Tag Archives: klb chemistry notes form 3

Chemistry Form 1-4 simplified Notes

Download the latest Chemistry notes for form 1 to 4 here. These notes are available at no cost.

FREE CHEMISTRY NOTES FOR FORM ONE TO FOUR.

FREE LATEST CHEMISTRY NOTES FOR FORM 2

FREE LATEST CHEMISTRY NOTES FOR FORM 1

FREE LATEST CHEMISTRY NOTES FOR FORM 1 UPDATED

FREE LATEST CHEMISTRY NOTES FOR FORM

FREE LATEST CHEMISTRY NOTES FOR FORM 2 UPDATED

FREE LATEST CHEMISTRY NOTES FOR FORM 1 NEW

FREE LATEST CHEMISTRY NOTES FOR FORM 1-4

FREE LATEST CHEMISTRY NOTES FOR FORM 1 UPDATED COPY

FREE LATEST CHEMISTRY NOTES FOR FORM 1-4 BOOKLET

FREE LATEST CHEMISTRY NOTES FOR FORM 1-4 UPDATED

FREE LATEST CHEMISTRY NOTES FOR FORM PRACTICALS GUIDE

FREE LATEST CHEMISTRY NOTES FOR FORM 3

FREE LATEST CHEMISTRY NOTES FOR FORM ONE TO FOUR

FREE LATEST CHEMISTRY NOTES FOR FORM 1 SIMPLIFIED

FREE LATEST CHEMISTRY NOTES FOR FORM 2 SIMPLIFIED

Download more notes and other teachers’ resources here; Teachers’ Free Resources Portal

Hot Downloads!!

Chemistry Topic By Topic Questions And answers (All Topics)

CHEMISTRY FORM ONE NOTES FREE

CHEMISTRY PRACTICALS GUIDE

CHEMISTRY FORM ONE NOTES: NEW

CHEMISTRY FORM FOUR NOTES: NEW

Free Chemistry notes, revision questions, KCSE past Papers, Exams, Marking Schemes,…

Chemistry High School notes for form 1-4 (Free updated pdf downloads)

CHEMISTRY NOTES F1-4: LATEST NOTES

Free updated schemes of work for all subjects (Secondary; Form 1-4)

Schemes of work for all subjects, free updated downloads- Secondary schools

KCSE Topical Revision Resources For All Subjects (Topic By Topic Questions…

CHEMISTRY NOTES FORM 3 PDF

 UNIT 2: NITROGEN AND ITS COMPOUNDS.

Unit checklist.

  1. Introduction
  2. Preparation of nitrogen
  • Isolation from air
  • Isolation from liquid air
  • Laboratory preparation
  • Preparation from ammonia
  • Properties of nitrogen
  • Oxides of nitrogen
    • Nitrogen (I) oxide
    • Nitrogen (II) oxide
    • Nitrogen (IV) oxide

GET THE FREE PDF NOTES HERE; Free Chemistry notes, revision questions, KCSE past Papers, Exams, Marking Schemes, Topical revision materials, Syllabus and Many more

  1. Action of heat on nitrates.
  2. Ammonia gas
  • Preparation
  • Laboratory preparation
  • Preparation from caustic soda
  • Test for ammonia
  • Fountain experiment
  • Properties and reactions of ammonia
  • Large scale manufacture of ammonia gas: the Haber process
  • Uses of ammonia
  1. Nitric (V) acid
  • Laboratory preparation
  • Industrial manufacture of nitric (V) acid: The Otswald’s process.
  • Reactions of dilute nitric acid
  • Reactions of concentrated nitric acid
  • Uses of nitric acid
  1. Test for nitrates.
  2. Pollution effects of nitrogen and its compounds
  3. Reducing pollution environmental pollution by nitrogen compounds.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Introduction:

– About 78% of air is nitrogen, existing as N2 molecules.

– The two atoms in the molecules are firmly held together.

– Nitrogen does not take part in many chemical reactions due to its low reactivity.

– Its presence in air dilutes oxygen and slows down respiration, burning and rusting.

 

Preparation of nitrogen.

(a). Isolation from air.

(i). Apparatus.

 

 

 

 

 

 

 

 

 

 

 

(ii). Procedure.

– Air is driven out of the aspirator by passing water into the aspirator from a tap.

– The air is the passed through a wash bottle containing concentrated potassium hydroxide solution.

Reason:

– To remove carbon (IV) oxide from air.

 

Equations:

2KOH(aq) + CO2(g)                   K2CO3(aq) + H2O(l)

 

Then

K2CO3(aq) + H2O(l) + CO2(g)                     2KHCO3(aq)

 

Thus;

KOH(aq) + CO2(g)                     KHCO3(aq)

 

– The carbon (IV) oxide-free air is then passed into a combustion tube with heated copper metal.

Reason:

– To remove oxygen from the air.

 

Note:

In this reaction the brown copper metal is oxidized to black copper (II) oxide.

 

Equation:

2Cu(s) + O2(g)                  2CuO(s)

Brown                                                Black

 

Note:

– Alternatively oxygen can be removed by passing the carbon (IV) oxide-free air through pyrogallic acid.

– The remaining part of air is mainly nitrogen and is collected over water.

 

Note:

– Nitrogen obtained by this method contains noble gases like xenon, argon etc as impurities.

– Purer nitrogen may be obtained by heating ammonium nitrite.

 

Equation:

NH4NO3(s)       Heat       N2(g) + 2H2O(g)

 

Summary.

 

 

 

 

 

 

 

 

 

(b). Removal from liquid air.

– Liquid air is primarily a mixture of nitrogen and oxygen with small amounts of noble gases.

– This method involves manufacture of liquid air and consequent fractional distillation.

 

The chemical process.

Step 1: removal of dust particles.

– Dust particles are first removed by either of the two processes:

  • Electrostatic precipitation

(i). Electrostatic precipitation:

– Air is passed through oppositely charged plates hence an electric field.

– Dust particles (charged) are consequently attracted to plates of opposite charges.

 

Diagram: electrostatic precipitation:

 

 

 

 

 

 

 

 

(ii). Filtration:

– The air is passed through a series of filters which traps dust particles as the air is forced through.

 

Step 2: removal of carbon (IV) oxide.

– The dust-free air is passed through a solution of potassium hydroxide; to remove carbon (IV) oxide.

 

Equations:

2KOH(aq) + CO2(g)                        K2CO3(aq) + H2O(l)

 

Then:

K2CO3(aq) + H2O(l) + CO2(g)                 2KHCO3(aq)

(Excess)

– Alternatively, sodium hydroxide may be used in place of potassium hydroxide.

 

Step 3: Removal of water vapour.

– The dustless, carbon (IV) oxide-free air is next passed into a chamber with concentrated sulphuric acid or anhydrous calcium chloride in which water vapour is separated and removed.

 

Note:

To remove water vapour, air may be alternatively passed into a freezing chamber where it is condensed at -25oC.

– The water vapour solidifies and is then absorbed by silica gel and separated out.

– Air is freed from carbon (IV) oxide, water vapour and dust particles (before compression) so as to prevent blockage of the pipes caused by  solid materials at liquefaction temperatures i.e. carbon (IV) oxide and water vapour form solids which may block the collection pipes.

 

Step 4: Liquification of air.

– The air free from dust, carbon (IV) oxide and water vapour is then compressed at about 200 atmospheres, cooled and allowed to expand through fine jet.

– This sudden expansion causes further cooling and the gases eventually liquefy.

– The liquid is tapped off through a valve while gas which has escaped liquefaction returns to the compressor.

– Liquid air is a transparent pale blue liquid.

– This liquid is then fractionally distilled.

 

Step 5: Fractional distillation of liquid air.

– The boiling point of nitrogen is -196oC (77K) and that of oxygen is -183oC (90K).

– Consequently when liquid air is allowed to warm up, the nitrogen boils off first and the remaining liquid becomes richer in oxygen.

– The top of the fractionating column is a few degrees cooler than the bottom.

– Oxygen, the liquid with the higher boiling point (-183oC) collects at the bottom as the liquid.

– The gas at the top of the column is nitrogen which ahs a lower boiling point.

– The more easily vapourised nitrogen is taken off.

– This way about 99.57% nitrogen is obtained.

 

Note:

– The separation of nitrogen and oxygen from air is a proof that air is a mixture and not a compound.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Summary: Fractional distillation of liquid air.

AIR

 

 

 

 

Step 1: Elimination of dust by Filtration

and electrostatic precipitation

 

 

 

 

Step 2: CO2 removal, pass dust free air

through KOH or NaOH

 

 

 

 

 

Step 3: Removal of water vapour; through

condensation -25oC) or conc. H2SO4

 

 

 

 

 

Recycling                        Step 4: Compression at approximately 200

atmospheres; cooling and expansion of air

 

 

 

 

 

Step 5: Fractional distillation

 

 

 

 

 

 

 

 

(c). Laboratory preparation method.

(i). Apparatus.

 

 

 

 

 

 

 

 

 

 

 

 

 

(ii). Procedure:

– Concentrated solutions of sodium nitrite and ammonium chloride are heated together in a round bottomed flask.

 

(iii). Observations.

– Colourless gas (nitrogen) is evolved rapidly and is collected over water.

 

(iv). Equation.

NaNO2(aq) + NH4Cl(aq)            heat       NaCl(aq) + N2(g) + 2H2O(l).

 

Note: the resultant gas is less dense than that isolated from air.

Reason:

– It does not contain impurities.

 

(d). Preparation from ammonia.

(i). Apparatus.

 

 

 

 

 

 

 

 

 

 

 

(ii). Procedure:

– Dry ammonia gas is passed over a heated metal oxide e.g. copper metal.

– The metal oxide is reduced to the metal while ammonia gas is itself oxidized to nitrogen and water.

– Water is condensed and collected in a u-tube immersed in ice cubes.

– Nitrogen produced is collected over water.

 

(iii). Observations and explanations.

  • Copper (II) oxide:

3CuO(s) + 2NH3(g)                           3Cu(s) + N2(g) + 3H2O(l)

(Black)                                                                            (Brown)    (Colourless)

 

  • Zinc (II) oxide

3ZnO(s) + 2NH3(g)                           3Zn(s) + N2(g) + 3H2O(l)

(Yellow-hot)                                                                  (Grey)    (Colourless)

(White-cold)

 

  • Lead (II) oxide

3PbO(s) + 2NH3(g)                           3Cu(s) + N2(g) + 3H2O(l)

(Red-hot)                                                                      (Grey)    (Colourless)

(Yellow-cold)

 

 

 

Properties of nitrogen.

(a). Physical properties.

  1. It is a colourless, odourless and tasteless gas; almost completely insoluble in water.
  2. Slightly lighter than air.

 

(b). Chemical properties.

  1. It is inert (unreactive)

Reason:

– The inert nature of nitrogen is due to the strong covalent bonds between the two nitrogen atoms in the molecule; N2.

 

Structurally;

 

 

 

 

 

 

– In air, it neither burns nor supports combustion and acts mainly as a diluent for the oxygen; slowing down the rate of burning.

 

Chemical test for nitrogen.

– A gas is proved to be nitrogen by elimination: –

  • It extinguishes a lighted splint and dos not burn; hence it is not oxygen, hydrogen or carbon (II) oxide.
  • It has neither smell nor colour; and therefore is not chlorine, ammonia, sulphur (IV) oxide or hydrogen chloride.
  • It does not form a white precipitate in lime water, and so it is not carbon (IV) oxide.
  • It is neutral to litmus and therefore cannot be carbon (IV) oxide, hydrogen sulphide, ammonia, hydrogen chloride

 

  1. Reaction with hydrogen.

– Under special conditions (i.e. high pressure, low temperatures and presence of iron catalyst), nitrogen combines with hydrogen to produce ammonia.

Equation:

N2(g) + 3H2(g)                2NH3(g)

 

– This reaction forms the basis of Haber process used in the manufacture of ammonia.

 

  1. Reaction with burning magnesium.

(i). Apparatus.

 

 

 

 

 

 

 

 

 

(ii). Procedure:

– A piece of burning magnesium ribbon is introduced into a gas jar full of nitrogen.

 

(iii). Observations:

– The magnesium ribbon continues to burn and a white solid; magnesium nitride is formed.

 

Equation:

3Mg(s) + N2(g)     Heat     Mg3N2(s)

 

Note:

– When magnesium nitride is treated with water or a solution of sodium hydroxide; the characteristic pungent smell of ammonia can be detected.

 

Equations:

In water

Mg3N2(s) + 6H2O(l)                   2NH3(g) + 3Mg(OH)2(aq)

 

In sodium hydroxide:

Mg3N2(s) + NaOH(aq)     

 

  1. Reaction with oxygen.

– When nitrogen and oxygen in air are passed through an electric arc small quantities of nitrogen (II) oxide are formed.

Equation:

N2(g) + O2(g)                      2NO(g)

 

Note:

– Nitrogen reacts with oxygen under various conditions to give different types of nitrogen oxides.

 

Uses of nitrogen

  1. Used in the Haber process in the manufacture of ammonia.
  2. Due to its inert nature, it is mixed with argon to fill electric bulbs (to avoid soot formation).
  3. In liquid state it is used as an inert refrigerant e.g. storage of semen for artificial insemination.
  4. Due to its inert nature, it is used in food preservation particularly for canned products i.e. it prevents combination of oxygen and oil which tends to enhance rusting.
  5. It is used in oil field operation called enhanced oil recovery where it helps to force oil from subterranean deposits.

 

 

 

 

 

 

 

 

 

 

 

 

Oxides of nitrogen.

– The three main oxides of nitrogen are:

  • Nitrogen (I) oxide, N2O
  • Nitrogen (II) oxide, NO
  • Nitrogen (IV) oxide, NO2

 

  1. Nitrogen (I) oxide.

Preparation of nitrogen (I) oxide, N2O

(i). Apparatus.

 

 

 

 

 

 

 

 

 

 

 

 

(ii). Procedure:

– Ammonium nitrate is gently heated in a boiling tube and gas produced collected over warm water.

– Heating is stopped while excess ammonium nitrate still remains.

Reason:

– To avoid chances of an explosion.

 

(iii). Observations:

– The solid (ammonium nitrate) melts and gives off nitrogen (I) oxide which is collected over warm water.

Reasons:

– Nitrogen (I) oxide is slightly soluble in cold water.

 

(iv). Equation:

NH4NO3(s)       Heat       NO2(g) + 2H2O(l)

 

Properties:

  1. It is a colourless gas, denser than air, fairly soluble in cold water and neutral to litmus.
  2. It supports combustion by oxidizing elements like sulphur, magnesium and phosphorus under strong heat.

Equations:

N2O(g) + Mg(s)    Heat     MgO(s) + N2(g)

 

2N2O(g) + S(s)     Heat     SO2(g) + 2N2(g)

 

2N2O(g) + C(s)     Heat     CO2(g) + 2N2(g)

 

5N2O(g) + 2P(s)   Heat     P2O5(g) + 5N2(g)

 

  1. Magnesium decomposes the gas and continues to burn in it.

Equation:

N2O(g) + Mg(s)    Heat     MgO(s) + N2(g)

 

  1. When exposed over red-hot finely divided copper it is reduced to nitrogen.

Equation:

N2O(g) + Cu(s)     Heat     CuO(s) + N2(g)

 

  1. Chemical test.
  • It relights a glowing splint.

Note:

  • It can be distinguished from oxygen by the following tests:
  • It has a sweet sickly smell; oxygen is odourless.
  • It will not give brown fumes (NO2) with nitrogen (II) oxide; oxygen does.
  • It is fairly soluble in cold water; oxygen is insoluble.
  • It extinguishes feebly burning sulphur; oxygen does not.

 

Uses of nitrogen (I) oxide.

– It was formerly used in hospitals as an aesthetic for dental surgery but has since been discontinued due to availability of more efficient anaesthetics.

 

Note:

– Nitrogen (I) oxide is also called laughing gas; because patients regaining consciousness from its effects may laugh hysterically.

 

  1. Nitrogen (II) oxide, NO.

Preparation:

(i). Apparatus.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(ii). Procedure:

– Action of heat on 50% concentrated nitric acid on copper turnings.

– Not any heat is required.

 

Equation:

3Cu(s) + 8HNO3(aq)                              3Cu(NO3)2(aq) + 4H2O(l) + 2NO(g)

 

(iii). Observations:

– An effervescence occurs in the flask; with brown fumes because the nitrogen (II) oxide produced reacts with oxygen of the air in the flask to form a brown gas, nitrogen (IV) oxide.

 

Equation:

2NO(g) + O2(g)              2NO2(g)

Colourless    Colourless                   Brown

 

– The brown fumes eventually disappear and the gas collected over water.

– The NO2 fumes dissolve in the water in the trough, resulting into an acidic solution of nitrous acid.

– The residue in the flask is a green solution of copper (II) nitrate.

– Industrially, the gas is obtained when ammonia reacts with oxygen in the presence of platinum catalyst.

– This is the first stage in the production of nitric acid.

 

(v). Properties.

  1. It is a colourless, insoluble and neutral to litmus. It is also slightly denser than air.
  2. Readily combines with oxygen in air and forms brown fumes of nitrogen (IV) oxide.
  3. Does not support combustion except in the case of strongly burning magnesium and phosphorus; which continues to burn in it, thereby reducing it i.e. it is an oxidizing agent.

 

Example:

2Mg(s) + 2NO(g)                       2MgO(s) + N2(g)

 

4P(s) + 10NO(g)             2P2O5(s) + 5N2(g)

 

  1. When passed over red-hot finely divided copper, it is reduced to nitrogen gas.

 

Equation:

2Cu(s) + 2NO(g)                        2CuO(s) + N2(g)

 

  1. Reaction with iron (II) sulphate.

– When iron (II) sulphate solution (freshly prepared) is poured into a gas jar of nitrogen (II) oxide, a dark brown colouration of Nitroso-iron (II) sulphate is obtained.

 

Equation:

FeSO4(aq) + NO(g)                                 FeSO4.NO(aq)

Green solution                                                                         Dark brown

                                                                                                    (Nitroso-iron (II) sulphate/ nitrogen (II) oxide iron (II) sulphate complex)

 

  1. It is also a reducing agent.

 

Equation:

Cl2(g) + 2NO(g)                       2ClNO(l)

Chloro nitrogen (II) oxide.

 

  1. Reaction with hydrogen.

– When electrically sparked with hydrogen, NO is reduced to nitrogen.

 

Equation:

2H2(g) + 2NO(g)          2H2O(l) + N2(g)

 

Chemical test:

– When exposed to air, nitrogen (II) oxide forms brown fumes of nitrogen (IV) oxide.

 

Uses of Nitrogen (II) oxide.

Note: –It is not easy to handle owing to its ease of oxidation.

  1. It is an intermediate material in the manufacture of nitric acid

 

  1. Nitrogen (IV) oxide.

Preparation:

(i). Apparatus.

 

 

 

 

 

 

 

 

 

 

(ii). Procedure:

– Action of conc. Nitric acid on copper metal.

 

Equation:

Cu(s) + 4HNO3(l)                              Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

 

Note:

– NO2 is also prepared by the action of heat on nitrates of heavy metals like lead nitrate.

– NO2 is given off together with oxygen when nitrates of heavy metals are heated.

– It is best prepared by heating lead (II) nitrate in a hard glass test tube.

  • Lead (II) nitrate is the most suitable because it crystallizes without water of crystallization (like other nitrates) which would interfere with preparation of nitrogen (IV) oxide that is soluble in water.

– The gas evolved passes into a U-tube immersed in an ice-salt mixture.

 

  • Apparatus:

 

 

 

 

 

 

 

 

 

 

 

  • Equation:

2Pb(NO3)2(s)                         2PbO(s) + 4NO2(g) + O2(g)

  • Observations:

– The heated white lead (II) nitrate crystals decompose and decrepitates (cracking sound) to yield red lead (II) oxide; which turns yellow on cooling.

– A colourless gas, oxygen is liberated, followed immediately by brown fumes of nitrogen (IV) oxide.

– Nitrogen (IV) oxide is condensed as a yellow liquid; dinitrogen tetroxide (N2O4); and is collected in the U-tube.

 

Note:

– At room temperature, nitrogen (IV) oxide consists of nitrogen (IV) oxide and dinitrogen tetroxide in equilibrium with each other.

 

Equation:
2NO2(g)                                    N2O4(g)

(Nitrogen (IV) oxide)                                       (Dinitrogen tetroxide)

 

– The oxygen being liberated does not condense because it ahs a low boiling point of -183oC.

 

Properties of nitrogen (IV) oxide.

  1. Red-brown gas with a pungent chocking smell
  2. It is extremely poisonous.
  3. It is acidic, hence turns moist litmus paper red.
  4. When reacted with water, the brown fumes dissolve showing that it is readily soluble in water.

 

Equation:

2NO2(g) + H2O(l)                           HNO3(aq) + HNO2(aq)

(Nitric (V) acid)     (Nitrous (III) acid)When liquid nitrogen

 

– Like carbonic (IV) acid, nitrous (III) acid could not be isolated. It is easily oxidized to nitric (V) acid.

 

Equation:

2NHO2(aq) + O2(g)                         2NHO3(aq)

(Nitric (III) acid)                                                (Nitrous (V) acid)

 

  1. Reaction with magnesium.

– Nitrogen (IV) oxide does not support combustion.

– However burning magnesium continues to burn in it.

Reason:

– The high heat of combustion of burning magnesium decomposes the nitrogen (IV) oxide to nitrogen and oxygen; the oxygen then supports the burning of the magnesium.

 

Equation:

4MgO(s) + 2NO2(g)                               4MgO(s) + 2N2(g)

 

Note:

– Generally nitrogen (IV) oxide oxidizes hot metals and non-metals to oxides and itself reduced to nitrogen gas.

Examples:

(i). Copper:

4Cu(s) + 2NO2(g)                                   4CuO(s) + N2(g)

 

(ii). Phosphorus:

8P(s) + 10NO2(g)                           4P2O5(s) + 5N2(g)

(iii). Sulphur:

2S(s) + 2NO2(g)                        2SO2(g) + N2(g)

 

Note:

– NO2 reacts with burning substances because the heat decomposes it to NO2 and O2.

 

Equation:

2NO2(g)      Heat        2NO(g) + O2(g)

 

– This is the oxidizing property of nitrogen (IV) oxide.

– The resultant oxygen supports the burning.

 

  1. Effects of heat:

– On heating, nitrogen (IV) oxide dissociates to nitrogen (II) oxide and oxygen and will support a burning splint.

 

Equation:

2NO2(g)       Heat          2NO(g) + O2(g)

 

  1. – When liquid nitrogen (IV) oxide or dinitrogen tetroxide is warmed, it produces a pale brown vapour.

– This is due to the reversible set of reactions:

Heat                                                                    Heat

N2O4(l)                                     2NO2(g)                                    2NO(g)     +    O2(g)

(Dinitrogen tetroxide)      Cool                (Nitrogen (IV) oxide)               Cool               (Nitrogen (II) oxide)   (Oxygen)

Pale yellow                                                               Brown                                                               

          Colourless

– Percentage of each in the equilibrium depends on temperature.

– At low temperatures, percentage of N2O4 is high and the mixture is pale yellow in colour.

– Percentage of nitrogen (IV) oxide increases with increase in temperature and the colour darkens till at 150oC when the gas is entirely NO2 and is almost black.

– Still at higher temperatures, nitrogen (IV) oxide dissociates into colourless gas (NO and O2).

 

  1. Reaction with alkalis.

– A solution of aqueous sodium hydroxide is added to a gas jar of nitrogen (IV) oxide and shaken.

 

Observation:

– The brown fumes disappear.

 

Explanation:

– Formation of sodium nitrate and sodium nitrite.

 

Equation:

2NaOH(aq) + 2NO2(g)                  2NaNO3(g) + NaNO2(aq) + H2O(l)

 

Ionically:

2OH(aq) + 2NO2(g)                        NO3(aq) + NO2(aq) + H2O(l)

 

Conclusion:

Nitrogen (IV) oxide is an acidic gas because it can react with an alkali.

 

 

 

Uses of nitrogen (IV) oxide.

  1. Mainly used in the manufacture of nitric (V) acid.

 

Summary on comparison between oxides of nitrogen.

 

  Nitrogen (I) oxide Nitrogen (II) oxide Nitrogen (IV) oxide
Colour – Colourless gas

– Sweet sickly smell

– Colourless; turns brown in air;

– Odourless

– Red brown gas;

– Choking pungent smell;

2. Solubility – Fairly soluble in cold water; but less soluble in hot water; – Almost insoluble in water – Readily soluble in water to form nitric (V) acid and nitrous (III) acid;
3. Action on litmus – Neutral to litmus – Neutral to litmus – Turns moist blue litmus paper red; i.e. acidic.
4. Combustion – Supports combustion; relights a glowing splint; – Does not support combustion; – Does not support combustion.
5. Density – Denser than air – Slightly denser than air – Denser than air;
6. Raw materials and conditions – Ammonium nitrate and heat; – Copper and 50% nitric acid; – Copper metal and concentrated nitric acid;

 

Action of heat on nitrates.

– All nitrates except ammonium nitrate decompose on heating tom produce oxygen gas as one of the products.

– Nitrates can be categorized into 4 categories based on the products formed when they are heated.

The ease with which nitrates decompose increases down the electrochemical series of metals.

 

  1. Nitrates of metals higher in the electrochemical series like sodium and potassium decompose on heating to give the corresponding metal nitrite and oxygen.

 

Examples:

2NaNO3(s)        Heat    2NaNO2(s) +  O2(g)

 

2KNO3(s)          Heat    2KNO2(s) +  O2(g)

 

  1. Nitrates of most other metals (heavy metals) that are average in the electrochemical series decompose on heating to give the metals oxide; nitrogen (IV) oxide and oxygen gas.

 

Example: action of heat on lead (II) nitrate.

(i). Apparatus:

 

 

 

 

 

 

 

 

 

 

 

 

 

(ii). Procedure:

– Solid white lead (II) nitrate crystals are strongly heated in a boiling (ignition) tube.

Products are passed into a U- tube immerse in ice.

– Excess gases are channeled out to a fume chamber.

 

(iii). Observations:

– The white crystalline solid decrepitates.

– A colourless gas (oxygen) is liberated and immediately followed by a red brown fumes/ gas (nitrogen (IV) oxide).

– A pale yellow liquid (dinitrogen tetroxide) condenses in the U-tube in the ice cubes.

– This is due to condensation of nitrogen (IV) oxide.

– A residue which is red when hot and yellow on cooling remains in the boiling (ignition) tube

 

Equation:

2Pb(NO3)2(s)      Heat       2PbO(s) + 4NO2(g) + O2(g)

(White crystalline solid)                 (Red-hot            Brown Fumes     Colourless gas
yellow-cold)

 

Further examples:

 

2Ca(NO3)2(s)      Heat       2CaO(s)  +   4NO2(g)   +  O2(g)

(White solid)                                   (White solid)      Brown Fumes     Colourless gas

2Mg(NO3)2(s)     Heat       2MgO(s) + 4NO2(g)   +  O2(g)

(White solid)                                  (White solid)       Brown Fumes     Colourless gas

2Zn(NO3)2(s)      Heat       2ZnO(s)  +  4NO2(g)    +   O2(g)

(White solid)                                    (Yellow-hot       Brown Fumes      Colourless gas
White-cold)

 

2Cu(NO3)2(s)      Heat       2CuO(s)   +  4NO2(g)   +   O2(g)

(Blue solid)                                     (Black solid)       Brown Fumes     Colourless gas

Note:

– Some nitrates are hydrated and when heated first give out their water of crystallization; and then proceed to as usual on further heating.

 

Examples:

 

Ca(NO3)2.4H2O(s)          Heat       Ca(NO3)2(s)  +   4H2O(g) 

(White solid)                                                       (White solid)             Colourless gas

On further heating;

 

2Ca(NO3)2(s)      Heat       2CaO(s)  +   4NO2(g)   +  O2(g)

(White solid)                                   (White solid)      Brown Fumes     Colourless gas

 

 

 

 

 

 

  1. Nitrates of metals lower in the reactivity series e.g. mercury and silver decompose on heating to give the metal, nitrogen (IV) oxide and oxygen.

 

Example:

 

Hg(NO3)2(s)       Heat       Hg(s)  +   2NO2(g)   +  O2(g)

(White solid)                                                          Brown Fumes     Colourless gas

2AgNO3(s)          Heat       2Ag(s) +  2NO2(g)   +  O2(g)

(White solid)                                                        Brown Fumes     Colourless gas

  1. Ammonium nitrate decomposes to nitrogen (I) oxide and water vapour.

 

Example:

NH4NO3(s)         Heat       N2O(g)   +  O2(g)

                                                     Colourless gas    Colourless gas
Note:

This reaction is potentially dangerous as ammonium nitrate explodes on strong heating.

 

Ammonia.

– Is a compound of nitrogen and hydrogen and is the most important hydride of nitrogen.

– It is formed when any ammonium salt is heated with an alkali whether in solid or solution form.

– It is a colourless gas with a pungent smell of urine.

– It is alkaline and turns moist red litmus paper to blue when introduced to it.

 

Laboratory preparation of ammonia.

(i). Reagents.

Base + ammonium salt                     NH3(g) + H2O(l)

 

(ii). Apparatus.

 

 

 

 

 

 

 

 

 

 

 

 

 

(iii). Procedure.

– Ammonium chloride (NH4Cl)/ sal-ammoniac is mixed with a little dry slaked lime i.e. Ca(OH)2 and the mixture thoroughly ground in a mortar.

Reason:

– To increase surface area for the reactions.

 

– The mixture is then heated in a round-bottomed flask.

Note:

– A round-bottomed flask ensures uniform distribution of heat while heating the reagents.

– The flask should not be thin-walled.

Reason:

The pressure of ammonia gas liberated during heating may easily crack or break it.

 

– The flask is positioned slanting downwards.

Reason:

– So that as water condenses from the reaction, it does not run back to the hot parts of the flask and crack it.

– The mixture on heating produces ammonia, calcium chloride and water.

 

Equation:

Ca(OH)2(s) + NH4Cl(s)                   CaCl2(aq) + 2NH3(g) + 2H2O(g)

(Slaked lime)

 

(iv). Drying:

– Ammonia is dried by passing it through a tower or U-tube filled with quicklime (calcium oxide) or pellets of caustic potash but not caustic soda which is deliquescent.

 

Note:

Ammonia cannot be dried with the usual drying agents; concentrated sulphuric acid and calcium chloride as it reacts with them.

  • With concentrated sulphuric acid.

2NH3(g) + H2SO4(l)                      (NH4)2SO4(aq)

 

  • With fused calcium chloride:

CaCl2(aq) + 4NH3(g)                  CaCl2.4NH3(s)

 

– i.e. ammonia reacts forming complex ammonium salt.

 

(v). Collection:

– Dry ammonia gas is collected by upward delivery.

Reasons:

– It is lighter than air.

– It is soluble in water.

 

 

 

 

 

 

 

 

 

 

 

 

 

Other methods of preparing ammonia.

 

(b). Ammonia from caustic soda (sodium hydroxide) or caustic potash (potassium hydroxide)

Note:

– The slaked lime is replaced by either of the above solutions.

– Thus the solid reactant is ammonium chloride and the liquid reactant is potassium hydroxide.

 

(i). Apparatus:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(ii). Procedure:

– The flask is not slanted. It is vertical and heated on a tripod stand and wire gauze.

Reason:

– No need of slanting since water produced is in liquid form and not gaseous. Thus there is no possibility of condensation of water on hotter parts.

 

Equations:

(i). With caustic soda:

NaOH(aq) + NH4Cl(s)                NaCl(aq) + H2O(l) + NH3(g)

 

Ionically;

Na+(aq) + OH(aq) + NH4Cl(s)               Na+(aq) + Cl(aq) + H2O(l) + NH3(g)

 

Hence; NH4Cl(s) + OH(aq)                Cl(aq) + H2O(l) + NH3(g)

 

(ii). With caustic potash:

KOH(aq) + NH4Cl(s)                  KCl(aq) + H2O(l) + NH3(g)

 

Ionically;

K+(aq) + OH(aq) + NH4Cl(s)                 K+(aq) + Cl(aq) + H2O(l) + NH3(g)

 

Hence; NH4Cl(s) + OH(aq)                Cl(aq) + H2O(l) + NH3(g)

 

 

 

 

 

Note:

Ammonium sulphate could be used in place of ammonium chloride in either case.

 

Equations:

 

(i). With caustic soda:

2NaOH(aq) + (NH4)2SO4(s)            Na2SO4(aq) + 2H2O(l) + 2NH3(g)

 

Ionically;

2Na+(aq) + 2OH(aq) + (NH4)2SO4(s)                   2Na+(aq) + SO42-(aq) + H2O(l) + NH3(g)

 

Hence; (NH4)2SO4(s) + 2OH(aq)               SO42-(aq) + 2H2O(l) + 2NH3(g)

 

(ii). With caustic potash:

2KOH(aq) + (NH4)2SO4(s)              K2SO4(aq) + 2H2O(l) + 2NH3(g)

 

Ionically;

2K+(aq) + 2OH(aq) + (NH4)2SO4(s)                     2K+(aq) + SO42-(aq) + H2O(l) + NH3(g)

 

Hence; (NH4)2SO4(s) + 2OH(aq)               SO42-(aq) + 2H2O(l) + 2NH3(g)

 

(iii). With calcium hydroxide:

Ca(OH)2(aq) + (NH4)2SO4(s)             CaSO4(aq) + 2H2O(l) + 2NH3(g)

 

Ionically;

Ca2+(aq) + 2OH(aq) + (NH4)2SO4(s)                    Ca2+(aq) + SO42-(aq) + H2O(l) + NH3(g)

 

Hence; (NH4)2SO4(s) + 2OH(aq)                 SO42-(aq) + 2H2O(l) + 2NH3(g)

 

Note:

Reaction with calcium hydroxide however stops prematurely, almost as soon as the reaction starts.

Reason;

– Formation of insoluble calcium sulphate which coats the ammonium sulphate preventing further reaction.

 

Preparation of ammonium solution.

(i). Apparatus.

 

(ii). Procedure:

– The apparatus is altered as above.

– The drying tower is removed and the gas produced is directly passed into water by an inverted funnel.

 

Reasons for the inverted broad funnel.

– It increases the surface area for the dissolution of thereby preventing water from “sucking back” into the hot flask and hence prevents chances of an explosion.

 

(iii). Equation.

NH3(g) + H2O(l)           NH4OH(aq)

 

Note:

– The solution cannot be prepared by leading the gas directly to water by the delivery tube.

Reason:

– Ammonia gas is very soluble in water and so water would rush up the delivery tube and into the hot flask causing it to crack.

– The rim of the inverted funnel is just below the water surface.

 

Tests for ammonia.

  1. It is a colourless gas with a pungent smell.
  2. It is the only common gas that is alkaline as it turns moist red litmus paper blue.
  3. When ammonia is brought into contact with hydrogen chloride gas, dense white fumes of ammonium chloride are formed.

 

Equation:

NH3(g) + HCl(g)                                 NH4Cl(s)

 

Fountain experiment.

(i). Diagram:

 

 

 

 

 

 

 

 

 

 

 

 

 

(ii). Procedure:

– Dry ammonia is collected in a round-bottomed flask and set up as above.

– The clip is open and solution let to rise up the tube.

– The clip is closed when the solution reaches the top of the tube after which it is again opened fro a while.

 

(iii). Observations and explanations.

– When a drop of water gets to the jet, it dissolves so much of the ammonia gas that a partial vacuum is created inside the flask.

– As the ammonia in the flask dissolves, the pressure in the flask is greatly reduced.

– The atmospheric pressure on the water surface in the beaker forces water into the flask vigorously.

– The drawn-out jet of the tube causes a fountain to be produced.

– The fountain appears blue due to the alkaline nature of ammonia.

 

(iv). Caution:

– Ammonia is highly soluble in water forming an alkaline solution of ammonium hydroxide.

 

Note:

1 volume of water dissolves about 750 volumes of ammonia at room temperature.

 

Properties and reactions of ammonia.

  1. Smell: has a characteristic pungent smell.
  2. Solubility: it is highly soluble in water. The dissolved ammonia molecule reacts partially with water to form ammonium ions (NH4+) and hydroxyl ions (OH)

 

Equation:

NH3(g) + H2O(l)                       NH4+(aq) + OH(aq)

 

– Formation of hydroxyl ions means that the aqueous solution of ammonia is (weakly) alkaline and turns universal indicator purple.

 

  1. Reaction with acids.

– Sulphuric acid and concentrated ammonia solution are put in a dish and heated slowly.

– The mixture is evaporated to dryness.

 

Observations:

– A white solid is formed.

 

Equation:

2NH4OH(aq) + H2SO4(aq)                             (NH4)2SO4(aq) + H2O(l)

 

Ionically:

2NH4+(aq) + 2OH(aq) + 2H+(aq) + SO42-(aq)                    2NH4+(aq) + SO42-(aq) + 2H+(aq) + 2OH(aq) + H2O(l).

 

Then;

2H+(aq) + 2OH(aq)                    2H2O(l)

 

– To some of the resultant white solid, a little NaOH(aq) was added and the mixture warmed.

– The gas evolved was tested fro ammonia.

 

Observation:

– The resultant gas tested positive for ammonia.

 

Equation:

(NH4)2SO4(s) + 2NaOH(aq)                   Na2SO4(aq) + 2NH3(g) + 2H2O(l).

 

 

Explanations:

– Evolution of ammonia shows that the white solid formed is an ammonium salt.

– The ammonia reacts with acids to from ammonium salt and water only.

 

Further examples:
HCl(aq) + NH4OH(aq)                      NH4Cl(aq) + H2O(l)

 

HNO3(aq) + NH4OH(aq)                   NH4NO3(aq) + H2O(l)

 

Ionic equation:

NH3(g) + H+(aq)                         NH4+(aq)

 

  1. Reaction of ammonia with oxygen.

– Ammonia extinguishes a lighted taper because it dos not support burning.

– It is non-combustible.

– However it burns in air enriched with oxygen with a green-yellow flame.

 

Experiment: Burning ammonia in oxygen.

(i). Apparatus.

 

 

 

 

 

 

 

 

 

(ii). Procedure:

– Dry oxygen is passed in the U-tube for a while to drive out air.

– Dry ammonia gas is then passed into the tube.

– A lighted splint is then passed into the tube.

 

(iii). Observations:

– A colourless gas is liberated.

– Droplets of a colourless liquid collect on cooler parts of the tube.

 

(iv). Explanations:

– The conditions for the reactions are:

  • Dry ammonia and oxygen gas i.e. the gases must be dry.
  • All air must be driven out of the tube.

– Ammonia burns continuously in oxygen (air enriched with oxygen) forming nitrogen and water vapour i.e. ammonia is oxidized as hydrogen is removed from it leaving nitrogen.

 

Equation:

4NH3(g) + 3O2(g)                                   2N2(g) + 6H2O(g)

 

 

 

Sample question:

Suggest the role of glass wool in the tube.

 

Solution:

– To slow down the escape of oxygen in the combustion tube, thus providing more time for combustion of ammonia.

 

  1. Ammonia as a reducing agent.

– It reduces oxides of metals below iron in the reactivity series.

 

Experiment: reaction between ammonia and copper (II) oxide.

(i). Apparatus.

Ice cubes

(ii). Procedure:

– Copper (II) oxide is heated strongly and dry ammonia is passed over it.

– The products are then passed through a U-tube immersed in cold water (ice cubes).

 

(iii). Observations.

– The copper (II) oxide glows as the reaction is exothermic.

– A colourless liquid collects in the U-tube.

– A colourless gas is collected over water.

– The black copper (II) oxide changes to brown copper metal.

 

(iv). Explanations.

– Ammonia gas reduces copper (II) oxide to copper and is itself oxidized to nitrogen and water.

 

Equation:

3CuO(s) + 2NH3(g)                                3Cu(s) + 3H2O(l) + N2(g)

Black                                                                                         red-brown                        (colourless)

 

– The water produced condenses in the U-tube immersed in cold (ice) water.

– The resultant nitrogen is collected by downward displacement of water.

– The nitrogen gas collected is ascertained indirectly as follows:

  • A lighted splint is extinguished and the gas does not burn; thus it is not oxygen, hydrogen, or carbon (II) oxide.
  • It has neither smell nor colour; it is not ammonia, chlorine, sulphur (IV) oxide or nitrogen (IV) oxide.
  • It is not carbon (II) oxide because it does not turn lime water into a white precipitate.

 

 

Note:

– This experiment proves that ammonia contains nitrogen.

 

  1. Reaction with chlorine.

(i). Procedure:

– Ammonia gas is passed into a bell jar containing chlorine.

 

(ii). Apparatus:

 

 

 

 

 

 

 

 

 

(iii). Observations:

– The ammonia catches fire and burns for a while at the end of the tube.

– The flame then goes out and the jar then gets filled with dense white fumes of ammonium chloride.

 

Equations:

2NH3(g) + 3Cl2(g)                      6HCl(g) + N2(g)

 

Then;

6HCl(g) + 6NH3(g)                     6NH4Cl(s)

 

Overall equation:

8NH3(g) + 3Cl2(g)                             6NH4Cl(s) + N2(g)

 

  1. Ammonia solution as an alkali.

– Solution of ammonia in water contains hydroxyl ions.

 

Equation:

NH3(g) + H2O(l)             NH4+(aq) + OH(aq)

 

– Thus it has many properties of a typical alkali.

– Ammonia salts are similar to metallic salts.

– The group (NH4+) precipitates in the reaction as a whole without splitting in any way.

– It exhibits unit valency in its compounds and therefore called a basic radical.

 

Note:

– It cannot exist freely as ammonia gas (NH3) which is a compound.

– Like other alkalis, ammonia solution precipitates insoluble metallic hydroxides by double decomposition when mixed with solution of salts of the metals.

 

 

 

 

  1. Reaction with air in the presence of platinum wire.

(i). Apparatus:

 

 

 

 

 

 

 

 

 

(ii). Procedure:

– Concentrated ammonia solution is put in a conical flask.

– The platinum (or even copper) wire is heated until white-hot.

– Oxygen gas or air is then passed through the ammonia solution.

– The red-hot platinum (copper) wire is then put into the flask containing the concentrated ammonia.

 

(iii). Observations:

– The hot platinum wire glows.

– Red-brown fumes are evolved.

 

(iv). Explanations:

– The hot platinum coil glows when it comes into contact with the ammonia fumes, which come from the concentrated ammonia solution.

– Reaction between ammonia and oxygen takes place on the surface of the platinum wire that acts a s a catalyst.

– A lot of heat is produced in the reaction that enables the platinum coil to continue glowing.

– Ammonia is oxidized to nitrogen (IV) oxide.

 

Equation:

4NH3(g) + 5O2(g)     Platinum catalyst   4NO(g) + 6H2O(l)

 

– Red-brown fumes of nitrogen (IV) oxide are produced due to further oxidation of the nitrogen (II) oxide to from nitrogen (IV) oxide.

 

Equation:

2NO(g) + O2(g)                     2NO2(g)

 

 

 

 

 

 

 

 

 

 

 

 

  1. Action of aqueous ammonia on solution of metallic salts

(i). Procedure:

– To about 2cm3 of solutions containing ions of calcium, magnesium, aluminium, zinc, iron, lead, copper etc in separate test tubes; aqueous ammonia is added dropwise till in excess.

 

(ii). Observations:

The various metal ions reacted as summarized in the table below.

 

Metal ions in solution Observations on addition of ammonia
Few drops of ammonia Excess drops of ammonia
Ca2+ White precipitate White precipitate persists;
Mg2+ White precipitate Precipitate persists;
Al3+ White precipitate Precipitate persists;
Zn2+ White precipitate Precipitate dissolves;
Fe2+ Pale green precipitate Precipitate persists; slowly turns red-brown on exposure to air;
Fe3+ Red-brown precipitate Precipitate persists;
Pb2+ White precipitate Precipitate persists;
Cu2+ Pale blue precipitate Precipitate dissolves forming a deep blue solution;

 

(iii). Explanations:

– Most metal ions in solution react with ammonia solution to form insoluble metal hydroxides.

– In excess ammonia, some of the so formed hydroxides dissolve forming complex ions.

 

(iv). Equations:

 

  1. Mg2+(aq) from MgCl2;

 

MgCl2(aq) + 2NH4OH(aq)                             Mg(OH)2(s) + 2NH4Cl(aq)

 

Ionically:

Mg2+(aq) + 2OH(aq)                            Mg(OH)2(s)

                                                                                        (White ppt)

 

  1. Fe2+ from Fe(NO3)2;

 

Fe(NO3)2(aq) + 2NH4OH(aq)                 Fe(OH)2(s) + 2NH4NO3(aq)

 

Ionically:

Fe2+(aq) + 2OH(aq)                    Fe(OH)2(s)

(Pale green ppt)

 

  1. Fe3+ from FeCl3;

 

Ionically:

Fe3+(aq) + 3OH(aq)                    Fe(OH)3(s)

(Red brown ppt)

 

 

 

 

 

Note:

Zn2+(aq) and Cu2+(aq) dissolve in excess ammonia solution forming complex ions.

 

  1. Zinc ions and ammonia solution.

 

  • With little ammonia:

ZnCl2(aq) + 2NH4OH(aq)                     Zn(OH)2(s) + 2NH4Cl(aq)

 

Ionically:

Zn2+(aq) + 2OH(aq)                   Zn(OH)2(s)

                                                                                (White ppt.)

 

  • In excess ammonia:

– The white precipitate of Zn(OH)2(s) dissolves in excess ammonia to form a colourless solution; proof that solution has Zn2+ ions;

– The colourless solution is a complex salt of tetra-amine zinc (II) ions.

 

Equation:

Zn(OH)2(s) + 4NH3(aq)                          [Zn(NH3)4]2+(aq) + 2OH(aq)

(White ppt.)                                                                              (Colourless solution-tetra amine zinc (II) ions)

 

  1. Copper (II) ions.

 

  • With little ammonia:

– A pale blue precipitate is formed.

 

Ionically:

Cu2+(aq) + 2OH(aq)                   Cu(OH)2(s)

(Pale blue ppt.)

 

  • In excess ammonia:

– The pale blue precipitate of Cu(OH)2(s) dissolves in excess ammonia to form a deep blue solution; proof that solution has Cu2+ ions;

– The deep blue solution is a complex salt of tetra-amine copper (II) ions.

 

Equation:

Cu(OH)2(s) + 4NH3(aq)                         [Cu(NH3)4]2+(aq) + 2OH(aq)

(Pale blue ppt.)                                                                       (Deep blue solution-tetra amine copper (II) ions)

 

Uses of ammonia gas and its solution:

  1. Ammonia gas is used in the manufacture of nitric acid and nylon.
  2. Ammonia gas is important in the preparation of ammonium salts used as fertilizers.
  3. It liquefies fairly easily (B.P is -33oC) and the liquid is used as a refrigerant in large cold storages and ice cream factories.
  4. Liquid ammonia is injected directly into the soil as a high nitrogen content fertilizer.
  5. Ammonia solution is used in laundry work as a water softener and a cleansing agent (stain remover)
  6. Ammonia is used in the manufacture of sodium carbonate in the Solvay process.
  7. Ammonia is used in “smelling salts”. It has a slightly stimulating effect on the action of the heart and so may prevent fainting

 

 

 

Qualitative analysis for cations using sodium hydroxide solution

(i). Procedure:

– To about 2cm3 of solutions containing ions of calcium, magnesium, aluminium, zinc, iron, lead, copper etc in separate test tubes; aqueous sodium hydroxide is added dropwise till in excess.

 

(ii). Observations:

The various metal ions reacted as summarized in the table below.

 

Metal ions in solution Observations on addition of ammonia
Few drops of ammonia Excess drops of ammonia
Ca2+ White precipitate White precipitate persists
Mg2+ White precipitate Precipitate persists;
Al3+ White precipitate Precipitate dissolves;
Zn2+ White precipitate Precipitate dissolves;
Fe2+ Pale green precipitate Precipitate persists; slowly turns red-brown on exposure to air;
Fe3+ Red-brown precipitate Precipitate persists;
Pb2+ White precipitate Precipitate dissolves;
Cu2+ Pale blue precipitate Precipitate dissolves forming a deep blue solution;

 

(iii). Explanations:

– Most metal ions in solution react with sodium hydroxide solution to form insoluble metal hydroxides.

– In excess sodium hydroxide, some of the so formed hydroxides (hydroxides of Zn, Al, Pb and Cu) dissolve forming complex ions.

 

(iv). Equations:

Ca2+(aq) + 2OH(aq)                             Ca(OH)2(s)

                                                                                        (White ppt)

 

Mg2+(aq) + 2OH(aq)                            Mg(OH)2(s)

                                                                                        (White ppt)

 

Al3+(aq) + 3OH(aq)                              Al(OH)3(s)

                                                                                        (White ppt)

 

Zn2+(aq) + 2OH(aq)                             Zn(OH)2(s)

                                                                                        (White ppt)

 

Pb2+(aq) + 2OH(aq)                              Pb(OH)2(s)

                                                                                        (White ppt)

 

Cu2+(aq) + 2OH(aq)                             Cu(OH)2(s)

                                                                                       (Pale blue ppt)

 

Fe2+(aq) + 2OH(aq)                    Fe(OH)2(s)

(Pale green ppt)

 

Fe3+(aq) + 3OH(aq)                    Fe(OH)3(s)

(Red brown ppt)

 

 

 

 

Note:

Hydroxides of Zn2+(aq) ; Pb2+(aq) ; and Al3+(aq) dissolve in excess ammonia solution forming complex ions.

 

  1. Zinc ions and sodium hydroxide solution.

 

  • With little sodium hydroxide:

 

Zn2+(aq) + 2OH(aq)                   Zn(OH)2(s)

                                                                                (White ppt.)

 

  • In excess sodium hydroxide:

– The white precipitate of Zn(OH)2(s) dissolves in excess sodium hydroxide to form a colourless solution;

– The colourless solution is a complex salt of tetra-hydroxo zinc (II) ions (zincate ion).

 

Equation:

Zn(OH)2(s) + 2OH(aq)                          [Zn(OH)4]2-(aq)

(White ppt.)                                                                              (Colourless solution-tetra hydroxo- zinc (II) ion/ zincate ion)

 

  1. Aluminium ions and sodium hydroxide solution.

 

  • With little sodium hydroxide:

 

Al3+(aq) + 3OH(aq)                    Al(OH)3(s)

                                                                                (White ppt.)

 

  • In excess sodium hydroxide:

– The white precipitate of Al(OH)3(s) dissolves in excess sodium hydroxide to form a colourless solution;

– The colourless solution is a complex salt of tetra-hydroxo aluminium (III) ions (aluminate ion).

 

Equation:

Al(OH)3(s) + OH(aq)                     [Al(OH)4](aq)

(White ppt.)                                                          (Colourless solution-tetra hydroxo- aluminium (III) ion/aluminate ion

 

  1. Lead (II) ions and sodium hydroxide solution.

 

  • With little sodium hydroxide:

 

Pb2+(aq) + 2OH(aq)                    Pb(OH)2(s)

                                                                                (White ppt.)

 

  • In excess sodium hydroxide:

– The white precipitate of Pb(OH)2(s) dissolves in excess sodium hydroxide to form a colourless solution;

– The colourless solution is a complex salt of tetra-hydroxo lead (II) ions (plumbate ions).

 

Equation:

Zn(OH)2(s) + 2OH(aq)                          [Zn(OH)4]2-(aq)

(White ppt.)                                                                              (Colourless solution-tetra hydroxo- lead (II) ion/ plumbate ion)

 

 

Summary and useful information on qualitative analysis:

Colours of substances in solids and solutions in water.

 

COLOUR IDENTITY
SOLID AQUESOUS SOLUTION

(IF SOLUBLE)

1. White Colourless Compound of K+; Na+, Ca2+; Mg2+; Al3+; Zn2+; Pb2+; NH4+
2. Yellow Insoluble Zinc oxide, ZnO (turns white on cooling); Lead oxide, PbO (remains yellow on cooling, red when hot)
Yellow Potassium or sodium chromate;
3. Blue Blue Copper (II) compound, Cu2+
4. Pale green

 

Green

Pale green (almost colourless)

Green

Iron (II) compounds,Fe2+

 

Nickel (II) compound, Ni2+; Chromium (II) compounds, Cr3+; (Sometimes copper (II) compound, Cu2+)

5. Brown Brown (sometimes yellow)

 

Insoluble

Iron (III) compounds, Fe3+;

 

Lead (IV) oxide, PbO2

6. Pink Pink (almost colourless)

Insoluble

Manganese (II) compounds, Mn2+;

Copper metal as element (sometimes brown but will turn black on heating in air)

7. Orange Insoluble Red lead, Pb3O4 (could also be mercury (II) oxide, HgO)
8. Black Purple

Brown

Insoluble

Manganate (VII) ions (MnO) as in KMnO4;

Iodine (element)-purple vapour

Manganese (IV) oxide, MnO2

Copper (II) oxide, CuO

Carbon powder (element)

Various metal powders (elements)

 

 

Reactions of cations with common laboratory reagents and solubilities of some salts in water

 

CATION SOLUBLE COMPOUNDS (IN WATER) INSUOLUBLE COMPOUNDS (IN WATER) REACTION WITH AQUEOUS SODIUM HYDROXIDE REACTION WITH AQUEOUS AMMONIA SOLUTION
Na+ All None No reaction No reaction
K+ All None No reaction No reaction
Ca2+ Cl; NO3; CO32-; O2-; SO42-; OH; White precipitate insoluble in excess White precipitate insoluble in excess, on standing;
Al3+ Cl; NO3; SO42- O2-; OH; White precipitate soluble in excess White precipitate insoluble in excess
Pb2+ NO3; ethanoate; All others; White precipitate soluble in excess White precipitate insoluble in excess
Zn2+ Cl; NO3; SO42- CO32-; O2-; SO42-; OH; White precipitate soluble in excess White precipitate soluble in excess
Fe2+ Cl; NO3; SO42- CO32-; O2-; OH; (Dark) pale green precipitate insoluble in excess (Dark) pale green precipitate insoluble in excess
Fe3+ Cl; NO3; SO42- CO32-; O2-; OH; (Red) brown precipitate insoluble in excess (Red) brown precipitate insoluble in excess
Cu2+ Cl; NO3; SO42- CO32-; O2-; OH; Pale blue precipitate insoluble in excess Pale blue precipitate soluble in excess forming a deep blue solution
NH4+ All None; Ammonias gas on warming Not applicable.

 

 

Qualitative analysis for common anions.

 

  SO42-(aq) Cl(aq) NO3(aq) CO32-(aq)
TEST Add Ba2+(aq) ions from Ba(NO3)2(aq); acidify with dilute HNO3(aq) Add Ag+(aq) from AgNO3(aq).

Acidify with dilute HNO3

Alternatively;

Add Pb2+ from Pb(NO3)2 and warm

Add FeSO4(aq);

Tilt the tube and carefully add 1-2 cm3 of concentrated H2SO4(aq)

Add dilute HNO3(aq); bubble gas through lime water;
OBSERVATION The formation of a white precipitate shows presence of SO42- ion; The formation of a white precipitate shows presence of Cl ion;

Formation of a white precipitate that dissolves on warming shown presence of Cl(aq) ions

The formation of a brown ring shows the presence of NO3 ions Evolution of a colourless gas that forma a white precipitate with lime water, turns moist blue litmus paper red; and extinguishes a glowing splint shows presence of CO32- ions
EXPLANATION Only BaSO4 and BaCO3 can be formed as white precipitates.

BaCO3 is soluble in dilute acids and so BaSO4 will remain on adding dilute nitric acid

Only AgCl and AgCO3 can be formed as white precipitates.

AgCO3 is soluble in dilute acids but AgCl is not;

– PbCl2 is the only white precipitate that dissolves on warming

Concentrated H2SO4 forms nitrogen (II) oxide with NO3(aq) and this forms brown ring complex (FeSO4.NO) with FeSO4; All CO32- or HCO3 will liberate carbon (IV) oxide with dilute acids

 

Checklist:

  1. Why is it not possible to use dilute sulphuric acid in the test for SO42- ions;
  2. Why is it not possible to use dilute hydrochloric acid in the test for chloride ions?
  3. Why is it best to use dilute nitric acid instead of the other two mineral acids in the test for CO32- ions?
  4. How would you distinguish two white solids, Na2CO3 and NaHCO3?

 

What to look for when a substance is heated.

 

1. Sublimation White solids on cool, parts of a test tube indicates NH4+ compounds;

Purple vapour condensing to black solid indicates iodine crystals;

2. Water vapour (condensed) Colourless droplets on cool parts of the test tube indicate water of crystallization or HCO3 (see below)
3. Carbon (IV) oxide CO32- of Zn2+; Pb2+; Fe2+; Fe3+; Cu2+;
4. Carbon (IV) oxide and water vapour (condensed) HCO3
5. Nitrogen (IV) oxide NO3of Cu2+; Al3+; Zn2+; Pb2+; Fe2+; Fe3+
6. Oxygen NO3 or BaO2; MnO2; PbO2;

 

 

 

 

 

 

Industrial manufacture of ammonia-The Haber process.

 

– Most of the world’s supply of ammonia is from the synthesis of Nitrogen and hydrogen in the Haber process.

 

(i). Raw materials

 

  • Nitrogen

– Usually obtained from liquid air by fractional distillation

 

  • Hydrogen

– Obtained from water gas in the Bosch process.

– Also from crude oil (cracking)

 

(ii). General equation

 

N2(g) + 3 H2(g)                        2NH3(g) + heat;

 

Note:

– Nitrogen and hydrogen combine in the ratio 1:3 respectively to form two volumes of ammonia gas plus heat.

-The reaction is exothermic releasing heat to the surrounding.

 

(iii). Conditions

 

  • High pressures

– The process is favoured by high pressures and thus a pressure of approximately 200 to 300 atmospheres is used.

 

Reason:

– The volume of gaseous reactants from equation is higher than volume of gaseous products. Thus increased pressure shifts the equilibrium to the right; favoring the production of more ammonia.

Note:

Such high pressures are however uneconomical.

 

  • Low temperatures

– Low temperatures favour production of ammonia;

Reason:

– The reaction is exothermic (releases heat to the surrounding) hence lower temperature will favour the forward reaction (shift the equilibrium to the right), producing more ammonia.

 

  • Catalyst

– The low temperatures make the reaction slow and therefore a catalyst is used to increase the rate of reaction

– The catalyst used is finely divided iron; impregnated with Aluminium oxide (Al2O3)

 

 

 

 

 

 

(iv). The chemical processes

 

Step 1: Purification

-The raw materials, nitrogen and hydrogen are passed through a purification chamber in which impurities are removed.

-The main impurities are CO2, water vapour,  dust particles, SO2, CO2 and O2;

 

Reason:

The impurities would poison the catalyst

 

Step 2: Compression

– The purified Nitrogen and Hydrogen gases are compressed in a compressor at 500 atmospheres.

 

Reasons:

  • To increase chances of molecules reacting;
  • To increase rate of collision of the reacting particles.
  • To increase pressure (attain desired pressures); and hence increase concentration of reacting particles.

 

Step 3: Heat exchanger reactions

– Upon compression the gaseous mixture, nitrogen and hydrogen are channeled into a heat exchanger; which heats them increasing their temperature.

– This enables the reactants (hydrogen and nitrogen) to attain the optimum temperatures for the succeeding reactions (in the catalytic chamber)

– From the heat exchanger the gases go to the catalyst chamber.

 

Step 4: Catalytic chamber.

– The gases then combine in the ratio of 1:3 (N2:H2 respectively), to form ammonia.

– This reaction occurs in presence of a catalyst; which speeds up the rate of ammonia formation;

– The catalyst is finely divided iron impregnated with aluminium oxide (Al2O3 increases the catalytic activity of iron).

 

Equation in catalytic chamber

 

N2(g) + 3H2(g)                          2NH3(g) + Heat (-92kjmol)

 

– Only about 6-10% of the gases combine.

– Due to the high heat evolution involved, the products are again taken back to the heat exchanger; to cool the gases coming from the catalytic chamber.

 

Step 5: Heat exchanger

– The gases from the catalytic chamber enter the heat exchanger a second time.

 

Reason:

– To cool the gases coming from the catalytic chamber, thus reduce cost of condensation.

-The gaseous mixture; ammonia and uncombined nitrogen and hydrogen are the passed through a condenser.

 

Step 6: The condenser reactions (cooling chamber)

– The pressure and the low temperatures in this chamber liquefy ammonia, which is then drawn off.

– The uncombined (unreacted) gases are recirculated back to the compressor, from where they repeat the entire process.

Summary: flow chart of Haber process.

Fractional distillation of air
Nitrogen
Hydrogen
Crude oil cracking; or water gas in Bosch process

 

 

 

 

 

 

 

Purifier: removal of duct particles; CO2; H2O vapour etc

 

 

 

 

 

 

Unreacted gases

(recycling)

 

 

 

 

 

 

6-10% ammonia + air;

 

 

 

 

LIQUID AMMONIA

 

 

Citing a Haber process plant

– When choosing a site for this industrial plant, the following factors are considered:

  1. Availability of raw materials (natural gas and crude oil)
  2. Presence of cheap sources of energy.
  3. Availability of transport and marketing.
  4. Availability of appropriate technology and labour force.

 

Ammonium salts as fertilizers

– Ammonium salts are prepared by the reaction between ammonia and the appropriate acid in dilute solution followed by evaporation and crystallization

 

(a). Ammonium sulphate

– Is prepared by absorbing ammonia in sulphuric acid.

 

Equation:

 

2NH3(g) + H2SO4(aq)                          (NH4)2SO4(aq)

 

Note: It is a cheap fertilizer.

 

(b). Ammonium nitrate

– Is prepared by neutralization nitric acid by ammonia.

 

Equation:

 

NH3(g) + HNO3(aq)                    NH4NO3(aq)

 

– As there is some danger of exploding during storage, ammonium nitrate is mixed with finely powdered limestone (CaCO3).

-The mixture, sold as nitro-chalk is much safer.

(c). Ammonium phosphate

– It is particularly useful as it supplies both nitrogen and phosphorus to the soil.

– It is prepared by neutralizing othophosphoric acid by ammonia

 

Equation:

 

3NH3(g) + H3PO4(aq)                       (NH4)3 PO4(aq)

 

(d) Urea CO (NH2)2

– Is made from ammonia and carbon (IV) oxide

– Its nitrogen content by mass is very high; nearly 47%

 

Equation:

 

NH3(g) +CO2(g)                       CO (NH2)2(aq)  + H2O(l)

   

Nitric (V) acid

– Is a monobasic acid (has only one replaceable Hydrogen atom); and has been known as strong water (aqua forty).

– It is a compound of hydrogen, oxygen and nitrogen.

 

Laboratory preparation of nitric (V) acid

(i). Apparatus

           

(ii). Reagents

– Nitric acid is prepared in the laboratory by action of concentrated sulphuric acid on solid nitrates e.g. potassium nitrate (KNO3) and sodium nitrate (NaNO3)

 

(iii). Procedure

– 30-40 grams of small crystal of KNO3 are put in a retort flask.

– Concentrated sulphuric acid is added just enough to cover the nitrate; and then heated (warmed) gently.

– The apparatus is all glass.

Reason:

– Nitric (V) acid would attack rubber connections.

– The neck of the retort flask is inserted into a flask that is kept cool continually under running water; this is where nitric acid is collected.

 

Note:

The cold water running over the collection flask is meant to cool (condense) the hot fumes of nitric (V) acid.

 

(iv). Observations and explanations

– Fumes of nitric are observed in the retort;

 

Equation

 

KNO3(g) + H2SO4(aq)                      KHSO4(aq) +HNO3(g)

 

– If Lead (II) nitrate was used;

 

Pb(NO3)2(s) + H2SO4(aq)                  PbSO4(s) + 2HNO3(g)

 

Note: with lead (II) nitrate the reaction soon stops because the insoluble lead (II) sulphate coats the surface of the nitrate preventing further reaction; yield of nitric (V) acid is thus lower;

 

-These fumes of nitric acid appear brown.

Reason:

– Due to the presence of nitrogen (iv) oxide gas formed by thermal decomposition of nitric acid.

 

Equation:

4HNO3(aq)                       4NO2(g) + O2(g) + 2H2O(g)

 

– Pure nitric (V) acid is colourless but may appear yellow (brown) due to the presence of Nitrogen (IV) oxide.

– The brown colour can be removed by blowing air through the acid.

– Fuming nitric acid boils at 83oC and is 99% pure; while concentrated nitric acid is only 70% acid and 30% water.

 

Note: Nitric acid is usually stored in dark bottles.

Reason:

– To avoid its decomposition by light to nitrogen (IV) oxide, oxygen and water.

– The reaction in the retort flask is a typical displacement reaction; in which the more volatile nitric (V) acid is displaced from nitrates by the less volatile sulphuric acid.

– The nitric acid distills over because it is more volatile than sulphuric acid.

 

 

 

 

 

 

 

 

 

 

 

 

Properties of concentrated nitric acid

– Nitric (V) acid readily gives oxygen and therefore is called an oxidizer.

– The acid is usually reduced to nitrogen (IV) oxide and water.

 

  1. Effects of heat on concentrated nitric acid

(i) Apparatus

 

 

 

 

 

 

 

 

 

 

 

 

(ii) Observations

– Brown fumes are seen in the hard glass tube.

– Colourless gas is collected over water.

 

(ii). Explanations

– Sand soaked in concentrated nitric acid produces nitric solid vapour on heating.

– The hot glass wool catalyzes the decomposition of nitric acid to nitrogen (IV) oxide (brown fumes), water vapour and oxygen.

 

Equation

 

4HNO3(l)                    4NO2(g) + 2H2O(l) + O2(g)

                                            (Brown fumes)

 

– The so formed nitrogen (IV) oxide dissolves in water forming both nitric and nitrous acids.

 

Equation:

 

2NO2(g) + H2O(l)                    HNO2(aq) + HNO3(aq)

 

– The oxygen gas is collected over water; and with the solution becoming acidic.

 

  1. Reaction with saw dust

– Saw dust contains compounds of carbon Hydrogen and oxygen.

 

Procedure:

– Some saw dust is heated in an evaporating dish and some few drops of concentrated nitric (V) acid on it (this is done in a fume cupboard)

 

Observation:

– A violent reaction occurs, the saw dust catches fire easily and a lot of brown fumes of nitrogen (IV) oxide given off.

– Nitric (V) acid oxidizes the compounds in saw dust to CO2 and water; and itself it is reduced to nitrogen (IV) oxide and water.

 

Equation:

(C, H, O) n(s) + HNO3(l)                      NO2(g) + CO2(g) +H2O(g)

Saw dust

 

– Warm concentrated nitric (V) acid oxidizes pure carbon and many other compounds containing carbon.

 

Equation:

C(s) + 4HNO3(l)                         2H2O(l) + 4NO2(g) + CO2(g)

 

  1. Reaction with sulphur

Procedure:

– 2 cm3 of concentrated nitric (V) acid is added to a little sulphur in a test tube and warmed.

– The mixture is filtered to remove excess sulphur and the filtrate diluted with distilled water.

– Drops of barium chloride are then added to the resultant solution.

 

Observations:

– Red brown gas, nitrogen (IV) oxide (NO2) is evolved and the sulphur is oxidized to sulphuric acid.

 

Equation

S(s) + 6HNO3(l)                          H2SO4(aq) + 6NO2(g) +2H2O(l)

 

– On addition of barium chloride to the solution, a white precipitate is formed.

– This is due to formation of barium sulphate and is a confirmation for the presence of SO42– ions.

 

Equation:

 

Ba2+(aq) + SO42-(aq)                     BaSO4(s)

                                                (White precipitate)

 

  1. Reaction with metals

– Concentrated nitric (V) acid reacts with metals except gold and platinum.

– Actual reaction depends on the concentration of the acid and the position of the metal in the reactivity series.

– The reaction results in a metal nitrate, NO2 and water.

– Copper, which is low in the reactivity series, reduces conc. HNO3 to NO2.

 

Equation:

 

Cu(s) + HNO3(l)                   Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

 

– Metals more reactive than copper e.g. Magnesium may reduce nitric acid to dinitrogen monoxide (N2O) or Nitrogen (N2).

– Some metals like iron and aluminium form insoluble layers when reacted with nitric acid thus stopping any further reaction.

 

 

 

  1. Reaction with iron (II) salts

 

Procedure:

– Few crystals of iron (II) sulphate are dissolved in dilute sulphuric acid.

– A little concentrated nitric (V) acid is added to the solution and mixture warmed.

 

Observation:

– Green solution turns brown.

 

Equation:

 

6FeSO4(s) + 3H2SO4(aq) +3HNO3(l)                    4H2O(l) +2NO(g) + 3Fe2 (SO4)3(aq)

 

Explanation:

– Nitric acid oxidizes green iron (II) salts (Fe2+) to brown iron (III) salts (Fe3+) and itself is reduced to Nitrogen (II) Oxide.

 

Note:

– In air, nitrogen (II) oxide is readily oxidized to nitrogen (IV) oxide; resulting to brown fumes.

 

Equation:
2NO(g) + O2(g)                           2NO2(g)

 

  1. Reduction of nitric (V) acid by hydrogen sulphide.

Procedure

– A few drops of conc. nitric (V) acid are added to a gas jar full of hydrogen sulphide and the jar then covered.

 

Observations

– Fumes (brown) of Nitrogen (IV) oxide and yellow deposits of sulphur;

 

Equation

– It is a REDOX reaction.

Oxidation

 

 

H2S(g) + 2HNO3(l)                       2H2O(l) + 2NO2(g) +S(s)

 

 

Reduction

 

 

 

 

 

 

 

 

 

 

 

 

Properties of dilute nitric (V) acid

  1. Reaction with metals

– Dilute nitric (V) acid reacts with most metals to form nitrogen (II) oxide instead of hydrogen.

 

Example:

3Mg(s) + 8HNO3(aq)                  3Mg(NO3)2(aq) +2NO(g) + 4H2O(l)

 

– In fact HNO3 is reduced to NO and water but NO soon gets oxidized in air to form brown fumes of NO2.

– However very dilute HNO3 (cold) reacts with more active metals such as Magnesium to produce Hydrogen.

 

  1. Dilute nitric (V) acid as a typical acid

(a). It turns blue litmus paper red.

(b). It reacts with metal oxides and metal hydroxides to form a metal nitrate and water (Neutralization)

 

Examples

  • CuO(s) + 2HNO3(aq)          Cu (NO3)2(aq) + H2O(l)

                  (Black)                                                                                (Blue)

 

  • Zn(OH)2(s) + 2HNO3(aq)            Zn (NO3)2(aq) + 2H2O(l)

                   (White ppt)                                                                      (Colourless)

 

  • KOH(aq) + HNO3(aq)            KNO3(aq) + H2O(l)

                 (Alkali)                (Acid)                                                 (Salt)              (Water)

 

  1. Reaction with metal carbonates and hydrogen carbonates

– Dilute HNO3 reacts with metal carbonates and hydrogen carbonates to form a nitrate, CO2 and water.

 

Examples.

CuCO3(s) + 2HNO3(aq)                   Cu(NO3)2(aq) + CO2(g) + H2O(l)

(Green)                                                                      (Blue solution)

 

NaHCO3(s) + HNO3(aq)                    NaNO3(aq) + CO2(g) + H2O(l)

 

Test for nitrates/nitric acid

  1. Oxidation of iron (ii) sulphate

– Concentrated HNO3 oxidizes green Iron (II) sulphate in presence of sulphuric acid into Iron (III) sulphate (yellow/brown)

– However the solution turns dark brown due to formation of a compound, FeSO4.NO

– NO is produced by reduction of nitrate to nitrogen monoxide by Fe2+

 

Ionically;

Fe2+(aq)                       Fe3+(aq)   +   e (oxidized)

 

NO3(aq) + 2H+(aq) + e                   NO2(g) + H2O(l) (reduced)

 

 

 

 

 

  1. Brown ring test

Procedure.

– An unknown solid is dissolved then acidified using dilute H2SO4.

– Some FeSO4 solution is then added.

– The test tube is then held at an angle and concentrated sulphuric (V) acid is added slowly (dropwise) to the mixture.

 

Observations

– The oily liquid (conc. H2SO4) is denser than water hence sinks to the bottom.

– A brown ring forms between the two liquid layers if the solid is a nitrate.

 

Diagrams:

 

 

 

 

 

 

 

 

Explanations:

– Suppose the solution tested isKNO3, the conc. H2SO4 and the KNO3 reacted to produce HNO3.

 

Equation:

KNO3(aq) +H2SO4(aq)                  KHSO4(aq) + HNO3(aq)

 

– The NO3 from nitric acid oxidizes some of the FeSO4 to Fe2 (SO4)3 (Fe2+ toFe3+) and itself reduced to NO by the Fe2+

 

-The NO so formed reacts with more FeSO4 to give a brown compound (FeSO4 NO) which appears as a brown ring.

 

Equation:

FeSO4(aq) + NO(g)               FeSO4. NO(aq)

(Green)                                                         (Brown)

 

Ionically:

Fe2+(aq) + 5H2O(l) + NO(g)                   [Fe(H2O)5NO]2+(aq)

(Green)                                                                                      (Brown)

 

  1. Heat

– Nitrates of less reactive metals decompose easily with gentle heating; clouds of brown NO2 can be seen.

 

Equation:

2Cu(NO3)2        heat         2CuO(s) + 4NO2(g) +  O2(g)

                                                      (Brown, acidic)

– The nitrates of more reactive metals need much stronger heating and decompose in a different way.

 

Equation:

2Na NO3(s)       heat      2NaNO2(s)  +  O2(g)

 

 

Uses of nitric acid 

– Large quantities are used in fertilizer manufacture.

– Manufacture of explosives (TNT)

– Manufacture of dyes

– Making nitrate salts

– Etching of metals.

– Manufacture of nylon and terylene

– Refining precious metals

– An oxidizing agent.

 

Industrial manufacture of nitric acid

The Otswald’s process

 (a). Introduction

– Nitric acid is manufactured by the catalyst oxidation of ammonia and dissolving the products in water.

 

(b). Raw materials

– Atmosphere air

– Ammonia from Haber process.

 

(c). Conditions

Platinum-rhodium catalyst or platinum gauze.

– The ammonia-air mixture must be cleaned (purified) to remove dust particles which could otherwise poison the catalyst.

 

(d). Chemical reactions.

Step 1: Compressor reactions.

– Ammonia and excess air (oxygen) (1:10 by volume) is slightly compressed.

– The mixture is then cleaned to remove particles which would otherwise poison the catalyst.

– They are then passed to the heat exchanger.

 

Step 2: Heat exchanger and catalytic chamber.

– In the heat exchanger, the gaseous mixture is heated to about 900oC and then passed over a platinum-rhodium catalytic chamber.

– An exothermic reaction occurs and ammonia is oxidized to nitrogen (II) oxide and steam.

 

Equation:

4NH3(g) + 5O2(g)                       4NO(g) + 6H2O(g)  + Heat.

 

– The exothermic reaction once started, provides the heat necessary to maintain the required catalytic temperature.

-This is of economical advantage i.e. electrical heating of catalyst is not continued hence lowering production costs.

 

Step 3: Heat exchanger.

– The hot products from catalytic chamber are again passed back through the heat exchanger.

– The hot gases are cooled and then passed into the cooling chamber.

 

 

Step 4: Cooling chamber

Once cooled, the NO is oxidized to NO2 by reacting it with excess oxygen.

 

Equation:

 

2NO(g) + O2(g)                       2NO2(g)

 

Step 5: Absorption towers:

– The NO2 in excess air is then passed through a series of absorption towers where they meet a stream of hot water and form nitric (V) acid and nitrous (III) acid.

 

Equations:

2NO2(g) + H2O(l)                      HNO3(aq) + HNO2(aq) (blue solution)

                                                                   Nitric                Nitrous

 

– The so produced nitrous (III) acid is oxidized by oxygen in excess air to nitric (V) acid so that the concentration of nitric acid in the solution (liquid) gradually increases.

 

Equation:

2 HNO2(aq) +  O2(g)                    2HNO3(aq)

 

– The resultant HNO3 is only 55%-65% concentrated.

– It is made more concentrated by careful distillation of the solution.

 

The process of distillation (increasing the concentration).

– Concentrated sulphuric (VI) acid is added to the dilute nitric (V) acid.

– The heat produced (when dilute sulphuric acid reacts with water) vapourises the nitric (V) acid.

– The resultant nitric (V) acid vapour is condensed.

Note:

  • Nitric (V) acid is stored in dark bottles.

Reason:

– To prevent its decomposition since it undergoes slow decomposition when exposed to light.

 

  • Dilute nitric (V) acid has higher ions concentration than concentrated nitric (V) acid.

Reason.

– Dilute nitric (V) acid is a stronger acid hence ionizes fully to yield more hydrogen ions than concentrated nitric (V) acid.

– Dilute nitric (V) acid is ionic whereas concentrated nitric (V) acid is molecular;

– Dilute nitric (V) acid is more (highly) ionized than concentrated nitric (V) acid.

 

 

 

 

 

 

 

 

 

 

 

 

Flow diagram for the otswald’s process.

                                       Ammonia

HEAT EXCHANGER
CATALYTIC CHAMBER

 

 

 

 

Air

 

 

 

Water                     Unreacted                NO(g)

                                                 NO + air;

 

 

 

 

 

 

 

 

            Nitric (V) acid

 

Pollution effects of nitrogen compounds.

  1. Acid rain

– Nitrogen (II) oxide is produced in internal combustion engines on combination of nitrogen and oxygen.

– Nitrogen (II) oxide oxidized to nitrogen (IV) oxide which dissolves in water to form nitric (III) and nitric (V) acids.

– Nitric (v) acid eventually reaches ground as acid rain and causes:

  • Loss of chlorophyll (chlorosis) from leaves
  • Corrosion of stone buildings and metallic structures, weakening them and destroying beauty.
  • Leaching of vital minerals from soils. These are converted into soluble nitrates and washed away from top soil. This leads to poor crop yields.

 

  1. Smog formation.

– Nitrogen (IV) oxide also undergoes series of chemical reactions in air to produce one of the major components of smog.

– Smog reduces visibility for motorists, irritates eyes and causes breathing problems.

 

  1. Eutrophication:

– Refers to enrichment of water with excess nutrients for algal growth.

– Presence of nitrate ions from nitrogen fertilizers in a water mass encourages rapid growth of algae.

– This eventually leads to reduction of dissolved oxygen in water, killing aquatic animals like fish.

– Presence of nitrate ions in drinking water may also cause ill health to humans. This is because they are converted into carcinogenic compounds.

 

Prevention.

  1. Recycling unreacted gases in manufacture of nitric acid to prevent release into environment.
  2. Treating sewage and industrial effluents to remove nitrogen compounds before releasing to rivers and lakes.
  3. Fitting exhausts systems of vehicles with catalytic converters which convert nitrogen oxides into harmless nitrogen gas.
  4. Adding lime to lakes and soils in surrounding regions to reduce acidity.
  5. Applying fertilizers at right and in the correct proportion to prevent them from being washed into water masses.

 

UNIT 3: SULPHUR AND ITS COMPOUNDS

Checklist:

 

  1. Occurrence of sulphur
  2. Extraction of sulphur
  • The Frasch pump
  • Extraction process
  1. Properties of sulphur
  • Physical properties
  • Chemical properties
  1. Uses of sulphur
  2. Allotropes of sulphur
  • Rhombic sulphur
  • Monoclinic sulphur
  1. Compounds of sulphur
  • Sulphur (IV) oxide
  • Laboratory preparation
  • Other preparation methods
  • Properties of sulphur (IV) oxide
    • Physical properties
    • Chemical properties
    • Uses of sulphur (IV) oxide
  1. Sulphur (VI) oxide
  • Laboratory preparation
  • Properties of sulphur (VI) oxide
  1. Sulphuric (VI) acid
  • Large scale manufacture
    • Raw materials
    • The chemical process
    • Pollution control
  • Properties of concentrated sulphuric (VI) acid
    • Physical properties
    • Chemical properties
  • Properties of dilute sulphuric (VI) acid
  • Uses of sulphuric (VI) acid
  1. Hydrogen sulphide gas
  • Laboratory preparation
  • Properties of hydrogen sulphide
  • Physical properties of hydrogen sulphide
  • Chemical properties of hydrogen sulphide
  1. Atmospheric pollution by sulphur compounds

 

 

 

 

 

Occurrence

– Occurs naturally as s free element in the underground deposits in Texas and Louisiana (USA) and Sicily (ITALY).

– It also occurs as a sulphate and sulphide ores.

 

Examples;

Metallic sulphides: iron pyrites (FeS2); Zinc blende (ZnS) Copper pyrites (CuFeS2)

Metallic sulphates e.g. Gypsum, CaSO4

Hydrogen sulphide e.g. H2S present in natural gas.

 

Extraction of sulphur: The Frasch process

– Is done using a set of 3 concentric pipes called Frasch pump; hence the name Frasch process.

 

(i). Apparatus: Frasch pump

Hot compressed air

 

 

Superheated water at 170oC
Froth of molten sulphur

 

Cross section of the Frasch pump

 

 

Outermost pipe: brings superheated water at 170oC

 

 

 

Innermost pipe: brings in hot compressed air;

 

Middle pipe: brings out a froth of molten sulphur

 

 

(ii). Chemical process

Note: Sulphur cannot be mined by conventional mining methods such as open cast, alluvial mining etc

Reasons:

– Sulphur deposits lie very deep under several layers of quicksand hence cannot be accessed easily.

– Sulphur deposits are associated with poisonous gases such as sulphur (IV) oxide gas which can cause massive pollution if exposed to open environment.

– Three concentric pipes, constituting the Frasch pump are drilled through the rock and soil down to the sulphur deposits.

 

 

(a). The outer tube (pipe)

– Is used to pump superheated water at 170o c and 10 atmospheres down the deposits.

– The heat of the water melts the sulphur.

– By the time the water reaches the sulphur, its temperature drops to 120oC, but this is enough to melt sulphur whose M.P is 114oC.

 

(b). The innermost tube

– Is the smallest pipe and is used to blow or force a jet of hot compressed air down the sulphur deposits.

– This produces a light froth of molten sulphur (mixture of air, water and sulphur) which is forced up the middle pipe.

 

(c). The middle pipe.

– Allows the sulphur froth (mixture of molten sulphur, water and air) into the surface; where mixture is run into large tanks.

– The forth usually settles in two layers, the bottom layer is mainly water while the upper layer is mainly molten sulphur; due to differences in density.

– Once in the settling tanks, sulphur solidifies and separates out; giving 99% pure sulphur.

– The sulphur is removed, melted again and poured into moulds, to form roll sulphur in which form it is sold.

 

Properties of sulphur

Physical properties

  1. – It is a yellow solid which exists in one amorphous form and 2 crystalline forms.

– A molecule of sulphur consists of a pluckered ring of 8 sulphur atoms covalently bonded.

 

Diagram: structure of a sulphur molecule.

 

 

 

 

 

 

 

  1. Solubility

– It is insoluble in water but soluble in organic solvents like carbon disulphide, xylene and toluene.

 

  1. It is a poor conductor of heat and electricity since it is a covalent element lacking free electrons or ions.

 

  1. Effects of heat

– When sulphur is heated out of contact with air, it melts at low temperatures of about 113oC to form an amber (orange) coloured mobile liquid.

Reason:

– The S8 rings open up to form chains of S8.

 

 

 

 

Diagrams:

The pluckered S8 ring of sulphur molecule                                           Chains of S8 molecule

 

 

 

 

 

 

– On further heating, the liquid darkens in colour.

– At 160oC, the liquid becomes much darker and very viscous (such that the test tube can be inverted without the sulphur pouring out.)

– The viscosity continues to increase until a temperature of about 1950C

Reason:

– The S8 rings of sulphur are broken and they then join to form very long chains of sulphur atoms, with over 100,000 atoms (S100 000).

 

Note: As the chains entangle with one another the viscosity increases and colour darkens.

 

– Near the boiling point, the liquid becomes less dark i.e. red-brown and more mobile (runny).

Reason

– The long chains are broken to shorter chains.

 

– At 444oC (boiling point), sulphur vapourises to form a red-brown vapour consisting of S8, S6, S4 and S2 molecules.

Reason

– The sulphur liquid changes state to form sulphur vapour.

– The vapour is light brown in colour, and consists of a mixture of molecules of formula S2-S10

 

Note

If heated further the larger sulphur vapour molecules (S8, S6 etc) dissociate and at 750oC the vapour is mostly constituted of diatomic molecules (S2)

On exposure to cold surfaces the light brown vapour condenses to a yellow sublimate. The yellow sublimate is called flowers of sulphur.

 

Chemical properties

  1. Burning in air

– It burns in air with a bright blue flame forming a misty gas with a choking smell.

– The gas is sulphur (IV) oxide, with traces of sulphur (VI) oxide, both of which are acidic.

 

Equation:

S(s) + O2(g)                        SO2(g)

 

Note:

The SO3 is formed due to further oxidation of some of the SO2 gas

 

Equation:

2SO2(s) + O2(g)                   2SO3(g)

 

 

 

  1. Reaction with acids.

– Dilute acids have no effect on sulphur.

– It is however easily oxidized by concentrated (VI) sulphuric acid and Nitric (VI) acid.

 

  • With conc. H2SO4

– When warmed with conc. H2SO4, sulphur is oxidized to sulphur (IV) oxide while the acid is reduced to the same gas.

 

Equation:

S(s) + 2H2SO4(l)                        3SO2(g) + 2H2O(l)

 

  • With conc. HNO3

– Sulphur is oxidized to sulphuric (VI) acid while acid itself is reduced to red-brown Nitrogen (IV) oxide.

 

Equation:

S(s) + 6HNO3(l)                   H2SO4(aq) + 6NO2(g) + 2H2O(l)

 

Note:

– The resultant solution gives a white precipitate with a solution of Barium chloride.

Reason

– Due to presence of sulphate ions which combine with Ba2+ to form insoluble BaSO4(s)

 

Ionically;

Ba2+(aq)  + SO42-(aq)                      BaSO4(s)

 

  1. Reaction with other elements.

– It combines directly with many other elements to form sulphides.

– With metals, sulphur forms metal sulphides, most of which are black.

 

Examples.

(a). With metals

 

  • Iron metal

Fe(s) + S(s)                         FeS(s) + Heat

(Grey)   (Yellow)                                  (Black)

 

Note:

– During the reaction, the mixture glows spontaneously; immediately the reaction has started.

 

  • Copper

2Cu(s) +   S(s)                  Cu2S

(Red-brown)  (Yellow)                    (Black copper (I) sulphide))

 

(b). Non-metals

 

  • Carbon

C(s)  +  2S(s)                     CS2(s)

(Black) (Yellow)                                (Black Carbon disulphide)

 

 

Note.

– Carbon (IV) sulphide has a distinct smell.

– It is an excellent solvent and is used as a pesticide due to its poisonous nature.

 

  • Hydrogen

H2(g) +  S(s)                     H2S(g)

 

  • Fluorine

S(s) + F2(g)                 SF2(g)

 

  • Chlorine

S(s) + Cl2(g)                SCl2(g)

 

  • Bromine

2S(s) +Br2(g)               S2Br2(g)

 

  • Phosphorous

10S(s) + 4P(s)                P4S10(s)

 

Note:

– Sulphur does not react with inert gases, nitrogen and iodine.

 

Uses of sulphur

  1. Industrial manufacture of sulphuric (VI) acid in the contact process.
  2. It is used as a fungicide for treatment of fungal skin diseases.
  3. It is used for vulcanization (hardening) of rubber
  4. Manufacture of calcium hydrogen sulphite (Ca(HSO3)2 used for bleaching in paper and textile industries.
  5. Manufacture of matches and fireworks.
  6. Manufacture of dyes e.g. sulphur blacks that gives paint smooth texture.
  7. Manufacture of sulphur ointments and drugs e.g. sulphur-guanidine for dysentery.
  8. Manufacture of hair oil.
  9. Small amounts of sulphur are added to concrete to prevent corrosion by acids.
  10. Manufacture of fungicides for spraying crops against fungal infections e.g. ridomil, dithane for potato and tomato blights

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Allotropes of sulphur

– Allotropy is the existence of an element in more than one form without change of state.

– Sulphur has 2 allotropes

  • Rhombic sulphur/ octahedral/ alpha-sulphur
  • Monoclinic/ prismatic sulphur/ beta-sulphur.

 

-Unlike carbon only the rhombic sulphur occurs naturally.

 

Comparison of rhombic and monoclinic sulphur.

 

                     Allotrope 

Characteristic

Rhombic sulphur Monoclinic sulphur
Stability – Stable below transitional temp. of 96oC – Stable above 96oC
Colour – Bright yellow crystalline solid – Pale yellow crystalline solid
Melting point – Melts at 113oC; – Melts at 119oC;
Density – About 2.06gcm-3(heavier than monoclinic Sulphur) – Lighter than 1.98gcm-3 (lighter than rhombic sulphur)
Shape – Octahedral shape

Diagram:

 

– Needle-like/ prismatic

Diagram:

 

Note.

96oC is called transitional temperature; because both allotropes are stable.

 

Compounds of sulphur

 

Oxides of sulphur.

 

Sulphur (IV) oxide

Laboratory preparation of sulphur (IV) oxide

(i). Apparatus:

Dry sulphur (IV) oxide gas
Sodium sulphite
Dilute HCl

 

Conc. H2SO4(l)

 

 

(ii). Procedure

– Dilute HCl or H2SO4 is poured into sodium sulphite crystals in the flask.

– The gas produced is passed through conc. Sulphuric acid to dry it.

– If the reaction is slow, the round-bottomed flask is heated (warmed) gently.

– Dry gas is collected by downward delivery as it is denser than air.

 

(ii). Equation.

Na2SO3(aq) + 2HCl(aq)                        H2O(l) + SO2(g) + 2NaCl(aq)

 

Ionically;

2H+(aq) + SO32-(aq)                         H2O(l) + SO2(g)

 

Note:

– Nitric (V) acid should not be used.

Reason:

– It is a strong oxidizing agent and cannot therefore reduce the metal sulphites.

– Instead it will oxidize the SO2 produced to sulphuric (VI) acid

 

Equation:

2HNO3(aq) + SO2(g)                      2NO2(g) +  H2SO4(l)

 

Other methods of preparing sulphur (IV) oxide.

(b). Preparation from concentrated sulphuric (VI) acid

(i). Apparatus

– As in (a) above

 

(ii). Procedure

– Copper turnings are covered with concentrated sulphuric (VI) acid and the mixture heated (a must in this case).

Note:

– Dilute sulphuric (VI) acid does not react with copper hence the need for concentrated acid.

– Cold concentrated sulphuric (VI) acid does not also react with copper hence warming.

 

(iii). Observation.

– When the solution becomes hot, there is evolution of sulphur (IV) oxide gas.

 

Equation.

Cu(s) +2H2SO4(l)                    CuSO4(aq) + 2H2O(l) + SO2(g)

 

Note:

– This reaction is in two stages.

  • Oxidation of Cu to CuO

– Concentrated sulphuric (VI) acid oxidizes copper to Copper (II) oxide

 

Equation:

Cu(s) + H2SO4(l)                     CuO(s) + H2O(l) + SO2(g)

 

  • CuO further reacts with the acid to form salt and water.

 

Equation:

CuO(s) + H2SO4(l)                  CuSO4(aq)  + H2O(l)

 

Overall equation:

Cu(s) + H2SO4(l)                      CuSO4(aq) + 2H2O(l) + SO2(g)

 

(c). Roasting sulphur in air

– When sulphur is burnt in air, SO2 is produced.

 

Equation:

S(s) + O2(g)                     SO2(g)

 

Note:

This reaction is not suitable for preparing a pure sample of the gas in the lab.

Reason

– The gas is contaminated with traces of O2; N2; CO2 and inert gases.

– There are higher chances of environmental pollution, due to escape of some of the gas into the atmosphere.

 

(d). Roasting metal sulphides in air

Examples:

2FeS(g) + 3O2(g)                    2FeO(s) + 2SO2(g)

2ZnS(g) + 3O2(g)                    2ZnO(s) + 2SO2(g)

 

Preparation of sulphur (IV) oxide solution.

(i). Apparatus

                       

(ii). Procedure

– Gas is directly passed into water using an inverted funnel; to prevent “sucking back” by increasing surface area for dissolution.

 

Properties of sulphur (IV) oxide gas

Physical properties

  1. It is a colourless gas with an irritating (pungent) characteristic smell.
  2. It neither burns nor supports combustion i.e. when a lighted splint is introduced into a gas jar full of sulphur (IV) oxide, the splint is extinguished.
  3. It has a low PH.

 

 

 

 

 

 

 

 

Chemical properties.

– It is a strong reducing agent.

– An aqueous solution of sulphur (IV) oxide, sulphurous acid is strong reducing agent.

– The sulphite radical, SO32-, acts as a supplier of electrons; the overall reaction results into formation of sulphate ions.

 

Equations:

H2SO3(aq)                       2H+(aq) + SO32-(aq) then;

 

SO32-(aq) + H2O(l)                SO42-(aq) + 2H+(aq) + 2e-

 

– The resultant electrons supplied are accepted by an oxidizing agent, which consequently gets reduced.

 

Examples:

(i). Reduction of acidified potassium manganate (VII).

Procedure.

-To about 2 cm3 of sulphur (IV) oxide solution, 2 cm3 of dilute H2SO4 was added followed by an equal volume of potassium manganate (VII) solution.

 

Observations

– Purple solution changes to colourless.

 

Explanation

– Purple manganate (VII) ions are reduced to colourless manganate (II) ions, while H2SO3 (sulphurous (IV) acid) is reduced to sulphate ions and water.

 

Equation:

 

5SO2(g) + 2KMnO4(aq) + 2H2O                       K2SO4(aq) + 2MnSO4(aq)+ H2SO4(aq)

 

 

Ionically;

2MnO4(aq) + 5SO32-(aq) + 6H+(aq)                      2Mn2+(aq) + 5SO42-(aq) + 3H2O(l)

 

(ii). Reduction of potassium chromate (IV) solution

 

Procedure

– To 2 cm3 of Sulphur (IV) oxide solution, 2 cm3 of dilute H2SO4 was added followed by an equivalent volume of potassium chromate (VI) solution.

 

Observation

– Acidified potassium chromate (VI) solution change from orange to green.

 

Equation

K2Cr2O7(aq) + 3SO2(aq) + H2SO4(aq)                    K2SO4(aq) + H2O(l) + Cr2(SO4)3(aq)

(Orange)                                                                                                                                                                    (Green)

 

Ionically:                                Oxidation

 

 

Cr2O72-(aq) + 3SO32-(aq) + 8H+(aq)                      2Cr3+(aq) + 3SO42-(aq)

 

 

Reduction

Note:  this is the usual chemical test for sulphur (IV) oxide.

 

(iii). Reduction of Iron (III) ions to Iron (II) ions (Fe3+ to Fe2+)

 

Procedure

– About 3 cm3 of Iron (III) chloride solution are heated in a test tube and sulphur (IV) oxide gas bubbled into it.

 

Observations

– The brown solution turns green.

 

Explanation

– Aqueous sulphur (IV) oxide reduces to Fe3+ in FeCl3 which are brown to green Fe2+ in FeCl2(aq).

 

Ionically

2Fe3+(aq) + SO32-(aq) + H2O(l)                               Fe2+(aq) + SO42-(aq) + H+(aq)

 

 

(iv). Reduction of bromine water

 

Procedure

– Bromine water (red brown) is added to a solution of sulphur (IV) oxide followed by HCl and BaCl2 solution.

 

Equation

Br2(aq) + SO2(g) + 2H2O(l)                   2HBr(aq) + H2SO4(aq)

 

Ionically:                                         Oxidation

 

 

Br2(aq) + H2O(l) + SO32-(aq)                    2HBr(aq) + SO42-(aq)

(Red-brown)                                                                             (Colourless)

 

 

Reduction

On addition of barium chloride

– A white precipitate is formed, due to the formation of insoluble barium sulphate.

 

Equation:

Ba2+(aq) + SO42-(aq)                   BaSO4(s)

Note

– This test confirms presence of SO42- since a white precipitate insoluble in dilute hydrochloric acid is formed.

– CO32-(aq) and SO32- also forms a white precipitate with BaCl2(aq) but the white precipitates dissolve in dilute HCl(aq)

 

 

 

 

 

 

(v). Reduction of hydrogen peroxide

 

Procedure

– To 2 cm3 of aqueous sulphur (IV) oxide, an equal volume of hydrogen peroxide is added followed by 1 cm3 of HCl, then a few drops BaCl2 solution.

 

Observation and explanations:

– Bubbles of a colourless gas; that relights a glowing splint.

– Hydrogen peroxide is reduced to water; while the sulphite ion in aqueous sulphur (IV) oxide (H2SO3(aq)) is oxidized to SO42-(aq)

 

Equation

H2O2(l) +SO32-(aq)                         H2O(l) + SO42-(aq)

 

– On addition of BaCl2, a white precipitate insoluble in dilute HCl.

– This confirms presence of sulphate ions.

 

Equation:

Ba2+(aq) + SO42-(aq)                   BaSO4(s)

 

(vi). Reduction of concentrated nitric (V) acid

 

Procedure

– Sulphur (IV) oxide is bubbled through (into) a solution of concentrated nitric (v) acid.

 

Observation

– Brown fumes (of NO2) are liberated.

 

Explanation

– Sulphur (IV) oxide reduces nitric (V) acid to nitrogen (IV) oxide (brown) while it is itself oxidized by HNO3 to form H2SO4.

– Thus while SO2 is the reducing agent; HNO3 is the oxidizing agent.

 

Equation:

2HNO3(l) + SO2(g)                               2NO2(g) + H2SO4(aq)

                                                                                               (Brown fumes)

 

(vii). Reaction with atmospheric oxygen in light.

 

Procedure:

– About 2 cm3 of Sulphur (IV) oxide solution is left in a test tube in light for 24 hours, dilute HCl is then added, followed by barium chloride.

 

Observations and explanations:

– Atmospheric oxygen in light oxidizes sulphite ion (SO32-) into sulphate (SO42-)

 

Equation:

2SO32-(aq) + O2(g)                          2SO42-(aq)

 

– On adding barium chloride, a white precipitate insoluble in dilute HCl results; confirming presence of sulphate ion.

Equation:

Ba2+(aq) + SO42-(aq)                   BaSO4(s)

                                                (White ppt)

 

  1. Sulphur (IV) oxide as oxidizing agent

– It reacts as an oxidizing agent with reducing agents more powerful than itself.

 

Examples

 

(a). Reaction with hydrogen sulphide

 

Procedure

– A test tube of dry hydrogen sulphide gas is inverted into a gas jar full of moist sulphur (IV) oxide, and the gases allowed to mix.

 

Observation

Yellow deposits of sulphur is produced.

 

Examples:     

                         Oxidation

 

 

2H2S(g) + SO2(g)                       2H2O(l) + 3S(s)

 

 

                  Reduction

Explanations:

– H2S is a stronger reducing agent than sulphur (IV) oxide.

– Thus sulphur (IV) oxide acts as an oxidizing agent supplying oxygen to the hydrogen sulphide.

 

Note

– Dry gases do not react and for this reaction to occur, the gases must be moist or at least one of them.

 

(b). Reaction with burning magnesium

 

Procedure

– Burning magnesium is lowered into a gas jar full of sulphur (IV) oxide.

 

Observation

White fumes of magnesium oxide and yellow specks of sulphur.

 

Equation

 

2Mg(s) + SO2(g)                        2MgO(s) + S(s)

 

  1. Sulphur (IV) oxide as bleaching agent.

 

Procedure

– Coloured flower petals are placed in a test-tube full of sulphur (IV) oxide.

 

Observation

– The coloured (blue or red) petals are bleached (turned colorless);

 

Explanations:

– In presence of water, sulphur (IV) oxide acts as a bleaching agent.  It bleaches by reduction (removal of oxygen form the dye)

– It first combines with water forming the sulphurous acid; which then reduces the dye to form a colourless product.

 

Equations:

SO2(g) + H2O(l)                  H2SO3(aq)

 

H2SO3(aq)                               2H+(aq) + SO32-(aq)

 

Then;

SO32-(aq) + [O]             SO42-(aq)

               From dye

 

General equation

SO2(g) + H2O(l) + [Dye + (O)]                      Dye + H2SO4(aq)

                                                    Coloured                                       Colourless

Note

– The original colour may be restored by oxidation or prolonged exposure to air.  This explains why old newspapers which were originally bleached white by sulphur (IV) oxide turn brown with time.

– Chlorine bleaches by oxidation hence its oxidation is permanent; SO2 is however preferred because it is milder in action.

 

  1. Reaction with sodium hydroxide (alkalis)

 

Procedure

– A gas jar full of sulphur (IV) oxide is inverted over sodium hydroxide solution in a trough and shaken.

Observations

– Solution seen rises up in the jar.

 

Explanation

– Sulphur (IV) oxide is acidic, hence easily absorbed by alkaline solutions such as sodium hydroxide solution.

– Sodium sulphite and sodium hydrogen sulphites are formed depending on amount of sulphur oxide.

 

Equations

  • With limited sulphur (IV) oxide:

 

2NaOH(aq) +  SO2(g)                             Na2SO3(aq) + H2O(l)

 

  • With excess sulphur (IV) oxide:

 

NaOH(aq) + SO2(g)                                NaHSO3(aq)

 

Reaction with chlorine:

– Sulphur (IV) oxide reacts with moist chlorine to form an acidic mixture of sulphuric (VI) acid and hydrochloric acid.

 

Equation:

SO2(g) + SO2(g) H2O(l)                             H2SO4(aq) + 2HCl(aq)

Explanation:

– Sulphur (IV) oxide serves as the reducing agent reducing chlorine into hydrochloric acid;

– Chlorine acts as the oxidizing agent; oxidizing the sulphur (IV) oxide into sulphuric (VI) acid

 

Tests for sulphur (iv) oxide

  1. Characteristic pungent smell.
  2. Bleaches flower petals.
  3. Decolourises purple potassium manganate (VII)
  4. Turns filter paper soaked in acidified orange potassium dichromate (VI) solution to green

 

Sulphur (IV) oxide as a pollutant

– It is industrial waste in some chemical processes.

– The emission to the air it dissolves forming sulphurous acid.

 

Equation:

SO2(g) + H2O(l)                          H2SO3(aq)

 

– Sulphurous acid is readily oxidized to sulphuric (VI) acid; which attacks stonework and metal structures causing them to corrode.

– If breathed in, SO2 causes lung damage.

 

Uses of sulphur (VI) oxide

– Industrial manufacture of sulphuric (VI) acid.

– Fumigation in green houses for purposes of pest and disease control.

– Preservative in jam and fruit juices.

– Bleaching agent for wool, straw, paper pulp etc.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Sulphuric (VI) acid

 

Industrial manufacture of sulphuric (VI) acid: The contact process

 

Raw materials

– Sulphide ores or sulphur.

– Water

– Oxygen (air)

– Concentrated sulphuric (VI) acid.

 

The chemical process

 

Step 1: Production of sulphur (VI) oxide

– Sulphur (IV) oxide is obtained b burning the metal ores of sulphides or elemental sulphur in air.

 

Equation:

 

S(s) + O2(g)                    SO2(g)

 

– Obtaining sulphur (IV) oxide form pyrites is cheaper than form sulphur.

– Flowers of sulphur form pyrites is impure and contains dust; which involves extra expenses and time in purification.

 

 

Step 2:            Purification and drying

– The Sulphur (IV) oxide and excess air are passed through a series of driers and purifiers.

– Purifiers remove dust particles, which would otherwise poison the catalyst used in this process by taking up the catalytic surface thus impairing the catalytic efficiency.

– Purification (removal of dust) is by electrostatic precipitation.

– Are dried through concentrated sulphuric acid then passed through heat exchanger.

 

Step 3:  Heat exchanger reactions

– The pure dry SO2 and excess air mixture are passed into heat exchanger reactions.

 

Reason:

– To lower their temperatures since reaction in the proceeding chamber (catalytic chamber) are exothermic hence requiring lower temperatures.

 

Step 4:            Catalytic chamber

– Dry dust-free SO2 is mixed with clean excess air, heated and passed into a catalytic chamber containing vanadium (V) oxide catalyst.

 

Equation                 V2O5

2SO2(g) + O2(g)                          2SO3(g) + Heat

450oC

 

– The product is sulphur (VI) oxide, SO3.

– Formation of sulphur (VI) oxide is accompanied by evolution of heat (exothermic reaction) and a reduction in volume.

 

Note:

– A good yield of SO3 is favoured by the following conditions.

 

  1. Temperature

– The forward reaction is exothermic hence the yield can be favourable in low temperatures.

– However, at such low temperatures the equilibrium is attained very slowly.

– At high temperatures, equilibrium is achieved very quickly but sulphur (VI) oxide decomposes considerably.

– Thus a compromise optimum temperature of about 450oC is used in order to enable as much sulphur (VI) oxide as possible to be made in a reasonable time.

– From the graph, high SO3 yield is favoured by relatively low temperatures.

 

Graph: %age yield of sulphur (VI) oxide against temperature.

 

 

 

 

 

 

 

 

 

 

 

  1. Pressure

– High pressures favour production of more sulphur (VI) oxide.

 

Reason

– The volume of gaseous reactants is higher than volume of gaseous products.

– Since reaction involves reduction in volume, theoretically pressure used should be as high as is economically convenient.

 

Note:

– High pressures are however disadvantageous.

 

Reason

– The equipment required to generate high pressure would be expensive to maintain.

– The high pressure could also liquefy sulphur (VI) oxide.

– A pressure slightly above atmospheric pressure is used providing 98% conversion at low maintenance costs.

 

  1. Catalyst

– A catalyst neither takes part in a reaction nor increases the yield.

– It merely speeds up the reaction i.e. reduces the time taken to react at equilibrium of 450oC.

– Main catalyst is vanadium (V) oxide (V2O5).

– It is spread out (in trays) on silica gel to increase the surface area for combination of reactants.

– Dust settled in the catalyst may reduce its effective area.

– Dust may also react with the catalyst, “poison” it and further reduce its efficiency.

– This explains need to purify gases thoroughly.

– An effective catalyst is platinised asbestos.

– However, vanadium (V) oxide is preferred.

 

Reasons:

– It is not easily poisoned by dust particles.

– It is cheaper and readily available.

 

Note:

– The highest yield of sulphur (VI) oxide is obtained at optimum conditions of 4500C and pressure 2-3 atmospheres in presence of vanadium (V) oxide or platinised asbestos.

 

Step 5:            Heat exchanger reactions

– Hot SO3 gas from catalytic chamber is again passed through heat exchanger for cooling after which the cooled gas is taken into an absorption chamber.

 

Step 6: Absorption chamber

– The SO3 is not dissolved (passed) into water directly.

 

Reason

– It dissolves in water exothermically with a loud, hissing sound giving off corrosive vapour resulting into harmful sulphuric acid “sprays” or mist all around.

 

– The SO3 is dissolved in conc. H2SO4 forming oleum (pyrosulphuric acid/ fuming sulphuric acid).

 

Equation:

 

SO3(g) + H2SO4(l)                      H2S2O7(l)

 

– Resultant “Oleum” is then channeled into a dilution chamber.

 

Step 7:  Dilution chamber.

– Oleum is diluted with correct amounts of water to form concentrated sulphuric acid.

 

Equation:

 

H2S2O7(l) + H2O(l)                          2H2SO4(aq)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Summary: flow diagram for the contact process:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Pollution control in contact process.

– Main source of pollution is sulphur (IV) oxide.

– In catalyst chamber, SO2 reacts with oxygen forming SO3.

Equation:                V2O5

2SO2(g) + O2(g)                          2SO3(g) + Heat

450oC

– This is a reversible reaction and upto 98% conversion is possible and excess (unreacted) SO2 warmed and released into atmosphere via long chimneys.

– However, SO2 being a pollutant, little or none should be released into atmosphere.

– This is done by scrubbing the gas.

– This involves neutralizing the chimney gas by a solution of Calcium hydroxide forming a salt (calcium sulphite) and water.

 

Equation:

Ca(OH)2(aq) + SO2(g)                   CaSO3(aq) + H2O(l)

 

Note:

– In certain cases, filters are also installed to remove any traces of acid spray or mist form the exhaust gases.

– The unreacted gases (SO2 and SO3) may also be recycled within the process.

Properties of concentrated sulphuric (VI) acid

Physical properties

  1. – Colourless, odourless, oily liquid.
  2. – Very dense; with density 1.84 gcm-3.
  3. – Soluble in water and gives out considerable heat when a solution is formed.
  4. – It is hygroscopic – absorbs atmospheric moisture to become wet.

 

Experiment: To show hygroscopic nature of conc. H2SO4.

(i). Procedure

– A small beaker half full of conc. H2SO4 is weighed.

– Level of acid in beaker is marked to the outside using gummed paper.

– Acid is left exposed to air for a week or so then weighed again and level also noted.

 

(ii). Observations

– There is an increase in weight of acid.

– Level of acid in beaker is now above the paper mark.

 

(iii). Explanations

– The increase in weight and size is due to water absorbed form the air by the conc. sulphuric (VI) acid.

 

Note:

– This explains why sulphuric (VI) acid is used as a drying agent.

 

Chemical properties

 

  1. – It is a dehydrating agent.

Examples:

 

(a). Action on blue hydrated copper (II) sulphate (CuSO4.5H2O) crystals.

 

(i). Procedure

– A few crystals of hydrated CuSO4.5H2O were put in a test tube and enough concentrated sulphuric (VI) acid added, to cover them completely.

 

(ii). Observation:

– Blue copper (II) sulphate pentahydrate crystals turn to white powder of anhydrous CuSO4.

 

Equation

 Conc. H2SO4

 

 

CuSO4.5H2O(s)               CuSO4(s) + 5H2O(l)

(Blue crystals)                                                      (White crystals)

 

Explanations:

– Conc.H2SO4 has a very strong affinity for water and hence removes water of crystallization from crystals hence dehydrating them.

 

 

 

 

 

 

(b). Action on white sugar (C12H22O11)

 

(i). Procedure:

– A tablespoonful of sugar is put in an evaporating dish form a beaker and adequate volume of conc. H2SO4 is added.

 

(ii). Observations:

– Sugar turns form brown then yellow and finally to a charred black mass of carbon.

– A spongy black mass of charcoal (carbon) rises almost filling the dish.

– Steam is also give off and dish becomes very hot since reaction is exothermic.

 

Equation

 Conc. H2SO4

 

 

C12H22O11(s)                       12C(s) + 11H2O(l)

(White crystals)                                       (Black solid)

 

Explanation

– The acid removed from the sugar elements of water (hydrogen and oxygen, ratio 2:1) to form water, leaving behind a black charred mass of carbon.

 

(c). Action on oxalic acid (ethanedioic acid (H2C2O4)

– Conc. H2SO dehydrates oxalic acid on heating to a mixture of carbon (II) oxide and carbon (IV) oxide.

 

 Conc. H2SO4

Equation

 

H2C2O4(s)                         CO(g) + CO2(g) + H2O(l)

 

Note: 

– Conc. H2SO4 acid gives severe skin burns because it removes water and elements of water from skin tissue.

– Should the acid spill on skin, it is washed immediately with plenty of water followed with a solution of sodium hydrogen carbonate.

– Holes appear where the acid spills on clothes for same reason.

 

(d). Action on alcohols (alkanols)

– Conc. sulphuric (VI) acid dehydrates alcohols to corresponding alkenes.

 

Example: dehydration of ethanol to ethene

Equation:

 Conc. H2SO4

 

 

CH3CH2OH(s)                        C2H4(g) + H2O(l)

(Ethanol)                                                        (Ethene)

 

(e). Action on methanoic acid.

– Conc. sulphuric (VI) acid dehydrates methanoic acid to form CO.

 Conc. H2SO4

Equation:

 

HCOOH(s)                       CO(g) + H2O(l)

 

 

  1. Further reactions of conc. H2SO4 as an oxidizing agent.

– Hot concentrated Sulphuric acts as an oxidizing agent in which cases it is reduced to sulphur (IV) oxide and water.

 

Examples:

 

(a). Reaction with metals.

  • Copper

Cu(s) + 2H2SO4(l)                      CuSO4(aq) + SO2(g) + 2H2O(l)

 

Note: the copper (II) sulphate formed is white since the conc. H2SO4 further dehydrates the hydrated CuSO4.

 

  • Zinc

Zn(s) + 2H2SO4(l)                      ZnSO4(aq) + SO2(g) + 2H2O(l)

                    (Hot acid)

 

Zn(s) +  H2SO4(l)                       ZnSO4(aq) + H2(g)

                    (Cold acid)

 

  • Lead

Pb(s) + 2H2SO4(l)                      PbSO4(aq) + SO2(g) + 2H2O(l)

                    (Hot; conc.)                                      (Insoluble)

 

Note: 

– Dilute sulphuric (VI) acid doesn’t  have any action on copper.

Reason:

– Copper is below hydrogen in reactivity series hence cannot displace it from the acid.

 

– This acid (H2SO4) has very little effects on lead, and usually the amount of SO2 liberated is very little.

Reason:

– Formation of an insoluble lead sulphate layer that forms a protective coating on the metal stopping further reaction.

 

(b). Reaction with non-metals.

– Concentrated sulphuric acid oxidizes non-metals such as sulphur and carbon to their respective oxides.

 

Equations:

Ø  With carbon

C(s) + 2H2SO4(l)                        CO2(g) + 2SO2(g) + 2H2O(l)

Ø  With sulphur

S(s) + 2H2SO4(l)                      3SO2(aq) +  2H2O(l)

 

  1. It is a less volatile acid; and displaces more volatile acids (refer to lab preparation of HNO3)

 

 

 

 

 

Reactions of dilute sulphuric acid

  1. Reaction with metals

– It reacts with metals above hydrogen in the reactivity series to produce a salt and hydrogen.

– With potassium and sodium, reaction is violent.

 

Equations:

  • With magnesium:

Mg(s) + H2SO4(aq)                     MgSO4(aq) + H2(g)

 

  • With zinc:

Zn(s) + H2SO4(aq)                      ZnSO4(aq) + H2(g)

 

Note:

– Copper is below hydrogen in reactivity series hence can’t displace hydrogen form dilute sulphuric (VI) acid.

 

  1. Reaction with carbonates and hydrogen carbonates

– Dilute H2SO4(aq) reacts with carbonates and hydrogen carbonates to produce a salt, carbon (IV) oxide and water.

 

Equations

  • With sodium carbonate:

Na2CO3(s) + H2SO4(aq)                      Na2SO4(aq) + CO2(g) + H2O(l)

 

  • With calcium hydrogen carbonate:

CaHCO3(s) + H2SO4(aq)                    CaSO4(aq) + CO2(g) + H2O(l)

 

Note:

– Reaction with lead carbonate however stops soon after the reaction.

 

Reason:

– Formation of an insoluble coating of the lead (II) sulphate on the lead (II) carbonate which prevents further contact between acid and carbonate.

– The same logic applies for calcium carbonate.

 

  1. Reaction with oxides and hydroxides

– Reacts to form salt and water.

– However, those metal oxides whose sulphates are insoluble react only for a while.

– Thus reaction between dilute sulphuric (VI) acid and lead (II) oxide stops almost immediately.

– This is due to formation of an insoluble layer of lead (II) sulphate which effectively prevents further contact between acid and oxide.

 

Equations:

  • With magnesium oxide:

MgO(s) + H2SO4(aq)                  MgSO4(aq) + H2O(g)

(White)                                                               (Colourless solution)

 

  • With copper (II) oxide:

CuO(s) + H2SO4(aq)                   CuSO4(aq) + H2O(g)

(Black)                                                               (Blue solution)

  • With sodium hydroxide:

NaOH(s) + H2SO4(aq)                Na2SO4(aq) + 2H2O(g)

(White)                                                               (Colourless solution)

 

  • With lead (II) oxide:

PbO(s) + H2SO4(aq)                   PbSO4(aq) + H2O(g)

(Red)                                                                   (White ppt; reaction stops immediately)

 

Uses of sulphuric (VI) acid

  1. Manufacture of fertilizers.
  2. Processing of metal ores.
  3. Manufacture of detergents.
  4. Manufacture of plastics.
  5. Manufacture of dyes and paints.
  6. Manufacture of lead and accumulators.
  7. Manufacture of polymers.
  8. Manufacture of petroleum (petroleum refinery).
  9. Drying agent in industrial processes.

 

 

Hydrogen sulphide gas

– It is a colourless gas with a characteristic “rotten egg” smell; and is usually given out by rotting cabbage and eggs.

 

Laboratory preparation

 

(i). Apparatus:

    Warm water

 

H2S(g)
Iron (II) sulphide
     Dil. HCl
                                           Anhydrous                       Dry H2S gas

Calcium chloride

    Iron (II) sulphide
Dil HCl
 

 

 

 

Or

(ii). Procedure:

– Dilute hydrochloric acid is poured into Iron (II) sulphide in a round-bottomed flask.

– Resultant gas is passed through U-tube with anhydrous calcium chloride to dry the gas.

– This can also be done with phosphorous (V) oxide.

 

Equation:

FeS(s) + 2HCl(aq)                      H2S(g) + FeCl2(aq)

 

Ionically:

S2-(aq) + H+(aq)                H2S(g)

 

(iii). Collection of gas

– When dry, the gas is collected by downward delivery because it is denser than air.

– When wet is collected over warm water because it is more soluble in cold water.

 

  • Hydrogen sulphide test.

– When a strip of filter paper soaked in aqueous lead (II) ethanoate is put in hydrogen sulphide, the paper turns black or dark brown.

Reason:

– Due to the formation of lead (II) sulphide which is black.

 

Equation

H2S(g) + (CH2COOH)2Pb(aq)                     PbS(s) + 2CH3COOH(aq)

 

 

 

 

 

 

 

Properties of hydrogen sulphide gas

Physical properties

  1. Colourless and very poisonous gas (similar to hydrogen cyanide)
  2. Has a repulsive smell (similar to that of rotten eggs or decaying cabbages)
  3. Soluble in water giving a weak acid (only slightly ionized)

 

Equation:

H2S(g) + H2O(l)                          H2S(aq)

 

Then:

H2S(aq)                                     H+(aq) + HS(aq)                         2H+(aq) + S2-(aq)

 

– The acid is dibasic hence forms hydrogen sulphides.

 

Equation:

2NaOH(aq) + H2S(g)                         NaHS(aq) + 2H2O(l)

 

Note: 

Potassium hydroxide reacts similarly like sodium hydroxide.

 

Chemical properties

  1. Combustion

– Burns in a blue flame in a limited supply of oxygen (air) forming a yellow deposit of sulphur and steam.

 

Equation:

2H2S(g) + O2(g)                      2SO2(s) + 2H2O(g)

 

– In plentiful supply (excess) of Oxygen (air) it burns with a blue flame forming SO2 and steam.

 

Equation:

2H2S(g) +3O2(g)       2S(s) + 2H2O(g)

 

  1. It is a reducing agent

– It supplies electrons which are accepted by the oxidizing agent and forms sulphur.

 

Ionically:

H2S(aq) + 2H+(aq) + S2-(aq)

 

Then

S2-(aq)                   S(s) + 2e(aq)

 

H2S(aq) + [O]                     S(s) + H2O(l); in terms of addition of oxygen.

 

 

 

 

Examples

(i). With acidified K2Cr2O7 solution (potassium dichromate VI)

 

Equation:

Reduction:

 

 

Cr2O72-(aq) + 3H2S(g) + 8H+(aq)                          2Cr3+(aq) + 7H2O(l) + 3S(s)

(Orange)                                                                                                        (Green)

 

                                                                                                         Oxidation

 

Observation: The orange solution turns green and H2S oxidized to yellow sulphur.

 

(ii). Potassium manganate (VII) (KMnO4)

Equation:

Reduction:

 

 

2MnO4(aq) + 5H2S(g) + 6H+(aq)                         2Mn2+(aq) + 8H2O(l) + 5S(s)

(Purple)                                                                                                         (Colourless)

 

                                                                                                         Oxidation

Observation:

– The Purple solution turns colourless

– Manganate (VII) ions are reduced to manganate (II) ions; H2S oxidized to yellow sulphur.

 

(iii). Action on Iron (III) chloride ions

Equation:

FeCl3(aq) + H2S(g)                                       2FeCl2(aq) + 2HCl(aq) + S(s)

 

Ionically:

Reduction:

 

 

Fe3+(aq) + S2-(g)                                                     Fe2+(aq) + 3S(s)

(Brown)                                                                                                         (Pale green)

 

                                                         Oxidation

 

Observation:

– The brown solution turns pale green;

– The Fe3+(aq) are reduced to Fe2+(aq); while the S2-(aq) are oxidized to yellow sulphur.

 

(iv). Action with Conc. HNO3

Equation:

2HNO3(aq) + H2S(g)                                   2H2O(aq) + 2NO2(aq) + S(s) + Heat

 

Ionically:

Reduction:

 

 

2H+(aq) + 2NO3(aq) + H+(aq) + S2-(aq)                   2H2O(l) + 2NO2(g) + S(s) + Heat

(Colourless solution)                                                                                                              (Brown)          (Yellow)

 

                                                                                                         Oxidation

Observation:

– Evolution of brown fumes; and deposits of a yellow solid;

– HNO3(aq) is reduced to brown NO2(g); while S2-(aq) are oxidized to yellow sulphur;

Note: The solution also contains H2SO4 produced by the reaction:

Reduction

 

 

2HNO3(aq) + H2S(g)                                 H2SO4(aq) + 8NO2(aq) + 4H2O(l) ;

 

 

Oxidation

 

(v). Action of air on H2S

– The gas is dissolved in distilled water in a beaker and exposed to air; after a few days, a white disposal is formed.

 

Equation:

H2S(g) + O2(g)                2H2O(l) + 2S(s)

 

(vi). Action with concentrated sulphuric (VI) acid.

 

Equation

Reduction

 

 

H2SO4(aq) + 3H2S(g)                               4S(s) + 4H2O(l)

 

 

Oxidation

 

(vii). Action with halogen elements

  • Red-brown bromine water

– Red-brown bromine water is reduced forming colourless hydrogen bromide (Hydrobromic acid) and yellow deposits (suspension) of sulphur.

 

Equation:

Reduction

 

 

Br2(aq) + H2S(g)                         2HBr(aq) + S(s)

(Red-brown)                                                       (Colourless)    (Yellow suspension)

 

Oxidation

 

(viii). Action with hydrogen peroxide.

Equation:

Reduction

 

 

H2O2(aq) + H2S(g)                      2H2O(l) + S(s)

(Red-brown)                                                       (Colourless)    (Yellow suspension)

 

Oxidation

 

 

 

 

Preparation of metallic sulphides

– Hydrogen sulphide reacts with metal ions in solution to form precipitates of metal sulphides; majority of which are black in colour.

 

(i). Procedure

– The gas is bubbled through solutions of the following salts: Pb (NO3)2, CuSO4, FeSO4 etc.

 

(ii). Observations and equations

  • Lead ions:

Pb(NO3)2(aq) + H2S(aq)                         PbS(s) + 2HNO3(aq)

(Colourless)                                                                              (Black)

 

Ionically:

Pb2+(aq) + S2-(aq)                        PbS(s)

 

  • Copper (II) ions:

CuSO4(aq) + H2S(aq)                         CuS(s) + H2SO4(aq)

(Blue)                                                                          (Black)

 

Ionically:

Cu2+(aq) + S2-(aq)                       CuS(s)

 

  • Iron (II) ions:

FeSO4(aq) + H2S(aq)                                  FeS(s) + H2SO4(aq)

(Pal green)                                                                                (Black)

 

Ionically:

Fe2+(aq) + S2-(aq)                      FeS(s)

 

  • Zinc ions:

Zn(NO3)2(aq) + H2S(aq)                         ZnS(s) + 2HNO3(aq)

(Colourless)                                                                              (Black)

 

Ionically:

Zn2+(aq) + S2-(aq)                     ZnS(s)

 

Note:

– Most metal sulphides are insoluble in water except those of sodium, potassium and ammonium.

 

 

 

Sulphites

– Are compounds of the sulphite radical (SO32-) and a metallic or ammonium cation

 

Effects of heat

– They decompose on heating, forming SO2;

 

Example:

CuSO3(s)       Heat         CuO(s) + SO2(g)

 

Test for sulphites

 

(i). Procedure

– To 2cm3 of the test solution, ad 2 cm3 of BaCl2 or Ba (NO3)2; i.e. addition of barium ions.

– To the mixture add 2 cm3 of dilute HCl or HNO3.

 

(ii). Observation

– A white precipitate (BaSO3) is formed which dissolves on addition of acid.

– Production of a colourless gas that turns filter paper soaked in acidified orange potassium dichromate (VI) to green.

 

(iii). Explanations

– Only BaSO3; BaCO3 and BaSO4 form white precipitates;

– The precipitates of BaSO3 and BaCO3 dissolve on addition of dilute acids; unlike BaSO4;

– BaSO3 produces SO2(g) as it dissolves on addition of a dilute acid; SO2 turns orange acidified potassium dichromate (VI) to green;

– BaCO3 of the other hand dissolves in dilute acids producing CO2; which has no effect on K2Cr2O7; but forms a white precipitate in lime water;

 

Equations:

  • On addition of Ba2+:

Ba2+(aq) + SO32-(aq)                        BaSO3(s)

(White precipitate)

 

  • On addition of dilute HCl(aq):

BaSO3(s) + 2HCl(aq)                     BaCl2(aq) + SO2(g) + H2O(l)

(White precipitate)                                                                              (Colourless)

 

Ionically:

BaSO3(s) + 2H+(aq)                        Ba2+(aq) + SO2(g) + H2O(l)

 

 

Sulphates

– Are compounds of the sulphate radical (SO42-) and a metallic or ammonium cation.

 

Effects of heat.

– Decompose on heating and liberate SO2 and SO3 or SO3 alone;

– However quite a number of sulphates do not decompose on heating; and thus require very strong heating in order to decompose.

 

Examples:

2FeSO4(s)         Heat       Fe2O3(s) + SO2(g) + SO3(g)

(Pale green)                                      (Brown)            (Colourless gases)

 

CuSO4(s)          Heat       CuO(s) + SO3(g)

(Blue)                                                 (Black)         (Colourless)

Action of acids

Test for sulphates

– To about 2 cm3 of the test solution, 2 cm3 of BaCl2 or Ba (NO3)2 solution is added.

– To the mixture, 2 cm3 of dilute HCl or HNO3 is added.

 

Observation

– A white precipitate is formed when Ba (NO3)2 is added; which is insoluble in excess acid.

 

Explanations.

– Only BaSO3; BaCO3 and BaSO4 form white precipitates;

– The precipitates of BaSO3 and BaCO3 dissolve on addition of dilute acids; unlike BaSO4;

– Thus the white precipitate insoluble in dilute HCl or HNO3 could only be a sulphate; in this case barium sulphate.

 

Equations:

  • On addition of Ba2+:

 

Ba2+(aq) + SO42-(aq)                        BaSO4(s)

                                                                                (white precipitate)

 

  • On addition of dilute acid:

BaSO4(s) + 2HCl(aq)                     BaSO4(s) + 2HCl(aq); i.e. no effect;

(White precipitate)                                                                    (White precipitate)

 

 

 

Pollution by sulphur compounds.

– Main pollutants are sulphur (IV) Oxide and hydrogen sulphide.

 

(a). Sulphur (IV) oxide.

– SO2 is emitted when sulphur-containing fuels are burnt; during extraction of metals like copper and in manufacture of sulphuric (VI) acid.

– SO2 is oxidized to SO3;

– SO3 reacts with water in atmosphere to form sulphuric (VI) acid which comes down as acid rain or acid fog.

Acid rain (fog) has environmental effects:

  • Leaching of minerals in soil;
  • Erosion of stone work on buildings;
  • Corrosion of metallic structures;
  • Irritation of respiratory systems thus worsening respiratory illnesses;
  • Death of plants as a result of defoliation (falling of leaves);
  • Destruction of aquatic life in acidified lakes;
  • Stunted plant growth due to chlorosis;

 

(b). H2S is very poisonous.

 

 

 

 

UNIT 5: CHLORINE AND ITS COMPOUNDS.

Unit Checklist:

  1. About chlorine.
  2. Preparation of chlorine.
  3. Properties of chlorine.
  • Colour and smell
  • Solubility in water
  • Action on litmus paper
  • Bleaching action
  • Action on hot metals
  • Reaction with non-metals
  • Oxidation reactions
  • Reaction with alkalis
  • Effect of sunlight on chlorine water.
  1. Industrial manufacture of chlorine (The mercury cathode cell)
  2. Uses of chlorine and its compounds
  3. Hydrogen chloride gas
  • Preparation
  • Properties
  1. Test for chlorides.
  2. Hydrochloric acid
  • Large scale manufacture
  • Uses of hydrochloric acid
  1. Environmental pollution of chlorine and its compounds

 

Introduction:

– Chlorine is a molecular non-metallic element made up of diatomic molecules.

– Its electron arrangement is 2.8.7 and it belongs to the halogen family.

 

Preparation of chlorine.

Note: It is usually prepared by oxidation of concentrated hydrochloric acid by removal of hydrogen.

 

Equation:

2HCl(aq) + [O]                Cl2(g) + H2O(l)

– The [O] is from a substance containing oxygen.

 

(a). Preparation of chlorine from MnO2 and HCl.

(i). Apparatus:

 

 

 

 

 

 

 

 

 

 

 

(ii). Conditions:

– Heating;

– Presence of an oxidizing agent; in this case it is manganese (IV) oxide.

 

(iii). Procedure:

– Hydrochloric acid is reacted with manganese (IV) oxide (dropwise);

Equation:

MnO2(s) + 4HCl(aq)     Heat        MnCl2(aq) + 2H2O(l) + Cl2(g)

 

(iv). Explanation:

– Manganese (IV) oxide oxidizes hydrochloric acid by removing hydrogen resulting into chlorine.

– The manganese (IV) oxide is reduced to water and manganese chloride.

– The resultant chlorine gas is passed through a bottle containing water.

Reason:

– To remove hydrogen chloride fumes (gas) which is very soluble in water.

– Next it is passed through concentrated sulphuric acid or anhydrous calcium chloride; to dry the gas.

 

(v). Collection:

(a). Wet chlorine is collected over brine (saturated sodium chloride solution) or hot water.

Reason:

– It does not dissolve in brine and is less soluble in water

 

(b). Dry chlorine is collected by downward delivery (upward displacement of air)

Reason:

– It is denser than air (2.5 times).

Note:

– Chlorine may also be dried by adding calcium chloride to the jar of chlorine.

 

(c). The first bottle must contain water and the second concentrated sulphuric acid.

Reason:

– If the gas is first passed through concentrated sulphuric acid in the first bottle then to the water; it will be made wet again.

 

Properties of chlorine gas.

  1. Colour and smell.

Caution: Chlorine is very poisonous.

– It is a green-yellow gas with an irritating pungent smell that attacks the nose and the lungs.

– It is 2.5 times denser than air, hence can be collected by downward delivery.

 

  1. Solubility in water.

– It is fairly soluble in water forming green-yellow chlorine water.

 

Equation:

Cl2(g) + H2O(l)                           HCl(aq) + HOCl(aq)

 

– Chlorine water is composed of two acids; chloric (I) acid (hypochlorous acid) and hydrochloric acid.

 

  1. Action on litmus paper.

– Moist chlorine turns litmus paper red then bleaches it.

– Dry chlorine turns damp blue litmus paper red then bleaches it.

– Moist chlorine bleaches red litmus paper; dry chlorine bleaches damp red litmus paper.

– Dry chlorine has no effect on dry litmus paper.

Reasons:

(i). In presence of moisture chlorine forms chlorine water which is acidic and hence turns blue litmus paper red.

(ii). Hypochlorous acid in the chlorine water is an oxidizing agent; thus adds oxygen (oxidizes) to the colour of most dyes; hence bleaching it.

 

Equations:

Cl2(g) + H2O(l)                           HCl(aq) + HOCl(aq)

 

 

Acidic solution

Then:

Dye + HOCl(aq)                      HCl(aq) + {Dye + [O]}

Coloured                                                                                              Colourless

 

  1. Bleaching action.

– Moist chlorine bleaches dyes but not printers ink which is made of carbon.

– The colour change is due to oxidation by hypochlorous acid.

 

Equations:

Cl2(g) + H2O(l)                           HCl(aq) + HOCl(aq)

 

 

Acidic solution

Then:

Dye + HOCl(aq)                      HCl(aq) + {Dye + [O]}

Coloured                                                                                            Colourless

  1. Action on a burning splint.

– The gas put out a glowing splint. It does not burn.

 

  1. Action on hot metals.

(a). Preparation of iron (III) chloride.

(i). Apparatus.

 

 

 

 

 

 

 

 

 

(ii). Precaution.

– Experiment should be done in a fume cupboard or in the open.

Reason:

– Chlorine gas is poisonous and will thus be harmful to the human body.

 

(iii). Procedure:

– Dry chlorine gas is passed over iron wool as per the diagram.

 

(iv). Conditions.

  • Chlorine gas has to be dry (done by the anhydrous calcium chloride in the U-tube)

Reason:

To prevent hydration hence oxidation of iron (which will then form Fe2O3.5H2O) hence preventing reaction between iron and chlorine.

 

  • Iron metal must be hot; and this is done by heating.

Reason:

To provide activation energy i.e. the minimum kinetic energy which the reactants must have to form products.

 

  • Anhydrous calcium chloride.

– In the U-tube; to dry the chlorine gas.

– In the thistle funnel; to prevent atmospheric water vapour (moisture) from getting into the apparatus and hence reacting with iron (III) chloride.

 

(v). Observations:

– Iron metal glows red-hot.

– Red brown fumes (FeCl3(g)) are formed in the combustion tube.

– A black solid (FeCl3(s)) is collected in the flask.

Note:

– Iron (III) chloride cannot be easily collected in the combustion tube.

Reason:

– It sublimes when heated and hence the hotter combustion tube causes it to sublime and its vapour is collected on the cooler parts of the flask.

 

(vi). Reaction equation.

2Fe(s) + 3Cl2(g)                  2FeCl3(g)

 

(vii). Conclusion.

– Iron (III) chloride sublimes on heating; the black solid changes to red-brown fumes on heating.

Equation:

FeCl3(s)                                  FeCl3(g)

(black)                                        (Red-brown)

 

 

(b). Aluminium chloride.

2Al(s) + 3Cl2(g)             2FeCl2(s)

2Al(s) + 3Cl2(g)             Al2Cl6(s)

 

Note:

– Aluminium chloride also sublimes on heating.

Equation:

AlCl3(s)                                  AlCl3(g)

(White)                                        (White)

 

(c). Reaction with burning magnesium.

(i). Procedure:

– Burning magnesium is lowered into a gar jar of chlorine gas.

 

(ii). Observations:

– The magnesium continues to burn with a bright blinding flame;

– Formation of white fumes (MgCl2); which cools into a white powder.

 

(iii). Equation:

Mg(s) + Cl2(g)                  MgCl2(s)

 

– Generally chlorine reacts with most metals when hot top form corresponding chlorides.

Note:

Where a metal forms two chlorides when it reacts with chlorine, the higher chloride is usually formed.

Reason:

The higher chloride is stable. This explains why reactions of chlorine with iron results into iron (III) chloride and not iron (II) chloride.

 

 

  1. Reaction with non-metals.

– It reacts with hot metals; forming covalent molecular compounds.

 

(a). Reaction with phosphorus.

(i). Procedure:

– A piece of warm phosphorus is lowered into a gas jar of chlorine.

 

(ii). Observations:

– Phosphorus begins to smoulder and then ignites spontaneously.

– Evolution of white fumes (PbCl3 and PCl5)

 

(iv). Explanation.

– Chlorine reacts with warm dry phosphorus to form white fumes of phosphorus (III) and (V) chlorides.

 

Equations:

P4(s) + 6Cl2(g)                  4PCl3(s)

(With limited chlorine)

P4(s) + 10Cl2(g)              4PCl5(s)

(With excess chlorine)

 

(b). Reaction with hydrogen.

(i). Conditions:

– Heating or presence of light; since chlorine and hydrogen do not react with each other at room temperature.

 

(ii). Precaution:

– The experiment is performed in a fume chamber (cupboard); since the reaction is explosive;

 

(iii). Procedure:

– Chlorine gas is mixed with hydrogen gas and the mixture heated or exposed to direct light; then aqueous ammonia brought near the mouth of the jar.

 

(iv). Observations:

White fumes at the mouth of the jar.

 

(v). Explanations:

– Chlorine reacts explosively with hydrogen to form hydrogen chloride gas.

Equation:

Cl2(g) + H2(g)    Heat/ Light     2HCl(g).

 

– The hydrogen chloride gas diffuses upwards and reacts with ammonia at the mouth of the test tube to form white fumes of ammonium chloride; NH4Cl.

Equation:

HCl(g) + NH3(g)       NH4Cl(g)

White fumes.

 

  1. Chlorine as an oxidizing agent.

– Chlorine is a strong oxidizing agent and oxidizes many ions, by readily accepting electrons.

– During the process, chlorine itself undergoes reduction.

 

(a). Reaction with hydrogen sulphide gas.

(i). Procedure:

– A gas jar full of chlorine gas is inverted into another containing hydrogen sulphide gas.

 

 

 

 

 

 

 

(ii). Apparatus:

 

 

 

 

 

 

 

 

 

(iii). Observations:

Yellow deposits (of sulphur)

Misty fumes (hydrogen chloride gas)

 

(iv). Explanations:

– Chlorine oxidizes hydrogen sulphide gas to sulphur solid, while itself is reduced to hydrogen chloride gas.

Equation:              Oxidation

 

 

Cl2(g) + H2S(g)               2HCl(g) + S(s)

 

 

Reduction

(v). Conditions:

– At least one of the gases must be moist; they do not react with each other in absence of moisture.

Note:

– In absence of moisture both gases are still in molecular form and hence cannot react; water facilitates their ionization hence ability to react.

 

– If aqueous hydrogen sulphide is used, then sulphur forms as a yellow suspension on the acidic solution.

Equations:

Stoichiometric:

Cl2(g) + H2S(aq)             2HCl(aq) + S(s)

 

Ionic:

Cl2(g) + S2-(g)                 2Cl(g) + S(s)

 

(b). Reaction with sodium sulphite.

Procedure:

– Chlorine gas is bubbled through sodium sulphate in a beaker.

– Resulting solution is then divided into two portions.

– To the first portion, drops of dilute nitric acid are added followed by few drops of barium nitrate solution.

– To the second portion, few drops of lead (II) nitrate are added and the mixture warmed then cooled.

 

(ii). Observations:

1st portion: White precipitate formed indicating presence of SO42-;

 

 

Explanations:

– The white precipitate indicate presence of SO42-; the precipitate is barium sulphate Ba(SO4)2;

– Chlorine oxidizes SO32- in Na2SO3 to SO42- while itself is reduced to chloride ions;

 

Equations:

H2O(l) + Cl2(g) + Na2SO3(aq)                  Na2SO4(aq) + 2HCl(aq)

 

Ionically:

Cl2(g) + SO32-(aq) + H2O(l)                        SO42-(aq) +  2H+(aq) + 2Cl(aq)

 

– On adding barium nitrate (Ba(NO3)2); the Ba2+ ions react with the SO42-  to form insoluble BaSO4; the white precipitate.

 

Ionically;

Ba2+(aq)   +  SO42-(aq)                              BaSO4(s)

(White precipitate)

Note:

– The solution is first acidified (with HNO3) before addition of Ba(NO3)2 to prevent precipitation of BaSO3(s) and BaCO3(s).

 

2nd portion:

Observation:

– Formation of a white precipitate on addition of Pb(NO3)2 solution.

– On warming the white precipitate dissolves then recrystalizes back on cooling.

 

Explanations:

– The white precipitate shows presence of either Cl; SO32- or  SO42-

– However the fact that it dissolves on warming confirms the presence of Cl(aq) and not SO32-(aq) and SO32-(aq)

 

Equation:

Pb2+(aq)   +  Cl(aq)                                     PbCl2(s)

(White precipitate soluble on warming)

 

(c). Reaction with ammonia.

(i). Procedure:

Chlorine gas is bubbled through aqueous ammonia.

 

(ii). Observations:

– Evolution of white fumes.

 

(iii). Explanation.

– Chlorine gas oxidizes ammonia to nitrogen, while is itself reduced to white fumes of ammonium chloride.

 

Equation:              Reduction

 

 

8NH3(g) + 3Cl2(g)                      6NH4Cl(g) + N2(s)

 

 

Oxidation

 

(d). Displacement reactions with other halogens.

(i). Procedure:

– Chlorine is bubbled through aqueous solutions of fluoride, bromide and iodide ions contained in separate test tubes.

 

(ii). Observations and explanations:

  • With fluoride ions.

– No observable change or no reaction; because chlorine is a weaker oxidizing agent than fluorine.

 

  • With bromide ions:

– If potassium bromide was used, the colourless solution turns red-brown.

Reason:

– Chlorine has a higher tendency to gain electrons than bromine.

– It readily oxidizes bromide ions (in KBr) to form potassium chloride and bromine which immediately dissolves to make the solution red-brown.

 

Equation:              Reduction

 

 

2KBr(aq) + Cl2(g)                       2KCl(aq) + Br2(l)

 

 

Oxidation                                         Red brown

Ionically;

2Br(aq) + Cl2(g)                              2Cl(aq) + Br2(l)

 

With iodide ions.

– Using potassium iodide the colourless solution would turn black.

Reason:

– Chlorine has a higher tendency to gain electrons that iodine.

– It readily oxidizes the I (in KI) to form iodine and potassium chloride.

– Iodine solid in the resulting solution makes it black.

 

Equation:          Reduction

 

 

2KI(aq) + Cl2(g)             2KCl(aq) + I2(l) (black)

 

 

Oxidation                                                

Ionically;

2I(aq) + Cl2(g)                    2l(aq) + Br2(l)

 

  1. Reaction with alkalis.

(a). Reaction with sodium hydroxide solution.

(i). Procedure:

– Bubble chlorine slowly through cold dilute sodium hydroxide solution.

– Dip litmus paper.

 

(ii). Observation:

– Litmus paper is bleached; the product has the colour and smell of chlorine.

 

 

(iii). Explanation:

– Chlorine dissolves in sodium hydroxide to form a pale yellow solution of sodium chlorate (I) or sodium hypochlorite (NaClO);

– The sodium chlorate (I) bleaches dyes by oxidation.

Equation:

Cl2(g)+ 2NaOH(l)                      NaCl(aq) + NaClO(aq) + H2O(l)

 

 

Pale yellow solution

Bleaching action of NaClO:

– The NaClO donates oxygen to the dye making it colourless; and thus it bleaches by oxidation.

Equation:

Dye + NaClO(aq)                     NaCl(aq) + {Dye + [O]}

Coloured                                                                                                Colourless

 

Note:

With hot concentrated sodium hydroxide, the chlorine forms sodium chlorate (III); NaClO3.

Equation:

3Cl2(g)+ 6NaOH(l)                          5NaCl(aq) + NaClO3(aq) + 3H2O(l)

 

(b). Reaction with potassium hydroxide

– Follows the trend of sodium.

 

(c). Reaction with slaked lime {Ca(OH)2(s)}

Equation:

Cl2(g)+ Ca(OH)2(l)                        CaOCl2(aq) + 3H2O(l)

Calcium chlorate I

 

Note:

Bleaching powder, CaOCl2 always smells of strongly of chlorine because it reacts with carbon (IV) oxide present in the atmosphere to form chlorine.

Equation:

CaOCl2(s) + CO2(g)                          CaCO3(s) + Cl2(g)

 

  1. Effects of chlorine gas on:

(a). A burning candle.

(i). Procedure:

– A burning candle is lowered into a gas jar of chlorine.

 

(ii). Observations:

– It burns with a small, red and sooty flame.

 

(iii). Explanations:

– Wax (in candles) consists of mainly hydrocarbons.

– The hydrogen of the hydrocarbon reacts with chlorine forming hydrogen chloride while leaving behind carbon.

 

(b). warm turpentine.

(i). Procedure:

– A little turpentine is warmed in a dish and a filter paper soaked (dipped) in it.

– The filter paper is then dropped into a gas jar of chlorine.

(ii). Observation:

– There is a red flash accompanied by a violent action whilst a black cloud of solid particles form.

 

(iii). Conclusion:

– Black cloud of slid is carbon.

– Turpentine (a hydrocarbon) consists of hydrogen and carbon combined together.

– The chlorine combines with hydrogen and leaves the black carbon behind.

 

Equation:

C10H16(l) + 8Cl2(g)                     16HCl(g) + 10C(s)

 

  1. Effects of sunlight on chlorine water.

(i). Procedure:

– Chlorine water is made by dissolving the gas in water.

– A long tube filled with chlorine water is inverted over a beaker containing water.

– It is then exposed to sunlight (bright light) as shown below.

 

(ii). Apparatus:

 

 

 

 

 

 

 

 

 

 

(iii). Observations:

– After sometime a gas collects in the tube and on applying a glowing splint, the splint is rekindles showing that the gas collected is oxygen.

 

(iv). Explanation:

– Chlorine water has two components.

Equation:

Cl2(g) + H2O(l)                           HCl(aq) + HOCl(aq)

 

– The HOCl being unstable will dissolve on exposure to sunlight, giving out oxygen.

Equation:

2HOCl(aq)                       2HCl(aq) + O2(g) (slow reaction)

 

Overall reaction:

2H2O(l) + 2Cl2(g)                       4HCl(aq) + O2(g)

 

 

 

 

 

 

Industrial manufacture of chlorine (the mercury cathode cell)

The electrolysis of brine

(i). Apparatus.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(ii). Electrolyte.

– Brine, concentrated sodium chloride solution, NaCl

 

(iii). Electrodes.

Anode: carbon (graphite)

Cathode: Flowing mercury;

 

(iv). Ions present:

NaCl(aq)                       Na+(aq) + Cl(aq)

 

 

H2O(l)                  H+(aq) + OH(aq)

 

(v). Reactions:

Anode:

– Cl and OH migrate to the anode.

– Because of high concentration of Cl(aq), they are discharged in preference to OH ions.

 

Equation:

2Cl(aq)                                     Cl2(g) + 2e

(Green-yellow)

 

Cathode:

– H+(aq) and Na+(aq) migrate to the cathode.

– Because the cathode is made of mercury, Na+(aq) is discharged in preference to H+(aq) ions;

 

Equation:

2Na+(aq) + 2e                         2Na(s)

 

Note:

– Sodium formed at the cathode dissolves in the flowing mercury cathode to form sodium amalgam (Na/Hg).

– Sodium amalgam is reacted with water to form sodium hydroxide and hydrogen.

– Mercury (in the sodium amalgam) remains unreacted.

 

Equation:

2Na/Hg(l) + 2H2O(l)                              2NaOH(aq) + H2(g) + 2Hg(l)

 

– The unreacted mercury is recycled.

 

(vi). Products:

Chlorine gas at the anode.

Hydrogen and sodium hydroxide at the cathode.

 

Uses of chlorine gas and its compounds.

  1. Manufacture of hydrochloric acid.
  2. Used in form of bleaching powder in textile and paper industries.
  3. For sterilization of water for both domestic and industrial use and in swimming pools.
  4. Used in sewage treatment e.g. NaOClO3 solution used in latrines.
  5. Manufacture of plastics (polyvinyl chloride; PVC)
  6. Manufacture of germicides, pesticides and fungicides e.g. DDT and some CFCs.
  7. CFCs are used to manufacture aerosol propellants.
  8. Manufacture of solvents such as trichloromethane and some chlorofluorocarbons (CFCs).
  9. CFCs are commonly freons are used as refrigerants in fridges and air condition units due to their low boiling points.
  10. Manufacture of chloroform, an aesthetic.

 

Hydrogen chloride gas.

Laboratory preparation of hydrogen chloride gas.

(i). Apparatus:

 

 

 

 

 

 

 

 

 

 

 

 

 

(ii). Procedure:

– Concentrated sulphuric acid is reacted with sodium chloride, and the mixture heated gently.

– Resultant gas is passed through conc. Sulphuric (VI) acid; to dry the gas.

 

(iii). Equation:

H2SO4(l) + NaCl(aq)                              NaHSO4(s) + HCl(g)

 

Ionically;

H+(aq) + Cl(aq)                          HCl(g)

Note:

– The reaction can proceed in the cold, but on large scale HCl(g) is produced by the same reaction but the heating is continued to re hot.

 

Properties of hydrogen chloride gas.

  1. Colourless gas with a strong irritating pungent smell.
  2. Slightly denser than air (1¼ times). This makes it possible to collect the gas by downward delivery.
  3. Very soluble in water; and fumes strongly in moist air forming hydrochloric acid deposits.

 

Diagram:

 

 

 

 

 

 

 

 

 

 

– The aqueous solution is known as hydrochloric acid.

– It is almost completely ionized (a strong acid) in aqueous solution.

Equation:

HCl(aq)                    H+(aq) + Cl(aq)

 

– This solution has the usual acidic properties:

Examples:

(i). turns blue litmus red.

(ii). Liberates hydrogen gas with certain metals e.g. zinc, Magnesium, iron etc.

Note:

Hydrochloric acid does not react with metals below hydrogen in the reactivity series.

Equations:

Zn(s) + 2HCl(aq)                        ZnCl2(aq) + H2(g)

Mg(s) + 2HCl(aq)                       MgCl2(aq) + H2(g)

Fe(s) + 2HCl(aq)                        FeCl2(aq) + H2(g)

 

(iii). Neutralizes bases to form salt and water.

Examples:

HCl(aq) + NaOH(aq)                          NaCl(aq) +H2O(l)

2HCl(aq) + CuO(s)                             CuCl2(aq) + H2O(l)

 

(iv). Liberates carbon (IV) oxide from carbonates and hydrogen carbonates.

Examples:

CaCO3(s) + 2HCl(aq)                    CaCl2(aq) + H2O(l) + CO2(g)

ZnCO3(s) + 2HCl(aq)                    ZnCl2(aq) + H2O(l) + CO2(g)

NaHCO3(s) + HCl(aq)                   NaCl(aq) + H2O(l) + CO2(g)

 

 

 

Note:

As the hydrogen chloride gas very soluble in water, the solution must be prepared using a funnel arrangement; to prevent sucking back and increase the surface area for the dissolution of the gas;

 

Diagram: dissolution of hydrogen chloride gas

 

 

 

 

 

 

 

 

 

 

 

 

  1. Dry hydrogen chloride is NOT particularly reactive at ordinary temperatures, although very reactive metals burn in it to form the chloride and hydrogen gas.

Equation:

2Na(s) + 2HCl(aq)                      2NaCl(s) + H2(g)

 

Metals above hydrogen in the reactivity series react with hydrogen chloride gas when heated.

Note:

If reacted with some metals it forms 2 chlorides e.g. iron where iron (II) and iron (III) chlorides exist.

 

  1. Hydrogen chloride gas forms white fumes of ammonium chloride when reacted with ammonia gas;

Equation:

NH3(g) + HCl(g)                         NH4Cl(s)

 

Note: This is the chemical test for hydrogen chloride gas.

 

  1. Hydrogen chloride is decomposed by oxidizing agents, giving off chlorine.

Examples:

PbO2(s) + 4HCl(g)                      PbCl2(s) + 2H2O(l) + Cl2(g)

MnO2(s) + 4HCl(g)                    MnCl2(s) + 2H2O(l) + Cl2(g)

 

Diagram: reacting hydrogen chloride with an oxidizing agent.

 

 

 

 

 

 

 

 

 

 

 

Test for chlorides.

Test 1: Using silver ions:

Procedure:

– To the test solution, add silver ions from silver nitrate.

– Acidify with dilute nitric acid.

 

 

(ii). Observations and inference:

– Formation of a white precipitate shows presence of Cl(aq)

 

(iii). Explanations:

– Only silver carbonate and silver chloride can be formed as white precipitates.

– Silver carbonate is soluble in dilute nitric acid but silver chloride is not.

 

Equations:

– Using Cl from NaCl as the test solution;

NaCl(aq) + AgNO3(aq)               NaNO3(aq) + AgCl(s)

White ppt.

 

Ionically;

Ag+(aq) + Cl(aq)                          Ag(s)

White ppt.

 

Note:

– This precipitate dissolves in excess ammonia.

– The white precipitate of silver chloride turns violet when exposed to light.

 

Test 2: Using lead ions

(i) Procedure:

– To the test solution, add lead ions from lead (II) nitrate, then warm

 

(ii). Observations and inference:

– Formation of a white precipitate that dissolves on warming shows presence of Cl(aq)

 

(iii). Explanations:

– Only lead carbonate, lead sulphate, lead sulphite and lead chloride can be formed as white precipitates.

– Only lead chloride dissolves on warming; unlike the rest which are insoluble even on warming.

 

Equations:

Using Cl from NaCl as the test solution;

2NaCl(aq) + Pb(NO3)2(aq)                     2NaNO3(aq) + PbCl2(s)

White ppt.

Ionically;

Pb2+(aq) + Cl(aq)                       PbCl2(s)

White ppt.

 

 

 

Hydrochloric acid.

Large scale manufacture of hydrochloric acid.

(i). Diagram:

 

 

 

 

 

 

 

 

 

 

 

 

 

(ii). Raw materials:

Hydrogen obtained as a byproduct of petroleum industry; electrolysis of brine or from water by Bosch process;

Chlorine obtained from the electrolysis of brine or as fused calcium chloride.

 

(iii). Procedure:

– A small sample of hydrogen gas is allowed through a jet and burnt in excess chlorine gas.

Equation:

H2(g) + Cl2(g)                 2HCl(g)

 

Precaution: A mixture of equal volumes of hydrogen and chlorine explodes when put in sunlight.

 

– The hydrogen chloride gas formed is dissolved in water over glass beads.

– The glass beads increase the surface area over which absorption takes place.

– Commercial hydrochloric acid is about 35% pure.

– Hydrochloric acid is transported in steel tanks lined inside with rubber.

– If the acid comes into contact with exposed parts of metal or with rust, it forms iron (III) chloride that makes the acid appear yellow.

 

Pollution in an industry manufacturing hydrochloric acid.

(i). Chlorine is poisonous.

(ii). Mixture of hydrogen and oxygen in air is explosive when ignited.

 

Uses of hydrochloric acid.

  1. Sewage treatment.
  2. Treatment of water (chlorination) at the waterworks.
  3. Removing rust from metal e.g. descaling iron before it is galvanized or and other metals before they are electroplated.
  4. Making dyes, drugs and photographic materials like silver chloride on photographic films.

 

 

 

 

Environmental pollution by chlorine and its compounds.

  1. Chlorine may dissolve in rain and fall as acid rain, which has adverse effects on plants and animals, buildings and soil nutrients.
  2. CFCs are non-biodegradable. Over time, they diffuse into the atmosphere breaking down to free chlorine and fluorine atoms. These atoms deplete the ozone layer. Chlorine is thus one of the greenhouse gases.
  3. PVCs are non-biodegradable.
  4. DDT is a pesticide containing chlorine and has a long life span, affecting plants and animal life.

Note: DDT is banned in Kenya; NEMA advises increased use of pyrethroids in mosquito control.

 

ORGANIC CHEMISTRY I

Contents checklist.

 

ORGANIC CHEMISTRY

Definition

– The chemistry of hydrogen carbon chain compounds.

– It the study of carbon compounds except the oxides of carbon i.e.  CO, CO2 and             Carbons.

 

ORGANIC CHEMISTRY I: THE HYDROCARBONS

 

Hydrocarbons

Are compounds of hydrogen and carbon only; and are the simplest organic compounds.

 

Main groups of hydrocarbons

Are classified on the basis of the type of bonds found within the carbon atoms.

  • Alkanes: Are hydrocarbons in which carbon atoms are linked by single covalent bonds.
  • Alkenes: Carbon atoms are held by at least one double bond.
  • Alkynes: Have at least one triple bond between any tow carbon atoms.

 

Saturated and unsaturated hydrocarbons

(a). Saturated hydrocarbons

– Are hydrocarbons which the carbon atoms are bonded to the maximum number of other             atoms possible.

– hydrocarbons which don’ react and hence cannot decolourise both Bromine water and acidified potassium manganate (VII).

– They are compounds in which each carbon atom has only single covalent bonds, throughout the structure.

 

(b). Unsaturated hydrocarbons

– Are hydrocarbons which contain at least one double or bond, between any two adjacent carbon atoms.

– The carbon atoms do not have maximum covalency.

– They can decolourise both bromine water and acidified potassium manganate (VII).

 

Examples: All alkenes and Alkynes.

 

Experiment:   To verify saturated and unsaturated hydrocarbons.

Procedure:
– 3 to 4 drops of bromine wate are added to about 1 cm3 of the liquid under investigation.

– The mixture is then shaken thoroughly and the observations recorded;

– For gases the gas under investigation is bubbled ito 1 cm3 of bromine water;

– The procedures are then repeated with acidified potassium manganate (VII);

 

Observations:

 

COMPOUND

OBSERVATIONS
With potassium permanganate With Bromine water
Kerosene No observable colour change No colour change
Laboratory gas No observable colour change No observable colour change
Turpentine Purple colour turns colourless Solution is decolourised
Hexane No observable  colour change No observable colour change
Pentene Potassium permanganate is decolourised Solution is decolourised

 

Conclusion

– Kerosene, laboratory gas and hexane are saturate hydrocarbons

– Turpentine and pentane are unsaturated hydrocarbons.

 

Homologous series

– Refers to a group of organic compounds that have the same general formula, whose consecutive members differ by a similar unit, and usually have similar chemical properties.

 

Characteristics of a Homologous series.

(i). Can be represented by a general formula;

(ii). Have similar chemical properties

(iii). Have similar structures and names

(iv). They show a steady gradation of physical properties

(v). Can usually be prepared by similar methods.

 

Structural and molecular formula

  • Molecular formulae

– Simply shows the number and type of elements (atoms) in the compound.

 

  • Structural formula

Shows how the different atoms in the molecules (of a compound) are bonded or joined together.

 

Example:

Methane

Molecular formula CH4;

 

Structural formula

H

H – C – H

H

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. Alkanes

Are the simplest hydrocarbons with the general formula; CnH2n + 2 where n = number of carbon atoms in the molecule.

 

Examples:

– For compound with only 1 carbon atom, formula = CH4

– 2 carbon atoms; the formula = C2H6

 

Names and formulas of the first 10 Alkanes

 

Note:

Consecutive members of the alkane series differ by a CH2-unit, hence a homologous series.

 

(a). General  formula

– The Alkanes have a general formula CnH2n+2 where n is the number of carbon atoms in the molecule.

Example:       

When n = 3, (2n + 2) = 8, and the alkane has the formula C3H8 (Propane)

 

(b). Structure

– In all Alkanes the distribution of bonds around each carbon atom is tetrahedral.

 

Example: Methane

 

(c). Homologous series

– The Alkanes differ from each other by a –CH2-.

– Thus methane, CH4 differs from ethane, C2H6 by –CH2-, and ethane in turn differs from             propane C3H8 by – C 2 -.

– They therefore form a homologous series.

 

(d). Functional groups

– A functional group is a part of a compound which has a characteristic set of properties.

– Thus when a bromine atom replaces a hydrogen atom in an alkane, it imparts to the compound new chemical and physical properties.

 

Examples: six important functional groups.

 

(e). Isomerism

– Is a situation whereby two or more compounds have similar molecular formulae but different structural formula.

– Such compounds are called isomers, i.e compounds with the same molecular formula but different structural formula.

 

Examples: For Butane, (C4H10) there are two possible structures.

 

Isomers have different physical and chemical properties.

 

 

 

 

Example: Ethanol and dimethyl ether.

– Molecular formula: both have C2H6O

 

  • Structural formula:

(i). Ethanol                                                                 (ii). Dimethyl ether

 

 

 

 

 

Differences

Ethanol Dimethyl ether
– A liquid of boiling point 78.4oC

– Completely soluble in water

– Reacts with sodium  ethoxide and  liberates hydrogen gas

– A gas at room temperature (B.P – 240C).

– Slightly soluble in water.

– Does not react with sodium metal.

 

(f). Alkyl groups

– Is a group formed by the removal of a hydrogen atom form a hydrocarbon.

– Alkyl groups don’t exist on their own but are always attached to another atom or group.

 

Naming of alkyl groups

– Is done by removing the ending -ane from the parent alkane and replacing it with –yl.

 

Examples

Methane (CH4) gives rise to Methyl -CH3

Ethane (C2H6) gives rise to ethyl, – C2H5 i.e. -CH2CH3

Propane (C3H8) gives rise to Propyl, – C3H7 // -CH2CH2CH3;

 

(g). Nomenclature of Alkanes

– Generally all Alkanes end with the suffix -ane;

– Alkanes can either be straight chain or branched.

 

(i). Straight chain Alkanes

– The names of all Alkanes end with the suffix -ane;

Examples:

Methane, ethane, propane, butane.

 

– With the exception of the first 4 members of the series (i.e. the 4 listed above) the names of Alkanes begin with a Greek prefix indicating the number of carbon atoms in the main chain.

Examples: – Pentane – 5 carbon atoms

Hexane – 6 carbon atoms.

 

(ii). Branched Alkanes

The naming of branched chain Alkanes is based on the following rules:-

  1. The largest continuous chain of carbon atoms in the molecule is used to deduce the parent name of the compound.
  2. The carbon atoms of this chain are numbered such that the branching // substituents are attached to the carbon atom bearing the lowest number.
  3. The substituent // branch is named e.g. methyl, ethyl etc and the name of the compound written as one word.

 

Examples

Further examples

H   H    H                                         CH2CH2CHCH2CH3

│   │    │                                         │           │

H – C – C – C – H                                   CH3       CH2

│                                                       │

H – C – H                                                 CH3

│                                                     3-ethylhexane;

H

2-methylpropane;

 

Further examples.

  1. CH3CH2CH2CH3

CH3

3-methylpentane;

 

  1. CH3

H3C – C – CH3

CH3

2, 2-dimethylpropane;

 

Note: refer to course books and draw as many examples as possible.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Draw the structural isomers of:

  1. Butane.

 

  1. Pentane;

 

  1. Hexane;

 

(f). Occurrence of Alkanes

– There are 3 known natural sources:

(i). Natural gas: this consists of mainly of methane;

 

(ii). Crude oil:

– Consists of a mixture of many Alkanes

– It can be separated into its components by fractional distillation.

Reason:

– The different components have different boiling points.

 

(iii). Biogas: This contains about 60-75% of methane gas/marshy gas.

 

Separation of the components of crude oil.

(i). Apparatus

 

 

 

 

 

 

 

 

 

(ii). Procedure

– The apparatus is arranged as shown above.

– The first distillate appears at about 120oC and is collected, the of 40oC intervals thereafter until the temperatures reach 350oC.

 

(iii). Observations and explanations

– This method of separation is called fractional distillation, and depends on the fact that the various components of the mixture have different boiling points.

– The various fractions vary in properties as explained below.

 

(a). Appearance

– Intensity of the colour increases with increase in boiling point.

– Boiling point increases with increasing number of carbon atoms.

Reason:         

– The higher the number of carbon atoms, the higher the number of covalent bonds.

– Thus the first fraction to be distilled (lab gas) is colourless while the last           distillates (between) is dark black in colour.

 

(b). Viscosity

Increases with increasing boiling point;

– The fractions with low boiling points are less viscous while the fraction with the highest boiling point is semi-solid;

 

(c). Inflammability:

– Decreases with increasing boiling points.

– The gaseous fractions, with least boiling points readily catches fire // burn, while the semi-solid fractions with very high boiling points are almost non-combustible.

 

Note:  Some Hydrocarbons are found in more than one fraction of crude oil and more advanced chemical methods are necessary for complete separation.

 

Uses of the various fractions of crude oil.

No. f carbon atom per molecule Fractions Uses
1-4 Gases Laboratory gases and gas cookers
5-12 Petrol Fuel in petrol engines
9-16 Kerosene (paraffin) Fuel for jet engines (aeroplanes) and domestic uses
15-18 Light diesel oils Fuel for heavy diesel engines e.g. for ships
18-25 Diesel oils Fuel for diesel engines
20-70 Lubricating oils Used for smooth running of engine parts
>70 Bitumen Road tarmacking

 

Changes // gradation of physical properties across the alkane homologous series

 

Name of alkane Formula State of room temperature (208K) M.P (K) B.P (K) Density

(g cm-3)

Solubility Solubility
Methane

Ethane

Propane

Butane

Pentane

Hexane

Heptane

Octane;

Nonane

Decane

CH4

C2H6

C3H8

C4H10

C5H12

C6H14

C7H16

C8H10

C9H20

C10H22

 

Gaseous

 

Liquid

 

 

90

91

85

138

143

178

 

 

 

243

112

184

231

273

309

342

447

0.424

0.546

0.582

0.579

0.626

0.659

0.730

   

 

 

 

 

 

 

 

 

 

 

Preparation and chemical properties of Alkanes

Note:

– Alkanes, like any other Homologous series have similar chemical properties.

– Generally any alkane can be represented form the reaction represented by the following equation:

CnH2n + 1COONa + NaOH(aq) → CnH2n +2 + Na2CO3(aq);

 

Thus;

– Methane can be prepared form sodium ethanoate (CH3COONa)

– Ethane can be prepared form sodium propanoate (CH3CH2COONa)

– Propane can be prepared form sodium Butanoate (CH3CH2CH2COONa)

Laboratory Preparation of methane

(i). Apparatus

 

 

 

 

 

 

 

 

 

 

 

(ii). Procedure

– About 5g of odium ethanoate and an equal mass of soda lime is put in a hard glass test tube, upon mixing them thoroughly in a mortar.

– The mixture is heated thoroughly in the test-tube.

 

(iii). Observation

– A colourless gas collects over water

Reasons:

– Methane does not react with and is insoluble in water.

 

Equation

CH3COONa + NaOH(s) → CH4(g) + Na2CO3(aq)

Sodium ethanoate         sodalime            Methane       Sodium carbonate

 

Physical properties of methane

  1. It is a non-poisonous, colourless gas.
  2. It is slightly soluble in water, but quite soluble in organic solvents such as ethanol and ether.
  3. II is less denser than air and when cooled under pressure, it liquefies.

 

Chemical properties

  1. Burning

– It is flammable and burns in excess air // oxygen with a pale blue non-luminous flame to give carbon (IV) oxide ad water vapour.

Equation:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

 

Note:  In a limited supply of air, the flame is luminous.

Reason:

– This is due to incomplete combustion of the methane.

– A mixture of methane and air explodes violently when ignited if the volume ratio is approximately 1:10 and this is often the cause of fatal explosions in coal mines.

 

  1. Reaction with Bromine water and acidified potassium permanganate

– When methane is bubbled through bromine water the red brown colour of bromine persists; and when bubbled through acidified potassium manganate (VII) solution; the purple colour of the solution remains;

– Thus it has no effect on either bromine water or acidified potassium permanganate.

Reason: It is a saturated hydrocarbon.

 

  1. Substitution reactions

– A substitution reaction is one in which one atom replaces another atom in a molecule.

 

Example: The substitution of Bromine in methane.

Procedure:

– A sample of Methane (CH4) is placed in a boiling tube and to it is added some bromine gas.

– The tube is stoppered, and the mixture shaken, then allowed to stand and exposed to ultra-violet lamp.

 

Observations

– The red colour of Bromine begins to fade, and the pungent smell of hydrogen bromide (HBr) gas is detectable when the stopper is removed.

– A moist blue litmus paper also turns red on dipping into the resultant mixture.

Equation                                                                                                                                                                CH4(g) + Br2(g) → CH3Br(g) + HBr(g)

Explanation                                                                                                                                                                      – For a chemical reaction to occur, bonds must be broken.                                                                                       – The light energy (V.V. light) splits the Bromine molecule into free atoms, which are very reactive species.                                                                                                                                                                               – Similarly the energy breaks the weaker carbon – hydrogen bonds, and not the stronger carbon – carbon bonds.                                                                                                                                                                                  – The free bromine atoms can then substitute (replace one of the hydrogen atoms of methane, resulting unto bromomethane and hydrogen bromide gas.

Note:  This process can be repeated until all hydrogen atoms in CH4 are replaced.

Write all the equations to show the stepwise substitution of all hydrogen atoms in methane.

– The substitution reactions can also occur with chlorine, forming chloremethane dichloromethane, trichloromethane (chloroform) and tetrachloromethane (carbon tetrachloride) respectively.

Equations:

 

 

 

 

 

 

Uses of methane                                                                                                                                                    – It is used as a fuel                                                                                                                                                     – Used in the manufacture of carbon black which is used in printers ink and paints.                                                         – Used in the manufacture of methanol, methanal, chloromethane and ammonia.

Cracking of Alkanes                                                                                                                                                 – Is the breaking of large alkane molecules into smaller Alkanes, alkenes and often hydrogen.                                  It occurs under elevated temperatures of about 400-700oC

Equation

Example: Cracking of propane

 

  1. Alkenes

– Are hydrocarbons with at least one carbon-carbon double bond, and have the general formula CnH2n.

– They thus form a homologous series – with the simplest member behind ethane.

 

Names and formulae of the first six alkenes.

 

Name of alkene Formula
Ethene

Propane

Pbut-l-ene

Pent-lene

Hex-tene

 

 

 

NOMENCLATURE OF ALKENES

 

Rules

  1. The parent molecule is the longest carbon chain; and its prefix is followed by the suffix –ene.
  2. The carbon atoms in the chain are numbered such that the carbon atoms joined by the double bonds get the lowest possible numbers.
  3. The position of the substituent groups is indicated by showing the position of the carbon atom to which they are attached.
  4. In case of 2 double bonds in an alkene  molecule, the carbon atom to which each double bond is attached must be identified.

 

Examples

 

Questions:      For each of the following alkenes, draw the structural formula

 

  1. Hex- l – ene
  2. Prop-l-ene
  • Hex-2-ene

 

  1. Give the IUPAC names for:

 

Note:  Branched alkenes:

 

Event for branched alkenes, the numbering of the longest carbon chain is done such that the carbon atoms joined by the double bonds gets the smallest numbers possible.

 

 

 

 

Isomerism in alkenes

  • Alkenes show two types of isomerism:-
  1. Branching isomerism
  2. Positional isomerism

 

  1. i) Branching isomerism

Occurs when a substitutent groups is attached to one of the carbon atoms in the largest             chain containing the double bond.

 

Positional isomerism; in alkenes

 

Is a situation whereby two or more unsaturated alkenes have same molecular formular but different structural formula; due to alteration of the position of the double bond.

 

 

 

Question:       Draw all the possible  isomers of Hexene , resulting from positional and                           branching isomerism.

 

Gradation of physical properties of Alkenes

 

Name of alkene Formula (MP0C) B.P (0) Density g/cm3 solubility
Ethene

Propene

But-l-ene

Pent-l-ene

Hex-l-ene

  -169

-189

-185

-138

-98

-104

–47.7

-6.2

-3.0

-98

0.640

0.674

 

 

Note:   the double bond is the reactive site in alkenes

 

Preparation and chemical properties of Ethene

 

  1. i) Apparatus

 

 

  1. Procedure

A mixture of ethanol and concentrated sulfuric acid in the ratio 1:2 respectively  are heated in a flask to a temp. of 1600C – 1800C.

 

  • Observation

A colourless gas results; and is collected over water.

 

Reasons:         Its insoluble, unreactive and lighter than water.

 

  1. Equation

 

  1. Explanation

 

At 1600C – 1800C the conc. H2SO4 dehydrates the ethanol, removing a water molecule form it and the remaining C and H atoms  rearrange and combine to form Ethene which is collected as  colourless gas.

 

Note:  At temperature  below 1400C, a different  compound called ether is predominantly              formed.

 

Ethene  can also be prepared by passing hot aluminum  oxide over ethanol.  The later of which acts as a catalyst i.e.

 

Reactions of ethene/chemical properties

 

  1. Burning/combustion

Just like an alkenes and alkanes, ethene  burn in air, producing carbon dioxide and large quantities of heat.

 

Equation:

 

Caution:         Mixtures of air and ethene  can be explosive and must be handled                         very carefully.

 

  1. Additional reactions:

Is a reaction in which are molecule adds to another to form  a single product occur              in alkenes due to presence of a double bond.

 

  1. With oxidizing agents
  2. i) Reaction with acidified potassium permanganate.

Procedure:      Ethene is bubbled into a test tube containing acidified potassium                                       permanganate.

Observation:   The purple   colour  of the solution disappears.

Explanation:  Ethene reduces the potassium permanganate.

The  permanganate ion is reduced to Manganese (II) ion and water.

 

Equation

 

Note:   The net effect of the above reaction is the addition of two –OH groups to                                     the double bond forming ethan-1, 2-dio(ethylene glycol).

In cold countries ethylene glycol is used as an antifreeze in car radiators.

 

  1. Reaction with acidified potassium chromate (VI) (K2Cr2O7)

 

  1. Halogenations is the addition of halogen atoms across a double bond.
  2. i) Reaction with Bromine Br2(g)

 

Procedure:     Ethene is mixed with Bromine liquid/gas

Observation:  The reddish  brown bromine gas is decoloursed/becomes  colourless.

Explanation:   Bromine is decoloursed due to the addition of Bromine  atoms to the twocarbon atoms f the double bond forming 1.2 dibromethane.

 

  1. ii) Reaction with chlorine

The Chlorine  (greenish yellow) also gets decoloursied due the addition of its             atoms on the double bond.

 

 

Note: Alkenes react with and decolourise halogens and potassium permanganate  by             additional  reaction at room temperature and pressure.

 

The reaction site is the double bond  and hence/all alkenes will react in a similar             manner.

Example; Butene and Bromine

 

 

iii)       Reaction with Bromine water

Bromine  is dissolved in water and reacted with ethene.

 

Equation:

 

Further examples of additional reactions

 

  1. Addition of hydrogen halides

 

  1. With hydrobromic acid; HBr (aq)

 

With  sulphuric acid

 

  1. Addition of Ethene  with sulphuric acid

 

Note:  When ethylhydrogen sulphate is hydrolysed, ethanol is formed.

 

In this reaction, water is added to ehylhydrogen sulphate and the mixture warmed.

 

  1. Ethene with Hydrogen i.e. Hydrogenation.

 

Is commonly termed hydrogenation though just a typical addition reaction.

 

Ethene  is reacted with hydrogen, under special conditions.

 

Conditions;     moderate temperature and pressure.

Nickel catalyst/palladure catalyst.

 

Equation:

 

Application:    it is used industrially in the conversion f various oils into fats e.g. in the preparation of Margarine.

 

  1. Polymerization reactions.

Also called self-addition reactions

Alkanes have the ability to link together (polymerise) to though the double bond to give a molecule of larger molecular mass (polymers)

 

Polymers:       Are  very large molecules formed when 2 or more (smaller) molecules link                         together  to form a larger unit.

Polymers have properties different form those of the original constituent manners.

 

Examples:      Polymerisation of ethene

 

  1. i) Conditions
  • High temperatures of about 2000C
  • High/elevated pressures of approximately 1000 atmospheres
  • A trace of oxygen catalyst.

 

  1. ii) Procedure: Ethene is heated at 2000C and 1000 atm. Pressure over  a catalyst.

 

iii)        Observation:  Sticky white substance  which hardens  on cooling  is formed.                                            This solid is called  polythene, commonly reffered to as polythene.

 

  1. Equation:

 

 

Generally

 

Uses of polythene

 

  1. Used for the manufacture of many domestic articles (bowls, buckets, water cans, and cold water pipes) e.t.c.

 

Note:  Polythene pipes have a great advantage over metal pipes as they can be             welded quickly and do not burst in frosty weather.

 

  1. Manufacture of reagent bottles, droppers, stoppers etc. since polythene is unaffected by alkalis and acids.

 

Test for Alkenes

 

–           They decolourise bromine water, acidified potassium manganate VII.

i.e. These addition reactions show the presence of a double bond.

 

Uses of Alkenes

 

  1. Manufacture of plastics, through polymerization.
  2. Manufacture of ethanol; through hydrolysis reactions
  3. Ripening of fruits.
  4. Manufacture of ethan – 1, 2-diol(glyco) which is used as a coolant.

 

           

  1. ALYKYNES

 

Are unsaturated hydrocarbons which form a homologous series of a general formula CnH2n-2, where n = 2 or more.

 

The functional groups of the alkyne series is the carbon – carbon tripple bond.

 

They also undergo addition reactions because of High unsaturation and may be polymerised like the alkenes.

 

Examples

 

Name Molecular formula Structural formular
Ethyne

Propyne

But-l-yne

Pent-l-yne

C2H2

C3H4

C4H6

C5H8

CH     CH

CH3C     CH

CH3CH2C       CH

CH3(CH2)2C     CH

 

Nomenclature of alkynes

 

  • The largest chain with the tripple carbon – carbon bond forms  the parent molecule.
  • Numbering of the carbon atoms is done such that the carbon atom with the tripple bond acquires the lowest possible number.
  • The substituent branch if any is named, and the compound written as a single word.

Examples

 

 

  1. Draw the structures of the following hydrocarbons
  2. 2,2 dimethyl-but-2-yne
  3. propyne
  • 4,4 diethyl-hex-2-yne.

 

Isomerism in alkynes

 

  1. Positional isomerism

Isomerism commonly occurs in alkynes due to the fact that the  position of the tripple bond can be altered.

 

Such isomers, as usual have same molecular but different structural  formulas.

 

Examples

  1. i) Isomers of Butyne

 

 

  1. Branching isomerism – occurs when alkyl group is present in the molecule.

 

  1. Others

 

Gradation in physical properties of Alkynes

 

Name of Alkyne Formula M.P/0C B.P/0C Density/gcm-3
Ethyne

Propyne

Butyne

Pent-l-yne

Hex-l-yne

HC    CH

CH3    CH

CH3CH2CC    CH

CH3CH2CH2C   CH

CH3(CH2)3C  CH

-8108

-103

-122

-90

-132

-83.6

-23.2

8.1

39.3

71

0.695

0.716

 

 

Preparation and chemical properties of Ethyne.

 

  1. Preparation
  2. i) Apparatus

 

 

 

  1. ii) Procedure:

Water is dripped over calcium carbide and is collected over water.

Reasons for over-water collection:-

  • It’s insoluble in water
  • Unreactive and lighter than water.

 

  • Conditions
  • Room temperature

 

  1. Equation

 

 

  1. Properties of Ethyne
  2. i) Physical
  • Colourless gas, with a sweet smell when pure.
  • Insoluble in water and can thus be collected over water.
  • Solubility is higher in non- solvents    *  Draw table on physical properties.
  1. Chemical properties
  • Combustion

Ethyne burns with a luminous and very sooty  flame; due to the high percentage of carbon content, some of which remains unburnt.

  • In excess air, the products are carbon dioxide and water.

 

Equation

 

In limited air, they undergoes incomplete combustion, forming a mixture of carbon and carbon dioxide.

 

Note:  A sooty flame observed when a hydrocarbon burns in air is an indication of             unsaturation in the hydrocarbon.

 

Addition reactions

During  addition reactions of alkynes (Ethyne) the tripple bond breaks in stages;

 

  1. Reaction with hydrogen (Hydrogenation)

 

 

Note:  This reaction occurs under special conditions i.e. –  Presence of a Nickel catalyst

Temperatures about 2000C

 

  1. Reaction with halogens
  2. i) Reaction with chlorine

 

 

  1. With Bromine gas
  • The red-brown bromine vapour is decoloursed.

 

Equations

 

Note:   In this reaction Cl2 should be diluted with an inert.

 

Reason:           Pure Cl2 reacts explosively with Ethyne, forming carbon and HCl.

 

  1. Reaction with Bromine liquid

When  Ethyne reacts with Bromine water, the reddish – brown  colour of bromine water disappears.

 

Reason:          The Bromine adds to the carbon  tripple bond leading to the      …… of                              1;1,2,2 tetrabromoethane.

 

Equation

 

E; Ethyne  also decolorizes acidified potassium permanganate.

 

Note:  Decolourization of acidified potassium  permanganate  and bromine water are             tests for unsaturated hydrocarbons (alkanes and alkynes)

  1. Reaction with hydrogen halides

Uses of Ethyne

  1. Industrial manufacture of compounds like adhesives and plastics
  2. It’s used in the oxy-acetylene flame which is used for welding and cutting metals.

CHEMISTRY NOTES FORM 1-4 PDF

UNIT 1: INTRODUCTION TO CHEMISTRY.

Unit Checklist.

Meaning of chemistry

  • Matter
  • States of matter
  • Properties of the different states of matter.
  • Conductors and non-conductors.

DOWNLOAD FREE PDF NOTES HERE; Free Chemistry notes, revision questions, KCSE past Papers, Exams, Marking Schemes, Topical revision materials, Syllabus and Many more

Role of chemistry in society.

Laboratory rules.

  • Necessity for laboratory rules.
  • The laboratory rules and regulations.
  • Laboratory safety symbols.

Common apparatus used in a chemistry laboratory.

The Bunsen burner.

  • Structure
  • Functions of the various parts.
  • Types of Bunsen burner flames.

Methods of gas collection.

  • Upward delivery
  • Downward delivery
  • Over water
  • Using a collecting syringe.

Drying of gases

  • Using concentrated sulphuric acid.
  • Using anhydrous calcium chloride
  • Using calcium oxide.

Drugs and drug abuse.

 

Meaning of chemistry.

– It is a branch of science which deals with the composition and properties of matter.

 

Matter

– Matter is anything that has mass and occupies space.

 

States of matter.

– Matter exists in three different states: solids, liquids and gases.

 

Properties of the different states of matter.

 

  Solid Liquids Gases
Mass Definite Definite Definite
Shape Definite Indefinite: they take the volume of the container in which they are in. Indefinite: they take the shape of the container in which they are.
Volume Definite Definite Indefinite: volume may increase with increase in temperature; and decrease with decrease in pressure

 

Note: Conductors and non-conductors:

– The flow of electric current through materials is called electrical conductivity.

– Solid substances which allow electric current to flow through them are called conductors.

– Solid substances that do not allow electric current to flow through them are called non-conductors.

 

Role of chemistry in a society.

– Chemistry has enabled extraction of chemicals from plants.

– It is used in the manufacture of substances such as soap, glass, plastics, medicine, rubber, textiles etc from naturally occurring substances.

– Purification of substances from natural raw materials.

– It forms a basis for entry into careers e.g. teaching, medicine, chemical engineering etc.

 

Laboratory rules and safety symbols.

Necessity of laboratory safety rules.

  1. To avoid accidents and injuries during practical experiments in the laboratory.
  2. To avoid damage to and breakage of apparatus and laboratory fittings.
  3. To avoid wastage of laboratory chemicals.

 

The laboratory safety rules and regulations.

– Never run while in the laboratory;

Reason: You may injure others or yourself in the laboratory.

– Never taste or eat something in the laboratory;

Reason: to avoid poisoning.

– Always consult your teacher before trying out any experiment; so as to avoid accidents.

– Label all chemicals in use so as to avoid confusion.

– Always use a clean spatula for scooping a substance from a container to minimize contamination.

– Always hold test-tubes and boiling tubes using test tube holder when heating; to avoid being burned.

– When heating a substance never let the open end of the tube face yourself or anybody else, because the liquid may spurt out and cause injury.

– Never look directly into flasks and test tubes where reactions are taking place, because the chemicals may spurt into your eyes and cause injury.

– Never smell gases directly. Instead, waft the gaseous fumes near your nose with your hand.

– Experiments in which poisonous gases and vapours are produced must be carried out in a fume cupboard or an open space outdoors.

– Always keep flammable substances away from flames because they easily catch fire.

– Always report any accidents to the teacher or the laboratory technician immediately for necessary action.

– In case of an accident do not scramble for the same exit, because it may hinder easy escape.

– Always put off flames that are not in use in order to avoid accidents and minimize fuel wastage.

– If a chemical gets on your skin or mouth rinse it immediately with a lot of water.

– Always dispose off the chemicals already used safely to avoid explosions and contaminations.

– Always work on a clean bench. After completing your experiment, clean all the pieces of apparatus you have used and return them to their correct storage places.

– Always read the label of the reagents before using them.

 

Safety symbols.

– These are signs found on the labels of bottles or cartons containing dangerous chemicals.

– The common safety symbols are as follows:

Symbols Meaning.
 

 

 

 

 

Toxic: are very poisonous and can easily kill if swallowed, inhaled or on contact with the skin.

Examples: Chlorine and mercury;

  Harmful: Less harmful (dangerous) than the toxic substances; are only likely to cause pain and discomfort.

Examples: copper (II) sulphate, lead (II) oxide

  Highly flammable: are substances that catch fire easily and must not be handled near open fire.

Examples: ethanol, hydrogen

  Oxidizing: rapidly provide oxygen and can cause fire to burn more fiercely.

Examples: potassium manganate (VII), hydrogen peroxide

  Corrosive: are substances that cause burns to skin and fabric; and can also react with other substances such as metals

Examples: nitric (V) acid, conc. sulphuric acid, bromine.

  Irritant: can cause blisters or reddening of the skin; usually irritate the respiratory tract, skin, eyes etc.

Examples: calcium chloride and zinc sulphate

Common Chemistry laboratory apparatus and their uses.            

 

Name of apparatus Diagram. Use.
Test tubes  

 

 

 

 

 

 

– General laboratory experiments; like heating solids; qualitative analysis etc
Boiling tubes

 

 

 

 

 

 

– Mainly used for heating small amounts of solids and liquids.
Test tube holder

 

 

 

 

 

 

  Used for holding test tubes and boiling tubes during heating experiments.
Measuring cylinder

 

 

 

 

 

 

 

– Measuring accurate volumes of liquids in the laboratory
Beaker

– Lipped glass or plastic vessels of various capacities.

 

 

 

 

– Glass beakers are used for boiling liquid substances;

– Holding solutions during chemistry experiments.

Filtering funnel

 

 

 

 

 

 

 

– Directing liquids into containers with small narrow mouths;

– Holding filter papers during filtration;

Name of apparatus Diagram Use
Stirring rod

 

 

 

 

 

   
Watch glass

 

 

 

 

 

   
Thermometer

 

 

 

 

 

– Measuring temperatures during experiments.
Conical flask

 

 

 

 

 

 

– Normal laboratory experiments like titration.

– May be used for measuring volumes if graduated.

Round-bottomed flask

 

 

 

 

 

 

  – Used when heating liquid substances because heat is supplied uniformly.
Flat-bottomed flask

 

 

 

 

 

 

 

 

 

 

 

 

 

 

– Used for general laboratory experiments.
Evaporating dish

 

 

 

 

 

 

 

 

 

– Used when evaporating liquids.
Name of apparatus Diagram Use
Crucible

 

 

 

 

 

 

  – Use when heating solid substances that require strong heating.
Pestle and mortar

 

 

 

 

 

 

 

  – Crushing substances while the mortar holds the substances being crushed.
Pie clay (ceramic) triangle

 

 

 

 

 

  – Supporting crucibles during heating.
Tripod stand

 

 

 

 

 

 

 

  – Supporting beakers and flasks in which liquids are being heated.

 

Wire gauze

 

 

 

 

 

 

 

  – Used when glass apparatus are being heated; to facilitate even distribution of heat when heating substances in beakers or flasks
Clamp

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

– Supporting and holding pieces of apparatus during experiments.
Name of apparatus Diagram Use
Deflagrating spoon

 

 

 

 

 

 

  – Holding burning substances.
Spatula

 

 

 

 

 

 

  – Scooping solid substances from containers
Crucible tongs

 

 

 

 

 

 

 

  – Holding solid chemicals.
Condenser

 

 

 

 

 

 

 

   
Separating funnel

 

 

 

 

 

 

 

  – Separating immiscible liquids.
Thistle funnel

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

– Delivering liquid substances into other containers like flasks during reactions.
Name of apparatus Diagram Use
Wash bottle

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

– Holding water for rinsing apparatus
Dropping funnel

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 
Test tube rack

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

– Holding boiling tubes and test tubes.
Teat pipette (dropper)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

– Sucking liquid chemicals and placing them in another container dropwise.
Burette

– It consists of a long narrow tube with a tap and a jet at the bottom.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

– Delivering accurate volumes of liquids
Name of apparatus Diagram Use
Pipette

 

 

 

 

 

 

 

 

  – Delivering a specified volume of liquid accurately.
Gas jar

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

– Gas collection.
Trough

 

 

 

 

 

 

 

 

  – Holding some amount of water for some experiments e.g. gas preparation.
Reagent bottles

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

– Storing chemicals in liquid state.
Desiccator

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

– Drying substances or keeping substances free from moisture.
Name of apparatus Diagram Use
Spirit lamp

Note: less preferred for heating because their flames are not hot enough; and they deposit soot on apparatus making them dirty hence difficulty in observing changes during experiments

 

 

  – Heating substances in the laboratory.

 

 

 

 

 

 

 

 

Kerosene stove

Note: less preferred for heating because their flames are not hot enough; and they deposit soot on apparatus making them dirty hence difficulty in observing changes during experiments

 

 

 

 

 

 

 

 

 

 

 

 

 

– Heating substances in the laboratory.
Electric heater

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

– Heating substances in the laboratory.
Candles

Note: less preferred for heating because their flames are not hot enough; and they deposit soot on apparatus making them dirty hence difficulty in observing changes during experiments

 

 

 

 

 

 

 

 

 

 

 

 

 

 

– Heating substances in the laboratory.
Bunsen burner

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

– Heating substances in the laboratory.
Name of apparatus Diagram Use
Stop watch (clock)

 

 

 

 

 

 

 

 

  – Measuring time particularly in determination of reaction rates.

 

 

 

Beam balance

 

 

 

 

 

 

 

  – To take accurate weight measurements
Electronic balance  

 

 

 

 

 

 

 

– Take accurate weight measurements; and can take extremely low weight measurements.
Volumetric flask

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Note: most chemistry apparatus are made of glass:

Advantages of glassware:

– It does not react with most chemicals

– Glass is transparent and hence reactions can easily be observed as they progress.

– Glass materials are easy to wash and rinse after experiments.

– They can be used comfortable in heating experiments.

 

Disadvantages:

– Have higher chances of breakages in case they fall during experiments.

– They are comparatively expensive to plastics

 

 

 

– Some materials like beakers may however be made of plastics.

Advantages of plastic apparatus.

– Have lower chances of breaking.

– They are relatively cheaper to buy.

 

Disadvantages:

– Plastics tend to react with some laboratory chemicals

– may not be transparent and hence reactions cannot easily be observed as they progress.

– Glass materials are difficult to wash and rinse after experiments.

– They cannot be used in heating experiments.

 

The Bunsen burner.

– is the most common heating apparatus in the laboratory.

– Was invented by a German scientist known as Wilhelm Bunsen hence the name.

– It uses natural (cooking) gas for heating.

 

Parts of a bunsen burner

– Chimney

– The air hole

– Collar

– The jet

– Base

– Gas inlet

 

Diagram: structure of a bunsen burner.                                                                  Diagram: Bunsen burner-components detached

 

 

 

 

 

 

 

 

 

 

 

 

 

Functions of the various parts:

  • The chimney:

– It is a hollow metallic cylinder with an air hole near its lower end.

Function:

– allows air and the laboratory gas from the jet mix before they start burning at the upper end of the chimney.

– Its upper opened end provided a site where the gas burns.

 

  • The air hole:

– Is a small aperture found at the lower end of the chimney and smaller than the collar.

– Its diameter (size) is regulated by the collar.

Function:

– Allows air (oxygen) to enter and mix with the laboratory gas in the chimney.

 

  • The collar:

– Is a metallic ring with an air hole whose diameter is the same size as that of the air hole in the chimney.

– It fits into the lower part of the chimney; and can rotate around the chimney opening or closing the air hole.

Function:

– Regulates the amount of air entering the chimney.

 

  • The jet:

– It is a very tiny opening just below the air hole, that connects the gas inlet to the chimney.

Function:

– allows the laboratory gas (methane) into the chimney at high pressure.

 

  • The gas inlet:

– It is a hollow metal connected to the base and extending into the jet.

– Its external opening is usually connected to a rubber tubing that is connected to a gas tap.

Function:

– Allows laboratory gas from the cylinders (reservoirs) in the lab; through the gas taps into the Bunsen burner.

 

  • The base:

– A thick heavy metal, that is usually circular or oval.

Function:

– It supports the Bunsen burner on the bench.

 

The Bunsen burner flames.

– A flame is a mass of burning gases.

– A bunsen burner can produce two types of flames depending on the size of the air hole and hence amount of air entering the chimney.

 

Types of flames.

  • Luminous flame.
  • Non-luminous flame.

 

(a). The luminous flame.

– It is a large bright yellow flame produced when the air hole is closed and hence no air enters the chimney.

 

Characteristics of a luminous flame.

– It is large, quiet and bright yellow.

– Colour is not uniform and it ahs four zones.

– It produces less heat.

– It gives a lot of light to the glow of unburnt hot carbon particles

– It produces soot.

 

 

Diagram: the luminous flame.

 

 

 

 

 

 

 

 

 

 

Parts of a luminous flame.

  • The thin outer zone:

– Is a fairly visible, narrow zone on the outer surface of the flame.

– At this point methane (lab gas) mixes with sufficient air from the outside and burns completely to carbon (IV) oxide and water.

 

  • The inner bright yellow zone:

– It is a large bright yellow zone that lies beneath the thin outer zone.

– Here, air supply is insufficient resulting to incomplete combustion of the gas.

– Consequently the gas burns producing tiny carbon particles instead of carbon (IV) oxide.

– The white hot carbon particles glow brightly and are responsible for the yellow colour and the emission of light.

– On cooling the carbon particles form soot, which blackens the bottom of the apparatus being heated.

 

  • The almost colourless inner zone.

– Is found below the yellow inner zone; and consists mainly of unburnt gases.

 

  • The blue zone (region)

– Is found on the outer side of the base of the flame.

– Here, air near the flame rises rapidly due to convection currents and mixes with the burning gas.

– This makes burning more complete than in the two upper parts above it.

 

Advantages of the luminous flame:

– Can be used for lighting purposes; because it produces more light.

 

Disadvantages.

– Produces less heat hence inefficient in heating.

– Due to production of soot it blackens apparatus thus preventing better observations of experiments.

 

 

 

 

 

 

 

 

 

 

(b). The non-luminous flame.

– It is a small blue flame produced when the air hole is completely open and hence a lot of air enters the chimney.

 

Characteristics of a non-luminous flame.

– It is small, noisy and blue.

– Colour is uniform and it ahs three regions.

– It produces comparatively more (a lot of) heat.

– It does not produce soot, due to complete combustion hence no carbon particles remain.

– It produces less light due to lack of white-hot carbon particles.

 

Diagram: the luminous flame.

 

 

 

 

 

 

 

 

 

 

Parts of a non-luminous flame.

  • The outer pale blue region.

– It is a large light blue zone.

– Here, there is a lot of air coming up the chimney from the air hole and from the outside.

– The air gas mixture thus burns completely to carbon (IV) oxide and water.

– No soot formation because there are no carbon particles.

 

  • The middle greenish blue region.

– consists of partially burnt gas-air mixture, due to insufficient air supply.

– However as the mixture rises up the pale blue region, it undergoes complete combustion due to plenty of air (from outside)

 

  • The inner almost colourless region.

– Is located at the base of the flame.

– It consists of unburnt gas-air mixture.

 

Advantages of a non-luminous flame.

– Gives out a lot of heat hence very efficient in hating.

– It does not form soot hence will leave apparatus clean even after experiment (heating).

 

Disadvantages:

– It uses a lot of laboratory gas in burning.

– cannot be used for lighting purposes since it produces very little light.

 

 

 

 

Differences between a luminous and a non-luminous flame.

 

Luminous flam Non-luminous flame
Bright yellow in colour Blue in colour
Produces a lot of light Produces a lot of light.
Large and unsteady Small and steady
Produces soot Does not produce soot
Has four zones Has three zones
Burns quietly Burns noisily
Moderately hot Very hot

 

Experiments on Bunsen burner flames.

  1. To investigate the heating effects of the luminous and non-luminous flames.

Apparatus:

– Bunsen burners, 250 ml beakers, lighter, stopwatch, tripod stand, wire gauze.

 

Procedure

– 100 cm3 of water is put into ach of the two 250 ml beakers.

– One beaker is put over a luminous flame while the other is simultaneously put over a luminous flame

– Time taken for water to boil is noted for each set up.

– The bottom of ach beaker is observed for any changes.

 

Apparatus

 

 

 

 

 

 

 

 

 

Observations.

– Water heated over the non-luminous flame boiled ion a shorter time than the same amount of water heated over a non-luminous flame.

– The bottom of the beaker heated over the non-luminous flame remained clear but the one heated over the luminous flame was covered with black deposits of soot.

 

Explanations.

– The non-luminous flame is hotter than the luminous flame; hence boils the water faster

– The hottest part of the luminous flam is the outer blue zone.

– Incomplete combustion in the luminous flame leads to production of carbon particles, which when hot glow yellow and on cooling forms black soot on the beaker;

– Incomplete combustion in a non-luminous flame leads to production of carbon (IV) oxide and steam only, hence no soot formation.

 

Conclusions.

– The non-luminous flame is hotter than the luminous flame.

– The non-luminous flame is cleaner than the luminous flame.

 

  1. To investigate the hottest part of a non-luminous flame.

Requirements

– Bunsen burner, stiff white paper (cardboard), wooden splint.

 

Procedure

– A bunsen burner is ignited with the air hole open to get anon-luminous flame.

– A piece of white paper (cardboard) is slipped into the flame in region marked X as shown below.

– The piece of paper is removed quickly before it catches fire.

– A fresh piece of paper is then slipped into region marked Y as shown below; then again quickly removed before it catches fire.

– The experiment for each of the regions marked X and Y is then repeated using wooden splints.

– The splints should be held long enough for some of their parts to get charred

Apparatus

 

 

 

 

 

 

 

 

Observations.

  • Using pieces of paper.

– In region X, the part of the paper that was in contact with the flame was charred uniformly as shown below.

– In region Y, the part of the paper in contact with the flame had a charred ring with an unburnt part in the middle of the ring as shown below

 

Diagrams

 

 

 

 

 

  • Using wooden splints.

– In region X, the part of the splint in contact with the flame was charred uniformly as shown below.

– In region Y, the part of the splint in contact with the flame had an unburnt part in between two charred regions as shown below.

 

Diagrams:

 

 

 

 

 

 

 

 

 

 

Explanations.

– Regions which become charred indicate that they are the hottest part of the flame.

– Region X corresponds to the outermost blue region of a non-luminous flame.

– Region Y is the almost colourless region of the non-luminous flame, which is however surrounded by the middle greenish blue and the outer pale blue zones.

–  Thus in region X, the uniform charring of the paper and splint indicate that the outer pale blue zone is the hottest pat of the flame.

– Similarly the charred ring for experiment in region Y show that the parts in contact with the outer pale blue zone gets burnt faster before the parts in contact with the almost colourless or the greenish blue zones.

 

Conclusions.

– The hottest part of the non-luminous flame is the outermost pale blue zone.

– During heating the object being heated should not be placed nearer the chimney; these parts are less hot.

– For efficient heating the object being heated should be placed at the outermost region of the flame.

 

  1. To show the presence of unburnt gases in a Bunsen burner flame.

Apparatus:

– Bunsen burner, tongs, narrow hard glass tubing.

 

Procedure

– A bunsen burner is lit and adjusted to get a non-luminous flame.

– A narrow hard glass tubing is held with a pair of tongs and one of its end s is placed in the colourless zone of the flame.

– A match is lit and placed at the free end of the glass tubing.

 

Apparatus

 

 

 

 

 

 

 

Observations.

– A flame is obtained at the free end of the glass tubing.

 

Explanations.

– The tubing trapped unburnt gases at the almost colourless zone of the flame.

– The trapped gases combined with atmospheric air (oxygen) at the other (free) end of the tubing hence the flame.

 

Conclusions.

– The almost colourless region contains unburnt gases.

 

 

 

 

  1. To show the hottest part of the flame.

Apparatus:

– Bunsen burner, match stick

 

Procedure

– A matchstick is placed at the top of the bunsen burner chimney using a pin.

– A bunsen burner is lit and adjusted to get a non-luminous flame.

– The match stick is observed fro sometime for any changes.

– If no observable changes are made, the matchstick is then slowly raised towards the blue zone and observed keenly.

 

Apparatus

 

 

 

 

 

 

 

Observations.

– The matchstick did not ignite while it was at the bottom of the flame (resting on top of the chimney).

– It ignited as it was being raised towards the outer pale blue zone.

 

Explanations.

– The bottom of the flame (just on top of chimney) corresponds to the almost colourless zone.

– This zone contains unburnt gases, hence no burning occurs and is thus least hot to cause ignition of the matchstick.

– As the matchstick is raised upwards it moves past the greenish blue zone (where there is partial combustion) then to the outer pale blue zone where there is complete combustion and hence most heat.

– The heat in this region is adequate to cause ignition of the matchstick.

 

Conclusions.

– The outer pale blue zone is the hottest part of the non-luminous flame, and is thus the correct position to place an object during heating.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Methods of gas collection.

– Various chemical reactions produce gases; some of which are colourless while others are coloured.

– Additionally some gases are poisonous to the human body, while others are major causes of environmental pollution.

 

Examples:

  • Coloured gases:

Chlorine (green-yellow); nitrogen (IV) oxide (brown); bromine (red-brown)

 

  • Colourless gases:

Oxygen; carbon (II) oxide; carbon (IV) oxide; sulphur (IV) oxide; hydrogen; ammonia etc.

 

Factors affecting method used in collecting a gas.

– Density

– Solubility in water.

– Colour

– Toxicity

 

Summary on collection methods.

 

Method Apparatus Characteristic of gas
Upward delivery

– Also called downward displacement of air

 

 

 

 

 

 

 

 

– Must be less dense than air.

Examples: Hydrogen, ammonia gas.

Note: being lighter the gas is supported by the denser air from below;

– When used for colourless gases, it is not possible to know when the gas jar is full;

Downward delivery (upward displacement of air)   – Must be denser than air.

Examples: carbon (IV) oxide; nitrogen (IV) oxide; chlorine gas;

Note: – The gas displaces air and settle at the bottom of the collecting vessel

– Unless the gas is coloured, it is difficult to know when the container is full

Over water   – insoluble or only slightly soluble in water;

– does not react with water

Examples: carbon (IV) oxide; hydrogen; carbon (II) oxide;

Note: – with this method it is easy to tell when the gas jar or collecting tube is full of gas;

– This method cannot be used when the gas is required dry;

 

Collecting syringe

– the gas produced is collected in a syringe;

  – Mainly for poisonous gases; since the gases are confined and leakages are limited;

Note: this method allows collection of small volumes of gases;

– It also allows direct measurement of volume of gas produced;

Drying of gases.

– Is the process by which the moisture in a gas being prepared is removed prior to collection.

– This is done by passing the gas through chemicals that absorb moisture.

– Such chemicals are called drying agents.

– The drying agents should not react with the gases being dried.

 

Examples of drying agents.

– Anhydrous calcium chloride

– Concentrated sulphuric acid.

– Calcium oxide.

 

Apparatus and drying agents

 

 

 

 

 

 

 

 

 

 

 

Collection and drying of some gases

 

Gas Collection method Drying agent
Oxygen Over water Concentrated sulphuric (VI) acid; anhydrous calcium chloride.
Hydrogen Over water, upward delivery Concentrated sulphuric (VI) acid; anhydrous calcium chloride
Nitrogen Over water Concentrated sulphuric (VI) acid; anhydrous calcium chloride
Carbon (IV) oxide Over water, downward delivery Concentrated sulphuric (VI) acid; anhydrous calcium chloride
Ammonia Upward delivery Calcium oxide

 

Drugs and drug abuse.

Drug: is a chemical substance that alters the functioning of the body.

 

Types of drugs

(i). Medicinal drugs (medicines):

– Are drugs mainly used for treatment and prevention of diseases.

– Are also classified into two: over the counter drugs and prescription drugs.

 

  • Over-the-counter drugs.

– Are medicinal drugs that can be bought at a pharmacy or retail shop without written instructions from a doctor.

Examples: Mild painkillers like aspirin, panadol, paracetamol, drugs for flu etc.

 

  • Prescription drugs:

– Are strong medicines which should only be taken upon a doctors instruction (prescription).

– In this prescription, the doctors give a dosage, which indicates the amount and the rate at which it should be taken.

 

(ii). Leisure drugs.

– Are drugs that are usually taken for pleasure.

– Are classified into two:

  • Mild drugs: alcohol, tobacco;
  • Narcotic drugs: marijuana, cocaine, heroin, mandrax etc.

 

 

Drug abuse:

– Is the indiscriminate use of a drug for purposes which it is meant for; or administration of an overdose or underdose of a drug; as well as use of drugs for leisure purposes.

Note:

– The worst form of drug abuse is the taking of drugs for leisure purposes; and the most commonly abused drugs are the leisure drugs.

 

Effects of commonly abused drugs.

  • Alcohol:

– Affects the brain and the nervous system

– Damages the liver, and is a common cause of liver cirrhosis.

– Poor health due to loss of appetite.

– Time for working is wasted in drinking and hence less productivity and even lose of jobs; which results to poverty and family disintegrations.

 

– Bad breath, discoloured fingers and teeth

– Cause diseases such as bronchitis and tuberculosis.

– damages the lungs and is a common cause of lung cancer due to chemicals found in the cigarettes.

– Smoking during pregnancy is a common cause of miscarriages or still births.

– It is expensive: money used for other better uses is wasted in cigarette smoking.

 

  • Narcotic drugs:

– Interferes with the functioning of the brain.

– Results to addiction and drug dependency.

– Some are administered directly into the blood through syringes and hence common routes of transmission of HIV/AIDS.

 

General effects of drug abuse on the society.

– Drug abuses spend most of their money on drugs and hence neglect their family leading to misery and societal breakdown.

– Drunk drivers cause accidents.

– People who are drunk with a drug are unreasonable and cannot make logical decisions; and hence cannot be productive at that time.

– Drug abuse has resulted into loss of morals leading to higher rates of rapes, violent crimes, murders, prostitution etc.

– Drug abuse has fueled the spread of sexually transmitted diseases and HIV/AIDS.

 

Note:

Drug addiction:

– Is a situation in which an individual becomes dependent on a particular drug such that he cannot function normally without it; and lack of it result to some discomfort.

UNIT 2: SIMPLE CLASSIFICATION OF SUBSTANCES

 

Unit Checklist:

  1. Elements compounds and mixtures.
  2. Mixtures
  • Types of mixtures
  • Separation of mixtures
  • Basic concepts
  • Method of separation of mixtures.
    • Decantation
    • Evaporation
    • Condensation
    • Filtration
    • Crystallization
    • Separating funnel separation
    • Distillation
    • Sublimation
    • Chromatography
    • Solvent extraction
  1. Criteria for purity
  • Effects of impurity on melting point
  • Effects of impurity on melting point
  1. Nature of matter and kinetic theory of matter.
  • Effects of heat on matter
    • Melting
    • Evaporation
    • Condensation
    • Freezing
    • Freezing
    • Sublimation
  1. Permanent and non-permanent changes
  2. Constituents of matter
  • Atoms
  • Elements
  • Molecules
  • Compounds
  1. Names and symbols of common elements
  2. Simple word equations.

 

 

 

 

 

 

 

 

 

 

Elements compound and mixtures.

(a). Element:

– Is a pure substance that cannot be split up into simpler substances by chemical means.

Examples: copper, hydrogen, carbon.

 

(b). Compound:

– A pure substance that consists of two or more elements that are chemically combined.

Examples:

 

Compound Elements in the compound
Calcium carbonate Calcium, carbon and oxygen
Sodium chloride Sodium and chlorine
Ammonium nitrate Nitrogen, hydrogen, oxygen
Iron (II) sulphate Iron, sulphur, oxygen

 

(c). Mixture:

– A substance that consists of two or more elements or compounds that are not chemically combined

– Some mixtures can be naturally occurring while some are artificial.

 

Examples

Naturally occurring mixtures.

 

Mixture Components
Air Nitrogen, oxygen, carbon (IV) oxide, water vapour, noble gases etc
Sea water Water and various salts like chlorides of sodium, potassium and magnesium
Crude oil A mixture of hydrocarbons like methane, petrol, bitumen, etc
Magadi soda Sodium carbonate, sodium hydrogen carbonate and sodium chloride

 

Artificial mixtures.

 

Mixture Components
Soft drinks Water, citric acid, sugar, carbon (IV) oxide, stabilizers, sodium benzoate
Black ink Blue, black, yellow dyes and solvent
Cement Oxides of aluminium, iron, silicon, calcium and calcium carbonate.

 

Types of mixtures:

– There are two types of mixtures;

  • Homogenous mixtures
  • Heterogenous mixtures

 

(i). Homogenous mixtures.

– Is a mixture with a uniform composition and properties throughout its mass.

– The parts (components) of the mixture are uniformly distributed throughout the mixture

Examples:

Tea with sugar solution.

 

(ii). Heterogenous mixture:

Is a mixture without uniform composition throughout its mass.

Examples:

– Soil, rocks and sand mixture.

 

 

Separating mixtures.

A mixture can be separated into its various components (constituents) by appropriate physical means, depending on type of mixture.

 

Basic concepts:

Residue: solid that remains on the filter paper during filtration

 

Filtrate: liquid that passes past the filter paper during filtration

 

Solute: a solid that dissolves in a particular liquid

 

Solvent: the liquid in which a solute dissolves.

 

Saturated solution: a solution in which no more solute can dissolve at a particular temperature

 

Unsaturated solution: a solution that can take more of the solute (solute) at a particular temperature.

 

Miscible liquids: liquids that can mix together completely.

 

Immiscible liquids: liquids that cannot mix together completely.

 

There are various methods that can be used to separate mixtures.

These include:

  • Decantation
  • Evaporation
  • Condensation
  • Filtration
  • Crystallization
  • Separating funnel separation
  • Distillation (simple and fractional)
  • Sublimation
  • Chromatography
  • Solvent extraction
  1. Decantation:

– Is a method used to separate insoluble solids from liquids; a heterogenous mixture.

 

Procedure:

– The solid-liquid mixture is allowed to stand in a container.

– The insoluble solid settles at the bottom and the upper liquid portion poured out with care.

 

Apparatus.

 

 

 

 

 

 

 

 

Examples:

– Separation of sand-water mixture

– Separation of maize flour-water mixture.

 

Limitations (disadvantages) of decantation.

– It is not efficient as some fine suspended solids may come long with the liquid during pouring.

 

  1. Filtration.

– Is the separation of an insoluble solid from a heterogenous mixture (liquid) using a porous filter that does not allow the solids to pass through.

– Upon filtration the undissolved solid is left on the filter paper and is called the residue.

– The liquid that passes the filter paper is called filtrate.

 

Examples: separation of sand from water.

(i). Procedure.

– The filter paper is folded into ¼ and opened to from a cone.

 

Diagram: folding a filter paper.

 

 

 

 

 

 

 

– It is carefully placed inside a filter funnel.

– The apparatus are then arranged as shown blow.

– The sand-water mixture is then poured into the filter paper in the filter funnel.

– The collecting liquid is directed into a conical flask.

 

(ii). Apparatus.

 

 

 

 

 

 

 

 

 

 

Applications of filtration.

– Filtration of domestic water.

– Extraction of medicinal substances from plants.

– Extraction of sugar from sugarcane.

– Operation of a vacuum cleaner.

– Fuel filters in automobile engines.

 

 

  1. Evaporation.

– Is used to separate a soluble solid from its solution.

– Such solutions are usually homogenous mixtures.

– The solid is called a solute while the liquid is called a solvent.

 

Example: separation of salt from salt solution.

(i). Procedure:

– The salt solution is poured in an evaporating dish.

– The set up is then arranged as in the apparatus shown below.

– The solution is boiled under steam or sand bath until all the water in the salt solution evaporates and salt crystals remain in the dish.

 

(ii). Apparatus.

 

 

 

 

 

 

 

 

 

(iii). Observations and explanations.

– Upon heating the solution, water evaporates because it has lower boiling point than the salt.

– The solution is boiled until salt crystals start appearing on a glass rid dipped into the solution.

– This shall indicate that the solution is saturated.

– The saturated solution is allowed to cool and crystallize.

– The mother liquor (liquid that remains with the crystals) is poured and the salt (solid) dried between absorbent papers.

 

Note:

A crystal: is a solid that consists of particles arranged in an orderly repetitive manner.

 

– It is advantageous to boil the solution under a steam or sand bath rather than directly.

Reason:

– The steam or sand bath prevents the mixture from splashing out (spitting) of the evaporating dish.

– It also reduces chances of the evaporating dish cracking.

 

Applications of evaporation:

– Extraction of soda ash from Lake Magadi.

 

 

 

 

 

 

 

 

 

  1. Crystallization and recrystallization.

(a). Crystallization:

– is the process of formation of crystals from a solution.

– It involves evaporation of the solution to form a concentrated solution.

 

Example: crystallization of potassium nitrate from its solution.

(i). Procedure:

– About 5g of powdered potassium nitrate is added to 10cm3 of water in a boiling tube.

– The solution is heated until all the solid dissolves and then allowed to cool and crystallize.

Note:

– More potassium nitrate dissolves in hot water than in cold water.

– The resultant solution is then heated until crystals start appearing; and this can be confirmed by dipping a glass rod into the solution and feeling for crystals.

– This is called a saturated solution i.e. a solution that cannot take in any more of the solute at a given temperature.

– The saturated solution is then allowed to cool and crystallize.

 

(ii). Observations:

– The resultant solid particles have definite shapes.

– Some are needle-like while others are flat and sharp-edged.

– These are the potassium nitrate crystals.

 

(b). Recrystallization:

– Is used in obtaining pure crystals from a soluble solid containing impurities.

– Involves filtration and evaporation.

 

Examples:

Obtaining pure copper (II) sulphate crystals from impure copper (II) sulphate.

Purification of rock salt.

 

Note: The process can be enhanced by suspending a small piece of pure crystal into the saturated solution.

 

Diagram: recrystallization of copper (II) sulphate.

 

 

 

 

 

 

 

Applications of crystallization.

– Separation of Trona from sodium chloride in Lake Magadi.

 

  1. Distillation.

– Is the vapourisation of a liquid from a mixture and then condensing the vapour.

– Is used in the purification of liquids and separation of liquids from a mixture.

– It utilizes the differences in boiling points of the components of the mixture.

– Are of two types:

 

  • Simple distillation
  • Fractional distillation.

(i). Simple distillation.

– Is mainly used for purification of liquids containing dissolved substances.

– It is also useful in separating two miscible liquids with widely differing boiling points

 

Note:

Miscible liquids: Liquids that mix to from a uniform a uniform homogenous solution

– The liquid with the lower boiling point usually distills over first, and is collected.

 

Example: To obtain pure water from sea water.

(i). Procedure:

– Salty sea water is poured into a distillation flask.

– A few pieces of pumice or porcelain is added to the solution.

Reason:

– To increase the surface area fro condensation and evaporation.

 

– The solution is heated until it starts boiling, then the burner removed so that the liquid boils gently.

– The boiling goes on until the liquid (distillate) starts collecting in the beaker.

 

(ii). Apparatus.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(iii). Observations and explanations:

– The water boils and the resultant steam is passed through the Liebig condenser.

– As the vapour passes through the condenser, it is cooled by circulating cold water through the jacket of the condenser.

– The cold water enters through the lower bottom and leaves through the top upper part.

Reason:

– To provide more time for the cold water to condense the vapours.

 

– The distillate is collected in the beaker while the residue remains in the distillation flask.

 

Applications of simple distillation.

– Manufacture of wins and spirits.

– Desalinization of sea water to obtain fresh water.

(ii). Fractional distillation:

– Is a method used fro the separation of miscible liquids with very close boiling points.

Examples:

Ethanol and water.

 

– It is a modification of simple distillation in which the fractionating column is inserted on top of the distillation flask.

– All the components must be volatile at different extents in order for separation to be possible.

 

The fractionating column.

– Is usually an elongated (glass) tube, packed with pieces of glass beads or pieces of broken glass.

 

Role of glass beads.

– To increase the surface area for vapourisation of the various components of the mixture and allow the separation of the vapours to occur.

– Thus the more the glass beads in the fractionating column, the higher the efficiency of separation.

Note:

– The efficiency of the fractional distillation so s to get more pure components can also be done by:

  1. Increasing the length of the fractionating column (making it longer)
  2. Making the fractionating column narrower (decreasing the diameter)

 

Volatile liquids:

– Are liquids with the ability to change into vapour.

– More volatile liquids vapourize and condense faster than the less volatile liquids.

 

Note:

– During fractional distillation, the components of a mixture are collected at intervals, one at a time with the most volatile (lowest boiling point) coming out first.

– Each component collected in the receiver is called a fraction.

 

Example: separation of ethanol and water.

(i). Apparatus.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(ii). Procedure:

– Water-ethanol mixture is poured into a round-bottomed flask.

– The apparatus is then connected and set up as shown below.

 

Note:

– The thermometer bulb must be at the vapour outlet to the condenser.

Reason:

– For accurate determination of the vapourisation temperature for each fraction.

 

– The mixture is then strongly heated until the first fraction comes out of the distillation flask into the conical flask.

– Collection of the fractions should be done in a conical flask other than in a beaker.

Reason:

– To reduce the rates of evaporation of the fractions, especially the highly volatile ones (in this case ethanol)

 

– For this particular separation the first temperatures recorded by the thermometer should not exceed 80oC; to ensure that the first fraction is only ethanol.

 

(iii). Discussion.

– Ethanol boils at 78oC and water boils at 100oC.

– When the mixture is heated, ethanol and water evaporate and pass through the fractionating column which is filled with glass beads to offer a large surface area

– The large surface area encourages evaporation of ethanol and condensation of water vapour.

– Water can be seen dropping back into the distillation flask.

-Ethanol vapour passes through the condenser and warm liquid ethanol is collected in the conical flask.

 

Note:

– The first portion is almost pure ethanol (about 97%) and burns quietly with a blue flame.

– It also has the characteristic smell of alcohols.

 

Industrial applications of fractional distillation.

– Separation of air into various components in BOC gases Kenya limited.

– Separation of crude oil into paraffin, petrol, kerosene diesel and other components in the Kenya oil refinery.

– Distillation of ethanol from molasses at Muhoroni Agro-chemicals company.

 

  1. Sublimation.

– Is the process by which a solid changes directly to gaseous state upon heating.

– It is used to separate a mixture in which one of the components sublimes on heating.

Note:

Solid                           Gas

 

Solids that sublime have very weak forces of attraction between the atoms and hence are easily broken on slight heating.

 

 

 

 

Examples of solids that sublime on heating.

– Iodine; sublimes to from a purple vapour.

– Ammonium chloride; sublimes to from dense white fumes;

– Solid carbon (IV) oxide (dry ice);

– Anhydrous iron (III) chloride; sublimes to give red brown fumes.

 

Examples: separation of iodine from sodium chloride.

(i). apparatus:

 

 

 

 

 

 

 

 

 

 

 

(ii). Procedure:

– The iodine-common salt mixture is poured into a beaker and placed in a tripod stand.

– A watch glass full of cold water is placed on the beaker.

– The beaker is heated gently until some dense purple fumes are observed.

 

(iii). Observations:

– A purple vapour appears in the beaker.

– A dark-grey shiny solid collects on the bottom of the watch-glass.

– White solid remains in the beaker.

 

(iv). Explanations:

– Upon heating the mixture iodine sublimes and condenses on the cold watch glass to form a sublimate of pure iodine.

Note:

– Solid carbon (IV) oxide (dry ice) is used a s refrigerant by ice cream and soft drink vendors.

Reason:

– It sublimes on heating; as it sublimes it takes latent heat from ice cream (soft drinks) thus leaving it cold.

– It is also advantageous as it does not turn into liquid, which could be cumbersome to carry and would mess up the ice cream.

 

  1. Chromatography.

– Is the separation of coloured substances using an eluting solvent.

– It is also used to identify the components of a coloured substance.

– It involves the use of a moving liquid (eluting solvent) on a material that absorbs the solvent.

– It involves two major processes:

  • Solubility:

The tendency of a substance to dissolve in a solvent.

  • Adsorption:

The tendency of a substance to stick on an adsorbent material.

Examples:

  1. Separation of components of black ink.

(i). Procedure:

– A filter paper is placed on the rim of an evaporating dish or a small beaker.

– A drop of the black ink is placed at the centre of the filter paper; allowed to spread out and dry.

– A drop of water (ethanol) is then added to the ink and allowed to spread.

– After complete spread of the drop, a second drop is added.

– Water drops are added continuously until the disc of coloured substances almost reaches the edge.

 

(ii). Observations:

 

 

 

 

 

 

 

 

Note: The dry filter paper showing the separated components of a mixture is called a chromatogram.

 

(iii). Explanations:

– Water is the eluting solvent since ink is soluble in it.

– The various dyes in the black ink move at different distances from the black spot hence the bands.

Reasons:

– The dyes have different solubilities in the solvent; the more soluble the dye, the further the distance it travels on the absorbent paper

– They have different rates of adsorption i.e. the tendency of the dyes to stick on the absorbent material; dyes with low rates of absorption travel far from the original spot.

 

Note:

Solvent front.

– Is the furthest distance reached by the eluting solvent on the filter paper.

 

Baseline:

– The point at which the dye to be separated is placed; i.e. it is the starting point of separation.

 

  1. To verify contents of red, black and blue inks.

(i). apparatus:

 

 

 

 

 

 

 

 

 

 

 

(ii). Observations:

 

 

 

 

 

 

 

 

 

 

 

(iii). Explanations:

– The mixtures A to D have various components with varying solubilities in the solvent (ethanol)

– Mixture D is the most pure because it has only one spot.

– Mixture C is the least pure (most impure), as it has the highest number of spots indicating it is composed of so many dyes (four)

– Mixture C has the most soluble dye; its last component is the one nearest to the solvent front.

– Mixtures with similar dyes in their composition have spots at same levels; in this case A, B and D.

 

Applications of chromatography.

– Purification of natural products such as hormones, vitamins and natural pigments.

– Detection of food poisons e.g. in canned foods and soft drinks.

 

  1. Solvent extraction.

– Is the extraction of a solute from its original solvent by using a second solvent in which it has a higher solubility

 

Example: extraction of oil from nuts.

(i). Apparatus.

 

 

 

 

 

 

 

 

 

 

(ii). Procedure:

– Some nuts are crushed in a mortar using a pestle; to increase the surface area for solubility.

– A suitable solvent such as hexane or propanone (acetone) is added.

– The nuts are further crushed in the solvent.

– The resultant solution is decanted in an evaporating dish, and left in the sun to evaporate.

– The liquid remaining in the evaporating dish is smeared onto a clean filter paper.

 

(iv). Observations;

– A permanent translucent mark appears on the filter paper.

(v). Explanations:

– The nuts are crushed when in contact with the solvent to bring more of the oil in the nuts closer to the solvent.

– Upon evaporation oil is left behind because it has a higher boiling point than the solvent.

– A permanent translucent mark verifies the presence of oils.

 

Applications of solvent extraction.

– Used by dry-cleaners to remove dirt (grease) and stains from “dry-clean-only” clothes such as sweaters, suits, dresses etc.

 

  1. Separating funnel separation.

– Is used fro separating a mixture with two or more immiscible liquids.

– Such liquids do not mix but instead form layers based on their densities.

– The heaviest liquid layer is found at the bottom of the separating funnel; while the lightest liquid is found at the top of the separating funnel.

– The liquids are drained one after the other by opening and closing the tap of the separating funnel.

 

Example: Separation of oil from water

Apparatus:

Beakers, separating funnel, paraffin oil, distilled water, rubber stopper.

 

Apparatus.

 

 

 

 

 

 

 

 

 

 

Procedure.

– The tap of the separating funnel is closed.

– Equal volumes of water and paraffin are put in a separating funnel until it is half full.

– The mouth of the funnel is closed with a stopper and the mixture shaken.

– The mixture is allowed to stand until two distinct layers are formed.

– The stopper is removed and the tap opened to allow the bottom layer to drain into the beaker.

– The tap is closed after most of the bottom layer has drained off.

– The beaker is removed and the rest of the bottom layer is drained into a separate container and discarded; to ensure that no part of the top layer (paraffin) gets into the beaker containing the bottom (water) layer.

– The other (top) layer is then drained into another beaker.

 

Observations:

– After the mixture has settled oil and paraffin separate into two layers.

– The first beaker contains only water; while the second beaker contains only paraffin.

 

 

Conclusion.

– Paraffin and water are immiscible.

– The top layer contains water which is denser while the top layer contains oil (paraffin) which is lighter.

 

Practical application:

– Extraction of useful substances from complex mixtures.

 

Example: Separation of iodine from sodium chloride.

******************************

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. Use of magnets.

– Is used to separate solid mixtures, one of which is magnetic (usually iron).

– The iron is picked with a magnet leaving the other components of the mixture behind.

 

Examples:

– Separation of powdered iron from iron powder-sulphur mixture.

 

Practical applications:

-In the extraction (mining of iron); where magnetic iron ore is separated from other materials in the crushed ore.

– Separation of scrap iron from non-magnetic materials like glass and plastics in recycling plants.

 

Criteria for purity.

  • Pure substance:

– Is a substance that contains only one type of compound or element.

 

Determination of purity

(a). Solids:

– Purity of solids is determined by measuring the melting point.

– Pure solids melt sharply over a narrow temperature range.

 

Examples:

– Naphthalene melts at 80oC – 81oC.

– Water melts at 0oC.

 

Effects of impurity on melting point.

– Impurities lower the melting point of a substance making it melt over a wide range of temperatures i.e. the melting point is not sharp.

 

Applications:

  1. Extraction of metals.

– Impurities are added to purified metal ores to lower their melting points, hence save energy and extraction costs.

  1. To prevent knocking of engines due to freezing of water in car radiators during cold seasons, impurities like ethylene glycol or salt (NaCl) is added.
  2. Defrosting of frozen roads and sidewalks in temperate countries, by sprinkling a salt such as sodium chloride.

 

(b). Liquids:

– Purity of a liquid is determined by measuring its boiling point.

– A pure liquid has a sharp boiling point.

 

Examples:

Pure water boils at 100oC at 1 atmospheric pressure.

Pure ethanol boils at 78oC.

 

 

 

 

Effect of impurity on boiling point.

– Impurities raise the boiling point of a liquid.

Example:

– Sea water boils at a higher temperature than pure water due to the presence of dissolved salts.

 

States of matter.

  • Matter:

– Matter is anything that occupies space and has mass.

– It is composed of either pure substances or a mixture of substances.

 

States of matter:

Matter exists in three states:

  • Solid
  • Liquid

 

Kinetic theory of matter:

– States that matter is made up of small particles which are in continuous random motion.

– The continuous random motion of particles in matter is called Brownian motion.

– The rate of movement of particles in matter depends on the state

 

(a). Solid state:

– Have closely packed particles held by strong forces of attraction.

– Particles do not move from one point to another but vibrate about a fixed position.

– Upon heating, they start to vibrate vigorously.

Reason:

– Due to increase in their kinetic energy.

 

(b). Liquid state:

– Particles are further apart from one another than those in the solid state

– Forces of attraction between the particles are weaker than those in the solid.

– The particles remain close, but are free to move from one position to another.

– On heating they gain energy and move rapidly.

 

(c). Gaseous state:

– The particles are far apart and free to move randomly in all directions.

– Consequently they lack definite shape and volume, but occupy the whole space within a container.

– Forces of attraction between the particles are very weak.

 

The effect of heat on substances.

– A substance can change from one state to another upon heating or cooling.

– These are physical changes and can be reversed.

– There are five processes involved in changes of state:

  • Melting
  • Evaporation
  • Condensation
  • Freezing
  • Sublimation

 

The processes involved in change of states of matter.

(i). Melting:

– Is a change of state from solid to liquid.

  • Melting point:

– Is the constant temperature at which the melting takes place.

 

Examples:
Ice melts at 0oC.

– Sodium chloride melts at 800oC.

 

– During melting the energy supplied to the particles is used to weaken the forces of attraction so that particles can move about.

 

(ii). Vapourisation (evaporation)

– Is change of state from liquid to gas.

  • Boiling point:

– Is the constant temperature at which a liquid changes from solid to a gas.

 

– During boiling, the energy supplied is used to break the forces of attraction in the liquid thus moving the particles far away from each other.

 

Examples:

Water boils at 100oC, while ethanol boils at 78oC at one atmospheric pressure.

Reason:

The forces of attraction between the water particles are stronger than those of ethanol.

 

(c). Condensation:

– Is the change of state from a gas to a liquid.

– Is a change due to decrease in temperature.

– When the temperature of gas is decreased, the particles lose kinetic energy to the surroundings to move slowly.

– The attractive forces become stronger, and the sample changes to a liquid.

Note:

– The temperature at which condensation occurs is the same as the boiling point.

 

(d). Freezing:

– Is the change from a liquid to a solid.

– It is also due to decrease in temperature.

– When a liquid is cooled, the particles lose energy and move very slowly.

– They attract one another strongly, and ultimately remain in fixed positions.

Note:

– The freezing point is the same as the melting point.

 

(e). Sublimation.

– Is the process whereby a solid does not melt when heated, but changes directly to the gaseous state.

 

Example:

– Iodine solid changes to purple vapour when heated to 70oC.

– Dry ice (solid CO2), used to cool ice cream, evaporates without leaving a liquid.

Note:

– The reverse of sublimation, whereby a gas changes directly to solid is called deposition.

 

Summary on changes of state.

E

 

Solid                           Liquid                         Gas

 

 

 

F

Key: A: melting; B: vapourisation; C: Freezing; D: Condensation; E: Sublimation; F: Deposition;

 

Experiment: Investigating changes in temperature when ice is heated.

(i). Procedure:

– A 250ml beaker is half-filled with dry ice, and the initial temperature recorded.

– The ice is heated, while stirring with a thermometer and the temperature recorded every 30 seconds.

– Heating and recording is done until the resultant water starts to boil.

– A graph of temperature against time is plotted.

 

(ii). Results:

 

Temperatures (oC) -10 0 30 60 90 120 150 180 210
Time (seconds)                  

 

(iii). Graph:

Effect of heat on pure ice

 

 

 

 

 

 

 

 

 

 

 

 

 

(iv). Explanations:

  • Point A-B:

– As the ice is heated the temperature rises steadily from -10oC to 0oC.

Reason:

– The heat supplied increases the kinetic energy of the ice (solid water) molecules; collisions between them hence increased temperature.

 

  • Point B-C:

-The temperature of the ice remains constant even as heat is applied.

Reason:

– Heat supplied is used to break the forces of attraction between the water molecules in ice.

– This is the melting point hence at B-C the ice melts.

  • Points C-D:

– At C, all the ice has already melted (turned to water).

– Between C and D, the temperature of the water increases as heating continues.

Reason:

– The heat supplied increases the kinetic energy of the water molecules; their rate of collision increases hence increased temperatures.

 

  • Points D-E:

– Temperature of the water remains constant even as heat is being supplied.

– Point D-E is the boiling point i.e. 100oC.

Reason:

– The heat supplied is used to break the forces of attraction between water molecules in the liquid.

 

  • Points E-F:

– At point E, all the liquid water has turned into vapour.

– Thus between E and F, the temperature of the vapour rises as heat is applied.

 

Note: Heating curve for an impure solid.

Effect of heat on impure ice

 

 

 

 

 

 

 

 

 

 

 

 

 

The cooling curve.

– Is a curve that shows how the temperature of a substance changes with time as it is cooled from a gas into a solid.

– It is the opposite of a heating curve.

 

Example: the cooling curve of water

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Explanations:

  • Points A-B:

– Gaseous state; temperature is declining.

– Heat loss results into decrease in kinetic energy of the gaseous molecules.

 

  • Points B-C:

– This is the condensation point;

– The water vapour condenses to the liquid state; bonds are formed as the hest is lost.

 

  • Points C-D;

– Temperature of the liquid water is declining;

– Heat loss results into decrease in kinetic energy of the liquid molecules.

 

  • Points D-E;

– This is the freezing point;

– The temperature remains constant as the heat is being lost;

– The heat being lost results into bond formation; as the liquid forms a solid.

 

  • Points E-F;

– The water is now in solid state.

– The temperature of the solid declines as heat is being lost.

 

Permanent and temporary changes.

– Heat causes matter to change.

– Changes due to heat can either be permanent or non-permanent (temporary).

 

(a). Permanent changes.

– Are also called chemical changes.

– Involves substances that are relatively less stable to heat.

 

Characteristics of permanent (chemical) changes.

(i). New substances are formed.

(ii). Involves considerable heat changes; energy is either given out or absorbed.

(iii). The mass of the substance changes.

 

Types of permanent (chemical) changes.

– Chemical changes are of two main types:

  • Reversible permanent changes.
  • Irreversible (non-reversible) permanent changes.

 

(i). Reversible permanent changes.

– Are chemical changes in which the final new products can recombine to form the original substance, under certain conditions.

 

Generally:

A + B                          C + D

 

Reactants                                         Products

 

 

Examples:

  1. Effect of heat on hydrated blue copper (II) sulphate.

(i). Apparatus:

– Test tubes, Bunsen burner, test tube holder

– Hydrated copper (II) sulphate

 

(ii). Apparatus:

 

 

 

 

 

 

 

(ii). Procedure:

– Dry crystals of hydrated blue copper (II) sulphate are put in a clean dry test tube.

– The apparatus are arranged as above.

– The copper (II) sulphate is heated until no further change.

– The delivery tube is removed from the collected liquid while heating continues.

Reason:

– To avoid sucking back of the condensing liquid which would otherwise rehydrate the anhydrous copper (II) sulphate

 

– The test tube is allowed to cool and the remaining solid is divided into two portions.

– To one portion of the powder, add distilled water, while to the other potion add the condensed liquid.

 

(iii). Observations:
– A white solid/ powder remains in the test tube after heating.

– A colourless liquid condenses in the test tube dipped into the ice cold water.

– The colourless liquid turns the white solid into blue.

 

(iv). Explanations:
– Hydrated blue copper (II) sulphate have water of crystallization, giving it the characteristic blue colour.

– During heating, the heat energy supplied is used to drive out the water molecules (particles) out of the crystals;

– Without water, the copper (II) sulphate turns white and thus called anhydrous copper (II) sulphate;

– The water driven out of the crystals condenses in the test tube immersed in the ice cold water.

 

Heat

Equation

Cool

Hydrated copper (II) sulphate                                        anhydrous copper (II) sulphate + Water

                 Blue                                                                                                                  White

 

Heat

Cool

 

 

 

Chemically:
CuSO4.5H2O(s)            CuSO4(s) + 5H2O(g);

 

Confirmatory test for water.

  • It turns white anhydrous copper (II) sulphate to blue;
  • It changes blue cobalt chloride paper pink;

Confirmatory test that the water is pure:

  • It boils at 100oC;
  • It melts at 0oC;
  • It has a density of 1g cm-3;
  • It has a refractive index of 1.33;

 

 

Note:
– The delivery tube is removed from the collecting liquid while heating is continued; to ensure that no water condenses back into the copper (II) sulphate as this would cause rehydration;

– Addition of water to the anhydrous copper (III) sulphate changes its colour from white to blue;

 

Conclusion:

The effect of heat on copper (II) sulphate is a reversible chemical change;

 

  1. Effect of heat on ammonium chloride.

– When ammonium chloride is heated, it produces ammonia gas and hydrogen chloride gas.

– These are seen as dense white fumes.

– Reversible, when ammonia and hydrogen chloride are gases are reacted or allowed to cool, they produce ammonium chloride;

 

Heat

In summary:

Cool

Ammonium chloride solid                                        ammonium chloride + hydrogen chloride;

                         White                                                                                                                                 Dense white fumes

 

Heat

Cool

 

 

 

Chemically:
NH4Cl(s)                         NH3(g) + HCl(g);

     Ammonium chloride                        Ammonia gas       hydrogen chloride gas

 

  1. Decomposition of calcium carbonate.
Heat

In summary:

Cool

Calcium carbonate solid                               Calcium oxide + Carbon (IV) oxide;

 

(ii). Irreversible chemical change.

– Are chemical changes in which the resultant products cannot recombine to form the original substance (reactants);

– Majority of the chemical changes are irreversible;

 

Generally:
A + B                    C + D

Reactants                               Products

 

 

 

 

 

 

 

 

 

Examples:
1. Action of heat on potassium manganate (VII)

(i). Apparatus and chemicals.

– Bunsen burner, test tube, trough, wooden splint;

– Potassium managnate (VII);

 

(ii). Apparatus set up.

Heat
Potassium manganate (VII)

(iii). Procedure:
– 2 end-fulls of a spatula of potassium manganate (VII) are put in a hard glass test tube;

– The set up is assembled as shown above;

– The solid potassium manganate (VII) is heated, and the resultant gas collected over water;

– The resultant gas(es) is tested with a glowing splint;

 

(iv). Observations:

– The purple solid turns black;

– A colourless gas collects over water;

– The colourless relights a glowing splint;

 

(v). Explanations:
Potassium manganate (VII), a purple solid was decomposed (splint up) on heating to yield (give potassium manganate (III) and oxygen.

– The potassium manganate (III) is the black residue;

– The colourless gas is oxygen; and relighting a glowing splint is the confirmatory test;

 

In summary:
Potassium manganate (VII) → Potassium manganate (III) + oxygen

Purple solid                                                                       Black solid                                         Colourless gas

 

Note: It is not possible for oxygen and potassium manganate (III) to recombine back to potassium manganate (VII); hence the change is irreversible;

 

  1. Thermal decomposition of copper (II) nitrate.

– The blue solid decomposes to form a black solid; copper (II) oxide, red-brown fumes of nitrogen (IV) oxide and a colourless gas, oxygen;

 

In summary:
Copper (II) nitrate → Copper (II) oxide + Nitrogen (IV) oxide + oxygen

Blue solid                                           Black solid                          Brown (red-brown) fumes           Colourless gas

 

 

Note: Further examples of chemical changes

– The burning of any substance (except platinum);

– The rusting of iron;

– Addition of water to calcium oxide;

– Explosion of natural gas or hydrogen with air;

– Reacting of sodium in water;

 

Note: Exothermic and endothermic reactions:

(i). Exothermic reactions.

– Are reactions in which heat is released // given out to the surrounding;

– Usually the final temperature of the reaction vessel // mixture (e.g. beaker is higher than initially;

 

Examples:
– Freezing;

– Condensation;

– Deposition // sublimation of fumes to solid;

 

(ii). Endothermic reactions.

– Are chemical reactions in which heat is absorbed from the surrounding;

– The final temperature of the reaction vessel or reaction mixture is usually lower than the initial i.e. they are accompanied by a drop in temperature;

 

Examples.

– Melting;

– Vapourization;

– Sublimation (of solid to gas)

 

Summary on chemical changes.

Reaction Appearance of substance Changes during reaction New substance(s) Type of change
Heating hydrated copper (II) sulphate Blue Blue crystals turn into a white powder; colourless liquid condenses on cooling;

 

Anhydrous copper (II) sulphate and water Chemical
Heating potassium manganate (VII) Shiny purple crystals The purple solid turns black; evolution of a colourless gas;

 

Potassium manganate (III) and oxygen; Chemical;
Heating ammonium chloride White solid // powder; Dense white fumed that cools to a white solid;

 

Ammonia gas and hydrogen chloride gas; Chemical;
Heating lead (II) nitrate; White solid The white solid turns into a red solid during heating which on cooling turns yellow;

– Decrepitating sound;

– Brown fumes;

– colourless gas;

 

– Lead (II) oxide; nitrogen (IV) oxide and oxygen gas; Chemical;
Heating lead (II) nitrate; White solid The white solid turns into a yellow solid during heating which on cooling turns white;

– Decrepitating sound;

– Brown fumes;  colourless gas;

– Zinc (II) oxide; nitrogen (IV) oxide and oxygen gas; Chemical;
Heating copper turnings Brown turnings;

 

– Brown turnings // solid turn black; – Copper (II) oxide; Chemical;
Rusting of iron; Grey solid – Grey solid turns into a red brown solid;

 

– Hydrated iron (III) oxide; Chemical;
Heating Copper (II) nitrate; Blue solid The blue solid turns into a black solid;

– Brown fumes;

– colourless gas;

 

– Copper (II) oxide; nitrogen (IV) oxide and oxygen gas; Chemical;
Heating copper (II) carbonate Green solid The green solid turns into a black solid;

– colourless gas;

– copper (II) oxide and carbon (IV) oxide; Chemical;

 

(b). Temporary (non-permanent) changes.

– Are also called physical changes;

– They are changes that involve substances that are more stable to heat;

– On heating they do not decompose hence no new substances are formed;

 

Characteristics of permanent changes.

  1. All are reversible upon changes in temperature;
  2. No new substance is formed (instead there are only changes of state);
  3. The mass of the substances do not change;

 

Examples:

Solid Original appearance Observations during heating then cooling
1. Candle wax White sticky solid; – The solid melts into a colourless viscous liquid; and on cooling solidifies to the original solid wax again;
2. Iodine solid. Shiny dark-grey crystals; – The solid turns directly to purple vapour (sublimation);

– On cooling the purple iodine vapour (gas) changes directly to solid iodine (deposition);

 

ie. Iodine solid ═                iodine vapour;

Dark grey                          Purple

 

3. Zinc oxide White solid; – The white solid turns yellow on heating and upon cooling changes back to the original white colour;

 

ie. Zinc oxide ═                Zinc oxide;

White (cold)                          Yellow (hot)

 

4. Ice White – The solid water melts into liquid and on further heating the liquid vapourizes and turns into gas;

– On cooling the gas condenses to liquid which then freezes back into solid;

 

i.e Ice ═                  Water ═                  Gas

 

5. Platinum wire;   – A white glow of the metal is seen on heating, but on cooling the metal changes back to its original grey colour;

 

6. Lead (II) oxide; Yellow – The yellow solid turns red on heating and upon cooling changes back to the original yellow colour;

 

ie. Lead (II) oxide ═             Lead (II) oxide;

White (cold)                                   Yellow (hot)

 

Differences between Physical and Chemical changes.

 

Physical change Chemical change
1. Produces no new kind of substance; – Always produces a new kind of substance;
2. UIs usually (generally) irreversible; Are generally irreversible; with only few exceptions (i.e. most are irreversible);
3. The mass of the substance does not change; – The mass of the substance changes;
4. No energy is given out or absorbed i.e. are not accompanied by great heat changes; – Energy is usually given out or absorbed i.e are usually accompanied by great heat changes;

 

Constituents of matter

– A detailed examination of matter reveals that it is built of very tiny units called toms;

– Presently about 115 atoms have been identified;

– The arrangement and number of atoms in a substance will result into other much larger constituents of matter;

– These are:

  • Elements;
  • Molecules;
  • Compounds;
  • Mixtures;

 

  1. The atom;

– Is the smallest particle of matter that can take part in a chemical reaction;

– It is the smallest particle into which an element can be divided without losing the properties of the element;

– Atoms of various elements all differ from one another;

Examples:
– Copper is made up of many copper atoms;

– Sodium element is made up of many sodium atoms;

 

  1. Elements.

– An element is a substance that cannot be split into anything simple by any known chemical means;

– An element consists of a single type of atom;

– There are about 155 known elements, 90 of which occur naturally.

– Elements are classified into two main groups;

 

  • Metals:

– All are solids at room temperature (except mercury); and are good conductors of electricity;

 

  • Non-metals;

– Exists as solids and gases;

– All are poor electric conductors except graphite;

 

Examples of elements.

(i). Metals:

– Sodium, magnesium, potassium, aluminium, lead, iron, zinc, silver, gold, tin, platinum, uranium, calcium, manganese etc.

 

(ii). Non-elements.

– Carbon, nitrogen, sulphur, oxygen, chlorine, fluorine, argon, neon, bromine, iodine, silicon, boron, xenon, krypton.

  1. Molecule.

– Is the smallest particle of a substance that can exist independently;

– It is made when 2 or more atoms (similar or dissimilar) are chemically combined together;

– However atoms of noble/ inert gases exist as single atoms;

 

Note:

– Depending on number of atoms molecules can be categorized into:

(i). Monoatomic molecule;

– made up of only one atom;

Examples:
– Argon;

Argon atom     Argon molecule;

– Neon;

– Helium;

 

(ii). Diatomic molecules.

– Made up of 2 similar atoms; chemically combines;

Examples

– Oxygen gas;

  Oxygen atom                Oxygen molecule;

– Nitrogen gas;

– Hydrogen gas;

– Chlorine gas;

 

(ii). Triatomic molecules.

– Made up of 3 similar atoms; chemically combines;

Examples

– Ozone molecule;

 

  Oxygen atom                Ozone molecule;

 

Note:

– Other molecules are also made from atoms of different elements chemically combined together;

Examples:
(i). Hydrogen chloride;

+        →

Hydrogen   Chlorine        Hydrogen chloride molecule;

           Atom.        Atom;

 

(ii). Water molecule;

+        +                     →

               

          Hydrogen atoms     oxygen atom;                      Water molecule;

 

 

 

 

 

 

  1. Compounds.

– A compound is a pure substance consisting of two or more elements that are chemically combined.

– Compounds usually have different properties from those of its constituent elements;

– Properties of a compound are uniform throughout any given sample and from one sample to another;

 

Examples of compounds and their constituent elements.

 

Compound Constituent elements
Colourless water liquid; Oxygen and hydrogen;
Green copper (II) carbonate; Copper, carbon and oxygen;
White sodium nitrate; Sodium, nitrogen and oxygen;
Black copper (II) oxide; Copper and oxygen;
Blue copper (II) nitrate; Copper, nitrogen and oxygen;
Blue copper (II) sulphate; Copper, sulphur, oxygen and hydrogen;
Ammonium chloride; Nitrogen, hydrogen and chlorine;

 

Note:

– Carbonates are derivatives of (derived from or made of) carbon and oxygen;

– Nitrates are derivatives of nitrogen and oxygen;

– Sulphates are derivatives of sulphur and oxygen;

– Hydrogen carbonates are derivatives of hydrogen, carbon and oxygen;       

 

  1. Mixtures

– A substance that consists of two or more elements or compounds that are not chemically combined.

 

Characteristics of mixtures.

– It properties are the average of the properties of its elements;

– Its components can be separated by physical means e.g. filtration, magnetism, distillation etc.

– Its components are not necessarily if fixed positions;

– Are formed by physical means; i.e. there is usually no heat change during its formation.

 

Examples of mixtures:
1. Air:

– A mixture of oxygen, nitrogen, carbon (IV) oxide, water vapour, and noble gases.

 

  1. Sugar solution.

– A mixture of sugar and water.

 

  1. Sea water.

– Water, dissolved salts;

 

Experiment: To distinguish between an element and a compound.

(i). Apparatus.

– Watch glass, test tube, wooden splint, magnet, iron fillings, sulphur powder, dilute hydrochloric acid.

 

(ii). Procedure.

– Approximately 7g of iron fillings and 4g of sulphur are mixed in a test tube and the mixture strongly heated;

 

 

Observations:
– A red glow starts and spreads throughout the mixture forming a black solid.

– The black solid is iron (II) sulphide.

– The two products // substances in steps 1 and 2 are subjected to the following tests;

 

Test // Analysis Observations
Iron-sulphur mixture. Iron (II) sulphide
1. Colour: The colour of the substance is noted; – The resultant substance is yellow-grey due to the yellow sulphur and the grey iron powder; – The yellow-grey mixture changes to a black solid; iron (II) sulphide on heating;
2. Separation: A magnet is passed over the substances separately;

– Alternatively, water was added to each substance;

– Before heating the iron could be separated from sulphur by use of a magnet or sedimentation;

Note: These are physical methods;

– Magnetism and sedimentation have no effect on iron (II) sulphide;
3. Reaction with dilute hydrochloric acid: To each of the substances, a few drops of hydrochloric acid is added; – Iron reacted with dilute hydrochloric acid to form a colourless gas that burns with a pop sound.

– This is hydrogen gas;

– Sulphur is not affected;

– Iron (II) sulphide reacted with hydrochloric acid to produce a colourless gas with a characteristic pungent (rotten egg) smell;

– The gas is hydrogen sulphide;

4. Heat change. No heat was produced  or applied in mixing iron and sulphur; – After heating the mixture, the formation of the new substance, iron (II) sulphide produced enough heat hence the bright red glow;

 

Explanations:
– These four experiments summarize the four main differences between compounds and mixtures.

– From the results, iron and sulphur powder is a mixture; while iron (II) sulphide is a compound.

 

Differences between a mixture and a compound.

Compounds Mixtures
1. Components are in fixed positions; – Components are in any positions;
2. Components can only be separated by chemical means; which require large amounts of energy; – Components ca be separated by physical means;
3. The properties are different from those of the constituent substances; – The properties are the average of the properties of the constituent elements;
4. Are formed by chemical means // methods; i.e. a new substance is formed and there is evolution of heat; – Are formed by physical mans; no new substance is formed and there is no // negligible heat change;
5. Formation involves heat changes; either liberation or absorption; – No heat change in the formation of a mixture;

 

Names and symbols of common elements.

Chemical symbols.

– Are chemical short hands, written to refer to elements.

– They are usually based on the letters of the element;

– Chemical symbols consist of one or two letters which are usually derivatives of the Latin or English name of the element;

 

Rules in writing chemical symbols.

  1. The first letter must always be a capital letter;
  2. The letters should not be joined with each other, like in handwriting; they must be printed.
  3. If present, the second letter of a symbol must be a small letter;

 

Note: The abbreviations of the chemical symbols are mainly derivatives of English, Latin or German names.

Examples:

Copper is ymbolised as Cu; derived from Cuprum which is ltin;

  • Iron; → Fe (Ferrum –Latin)
  • Potassium; → K (Kalium –Latin)
  • Sodium; → Na (Natrium –Latin)
  • Lead; → Pb (Plumbum –Latin);

– The symbol of each element represents one atom of that element.

Example:
Ag represents one atom of silver;

– 2Ag represents 2 atoms of silver;

 

Some common elements and their symbols.

Element Latin // Greek // German name Symbol
Carbon

Fluorine

Hydrogen

Iodine

Nitrogen

Oxygen

Phosphorus

Sulphur

Aluminium

Argon

Barium

Calcium

Chlorine

Helium

Magnesium

Neon

Silicon

Zinc

Copper

Iron

Lead

Mercury

Potassium

Silver

Sodium

Gold

Tin

Manganese

Cuprum

Ferrum

Plumbum

Hydagyrum

Kalium

Argentum

Natrium

Aurum

C

F
H
I

N

O

P

S
Al

Ar

Ba

Ca
Cl

He

Mg

Ne

Si

Zn

Cu

Fe

Pb

Hg

K

Ag

Na

Au

Sn

Mn

 

Importance of chemical names and symbols over common names.

  1. Chemical names and symbols indicate the names in the compounds.

Examples:
– Compounds whose names end in  -ide; contains only two elements;

  • Iron (II) sulphide: iron and sulphur;
  • Magnesium nitride: magnesium and nitrogen;

– Compounds whose names end in –ate contain three elements and oe of them is oxygen;

  1. Chemical names are universally known and accepted; hence provide mean of easy communication among chemists all over the world;

Simple word equations.

Equations:

– Is a linear summary of a chemical reaction, showing the reactants and products.

Examples:
Copper (II) oxide + hydrogen → copper + water;

 

Explanations:
– Substances on the left hand side are called reactants;

– Substances on the right hand side are called products;

– The addition sign (+) on the left hand side means “reacts with”;

– The arrow (→) means to form;

– The addition sign (+) on the right hand side (products side) means “and”.

 

Conclusion:

Copper (II) oxide reacts with hydrogen to form copper and water;

 

Note:
– Some chemical reactions are reversible and hence have two opposite arrows (                  ) between reactants and products

– The arrows (                    ) in chemistry means a reversible chemical reaction;

 

Further examples:
1. Copper (II) carbonate → Copper (II) oxide + carbon (IV) oxide;

  1. Magnesium + oxygen → magnesium oxide;

 

 

UNIT 3: ACIDS BASES AND INDICATORS.

Unit checklist

  1. Acids:
  • Meaning of acids;
  • Organic acids
  • Mineral acids;
  1. Bases.
  • Meaning;
  1. Indicators.
  • Meaning;
  • Preparation of acid-base indicators;
  • Commercial acid-base indicators;
  • Colour of indicators in acids and bases;
  • Classifications of substances as acids or bases using acid base indicators;
  • The universal indicator;
  • The pH scale;
  • pH values of various solutions in universal indiactor;
  1. Properties of acids.
  • Physical properties;
    • Taste;
    • Effect on litmus papers;
    • Electrical conductivity;
  • Chemical properties.
    • Reaction with alkalis and bases;
    • Reaction with metals;
    • Reaction with carbonates and hydrogen carbonates;
  1. Properties of bases
  • Physical properties;
    • Taste;
    • Texture;
    • Effect on litmus papers;
    • Electrical conductivity;
  • Chemical properties.
    • Reaction with acids;
    • Precipitation of some hydroxides;
    • Effect of heat
  1. Uses of some acids and bases.

 

  1. Acids:

– Are substances that dissolve in water to release hydrogen ions.

– Acids can either be organic acids or mineral acids;

 

(i). Organic acids:

– Are acids found in plants and animals;

Examples:
– lactic acid in sour milk;

– Citric acid in citrus fruits like oranges;

– Ethanoic acid in vinegar;

– Tartaric acid in baking powder;

– Methanoic acid in bee and ant stings;

– Tannic acid in tea;

 

(ii). Mineral acids.

– Are acids made from minerals containing elements such as sulphur, chlorine, nitrogen etc.

– Are formed from reactions of chemicals;

– Main examples include:

  • Sulphuric (VI) acid (H2SO4); contains hydrogen, sulphur and oxygen;
  • Hydrochloric acid; contains hydrogen and chlorine;
  • Nitric (V) acid (HNO3); contains nitrogen, oxygen and hydrogen;

Note:
– Mineral acids are more powerful than organic acids; because they yield // release more hydrogen ions in water

– They are thus more corrosive.

  1. Bases.

– Are substances that dissolve in water to yield // release hydroxyl ions;

– Just like acids they are bitter to taste;

Examples:
– Sodium hydroxide;

– Ammonium hydroxide;

– Calcium hydroxide;

Note:
– Some bases insoluble in water while some are soluble in water;

– Soluble bases are called alkalis;

 

  1. Indicators.

– Are substances which give definite colours in acidic or basic solutions;

– Are substances which can be used to determine whether a substance is an acid or a base;

– Consequently they are called acid-base indicators;

– The determination is based on colour changes, where each indicator have particular colourations in acids and bases.

– Indicators can be commercially or locally prepared in the laboratory;

 

 

 

 

 

Indicators:

Experiment: preparation of simple acid-base indicators from flower extracts.

(i). Apparatus and chemicals.

– Test tubes;

– Pestle and mortar;

– Flower petals;

– Ethanol // propanone;

– Water;

– Various test solutions: sulphuric (VI) acid, hydrochloric acid, Ethanoic acid, sodium hydroxide, magadi soda, ammonia solution.

 

(ii). Procedure:
– Flowers from selected plants are collected and assembled e.g. bougainvillea, hibiscus etc;

– They are crushed in a mortar using a pestle and some ethanol added with continued crushing;

– The resultant liquid is decanted into a small beaker; and its colour recorded.

– Using a dropper, two to three drops of the resultant indicator are added to the test solutions.

 

(iii). Observations:
(a). Colour of extract in acids and bases

 

Plant extract Colour in hydrochloric acid Colour in dilute sodium hydroxide
1    
2    

 

(b). Result with various test solutions:

Test substance Colour (change) Type of substance (acid/base)
Lemon juice

Wood ash

Ammonia

Sour milk

Vinegar

Nitric (V) acid

Toothpaste

Lime water

Baking powder;

Sugar

Potassium hydroxide

   

 

Note:

– Plant extracts acid-base indicators are not normally preferred in Chemistry experiments.

Reason:

– They don’t give consistent (reproducible results because they are impure.

– Commercial indicators give more distinctive and reproducible results.

 

 

 

 

Commercial indicators.

– Are commercially prepared indicators which are sold in already purified forms.

 

Advantages of commercial indicators.

– They are relatively pure hence give consistent and reproducible results;

– They are readily available and easy to store in a Chemistry laboratory;

 

Main examples:
– Phenolpthalein;

– Methyl orange;

– Bromothymol blue;

– Litmus paper;

 

Note:
– Litmus is a blue vegetable compound which is extracted from “plants” called lichens;

– Litmus paper is an adsorbent paper which has been dipped in litmus indicator solution then dried;

 

Colours of various commercial indicators in acids and bases.

Indicator Colour in.
Neutral Base Acid
1. Litmus;

2. Phenolphthalein;

3. Methyl orange;

4. Bromothymol blue

Purple

Colourless;

Orange

Blue

Blue;

Pink;

Yellow

Blue

Red;

Colourless;

Pink;

Yellow;

 

Classification of various substances as acids or bases using indicators.

Substance Colour in Classification
Litmus Phenolphthalein Methyl orange Bromothymol blue
Hydrochloric acid

Sodium hydroxide

Omo (detergent)

Soda

Actal tablets

Lemon juice

Sour milk

Bleach (jik)

Fresh milk

Wood ash

         

 

The universal indicator.

– Is a full range indicator which gives range of colours depending on the strength of the acid or alkali.

– It is prepared by suitable mixing certain indicators;

– It gives a range of colour depending on the strength of acids and bases;

– Each universal indicator is supplied with a chart, to facilitate this classification.

 

 

 

 

 

 

The pH scale.

– Is a scale of numbers which shows the strength of acids or bases.

– It refers to the potential (power) of hydrogen;

– It ranges from 0 – 14;

– To determine the strength of an acid or base, the colour it gives in universal indicators solution is compared to the shades on the pH chart of the indicator;

 

Diagram: The pH scale.

 

 

 

 

 

Note:
– The strongest acid has a pH of 1;

– The strongest alkali has a pH of 14;

– Neutral substances have a pH of 7;

– Any pH less than 7 is acidic solution; while any pH above 7 is for a alkaline / basic solution;

 

Colour and pH of various solutions in universal indicator;

Substance Colour pH on chart Classification
Hydrochloric acid

Sodium hydroxide

Omo (detergent)

Soda

Actal tablets

Lemon juice

Sour milk

Bleach (jik)

Fresh milk

Wood ash

     

 

Properties of acids.

(a). Physical properties.

  1. They have a sour taste.

Examples:
– The sour taste of citric fruits is due to the citric acid in them.

– The sour taste in sour milk is due to lactic acid;

 

  1. They turn blue litmus to red;

– Red litmus will remain red in acidic solution; blue litmus will turn red;

 

  1. Electrical conductivity;

– Acids conduct electric current when dissolved in water;

– This is because they dissolve in water to release hydrogen ions; which are the ones that conduct electric current;

 

  1. Thy destroy clothing when strong; i.e. they at away clothing material leaving holes in it;

 

  1. Strong acids are corrosive; hence able to burn plant and animal tissues;

Chemical properties.

  1. Reaction with alkalis ad bases.

– Acids react with alkalis to form salt and water only;

– These types of reactions are called neutralization reactions;

– The hydrogen ions of the acid react with the hydroxyl ion of the alkali to form water;

– The name of the salt is usually derived from the acid;

Examples:

Acid Derivative salt
Sulphuric (VI) acid

Hydrochloric acid

Nitric (V) acid

Phosphoric acid

Sulphates;

Chlorides

Nitrates;

Phosphates;

 

Summary:

Acid + base (alkali) → salt + water; (a neutralization reaction);

Examples:
1. Sodium hydroxide + Hydrochloric acid → Sodium chloride + water;

  1. Calcium oxide + Sulphuric (VI) acid → calcium sulphate + water;

 

  1. Reaction with metals;

– Acids react with some metals to produce hydrogen;

 

Examples: Reaction with dilute hydrochloric acid and zinc metal;

Procedure:

– 2 cm3 of hydrochloric acid is put in a test tube;

– A spatula end-full of zinc powder is added.

– A burning splint is lowered in the test tube.

 

Observations.

– Effervescence of a colourless gas;

– The colourless gas burns with a pop sound;

 

Explanations.

– Zinc metal displaces the hydrogen ions in the acid which form the hydrogen gas;

– When a glowing splint is introduced into the hydrogen gas; it burns with a pop sound;

– This is the chemical test to confirm that a gas is hydrogen;

 

Conclusion;

– The gas produced is hydrogen gas;

– Thus, acids react with some metals to produce hydrogen gas, and a salt;

 

General equation:

Metal + Dilute acid → salt + Hydrogen gas;

 

Reaction equation:

Zinc + Hydrochloric acid → Zinc chloride + Hydrogen chloride;

 

Further examples:
i. Magnesium + Dilute sulphuric (VI) acid → magnesium sulphate + hydrogen gas;

  1. Magnesium + Dilute Hydrochloric acid → magnesium chloride + Hydrogen gas;

 

  1. Reaction with carbonates and hydrogen carbonates.

– Metal carbonates and hydrogen carbonates react with acids to form carbon (IV) oxide, water and a salt;

 

General equation:
Metal carbonate + Dilute acid → A salt + water + carbon (IV) oxide;

Metal hydrogen carbonate + Dilute acid → A salt + water + carbon (IV) oxide;

 

Examples: Reaction of sodium carbonate with dilute hydrochloric acid.

Procedure:

– About 2 cm3 of dilute hydrochloric acid is put in a test tube;

– A spatula end-full of sodium carbonate powder is then added;

– A burning splint is carefully lowered into the test tube.

 

Apparatus.

Observations.

– An effervescence occurs (bubbles); and a colourless gas is produced;

– The colourless gas does not relight a glowing splint; showing that it is carbon (IV) oxide;

 

Properties of bases.

Note:

– Bases are substances that release hydroxyl ions when added to water;

– Soluble bases are called alkalis;

 

Examples:
Sodium hydroxide + Water → Sodium ions + hydroxyl ions;

 

(a). Physical properties of bases

  1. They are bitter to taste;
  2. They are slippery or soapy to feel;
  3. They turn litmus blue;
  4. They conduct electricity / electric current. This is because when they are added to water they release hydroxyl ions which are the ones that conduct electricity;

 

(b). Chemical properties.

  1. Reaction with acids.

– They react with acids to form a salt and water as the only products;

– This is a neutralization reaction; and is used to cure indigestion;

 

Example:
– Actal tablets contain a base that neutralizes the stomach acid.

 

  1. They precipitate some metal hydroxides.

– Addition of some alkalis to salt solutions results in formation of solids;

– Most of these are normally hydroxides;

– A solid that is formed when two solutions are mixed is called a precipitate;

 

Example:

– Copper (II) sulphate + Sodium hydroxide → Copper (II) hydroxide + Sodium sulphate solution;

Blue solid;

 

  1. Effects of heat.

– Most metal hydroxides are decomposed by heat to form their oxides and water;

 

Heat

General equation:
Metal hydroxide                 Metal oxide + Water;

 

Example:
Zinc hydroxide → Zinc oxide + water;

 

Applications of acids and bases.

Application of acids

  1. Manufacture of aerated drinks;
  2. Cleaning metal surfaces to remove oxide layer;
  3. Sulphuric (VI) acid is used in car batteries to store and produce electricity;
  4. Treatment of some insect bites.

Examples:
– Wasp and bee stings can be treated by applying vinegar (Ethanoic acid) or lemon juice;

– These acidic substances neutralize the alkaline insect stings;

 

Uses of bases.

  1. Manufacture of anti-acid tablets to neutralize acid indigestions e.g. actal;
  2. Calcium oxide ad calcium chloride are used to dry gases in the laboratory;

 

UNIT 4: AIR AND COMBUSTION.

Checklist.

  1. Components
  2. Determination of percentage of the active part of air.
  • Burning candle.
  • Heating copper turnings;
  • Heating magnesium turnings.
  • Smouldering of white phosphorus.
  • Rusting of iron;
  1. Determination of presence of water and carbon (IV) oxide in water.
  2. Fractional distillation of liquid air.
  3. Rusting
  • Meaning and formula;
  • Conditions necessary for rusting;
  • Prevention of rusting;
  1. Oxygen
  • Laboratory preparation of oxygen gas;
  • Chemical test for oxygen gas;
  • Alternative methods of oxygen preparation;
    • Addition of water to solid sodium peroxide;
    • Heating potassium manganate (VII) solid.
  • Use of oxygen
  1. Burning substances in air;
  • Changes in mass;
  • Burning substances in oxygen;
    • Metals;
    • Non-metals;
  • Reactivity series;
  1. Competition for oxygen among metals;
  2. Applications for the competition for oxygen;
  3. Atmospheric pollution.

 

Introduction:

– Air is a gaseous mixture constituted of several gases, water vapour and pollutants.

 

Combustion:

– Is the burning of substances, usually in presence of air // oxygen;

– During combustion only the oxygen component of air is used; .e the active part of air.

 

Percentage composition of air.

Component Percentage volume.
Nitrogen 78%
Oxygen 21%
Carbon (IV) oxide 0.03%
Noble gases (argon) About 1%
Water vapour Variable
Smoke/dust particles Variable;
Others Trace

 

Note:

– From the noble gases argon is the most abundant, constituting about 0.93% of the entire 1%

 

Oxygen and combustion.

– When substances burn in air they consume oxygen.

– Thus the process of combustion utilizes mainly oxygen;

– The reactions in combustion are normally exothermic (give out heat) and often involve flame.

Note:

Combustion in which a flame is used is called burning;

– In combustion if all the oxygen in a given volume of air is used, the final volume of air reduces by about 21.0%;

– Since oxygen is the only constituent of air participating in combustion its termed the active part of air.

 

Experiments: Determination of the active part of air.

  1. Burning candle in air.

Apparatus and requirements.

– Candle;

– Cork / evaporating dish;

– Sodium hydroxide solution;

 

Procedure:

– A candle about 3cm long is put on a wide cork/ evaporating dish;

– It is then floated in a dilute solution of sodium hydroxide solution just above the beehive shelf;

– It is carefully covered with a dry 100cm3 measuring cylinder, during which the level of solution in the cylinder is noted and marked;

– The measuring cylinder is removed and the candle lit;

– The lighting candle is then covered with a measuring cylinder;

– The experiment is allowed to proceed until the candle goes off;

 

 

 

Observations:

– The candle went off after sometime;

– The sodium hydroxide level inside the gas jar rises;

– The sodium hydroxide level in the trough goes down;

 

Diagrams:

 

 

 

 

 

 

 

 

 

Explanations:
– The candle wax is made up of hydrogen and carbon, hence called a hydrocarbon;

– During burning it melts in air consuming oxygen and producing carbon (IV) oxide and water vapour;

– The water vapour condenses giving a negligible volume of water;

– The resultant carbon (IV) oxide is absorbed by the sodium hydroxide;

– Absorption of carbon (IV) oxide in the gas jar creates a partial vacuum within it;

– The sodium hydroxide in the trough rises to fill the resultant space; and hence a drop in the sodium hydroxide level in the tough;

 

Conclusion.

– Oxygen is the active part of air that is utilized during burning;

– Air is basically made up of 2 parts; an active part that supports burning and an inactive part that does not support burning;

 

  1. Quantitative determination of percentage of oxygen in air.

Apparatus and chemicals.

– Tough;

– Beehive;

– Candle and gas jar;

– A 30ml ruler;

– Sodium hydroxide solution;

 

Procedure:

– The entire apparatus is arranged as shown below;

– An empty gas jar is inverted over the candle before lighting it;

– The initial height A, is measured and recorded;

– The gas jar is then removed; the candle lit and covered with the gas jar again;

– The set up is allowed to run till the candle extinguishes (goes off); and the final height (B) of the air column measured.

 

 

 

 

 

Diagrams

 

 

 

 

 

 

 

 

 

 

Calculations:
– Amount of air in the gas jar initially = A cm3;

– Final amount of air remaining after burning; = Bcm3;

– Amount of oxygen used; = (A – B)

Thus;

Percentage of oxygen in air: = (A – B) x 10

A

= C%

Sample data:
Volume of air in the gas jar before burning =

Volume of air in the gas jar after burning =

Volume of air used during burning =

Percentage of air (by volume) used up = Volume used in burning x 100

Original (initial) volume

Substituting:

 

 

 

Conclusion:
– When candle burns in air, about 20% of air, which is oxygen used up;

 

  1. Determination of percentage of air used up in rusting.

Apparatus:
– Gas jar;

– Trough;

– Beehive;

– Iron fillings//powder

 

Procedure:
– The gas jar is divided into five equal portions by marking around it using a waterproof marker;

– The gas jar is wet near the bottom and some iron fillings sprinkled on it;

– Some water is put in a trough and the jar with iron fillings // wool // powder inverted over it;

– The initial colour of iron fillings is noted;

– The set up is left undisturbed for a few days until the water shows no further change in rising;

 

 

 

 

Diagrams of apparatus set up:

 

 

 

 

 

 

 

 

Observations:
– The iron fillings change colour from a grey to form a brown solid;

– Water level in the gas jar rises // increases until the first mark in the gas jar;

– The water level in the trough decreases;

 

Explanations:
– The gas jar is moistened to make the iron fillings stick onto its surface so that the fillings do not fall in the water when the gas jar is inverted;

– The brown substance formed is called rust and its chemical name is hydrated iron (III) oxide;

– During rusting, oxygen is utilized, thus creating a partial vacuum in the gas jar;

– This causes the water level in the gas jar to rise up and the water level in the trough to go down;

– The rise in water level is equivalent to about 1/5 of the original air volume, which translates to about 20%;

 

Conclusion:
– When rusting occurs about 20% of air, which is oxygen, is used up;

 

  1. Determination of percentage of air used up when air is passed over heated copper.

Apparatus and chemicals.

– Two 100cm3 syringes labeled Y and Y;

– Hard glass test tube;

– Glass wool’;

– Bunsen burner;

– Copper turnings;

 

Procedure:

– A small amount of copper turnings is put in a hard glass tube and glass wool put at both ends of the tube;

– All the air in the syringe Y is removed by pushing the plunger inside, upon which the syringe is tightly fixed at one end of the tube// glass tube;

– The plunger of syringe Z is pulled out to the 100cm3 mark; to fill it with air.

The apparatus is arranged as below.

 

Diagram:

 

 

 

 

 

 

– The tube containing copper turnings is strongly heated;

– Air is then passed over the hot copper turnings by slowly pushing the plunger Z to and fro for several times.

Reason:

– Ensure complete reaction between the hot copper turnings and oxygen (air);

 

– When no further air change in volume of air in the syringe occurs, the apparatus is allowed to cool;

– The volume of air left in syringe Z is recorded;

 

Observations:
– The brown solid (copper) turns into a black solid (copper (II) oxide);

– The plunger of syringe Z moves inwards to approximately 80 cm3 mark;

 

Explanations:
– The heated copper reacted with oxygen in air to form black copper (II) oxide;

– The percentage of oxygen that was in the air is approximately 20%, causing the plunger to move inwards to the 80 cm3 mark;

 

Equation:

Copper + Oxygen → Copper (II) oxide;

 (Brown)          (Colourless)              (Black)

 

2Cu(s) + O2(g) → 2CuO(s)

Brown      (Colourless)    (Black)

 

Conclusion:

– Burning of copper in air utilizes oxygen and produces black copper (II) oxide.

 

Sample results and calculations:

– Initial air volume in syringe Z = cm3;

– Final air volume in syringe Z = cm3;

– Volume of air used = (100 – 80) = 20 cm3;

Percentage of oxygen in air = 20 x 100 = 20%;

100

 

  1. Determination of percentage of air used up when air is passed over heated Magnesium.

– When the same set up is used to investigate the percentage of air used up in combustion of magnesium the volume of air used up is relatively higher than the 20%.

Reason:

Magnesium produces a lot of heat during combustion and thus reacts with both oxygen and nitrogen to form two products; magnesium oxide and magnesium nitride respectively;

 

Observations:

– Magnesium glows giving a bright blinding flame;

– Formation of a mixture of two white powders.

 

 

 

 

Equations:

Reaction with oxygen:

Magnesium + Oxygen → Magnesium oxide;

       (Grey)                (Colourless)                      (White)

 

2Mg(s) + O2(g) → 2MgO(s)

(Grey)         (Colourless)      (White)

 

Reaction with nitrogen:

Magnesium + Nitrogen → Magnesium nitride;

       (Grey)                (Colourless)                      (White)

 

3Mg(s) + N2(g) → Mg3N2(s)

(Grey)         (Colourless)      (White)

 

Note:

– Sodium metal will also react with both oxygen and nitrogen during combustion; forming sodium oxide and sodium nitride respectively;

 

  1. Smouldering of white phosphorus.

Apparatus and requirements.

– Graduated measuring cylinder;

– Water tough // pneumatic trough;

– Copper wire;

– White phosphorus;

 

Procedure:
– An empty measuring cylinder is inverted in a water trough and the water level noted;

– A small piece of white phosphorus is attached to the end of a piece of copper wire then put // inserted into the inverted measuring cylinder ensuring it is above the water;

– The set up is left undisturbed for 24 hours;

 

Precaution:

– Avoid contact with the phosphorus;

– Avoid inhalation of the fumes;

 

Observations:

– White fumes inside the cylinder at the start of the experiment;

– After 24 hours:
– water level inside the measuring cylinder rises;

– Water level in the trough drops;

 

Explanations:
– Yellow or white phosphorus smoulders in air; due to the fact that phosphorus reacts with oxygen to form phosphorus oxides;

-The phosphorus oxides are the white fumes;

– The phosphorus oxides then dissolves in water; forming acidic solutions of phosphoric acids;

– The water level rises inside the cylinder to occupy the volume of oxygen used up in reaction with phosphorus;

 

Equations:

Phosphorus + Oxygen → Phosphorus (V) oxide

White // yellow            Colourless              White fumes

 

P4(s) + 5O2(g) → 2P2O5(g);

White   Colourless              White fumes

Yellow

 

Phosphorus + Oxygen → Phosphorus (III) oxide

White // yellow            Colourless              White fumes

 

P4(s) + 3O2(g) → 2P2O3(g);

White   Colourless              White fumes

Yellow

 

Conclusion:

– Phosphorus smolders easily in air, reacting with oxygen (active part of air) to form phosphorus (III) or phosphorus (V) oxide;

– For this reason phosphorus is stored under water; to prevent it from reacting with atmospheric oxygen;

 

Note:
– This reaction can be made much faster by heating the copper wire; which will transmit heat to the piece of phosphorus at the tip, causing rapid burning of phosphorus to give dense white fumes of phosphorus (V) oxide // phosphorus (III) oxides;

 

Test for evidence of some components of air.

  1. Water vapour.

(i). Formation of dew;

 

(ii). When white anhydrous copper (II) sulphate is left in the open overnight; it forms a blue solid of hydrated copper (II) sulphate;

Reason:
– The white anhydrous copper (II) sulphate absorbs atmospheric water vapour;

– Upon hydration the copper (II) sulphate turns blue;

 

(iii). Sodium hydroxide pellets form a colourless solution when left in the open air overnight.

Reason:

– The sodium hydroxide pellets absorbs atmospheric water vapour and dissolves in it forming sodium hydroxide solution;

 

(iv). When air is passed through anhydrous calcium hydroxide solid in a U-tube for sometime; there is formation of a colourless solution in the U-tube.

Apparatus:

 

 

 

 

 

 

 

Reason:

– The anhydrous calcium chloride absorbs atmospheric water vapour forming a colourless solution of calcium chloride;

 

Equation:

Calcium chloride + water → calcium chloride solution;

White                                                                     Colourless solution;

 

Note:
– Substances that absorb moisture from the air to form a colourless solution are called deliquescent substances.

– Other examples of deliquescent substances include: iron (III) chloride, magnesium chloride and zinc chloride;

 

  1. Carbon (IV) oxide.

(i). Glass stoppers of reagent bottles containing sodium hydroxide solution tend to stick when left on for sometime in the laboratory;

Reason:

– The sodium hydroxide solution at the edges of the stopper is exposed to air; and thus reacts with atmospheric carbon (IV) oxide forming white sodium carbonate solid.

 

Equation:
Sodium hydroxide + Carbon (IV) oxide → Sodium carbonate + Water;

2NaOH(aq) + CO2(g) → Na2CO3(s) + H2O(l);

 

(ii). Bubbling atmospheric air through lime water (calcium hydroxide) to form a white insoluble salt of calcium carbonate.

 

Apparatus:

 

 

 

 

 

 

 

 

 

 

Reason:

– Atmospheric carbon (IV) oxide reacts with limewater (calcium hydroxide) to form a white insoluble precipitate of calcium carbonate salt;

 

Equation:
Calcium hydroxide + Carbon (IV) oxide → Calcium carbonate + Water;

 

Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l);

 

 

Note:

– When the air is bubbled on even after the formation of the white precipitate; the white precipitate dissolves after sometime to form a colourless solution;

Reason:

– Excess carbon (IV) oxide reacts with the calcium carbonate to form soluble calcium hydrogen carbonate.

 

Equation:

Calcium carbonate + Water + Carbon (IV) oxide → Calcium hydroxide solution;

CaCO3(s) + H2O(g) + CO2(g) → Ca(HCO3)2(aq);

 

Fractional distillation of Liquefied air.

– Air is a mixture of gases;

– It can be separated into its constituents by fractional distillation of liquid air.

– During the process air is passed through a series of steps during which it is purified, some components eliminated then it is compressed into liquid prior to fractional distillation.

– The process can be divided into two main stages;

Purification and liquefaction;

Fractional distillation of air;

 

(a). Purification and liquefaction.

Step 1: Purification:
– The air is purified by removal of dust particles ;

– This is done through the following ways:

  • Passage through filters; during which air is passed through a series of filters; the dust particles remain within the filters while dust free air passes on to the next stage;
  • Electrostatic precipitation where air is passed through charged electrodes which trap oppositely charged dust particles;

 

Step 2: Removal of carbon (IV) oxide.

– The dust-free air is passed through a chamber containing calcium hydroxide solution;

– The sodium hydroxide solution dissolves the carbon (IV) oxide present in the air;

– During the reaction, sodium carbonate and water are formed;

– Over a prolonged time; the sodium carbonate absorbs more (excess) carbon (IV) oxide forming sodium hydrogen carbonate;

 

Equations:

Sodium hydroxide + carbon (IV) oxide → Sodium carbonate + Water;

2NaOH(aq) + CO2(g) → Na2CO3(aq) + H2O(l);

 

In excess;

Sodium carbonate + Water + Carbon (IV) oxide → Sodium hydrogen carbonate;

Na2CO3(aq) + H2O(l) + CO2(g) → 2NaHCO3(aq);

 

 

 

 

 

Step 3: Removal of water vapour;

– The dust-free, CO2 – free air is then cooled to -25oC;

– This process solidifies the water vapour out as ice;

– This cooling process may be done at temperatures a s low as -80; so as to solidify any carbon (IV) oxide (freezing point -78oC) that may have escaped absorption by the sodium hydroxide;

– The removal of water vapour and carbon (IV) oxide are important because it prevents blockage of the pipes in the rest of the system;

 

Step 4: Liquefaction of air;

– The dry, dust-free and carbon (IV) oxide-free air is compressed to about 100 atmospheres of pressure; causing it to warm;

– The compressed air is cooled by refrigeration;

– The cold compressed air is made to expand rapidly by passage through a nozzle which cools it further;

– The repeated compression, cooling and expansion of air causes it to liquefy at about -200oC:

Note:
At this temperature only neon and helium whose boiling points re lower than -200oC remain in gaseous states;

 

(b). Fractional distillation;

– The liquid air now consists only of nitrogen, oxygen and noble gases (especially argon);

– The liquid air is fed at the bottom of a fractionating column;

– It is warmed to a temperature of -192oC;

-Nitrogen distils over fast at -196oC because it has a lower boiling point; and is collected at the top of the fractionating column;

Note:

– Any vapours of oxygen and argon which rise together with nitrogen vapour condense in the column and fall back as liquids;

– The nitrogen collected is 99% pure;

The small amounts of impurities include neon and helium;

– The liquids remaining at the bottom of the fractionating column after vaporization of all nitrogen is mainly oxygen and argon; with traces of krypton and xenon;

 

– The liquid is again warmed further to a temperature of -185oC; causing the vapourization of argon whose boiling point is -186oC;

– This is collected as a gas at the top of the fractionating column;

– The residue liquid is mainly oxygen with minute quantities of krypton and xenon which have even high boiling points;

– The oxygen is drained off and stored as pressurized oxygen in steel cylinders;

 

Uses of the products:

Oxygen;

– Used in hospitals with patients with breathing difficulties;

– It is used by mountain climbers and deep-sea divers for breathing;

– It is used to burn fuels;

– It is combined with acetylene to form oxy-acetylene flame which is used in welding;

– During steel making oxygen is used to remove carbon impurities;

 

 

Nitrogen:

– Manufacture of ammonia;

– Used in light bulbs; because of its inert nature it dies not react with the filament;

– As a refrigerant e.g. storage of semen for artificial insemination;

 

Rusting.

– Is the corrosion of iron in presence of oxygen and moisture to form brown hydrated iron (III) oxide;

– The chemical name rust is therefore hydrated iron (III) oxide with the formula Fe2O3.2H2O;

– Rust itself is a brown porous substance;

 

Disadvantage of rusting:

– It weakens the structure of the metal (iron) and hence eventually destroys them.

 

Experiment: To show the conditions necessary for rusting.

 

Experiment Procedure Observation Explanation
1 – Two clean iron nails are put inside the test tube;

– 10 cm3 of tap water are then added;

– Examine for two days;

 

– Iron nails turn brown implying there is rusting; – There is presence of both oxygen and water;
2 – Two clean iron nails are added to the test tube;

– 10 cm3 of boiled hot water is added followed by about 3 cm3 of oil;

– Examine for two days;

 

– No rusting occurs; – There is water but no oxygen so no rusting occur;

– Boiling the water removes any dissolved oxygen;

– Addition of the oil on top prevents dissolution // entry of any air containing oxygen into the water;

3 – Two clean iron nails are added to the test tube;

– Push a piece of cotton wool half way the test tube;

– Place some anhydrous calcium chloride on it and cork the tube tightly;

– Examine for two days;

 

– No rusting occurs; – There is no air // oxygen but no water;

– Anhydrous calcium chloride absorbs any moisture form the air in the test tube;

– Corking the tube tightly prevents more moisture from the atmosphere from getting into the tube as the calcium chloride may get saturated and allow moisture into the nails;

4 – Two clean iron nails are added to the test tube;

– Examine for two days;

 

– Some little rusting occurs; – Air contains and oxygen and some moisture that will facilitate rusting;
5 – Two clean iron nails are added to the test tube;

– Add salty water;

– Examine for two days;

Rusting occurs; and at a faster rate than the rest;

 

– Rusting occurs due to presence of both water and oxygen in the salty water;

– Rusting is faster because the salty water contains ions which gain electrons hence facilitate faster oxidation of iron;

 

 

 

 

 

Summary diagrams

 

 

 

 

 

 

 

 

 

 

 

Further explanations:
During rusting the first step is the oxidation of iron b7y xygen (in the air) to form anhydrous iron (III) oxide;

Equation:

Iron + Oxygen → Iron (III) oxide;

 

4Fe(s) + 3O2(g) → 2Fe2O3(s);

 

– The anhydrous ion (III) oxide then undergoes hydration with water to form brown hydrated iron (III) oxide;

Equation:
Anhydrous iron (III) oxide + Water → Hydrated iron (III) oxide.

Black                                                                                Brown;

 

– Rusting occurs faster in salty conditions;

Reason:

– The initial step is the oxidation of iron, from iron (II) ions (Fe2+) to iron (III) ions (Fe3+);

– During oxidation iron (II) ions give out electrons to undergo oxidation and form iron (III) ions;

– Salty water contains several dissolved salts whose ions easily accept electrons from the iron (II) ions and thus accelerating the oxidation of iron and hence rusting;

 

Prevention of rusting.

Note:
– Rusting destroys materials; equipment and roofs made of iron;

– Rust is porous and thus allows air and water to reach the iron beneath.

– Thus if not removed iron will continue corroding until it is all eaten up.

 

Methods of preventing rusting.

  1. Galvanizing.

– Is the coating of iron with a small layer of zinc;

– Can be done by either dipping the iron object in molten zinc, spraying with a spray of molten zinc, or by electroplating (electrolytic deposition);

– On exposure to air the zinc acquires an inert layer of zinc oxide that is impervious to both air and water;

– The iron beneath is thus prevented from air and water and thus rusting.

 

 

Note:
– The iron is protected even if the zinc coating is scratched.

Reason:
– Upon scratching both the iron and zinc get into contact with air and water;

– Since zinc is more reactive than iron, air and water reacts with zinc at the expense (instead) of iron;

 

  1. Electroplating:
    – Refers to electrolytic coating of metals (iron) with less reactive (less corrosive metals);

– This is done through the process of electrolysis where floe of electric current causes the less reactive metal to coat the metal being protected from rusting;

 

Example:

– Most tin cans are in fact made yup of steel coated with a thin layer of tin.

– Other than being non-toxic tin is unreactive and rarely reacts with the contents of the can or air;

Note:

– Unlike in galvanizing, when an electroplated material gets scratched, the metal underneath (iron) rusts, and very fast;

Reason:

– Both iron and the electroplating metal (tin) are exposed to air and water;

– Since iron is more reactive than tin (the less reactive electroplating material) it reacts with oxygen and water in preference to tin;

 

– This explain why galvanization is more durable than electroplating;

– Other less reactive metals that are used to coat iron objects include chromium, silver and gold;

– Some electroplating metals such as gold and silver also increase the aesthetic value of the electroplated object;

 

  1. Sacrificial protection.

– Blocks of a more reactive metal such as zinc or magnesium are attached to the iron structure;

– The more reactive metal will be corroded in preference to iron;

– To keep the iron structure from rusting, the block of reactive metal has to be replaced regularly;

– This metal is used for the protection of underground water pipes as well as ship hulls;

– The blocks of reactive metal are either attached directly to the iron structure or connected to it by a wire.

 

Diagrams:

 

 

 

 

 

 

 

 

 

  1. Painting.

– The paint coats the metal surface and thus prevents contact with air and water hence no rusting;

– However if the paint is scratched, rusting occurs quickly;

– It is used mainly in ion railings, gates, bridges, roofs, ships ad cars;

  1. Alloying.

– Alloys are mixtures of two or more metals;

– Thus to prevent iron from rusting it may be mixed with one o more metals resulting into a substance that does not rust;

Example:

– Stainless steel is an alloy of iron with chromium, nickel and manganese and it resistant to rusting.

 

  1. Oiling and greasing;

– Oil is used in moving engine parts while grease is used I other movable metal joints;

– The oil // grease forms a barrier that prevents water and air from coming into contact with the metal surface and hence preventing rusting;

– Oiling and greasing are unique in the sense that they are the only methods that can be used to prevent rusting in movable car parts;

 

Oxygen.

– A very important constituent of air;

– Lavoisier (1743 – 1794), A French Chemist showed that it is the component of air used in respiration and also in burning fuels;

– It is the most abundant of all elements; occurring both freely as well as in combination with other elements;

– Freely it constitutes about 21% by volume of atmospheric air;

 

Laboratory preparation of oxygen gas.

Apparatus.

– Zinc; round-bottomed // flask bottomed flask; thistle // dropping funnel; rubber stopper, deliver tubes, rubber tubings, beehive shelf, trough, gas jars, wooden splint, hydrogen peroxide (20% by volume), manganese (IV) oxide.

 

Diagram;

 

 

 

 

 

 

 

 

 

 

 

 

 

Procedure.

– Some manganese (IV) oxide is placed into a flat-bottomed flask;

– The apparatus is set up as shown in the diagram above;

– Add hydrogen peroxide from a thistle funnel into the flask dropwise;

– The gas is collected as shown;

 

 

Observations:

– Bubbles of a colourless gas are released from the flask through the water then into the gas jar;

– The colourless gas collects on top of the water;

 

Explanations:
– Hydrogen peroxide decomposes slowly to oxygen and water under normal conditions;

– This process is however slow to collect enough volumes of oxygen;

– On addition of manganese (IV) oxide the decomposition is speeded up;

– Thus manganese (IV) oxide speeds up the decomposition of hydrogen peroxide and thus acts a s a catalyst;

 

Equation:

  • Without a catalyst:

Hydrogen peroxide → Water + oxygen;

2H2O2(l) → 2H2O(l) + O2(g) (slow process)

 

  • With manganese (IV) oxide catalyst:
 Manganese (IV) oxide

 

 

Hydrogen peroxide                              Water + oxygen;

2H2O2(l) → 2H2O(l) + O2(g) (faster process)

 

Note:

The first few bubbles of oxygen gas are not collected.

Reason: The gas is mixed with air which was originally in air and hence impure.

 

Method of collection;

– Over water collection.

Reason:

– It is insoluble in water and less dense than water

 

Physical properties of oxygen gas.

– It is colourless;

– It is odourless;

– Has a low boiling point of about -183oC;

– Almost insoluble in water (hence collected over water);

 

Chemical test for oxygen gas.

– On inserting a glowing splint on a gas jar full of oxygen gas; it relights a glowing splint;

 

Drying of oxygen gas.

– The resultant oxygen is usually moist due to the fact that it is collected over water;

– If required dry the gas ca be died using either of the two methods:

(i). Using sulphuric (VI) acid.

– Bubbling the gas through a wash bottle containing concentrated sulphuric (VI) acid;

– The concentrated sulphuric (VI) acid absorbs moisture from the gas leaving it dry;

– The dry gas is then draw into collection syringe;

 

 

Diagram;

 

 

 

 

 

 

 

 

 

(ii). Using anhydrous calcium chloride.

– From the flask the gas is passed through a U-tube containing anhydrous calcium chloride;

– The anhydrous calcium chloride also absorbs moisture from the gas leaving it dry;

– The dry gas is then drawn into a collection syringe;

 

Diagram:

 

 

 

 

 

 

 

 

 

Alternative methods of oxygen preparation.

  1. Addition of water to sodium peroxide.

Apparatus:
– Sodium peroxide; round-bottomed // flask bottomed flask; thistle // dropping funnel; rubber stopper, deliver tubes, rubber tubings, beehive shelf, trough, gas jars, wooden splint, water;

 

Diagram of apparatus.

 

 

 

 

 

 

 

 

 

 

 

Procedure.

– Some sodium peroxide is placed into a flat-bottomed flask;

– The apparatus is set up as shown in the diagram above;

– Add water from a thistle funnel into the flask dropwise;

– The gas is collected as shown;

 

Observations:

– Bubbles of a colourless gas are released from the flask through the water then into the gas jar;

– The colourless gas collects on top of the water;

 

Explanations:
– Sodium peroxide reacts with water to liberate oxygen;

– A solution of sodium hydroxide remains in the flask;

– This solution will turn litmus paper blue showing it is alkaline.

 

Equation:

Sodium peroxide + water → Sodium hydroxide + oxygen;

2Na2O2(l) + 2H2O(l) → 4NaOH(aq) + O2(g)

 

Note:

The first few bubbles of oxygen gas are not collected.

Reason: The gas is mixed with air which was originally in air and hence impure.

 

Method of collection;

– Over water collection;

Reason:

– It is insoluble in water and less dense than water;

 

Chemical test for oxygen gas.

– On inserting a glowing splint on a gas jar full of oxygen gas; it relights a glowing splint;

 

 

 

 

 

  1. Heating potassium manganate (VII) solid.

Apparatus:
– Ignition tube // boiling tube; means of heating; solid potassium manganate (VII); rubber stopper, deliver tubes, beehive shelf, trough, gas jars, wooden splint, water;

 

Diagram of apparatus.

 

 

 

 

 

 

 

 

 

Procedure:

– The apparatus is set up as shown above.

– Some solid potassium manganate (VII) is put in a hard ignition// combustion tube and strongly heated as shown above.

– The resultant gas is collected over water as shown above.

Observations;

– The purple solid forms a black solid (potassium manganate (II) solid);

– Bubbles of a colourless gas are evolved and collect over water;

 

Explanations:
– Upon heating potassium manganate (VII) decompose to manganese (VI) oxide; potassium

 

Equation:
Potassium manganate (VII) → Potassium manganate (II) + Oxygen gas

KMnO4(s) → KMnO2(s) + O2(g);

 

Uses of oxygen.

  1. Used in hospitals for breathing by patients with breathing difficulties;
  2. It is used by mountain climbers and deep sea divers for breathing;
  3. It is used to burn fuels e.g. burning fuels for propelling rockets;
  4. Used in welding and cutting metals

Examples:
It combines with hydrogen to form a very hot oxy-hydrogen flame that is used in welding and cutting metals;

– It combines with acetylene to form oxy-acetylene flame which is also used in welding and cutting metals;

 

  1. During steel making, oxygen is used to remove iron impurities.

– During this process oxygen is blown over hot impure iron.

– The oxygen react with carbon impurities forming carbon (IV) oxide which escapes laving pure iron which is steel due to its higher purity;

 

Burning substances in air.

– When substances burn in air they mainly react mainly with oxygen (the active part of air);

– Some metals however also react with nitrogen;

– During burning there is usually change in mass;

 

Experiment: To investigate burning substances in air.

Requirements:
– Metal (magnesium ribbon); crucible; tripod stand; pipe clay triangle; means of heating;

 

Apparatus.

 

 

Procedure:

– About 1g of magnesium is put in the crucible;

– The crucible (with the magnesium is then weighed)

– The apparatus is set up as above;

– The crucible is heated with the lid lifted occasionally; so as to allow in air;

– No content of the crucible is allowed to escape; to ensure all products of the burning are retained;

– After all the magnesium has burned the crucible is allowed to cool;

– The crucible and its contents are weighed again;

 

Observations

Mass of crucible + magnesium before burning = xg

Mass of crucible + contents after burning = yg

Change in mass = (x –y) g;

Mass of product before burning is lower // less than the mass of the product after burning;

 

Explanations:

– When the magnesium is burned in a closed crucible in a closed container, most of the air is consumed;

– It is therefore necessary to allow in air so that the burning can continue;

– During burning the magnesium combines with air to form a new product;

– Magnesium combines with both oxygen and nitrogen in air to form magnesium nitride and magnesium oxide;

 

Equations

With oxygen:
Magnesium + oxygen → Magnesium oxide;

Mg(s) + O2(g) → MgO(s);

 

With nitrogen:

Magnesium + Nitrogen →Magnesium nitride;

3Mg(s) + N2(g) → Mg3N2(s);

 

Conclusion:

– Generally when metals burn in air, there is increase in mass;

– All metals react with oxygen to form metal oxides;

– Only more reactive metals react with nitrogen in air;

 

Note:

– During burning if the product(s) of the burning is gaseous, then there would be decrease in mass.

 

Examples:
Phosphorus → Phosphorus (V) oxide;

Lead (II) nitrate → Phosphorus (V) oxide + Nitrogen (IV) oxide + Oxygen gas;

Calcium carbonate → Calcium oxide + carbon (IV) oxide;

 

 

 

 

 

Burning metals in air and in oxygen.

Requirements:

– Metals; deflagrating spoon; gas jar; source of heat;

 

Diagram of apparatus;

 

 

 

 

 

 

 

 

 

 

 

Procedure:

– A piece of sodium is warmed on a deflagrating spoon until it begins to burn;

– It is then lowered into a gas jar of air as shown above;

– The flame colour is noted;

– The gas jar is allowed to cool; some water added to the product(s) in the gas jar and shook well;

– Any gases produced are tested by smell and also with litmus papers;

– The experiment is then repeated with pure oxygen;

– The whole procedure is repeated with other metals;

 

Observations;

– When substances burn in oxygen they form only oxides; as opposed to burning substances in air where some react with both air and nitrogen;

– Different substances produce different flame colours;

– Many metals burn in air and in oxygen at different speeds; with more reactive metals burning more vigorously than the less reactive metals;

– Burning is faster in oxygen than in air;

Reason:

– Oxygen is pure but in air there are other constituents such as nitrogen, carbon (IV) oxide and noble gases which slow down the burning;

– In air products are generally oxides and in some few cases (magnesium and sodium) nitrides as well;

– Metals that tend to be more reactive are the ones that react with both oxygen and nitrogen;

– In oxygen products are strictly oxides;

– Some of then products are soluble in water while others are not.

 

Sample equations:

Magnesium:

With oxygen:
Magnesium + oxygen → Magnesium oxide;

Mg(s) + O2(g) → MgO(s);

 

With nitrogen:

Magnesium + Nitrogen →Magnesium nitride;

3Mg(s) + N2(g) → Mg3N2(s);

Sodium:

With oxygen:
Sodium + oxygen → Sodium oxide;

4Na(s) + O2(g) → 2Na2O(s);

 

With nitrogen:

Sodium + Nitrogen →Magnesium nitride;

6Na(s) + N2(g) → 2Na3N(s);

 

Summary: burning metals in air.

Metal How it burns Appearance of product Name of products Solubility of product in water Effect of solution on litmus paper
Magnesium Burns with a bright white flame; White powder Magnesium oxide and magnesium nitride; Slightly soluble; alkaline gas (ammonia) is produced during the process; Turns blue;
Copper Burns with a blue flame; surface turns black; Black solid; Copper (II) oxide; Insoluble; No effect;
Iron. Glows to red hot; produces sparks; Brown black (dark brown) solid; Iron (II) oxide Insoluble; No effect;
Sodium Buns very vigorously with a golden yellow flame; White solid; Sodium oxide and sodium nitride Soluble; alkaline gas (ammonia) is produced in the process; Turns litmus blue;
Calcium Vigorous with a red flame; White solid; Calcium oxide and calcium nitride; Slightly soluble; alkaline gas evolved in the process; Turns blue;
Zinc   Yellow solid which cools to white; Zinc oxide Insoluble; No effect
Lead   Red solid which cools to yellow; Lead (II) oxide Insoluble; No effect;
Potassium Very vigorously with a lilac flame; White solid; Potassium oxide and potassium nitride; Soluble; alkaline gas evolved in the process; Turns blue;

 

Note:
– When metals combine with oxygen, it forms metal oxides. In these reactions oxygen is added to the metals; hence the reaction is called oxidation.

Oxidation refers to the addition of oxygen to a substance;

– The reactivity of various metals with oxygen differs.

– The arrangement of the metals in order of their activity forms the reactivity series;

– Metallic oxides generally turn litmus paper blue and are thus said to be basic oxides;

– Some metallic oxides however have both acidic and basic properties and are thus termed amphoteric oxides e.g. aluminium oxides;

 

 

 

 

 

The Reactivity series of metals;

Potassium;      Most reactive;

Sodium;

Calcium;

Aluminium;

Zinc;               increasing reactivity;

Iron;

Lead;

Copper;

Mercury;

Silver;

Gold;               Least reactive

 

Burning non-metals in oxygen

Requirements:

– Non-metals; deflagrating spoon; gas jar; source of heat;

 

Diagram of apparatus;

 

 

 

 

 

 

 

 

 

 

 

Procedure:

– A piece of sulphur is heated on a deflagrating spoon until it begins to burn;

– It is then lowered into a gas jar of oxygen as shown above;

– The flame colour is noted;

– The gas jar is allowed to cool; some water added to the product(s) in the gas jar and shook well;

– Any resultant solution is tested with litmus papers;

– Any gases produced are tested by smell and also with litmus papers;

– The experiment is then repeated with pure oxygen;

– The whole procedure is repeated with other non-metals such as carbon and phosphorus;

 

Explanations:
1. Sulphur.

– Burns in oxygen with a blue flame to give a colourless gas with a choking irritating smell;

– The gas is sulphur (IV) oxide;

Equation:
Sulphur + Oxygen → Sulphur (IV) oxide;

S + O2(g) → SO2(g);

– The sulphur (IV) oxide dissolves in water to form sulphurous acid, which turns litmus rd;

SO2(g) + H2O(l) → H2SO3(l);

 

  1. Carbon

– Glows red to give a colourless gas that forms a white precipitate in lime water;

– The gas is Carbon (IV) oxide;

Equation:
Carbon + Oxygen → Carbon (IV) oxide;

C + O2(g) → CO2(g);

– The Carbon (IV) oxide dissolves in water to form weak carbonic acid, which turns litmus rd;

CO2(g) + H2O(l) → H2CO3(l);

 

Note:
– In limited oxygen the carbon undergoes partial oxidation forming carbon (II) oxide;

Equation:
2C(s) + O2(g) → 2CO(g)

 

  1. Phosphorus.

– Burns in oxygen with a white flame to give dense white fumes;

– The white fumes are either phosphorus (V) oxide or phosphorus (III) oxide;

– Both solids // fumes dissolve in water to form phosphoric acid;

 

Equations:

With limited supply of oxygen:
Phosphorus + Oxygen → Phosphorus (III) oxide;

P4(s) + 3O2(g) →2P2O3(g);

 

With excess oxygen.

Phosphorus + Oxygen → Phosphorus (V) oxide;

P4(s) + 5O2(g) →P2O10(g);

 

Summary: effects of burning non-metals in air.

Non-metal How it burns in oxygen Name of products formed Appearance of the product Effect of solution on litmus paper
Sulphur Burns with a blue flame; Sulphur (IV) oxide White fumes; Turns ed
Carbon Glows red Carbon (IV) oxide Colourless gas; Turns red;
Phosphorus Burns with a white flame Phosphorus (V) oxide and phosphorus (III) oxide White fumes Turns red;

 

Precautions:
The experiment should be done in a fume cupboard.

Reason:
Products of burning sulphur and phosphorus in air are poisonous.

 

 

 

Note:
– Most non-metallic oxides are acidic in nature and therefore turn litmus red and are thus referred to as acidic oxides;

– Some non-metallic oxides form oxides which are neither acidic nor basic and are thus termed neutral oxides; e.g. carbon (II) oxide and water (hydrogen oxide);

 

Competition for combined oxygen among elements.

Apparatus:

– Metal oxides, source of heat, metals.

 

Procedure:

– A spatula end full of copper (II) oxide in a bottle top.

– Magnesium powder and mixed well;

– Record the observations;

– The experiment is repeated using other metal oxides with various other metals like zinc, iron etc.

 

Observations:

                Metal

Metal

oxide

Magnesium Zinc Iron Lead Copper
Magnesium oxide (white) No reaction No reaction No reaction No reaction No reaction
Zinc oxide (white) White magnesium oxide and grey zinc metal No reaction No reaction No reaction No reaction
Iron (III) oxide White magnesium oxide and grey iron metal;

 

White zinc oxide and iron; No reaction No reaction No reaction
Lead (II) oxide (yellow when White magnesium oxide and White zinc oxide and lead; Iron (III) oxide and lead; No reaction No reaction
Copper (II) oxide (Black White magnesium oxide and brown copper metal; White zinc oxide and brown copper metal; Brown iron (III) oxide and brown copper metal Yellow lead (II) oxide and brown copper metal No reaction

 

Explanations:

– A more reactive metal takes away oxygen from a less reactive metal;

– This is because a more reactive metal reacts more readily with a less reactive metal;

– These reactions are called displacement reactions;

– Some metals can displace other metals from their oxides upon heating;

– Metals which are higher in the reactivity series can displace metals which are lower in the reactivity series from their oxides;

– From the table none of the metals can displace magnesium from its oxide, while copper can be displaced from its oxides by all the metals.

– Thus from the list magnesium is the most reactive while copper is the least reactive.

– Such results of displacement reactions can also be used to develop a reactivity series of the metals (elements) concerned.

Selected equations:

  1. Copper (II) oxide + Magnesium → Magnesium oxide + Copper

CuO(s) + Mg(s) → MgO(s) + Cu(s);

       Black             Grey               White          Brown.

 

  1. Zinc (II) oxide + Magnesium → Magnesium oxide + Zinc

White when cold                           Grey                                  White                      Grey.

      Yellow when hot

ZnO(s) + Mg(s) → MgO(s) + Cu(s);

 

  1. Copper (II) oxide + Zinc → Zinc oxide + Copper

  Black                                       Grey          White when cold        Brown.

                                                                         Yellow when hot

CuO(s) + Mg(s) → MgO(s) + Cu(s);

 

  1. Lead (II) oxide + Magnesium → Magnesium oxide + Lead

Yellow when cold                           Grey                                  White                      Grey.

      Redwhen hot

CuO(s) + Mg(s) → MgO(s) + Cu(s);

 

  1. Copper (II) oxide + Zinc → Magnesium oxide + Copper

CuO(s) + Zn(s) → MgO(s) + Zn(s);

       Black             Grey               White         Brown.

 

Typical reactivity series from the results above:

Magnesium ↑ Most reactive

Zinc

Iron

Lead

Copper       ↓ Least reactive

 

Note:

– Removal of oxygen is called reduction;

– Addition of oxygen is called oxidation;

– A substance that loses oxygen during a reaction is said to be reduced while a substance that removes oxygen from another is called reducing agent;

– A substance that gains oxygen during a reaction is said to be oxidized while a substance that loses / donates oxygen to another is called an oxidizing agent;

 

Examples:

  1. Copper (II) oxide + Magnesium → Magnesium oxide + Copper

CuO(s) + Mg(s) → MgO(s) + Cu(s);

       Black             Grey               White          Brown.

 

Reducing agent: magnesium

Oxidizing agent: Copper (II) oxide

Oxidized species: Magnesium

Reduced species: Copper

 

 

 

 

  1. Copper (II) oxide + Zinc → Magnesium oxide + Copper

CuO(s) + Zn(s) → MgO(s) + Zn(s);

       Black             Grey               White         Brown.

Reducing agent: Zinc

Oxidizing agent: Copper (II) oxide

Oxidized species: Zinc

Reduced species: Copper

 

– In the above reactions both reduction and oxidation take place at the same time;

– A reaction in which both reduction and oxidation occur at the same time is called a redox reaction; “red” from reduction and “ox” from oxidation;

 

Application of Redox reactions:

  1. Extraction of metals;

– Ores of metals such as zinc, iron’ lead etc are roasted in air to form corresponding metal oxides;

– The metal oxides are then reduced to corresponding metals using common reducing agents like carbon and carbon (II) oxide.

 

Examples:

Zinc (II) oxide + Carbon (II) oxide → Zinc + carbon (IV) oxide;

 

Atmospheric pollution and percentage composition of air.

– Human activities have changes the normal composition of air in some places;

– This has not only altered the percentage composition of the main components but also added other components into the air.

 

Examples:

– Mining increases the amount of dust particles in the air;

– Geothermal power drilling may result into emission of gases like hydrogen sulphide, sulphur (IV) oxide into the air;

– Industrial processes like manufacture of nitric (V) acid, contact process etc may add gases ilke sulphur (IV) oxide, nitrogen oxides into the air;

 

These gases and emissions cause atmospheric pollution:

– Gases like sulphur (IV) oxide and nitrogen (IV) oxide dissolve I rain water to form acidic rain that causes corrosion of buildings, iron sheet roofing, bleaching of plants; irritation in bodies and respiratory surfaces in animals etc;

– Dust particles may block stomata in plants; cause smog formation hence reducing visibility (leading to more cases of road accidents).

 

Uses of oxygen:
1. Used for breathing in hospitals fro patients with breathing difficulties.

  1. Used for breathing by mountain climbers and deep sea divers.
  2. Used to burn fuels e.g. burning fuels to propel rockets.
  3. Manufacture of the oxy-acetylene flame that is used in welding and cutting of metals;
  4. Removal of iron impurities during steel making i.e. oxygen is blown through impure iron; the oxygen then reacts with carbon impurities forming carbon (IV) oxide which escapes leaving behind pure iron (steel).

 

UNIT 5: WATER AND HYDROGEN.

Checklist:
1. Introduction

  1. Burning candle wax in air.
  2. Reaction of water with metals
  • Sodium
  • Potassium
  • Calcium
  1. Reaction of metals with steam.
  • Calcium
  • Magnesium
  • Zinc
  • Iron
  • Aluminium
  1. Hydrogen
  • Hydrogen;
  • Reduction property of hydrogen
  • Burning hydrogen in air;
  1. Uses of hydrogen

 

Introduction:

– Water is the most abundant substances on earth;

– It covers about 71% of the earth’s surface;

– Main sources of water include seas, lakes, rivers, oceans.

 

Burning candle wax in air.

Apparatus:

 

 

 

 

 

 

 

 

 

 

Procedure.

– The candle is lit under the funnel and the suction pump turned on.

– The set up is left undisturbed for about 15 minutes.

 

Observations;

– The candle continues to burn.

– Droplets of a colourless liquid in the tube A;

– The colourless liquid turns white anhydrous copper (II) sulphate to blue and blue anhydrous cobalt (II) chloride into pink;

– A white precipitate forms in the calcium hydroxide in tube B;

– Deposits of a black solid on the inner sides of the funnel;

 

Explanations;

– The suction pump ensures continuous supply of air hence the candle continues to burn;

– Candle wax buns in oxygen to form carbon (IV) oxide and steam;

– The carbon (IV) oxide is sucked out through the apparatus by the suction pump;

– Carbon (IV) oxide forms a white precipitate of calcium carbonate when bubbled through lime water (calcium hydroxide)

– Incomplete combustion of the carbon in the candle wax produces carbon particles which cools and deposits as black solids;

 

Equations:

As the candle burns:
Carbon + Oxygen → Carbon (IV) oxide;

C(s) + O2(g) → CO2(g);

 

Hydrogen + Oxygen → Carbon (IV) oxide;

2H2(s) + O2(g) → 2H2O(g);

 

For the formation of the black deposits (soot)

Carbon + Oxygen → Carbon + Carbon (IV) oxide;

2C(s) + O2(g) → C(s) + CO2(g);

In the calcium hydroxide:

Carbon (IV) oxide + calcium hydroxide → Calcium carbonate + Water

Colourless                                                            Colourless                                        White precipitate                    Colourless

CO2(g) + Ca(OH)2(aq) → CaCO3(s) + H2O(l);

 

– The steam condenses into water in the boiling tube;

– Water turns white anhydrous copper (II) sulphate to blue and blue anhydrous cobalt chloride paper into pink;

 

General equation:
Hydrocarbon + Oxygen → Water + Carbon (IV) oxide;

 

Conclusion:

– Candle wax is a compound of carbon and hydrogen only; and such compounds are defined as hydrocarbons;

– When burned in air (oxygen) hydrocarbons produce carbon (IV) oxide ad steam (water);

– Other examples of hydrocarbons include: petrol; diesel; kerosene etc;

 

Note: Effects of repeating the same experiment without a suction pump.

Apparatus:

 

 

 

 

 

 

 

 

 

Observations:
– The candle went off;

– Deposition of black solid on the inner sides of the funnel;

– No colourless liquid in tube A;

– No white precipitate in tube B;

 

Explanations:
– The carbon (IV) oxide and steam produced would accumulate in the filter funnel hence making the flame to go off;

– Incomplete combustion of the candle would produce carbon particles which cool as soot;

– Only negligible amount of water and carbon (IV) oxide would pass through the apparatus;

 

 

 

 

 

 

 

 

Reactions of water with metals:
1. Potassium.

Procedure:
– A small piece of potassium metal is cut and dropped into a trough containing water;

– The resultant solution is tested with litmus paper;

 

Diagram of apparatus:

 

 

 

 

 

 

 

 

Observations and explanations:

– The metal floats on the water surface; because it is less dense than water;

– A hissing sound is produced; due to production of hydrogen gas;

– It explosively melts into a silvery ball then disappears because reaction between water and sodium is exothermic (produces heat). The resultant heat melts the potassium due to its low melting point.

– It darts on the surface; due to propulsion by hydrogen;

– The metal bursts into a lilac flame; because hydrogen explodes into a flame which then burns the small quantities potassium vapour produced during the reaction;

– The resultant solution turns blue; because potassium hydroxide solution formed is a strong base;

 

(b). Reaction equations.

Equation I

2K(s) + 2H2O(l) → 2KOH(aq) + H2(g);

 

Equation II

4K(s) + O2 (g) → 2K2O(s);

 

Equation III:

K2O(s) + H2O(l) → 2KOH(aq)

 

Effect of resultant solution on litmus paper;

– Litmus paper turns blue; sodium hydroxide formed is highly soluble in water; releasing a large number of hydroxyl ions which result into alkaline conditions // high pH;

 

 

 

 

 

 

 

 

 

 

 

  1. Sodium.

Procedure:
– A small piece of sodium metal is cut and dropped into a trough containing water;

– The resultant solution is tested with litmus paper;

 

Diagram of apparatus:

 

 

 

 

 

 

 

 

Observations and explanations:

– The metal floats on the water surface; because it is less dense than water;

– A hissing sound is produced; due to production of hydrogen gas;

– It vigorously melts into a silvery ball then disappears because reaction between water and sodium is exothermic (produces heat). The resultant heat melts the sodium due to its low melting point.

– It darts on the surface; due to propulsion by hydrogen;

– The metal may burst into a golden yellow flame; because hydrogen may explode into a flame which then burns the sodium;

– The resultant solution turns blue; because sodium hydroxide solution formed is a strong base;

 

(b). Reaction equations.

Equation I

2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g);

 

Equation II

4Na(s) + O2 (g) → 2Na2O(s);

 

Equation III:

Na2O(s) + H2O(l) → 2NaOH(aq)

 

Effect of resultant solution on litmus paper;

– Litmus paper turns blue; sodium hydroxide formed is highly soluble in water; releasing a large number of hydroxyl ions which result into alkaline conditions // high pH;

 

  1. Calcium.

Procedure:
– A small piece of calcium metal is cut and dropped into a trough containing water;

– A filter funnel is inverted over it;

– A test tube filled with water is then inverted over the funnel;

– The gas given out is collected as shown in the apparatus below.

– The resultant gas is then tested with a burning splint;

– The resultant solution in the trough is tested with litmus paper.

 

 

 

Diagram of apparatus:

Observations and explanations:

– Calcium sinks to the bottom of the beaker; because it is denser than water;

– Slow effervescence of a colourless gas; due to slow evolution of hydrogen gas;

– Soapy solution formed; due to formation of alkaline calcium hydroxide;

– A white suspension is formed; because calcium hydroxide is slightly soluble in water;

 

Reaction equation:

Ca(s) + H2O (l) → Ca (OH) 2(aq) + H2 (g);

 

Effect of resultant solution on litmus paper;

– Litmus paper slowly turns blue; calcium hydroxide formed is slightly soluble in water; releasing a small number of hydroxyl ions which result into alkaline conditions // high pH;

 

  1. Magnesium and other metals.

– Magnesium reacts with atmospheric oxygen to form magnesium oxide that coast the metal surface;

– Thus before reacting it with water this oxide layer has to be removed e.g. by polishing metal surface using sand paper;

– Reaction between magnesium and cold water is generally very slow; with very slow evolution of hydrogen gas;

– Zinc and iron metals do not react with cold water;

 

Reaction of metals with steam.

Note:

– Metals that react with cold water would react very explosively with steam and thus their reactions with steam should not be attempted in the laboratory;

– However some metals which react only sparingly with cold water or do not react with cold water at all react with steam to produce respective metal oxide and hydrogen gas;

 

  1. Magnesium

Procedure:
– A small amount of wet sand is put at the bottom of a boiling tube;

– A small piece of magnesium ribbon is cleaned and put in the middle of the combustion tube;

– The magnesium ribbon is heated strongly then the wet sand is warmed gently;

– The delivery tube is removed before heating stops; and the gas produced is tested using a burning splint;

 

 

 

 

Diagram of apparatus:

Observations and explanations.

– Magnesium burns with a bright blinding flame;

– Grey solid (magnesium) forms a white solid; due to formation of magnesium oxide;

– Evolution of a colourless gas that burns with a pop sound; confirming it is hydrogen;

 

Reaction equation.

Magnesium + Steam → Magnesium oxide + Hydrogen gas;

Mg(s) + H2O(g) → MgO(s) + H2(g);

 

  1. Zinc

Procedure:
– A small amount of wet sand is put at the bottom of a boiling tube;

– A small piece of zinc put in the middle of the combustion tube;

– The zinc is heated strongly then the wet sand is warmed gently;

– The delivery tube is removed before heating stops; and the gas produced is tested using a burning splint;

 

Diagram of apparatus:

Observations and explanations.

– Zinc metal does not burn but rather glows;

– Grey solid (zinc) forms a yellow solid which cools to a white solid (zinc oxide);

– Evolution of a colourless gas that produces a pop sound when exposed to a burning splint; confirming it is hydrogen;

 

Reaction equation.

Zinc + Steam → Zinc oxide + Hydrogen gas;

Grey        Colourless      Yellow when hot            Colourless

                                       White on cooling

Zn(s) + H2O(g) → ZnO(s) + H2(g);

 

 

 

  1. Iron

Procedure:
– A small amount of wet sand is put at the bottom of a boiling tube;

– A small piece of iron put in the middle of the combustion tube;

– The iron is heated strongly then the wet sand is warmed gently;

– The delivery tube is removed before heating stops; and the gas produced is tested using a burning splint;

 

Diagram of apparatus:

Observations and explanations.

– Iron metal does not burn but rather glows;

– Grey solid (zinc) forms a black solid; due to formation of tri-iron tetra-oxide;

– Evolution of a colourless gas that burns with a pop sound; confirming it is hydrogen;

 

Reaction equation.

Iron + Steam → Tri-iron tetra-oxide + Hydrogen gas;

Grey        Colourless                     Black                                     Colourless

 

3Fe(s) + 4H2O(g) → Fe3O4(s) + 4H2(g);

 

  1. Aluminium

Procedure:
– A small amount of wet sand is put at the bottom of a boiling tube;

– A small piece of aluminium put in the middle of the combustion tube;

– The aluminium is heated strongly then the wet sand is warmed gently;

– The delivery tube is removed before heating stops; and the gas produced is tested using a burning splint;

 

Diagram of apparatus:

Observations and explanations.

– Aluminium burns in steam but the reaction quickly stops; because the reaction forms a layer of aluminium oxide that coats the metal surface preventing further reaction;

– Grey solid (aluminium) forms a white solid of aluminium oxide;

– Slight evolution of a colourless gas that burns with a pop sound; confirming it is hydrogen;

– The production of the gas however stops soon after the reaction starts because the oxide layer stops further reaction;

 

Reaction equation.

Aluminium + Steam → Zinc oxide + Hydrogen gas;

Grey                        Colourless                White                    Colourless

 

2Al(s) + 3H2O(g) → Al2O3(s) + 3H2(g);

 

  1. Other metals.

– Lead and copper do not react with steam;

 

Summary of the reaction between metals with cold water and steam

Metal Action of metal on water Action of metal on steam
Potassium

Sodium

Calcium

Magnesium

Aluminium

Zinc

Iron

Lead

Copper

Violent

Violent

Moderate

Very slow

No reaction

No reaction

No reaction

No reaction

No reaction

Explosive

Explosive

Violent

Rapid

Slow

Slow

Slow

No reaction

No reaction

 

Note:

– Metals can thus be arranged in order of their reactivities with water; resulting to a reactivity series similar to that obtained form reaction between metals with oxygen;

 

Reactivity series of metals:

Potassium;           Most reactive;

Sodium;

Calcium;

Magnesium

Aluminium;

Zinc;                     increasing reactivity;

Iron;

Lead;

Copper;                Least reactive;

 

 

 

 

 

 

 

 

 

 

 

 

Hydrogen

– An element that does not exist freely in nature;

– Generally exists in compounds such as water, sugars, fuels etc;

 

Laboratory preparation of hydrogen gas.

Note:

– Hydrogen gas is generally prepared by the reaction between dilute acids and metals;

– Most suitable acids are dilute hydrochloric acid and dilute sulphuric (VI) acid;

– Most suitable metal is zinc metal;

 

Apparatus:

Procedure:
– Zinc granules are added to dilute sulphuric (VI) acid;

– Small amounts of copper (II) sulphate are added to the zinc – acid mixture;

Reason: To act as a catalyst hence speed up the reaction;

– The resultant colourless gas is collected over water;

Reason: The gas is insoluble in water;

– If the gas is required dry, the gas is passed through concentrated sulphuric (VI) acid or a U-tube containing calcium chloride;

 

Diagrams for drying the gas:

             Zinc granules

– The dry gas is collected by upward delivery (downward displacement of air);

Reason: It is less dense than air (note that hydrogen is the lightest gas known);

 

 

 

 

 

Note:

– Nitric (V) acid is not used in preparation of hydrogen gas; except very dilute nitric (V) acid and magnesium

Reason: Nitric acid is a strong oxidizing agent hence the hydrogen formed is immediately oxidized to water
– Potassium, sodium, lithium and calcium are not used in laboratory preparation of hydrogen gas;
Reason: They react explosively with acids;

– Magnesium is not usually used for laboratory preparation of hydrogen;

Reason: It is expensive;

– Prior to using it for preparation of hydrogen; aluminium should be washed with concentrated hydrochloric acid;

Reason: To remove the protective oxide layer that usually forms on the aluminium surface on its exposure to air;

– Hydrogen gas produced from iron metal tends to have a foul smell;

Reason: Iron gives a mixture of gases due to impurities in the iron; the foul smell is usually due to production of hydrogen sulphide that results from sulphide impurities in the iron metal;

 

Reaction equations:

  1. Iron + Sulphuric (VI) acid → Iron (II) sulphate + Hydrogen gas;

Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g)

 

  1. Iron (II) sulphide + Sulphuric (VI) acid → Iron (II) sulphate + Hydrogen sulphide gas;

FeS(s) + H2SO4(aq) → FeSO4(aq) + H2S(g);

 

Other sources of hydrogen.

– Cracking of alkanes;

– Fractional distillation of petroleum;

 

Properties of Hydrogen gas;

Physical properties.

  1. Colourless and odourless;
  2. It is insoluble in water;
  3. It is less dense than air (it is the lightest gas known);

 

Chemical properties.

  1. It has no effect on litmus paper implying that it is neutral;
  2. It burns with a pop sound when mixed with air;
  3. Pure hydrogen burns quietly with a blue flame producing water;
  4. Hydrogen does not support combustion but it burns in air;
  5. It is a reducing agent;

 

Chemical test for hydrogen;

– When a burning splint is introduced into a gas jar full of hydrogen gas; the gas buns with a “pop sound”

Note:
– The intensity of the “pop sound” diminishes as the purity of hydrogen increases;

 

 

 

Reactions of hydrogen:

  1. Passing hydrogen over heated copper (II) oxide;

Requirements:

Copper (II) oxide; porcelain boat; mean of heating; combustion tube; dry hydrogen gas;

 

Diagram of apparatus.

Procedure:

– Apparatus are arranged as shown above;

– Dry hydrogen gas is passed through the combustion tube for sometime prior to heating the oxide;

Reason:

  • To drive out all the air from the apparatus which may otherwise re-oxidize the metal after reduction.
  • A mixture of hydrogen and air will explode when the combustion tube is heated;

– The gas is continuously collected at the jet and tested; until the gas burns smoothly without a “pop” sound;

– The gas is then lit at the jet and the copper oxide heated;

– This is done until no further change;

– The apparatus is allowed to cool as hydrogen is still continuously allowed to pass through;

Reason:

  • To prevent re-entry of atmospheric oxygen which will re-oxidize the hot metal back to the metal oxide;

– Excess hydrogen must be burnt at the jet so that excess gas is not allowed to escape into the air;

Reason:

  • The mixture of hydrogen and oxygen is explosive when ignited;

 

Observations and explanations

– The black solid turns into a brown solid;

– Droplets of a colourless liquid on the cooler parts of the combustion tube;

Reason:

The hot black copper (II) oxide is reduced by hydrogen gas into brown copper metal while hydrogen gas is oxidized to water;

– The colourless liquid is confirmed to be water by:

  • Adding drops of it to anhydrous copper (II) sulphate which forms a blue solid;
  • Adding drops of it onto blue anhydrous cobalt (II) chloride which turns pink;

 

Reaction equations:

In the combustion tube:

Copper (II) oxide + Hydrogen → Copper + Water;

Black solid                                   Colourless gas    Brown solid     Colourless liquid

CuO(s) + H2(g) → Cu(s) + H2O(l);

 

At the jet:

– Hydrogen burns with a blue flame producing water;

Equation:

2H2(g) + O2(g) → 2H2O(l);

 

Conclusion:

– Hydrogen is a reducing agent and thus reduces the copper (II) oxide to copper metals while hydrogen itself undergoes oxidation to form water;

– Hydrogen is the reducing agent;

– Copper (II) oxide is the oxidizing agent;

 

  1. Passing hydrogen over heated Lead (II) oxide;

Requirements:

Lead (II) oxide; porcelain boat; mean of heating; combustion tube; dry hydrogen gas;

 

Diagram of apparatus.

Procedure:

– Apparatus are arranged as shown above;

– Dry hydrogen gas is passed through the combustion tube for sometime prior to heating the oxide;

Reason:

  • To drive out all the air from the apparatus which may otherwise re-oxidize the metal after reduction.
  • A mixture of hydrogen and air will explode when the combustion tube is heated;

– The gas is continuously collected at the jet and tested; until the gas burns smoothly without a “pop” sound;

– The gas is then lit at the jet and the copper oxide heated;

– This is done until no further change;

– The apparatus is allowed to cool as hydrogen is still continuously allowed to pass through;

Reason:

  • To prevent re-entry of atmospheric oxygen which will re-oxidize the hot metal back to the metal oxide;

– Excess hydrogen must be burnt at the jet so that excess gas is not allowed to escape into the air;

Reason:

  • The mixture of hydrogen and oxygen is explosive when ignited;

 

Observations and explanations

– The yellow solid turns red on heating then finally into a grey solid;

– Droplets of a colourless liquid on the cooler parts of the combustion tube;

 

 

 

Reason:

The hot red lead (II) oxide is reduced by hydrogen gas into grey lead metal while hydrogen gas is oxidized to water;

– The colourless liquid is confirmed to be water by:

  • Adding drops of it to anhydrous copper (II) sulphate which forms a blue solid;
  • Adding drops of it onto blue anhydrous cobalt (II) chloride which turns pink;

 

Reaction equations:

In the combustion tube:

Lead (II) oxide + Hydrogen → Lead + Water;

Yellow-cold; red – hot        Colourless gas       Grey solid     Colourless liquid

PbO(s) + H2(g) → Pb(s) + H2O(l);

 

At the jet:

– Hydrogen burns with a blue flame producing water;

Equation:

2H2(g) + O2(g) → 2H2O(l);

 

Conclusion:

– Hydrogen is a reducing agent and thus reduces the lead (II) oxide to lead metals while hydrogen itself undergoes oxidation to form water;

– Hydrogen is the reducing agent;

– Lead (II) oxide is the oxidizing agent;

 

  1. Passing hydrogen over heated Iron (III) oxide;

Requirements:

Iron (III) oxide; porcelain boat; mean of heating; combustion tube; dry hydrogen gas;

 

Diagram of apparatus.

Procedure:

– Apparatus are arranged as shown above;

– Dry hydrogen gas is passed through the combustion tube for sometime prior to heating the oxide;

Reason:

  • To drive out all the air from the apparatus which may otherwise re-oxidize the metal after reduction.
  • A mixture of hydrogen and air will explode when the combustion tube is heated;

– The gas is continuously collected at the jet and tested; until the gas burns smoothly without a “pop” sound;

– The gas is then lit at the jet and the copper oxide heated;

– This is done until no further change;

– The apparatus is allowed to cool as hydrogen is still continuously allowed to pass through;

Reason:

  • To prevent re-entry of atmospheric oxygen which will re-oxidize the hot metal back to the metal oxide;

– Excess hydrogen must be burnt at the jet so that excess gas is not allowed to escape into the air;

Reason:

  • The mixture of hydrogen and oxygen is explosive when ignited;

 

Observations and explanations

– The Brown solid turns into a grey solid;

– Droplets of a colourless liquid on the cooler parts of the combustion tube;

 

Reason:

The hot brown iron (III) oxide is reduced by hydrogen gas into grey iron metal while hydrogen gas is oxidized to water;

– The colourless liquid is confirmed to be water by:

  • Adding drops of it to anhydrous copper (II) sulphate which forms a blue solid;
  • Adding drops of it onto blue anhydrous cobalt (II) chloride which turns pink;

 

Reaction equations:

In the combustion tube:

Iron (III) oxide + Hydrogen → Iron + Water;

Brown solid                     Colourless gas      Grey solid     Colourless liquid

Fe2O3(s) + 3H2(g) → 2Fe(s) + 3H2O(l);

 

At the jet:

– Hydrogen burns with a blue flame producing water;

Equation:

2H2(g) + O2(g) → 2H2O(l);

 

Conclusion:

– Hydrogen is a reducing agent and thus reduces the iron (III) oxide to iron metal while hydrogen itself undergoes oxidation to form water;

– Hydrogen is the reducing agent;

– Iron (III) oxide is the oxidizing agent;

 

Note:

– Hydrogen does not reduce (remove oxygen) from oxides of metals above it in the reactivity series;

 

 

 

 

 

 

 

 

 

 

 

 

Products of burning hydrogen gas in air.

Apparatus and requirements:

– The apparatus is arranged as shown below.

Procedure:

– Apparatus is arranged as shown below.

– A stream of hydrogen is passed through anhydrous calcium chloride;

– The gas is tested for purity by collecting samples over the jet and testing with a burning splint.

Note: pure hydrogen gas should burn smoothly without the typical “pop sound”;

– The gas is then lit and the pump tuned on;

– The products of burning hydrogen is drown in through the apparatus using the pump for about 15 minutes;

– The product condensing in the test tube in cold water is tested with white anhydrous copper (II) sulphate and blue cobalt chloride paper;

 

Observations:
– Pure hydrogen burns with a blue flame;

– A colourless liquid condenses in the test tube immersed in cold water;

– The liquid turns white anhydrous copper (II) sulphate blue;

– The colourless liquid turns blue anhydrous cobalt chloride pink;

 

Explanations:

– The calcium chloride in the U-tube is used to dry the gas;

– The pure dry hydrogen gas burns with a blue flame to form steam which condenses into liquid water;

– Water turns anhydrous copper (II) chloride from white to blue; and turns blue anhydrous cobalt chloride paper into pink;

 

Reaction equations:

At the jet:

2H2(g) + O2(g) → 2H2O(g);

 

 

 

 

 

 

 

 

 

 

 

Uses of Hydrogen

  1. Large scale manufacture of ammonia in the Haber process;
  2. Hydrogenation for the manufacture of margarine.

– This refers to hardening of oils into fats.

– In this reaction Hydrogen gas is bubbled into liquid oil in presence o0f nickel catalyst;

– The oil takes up hydrogen and is converted into fat;

  1. Hydrogen is used in weather balloons because it is lighter than air;

– Usually a radio transmitter is connected to a weather balloon filled with air; as the balloon floats in air the transmitter collects information which is conveyed to weather stations for interpretation by meteorologists;

  1. A mixture of hydrogen and oxygen forms the very hot oxy-hydrogen flame (with temperatures up to 2000oC that is used in welding and cutting metals;
  2. It is used in rocket fuels for propulsion of rockets;
  3. Manufacture of hydrochloric acid; during which hydrogen is burnt in chlorine;

CHEMISTRY FORM 1, 2, 3 & 4 FREE PDF NOTES

UNIT 1: ATOMIC STRUCTURE AND THE PERIODIC TABLE.

Checklist

  1. Structure of the atom
  2. The subatomic particles;
  • Protons
  • Neutrons
  • Electrons;

GET THE FREE CHEMISTRY NOTES IN PDF BELOW;

Chemistry High School notes for form 1-4 (Free updated pdf downloads)

CHEMISTRY FORM ONE NOTES

CHEMISTRY FORM FOUR NOTES: NEW

  1. Atomic number and mass number
  2. Isotopes;
  3. Energy levels and electron arrangements.
  4. The periodic table;
  • Groups
  • Periods;
  1. Relative atomic mass and isotopes;
  2. Ion formation;
  • Ion;
  • Cations;
  • Anions;
  • Ionization energy;
  • Electron affinity;
  1. Valency and oxidation numbers;
  2. Chemical formula;
  3. Chemical equations
  4. Balancing chemical equations;

 

The atom:

– Refers to the smallest particle of an element hat can take part in a chemical reaction;

– It has an average diameter of 10-8 cm with a nucleus of about 10-13 cm;

 

Parts of an atom

– The atom is made of two main parts:

  • The nucleus
  • The energy levels;

 

  1. The nucleus:

– Is the positively charged part of an atom;

– The nucleus contains two subatomic particles; neutrons and protons;

– The positive charge is due to presence of protons;

– The nuclei of all atoms contain neutrons except the hydrogen atom;

– The protons and the neutrons are together referred to as the nucleons;

 

  1. The energy levels.

– They contain the electrons;

– Electrons are so small and move so fast that their path cannot be traced directly;

– Thus the energy level simple represents the region where the electrons are most likely to be found;

 

Structure of the atom.

 

 

 

 

 

 

 

 

Note:

– The atom can still however be split into smaller particles termed the sub-atomic particles;

 

The sub-atomic particles.

– Are generally three:

  • Protons;
  • Neutrons;
  • Protons;

 

  1. Protons.

– Are the positively charged sub-atomic particles;

– Are found in the nucleus and thus form part of the nucleons;

– The number of protons in the nucleus is equal to the number of electrons in the energy levels;

 

  1. Neutrons.

– Are neutrally charged sub-atomic particles found in the nucleus of the atom;

– They are thought to probably prevent the positively charged protons from getting too close to each other;

 

  1. Electrons.

– Are negatively charged sub-atomic particles found in the energy levels;

– The number of electrons in the energy levels is equal to the number of protons in the nucleus;

– This makes the atom to be electrically neutral;

 

Atomic number and mass number.

Atomic number.

– Refers to the number of protons in the nucleus of an atom;

Examples.

  • Sodium has 11 protons I the nucleus and thus said to have atomic number 11;
  • Chlorine has 17 protons in the nucleus and thus said to have atomic number 17;

 

Mass number;

– Refers to the sum of the number of protons and neutrons in an atom of an element;

Examples:

  • Sodium has 2 neutrons and 11 protons hence a mass number of 23;
  • Chlorine has 18 neutrons and 17 protons hence a mass number of 35.

 

Notation of atomic number and mass number;

– Both atomic number and mass number of an element can be written along with the symbol of an element;

 

Mass number;

– Is conventionally represented as a superscript to the left of the symbol;

Examples:

Sodium; 23Na;

Magnesium 24Mg;

 

Atomic number;

– Is conventionally represented as a superscript to the left of the symbol;

Examples:

Sodium; 11Na;

Magnesium 12Mg;

 

Thus the elements can be conventionally represented as:

Sodium 23Na

Magnesium 24Mg

 

Atomic properties of the first 20 elements.

Element Symbol Number

of electrons

Number

of  protons

Number

Of  neutrons

Atomic number Mass number
Hydrogen H
Helium He
Lithium Li
Beryllium Be
Boron B
Carbon C
Nitrogen N
Oxygen O
Fluorine F
Neon Ne
Sodium Na
Magnesium Mg
Aluminium Al
Silicon Si
Phosphorus P
Sulphur S
Chlorine Cl
Argon Ar
Potassium K
Calcium Ca

 

Isotopes.

– Are atoms of the same element with same atomic number but different mass number due to different number of neutrons.

 

Examples of isotopes.

Element Isotope Atomic No. Number of protons Number of neutrons Mass number Isotopic representation
Hydrogen
Carbon
Oxygen
Chlorine

 

Energy levels and electron arrangements.

  • Energy levels:

– Are definite orbits in an atom that the electrons occupy.

– The energy levels are numbered 1, 2, 3……starting with the one closest to the nucleus.

– Electrons occupying the same energy level have approximately the same amount of energy.

– Each energy level can only accommodate a given maximum number of electrons.

 

 

Maximum number of electrons per energy level

Energy level Maximum number of electrons
1st 2
2nd 8
3rd 8 (only for the first 20 elements)

 

Illustrations:

  • Hydrogen

– It has only one electron and thus this electron occupies the first energy level.

– Since the first energy level is not yet full, hydrogen does not have the second energy level;

– The electron arrangement of hydrogen is thus 1.

 

  • Helium:

– Helium is atomic number 2 and has only two electrons, which occupy the first energy level.

– The first energy level is thus completely full, but since there are no other r electrons lithium also has only one energy level;

– The electron arrangement is thus 2.

 

  • Chlorine:

– Chlorine has atomic number 17 and thus has 17 electrons;

– The first two electrons occupy the fist energy level which is thus completely filled up;

– The remaining 15 electrons occupy the second energy level, which can however accommodate only 8 to be completely filled up;

– Thus the remaining 7 electrons move to the third energy level; which needs 8 to be completely filled up;

– Since the third energy level is not yet full chlorine does not have a fourth energy level;

– The electron arrangement is thus 2.8.7.

 

Electron arrangement.

– Refers to the distribution of electrons in the energy levels of an atom.

 

Example: electron arrangement for the first 20 elements.

Element Symbol Atomic number No. of electrons Electron arrangement
Hydrogen
Helium
Lithium
Beryllium
Boron
Carbon
Nitrogen
Oxygen
Fluorine
Neon
Sodium
Magnesium
Aluminium
Silicon
Phosphorus
Sulphur
Chlorine
Argon
Potassium
Calcium

 

Dot and cross diagrams.

– Is a diagrammatic representation of the electron arrangements in an atom in which the energy levels are represented by concentric lines while electrons are represented by dots or crosses.

– However all electrons are the same regardless of whether they are represented as dots or crosses.

 

Examples:

Lithium Magnesium Aluminium Carbon
 

 

 

 

 

 

 

No. of electrons = No. of electrons No. of electrons No. of electrons
No. of protons   = No. of protons No. of protons No. of protons
No. of neutrons = No. of neutrons No. of neutrons No. of neutrons

 

Beryllium Nitrogen Chlorine Argon
 

 

 

 

 

 

 

No. of electrons = No. of electrons = No. of electrons = No. of electrons =
No. of protons   = No. of protons   = No. of protons   = No. of protons   =
No. of neutrons = No. of neutrons = No. of neutrons = No. of neutrons =

 

Potassium Boron Silicon Phosphorus
 

 

 

 

 

 

 

No. of electrons = No. of electrons = No. of electrons = No. of electrons =
No. of protons   = No. of protons   = No. of protons   = No. of protons   =
No. of neutrons = No. of neutrons = No. of neutrons = No. of neutrons =

The Periodic table

– Is a table showing the arrangements of elements in order of their relative atomic masses.

– Is based on the ideas of Dmitri Ivanovich Medeleev.

– The modern periodic table is based on Meneleev’s periodic law which states that:

                        “The properties of elements are a periodic functions of their relative atomic masses”

– The modern periodic law itself states that:

                        “The properties of elements are a periodic functions of their atomic numbers”

 

Design of the modern periodic table.

It has vertical columns called groups and horizontal rows called periods.

  • Groups:

– Are the vertical columns of a periodic table.

– Are eight in number; ad numbered in capital Roman numerals I all through to VIII.

Note: group VIII is also called group zero because the elements have little tendency to gain o lose electrons during chemical reactions

– Between group 2 and group 3 is a group of elements called the transition metals;

 

  • Periods:

– Are the horizontal rows in a periodic table.

– They are 8 in number in a modern periodic table.

 

  • Transition metals.

– Are elements that form a shallow rectangle between group II and group III.

– These elements are generally metallic and hence the name “transition metals

– They are not fitted in any group because they have variable valencies.

– Are sometimes called the “d-block” elements.

– They are much less reactive than the elements in groups I and II

– They have some unique characteristics that make them not fit in the 8 groups of the periodic table

– These include:

  • Have variable valencies; hence show different oxidation states in their compounds;
  • They form coloured compounds as solids and in aqueous solutions;
  • Have very high melting and boiling points (than metals in groups I and II).
  • They do not react with water;
  • Have very high densities (compared to metals in groups I ad II)

 

  • Lanthanides and Actinides.

– They form a block of elements within the transition metals.

– Are sometimes called the inner transition metals.

– The lanthanides consist of 14 elements from Cerium (Ce) to Lutetium (Lu).

– The Actinides are the 14 elements from Thorium (Tho) to Lawrencium (Lr).

 

Placing an elements in the periodic table

– The position of an elements in the periodic table is governed by the atomic number and hence the

electron arrangement.

 

  • The period:

– The period to which an element belongs is determined by the number of energy levels.

– The number of energy levels is equal to the period to which an elements belongs.

 

Examples:

Elements Symbol Atomic number Electron arrangement Number of energy levels Period
Lithium Li 3 2.1 2 2
Sodium Na 11 2.8.1 3 3
Calcium Ca 20 2.8.8.2 4 4
Nitrogen N 7 2.5 2 2
Helium He 2 2 1 1

 

  • Group:

– The group to which an element belongs to is governed by the number of electrons in the

outermost energy level.

– The number of electrons in the outermost energy level is equal to the group to which the element

belongs.

 

Examples:

Elements Symbol Atomic number Electron arrangement Outermost electrons Group
Potassium K 19 2.8.8.1 1 1
Aluminium Al 13 2.8.3 3 3
Silicon Si 14 2.8.4 4 4
Oxygen O 8 2.6 6 6
Chlorine Cl 17 2.8.7 7 7

 

Diagram of a full periodic table.

Note: In the modern periodic table atomic masses are used instead of mass numbers. The atomic

masses are preferable because they take care of elements with isotopes unlike mass numbers.

Diagram: part of the periodic table showing the first 20 elements

 

 

 

 

 

 

 

 

 

 

 

Relative Atomic Mass and Isotopes.

  • Introduction:

– The masses of individual atoms of elements are very negligible and thus quite difficult to weigh.

– On average the mass of an atom is approximately 10-22g which cannot be determined by an

ordinary laboratory balance.

– For this reason the mass of atom has been expressed relative to that of a chosen standard element

hence the term relative atomic mass.

– The initial reference element was hydrogen which was later replaced with oxygen.

– Later the oxygen scale was found unsuitable;

Reason:

Oxygen exists I several isotopes and thus led to problems when deciding the mass of an oxygen atom.

– For this reason oxygen was replaced with carbon as the reference atom and to date relative atomic

masses of elements are based on an atom of carbon-12 (note that carbon is isotopic and exists as

Carbon -12 or carbon-14).

 

  • Definition:

Relative atomic mass (R.A.M) of an element refers to the average mass of an atom of the element

compared with a twelfth (1/12) of an atom of carbon-12.

RAM = Average mass of one atom of an element

              1/12 mass of an atom of carbon-12

 

 

 

 

 

 

  • Measurement of Relative atomic mass

– RAM of elements is determined by an instrument called Mass Spectrometer.

– The instrument can also be used to determine the relative abundance of isotopes.

– The use of a mass spectrometer in determining the RAM of elements is called mass spectrometry.

 

  • How mass spectrometry works.

– In the mass spectrometer atoms and molecules are converted into ions.

– The ions are then separated as a result of the deflection which occurs in a magnetic field.

– Each ion (from an atom, isotope or molecule) gives a deflection which is amplified into a trace.

– The height of each peak measures the relative abundance of the ion which gives rise to that peak.

 

Note:

– Generally the relative atomic mass of an element is closest in value to the mass of the most

abundant isotope of the element.

 

Example: Diagram of a spectrometer trace for Lithium

  • Explanations:

– The trace has two peaks indicating that there are two isotopes for lithium.

– The fist peak occurs at a relative isotopic mass of 6 and the second at 7; these are the RAM of the two

isotopes respectively.

– The percentage abundance of the isotope with RAM of 6 (6Li) is 9 while the RAM of the isotope with

RAM 7 (7Li) is 91.

 

Calculating relative atomic masses of isotopic elements.

– Information form a spectrometer trace is usually extracted and used in calculation the relative atomic

mass of elements.

 

Worked examples.

  1. The mass spectrum below shows the isotopes present in a sample of lithium.

(i). Use this mass spectrum to help you complete the table below for each lithium isotope in the sample.                                                                                                                                   (3 marks)

 

Isotope Percentage composition Number of
Protons Neutrons
6Li
7Li

 

(ii). Calculate the relative atomic mass of this lithium sample. Your answer should be given to three significant figures.                                                                                                                      (3 marks)

 

  1. Element X with atomic number 16 has two isotopes. ⅔ of 33X and ⅓ of 30X. What is the relative atomic mass of element X? (2 marks)
  2. Calculate the relative atomic mass of an element whose isotopic masses and relative abundances are shown below.                                                       (2 marks)

 

 

  1. A neutral atom of silicon contains 14 electrons, 92% of silicon – 28, 5% silicon – 29 and 3% silicon – 30

(i). What is the atomic number of silicon?                                                                                    (1mark)

 

(ii). Calculate the relative atomic mass of silicon.                                                                       (1mark)

 

  1. Oxygen exists naturally as isotopes of mass numbers 16, 17 and 18 in the ratio 96:2:2 respectively. Calculate its R.A.M                                                            (2 marks)

 

  1. Calculate the relative atomic mass of potassium from the isotopic composition given below.

Isotope                                                Relative abundance

39K                                                      93.1

40K                                                      0.01

41K                                                      6.89

 

  1. Sulphur and sulphur compounds are common in the environment.

(a). A sample of sulphur form a volcano contained 88% by mass of 32S and 12% by mass of 34S.

(i). Complete the table below to show the atomic structure of each isotope of sulphur.

 

Isotope Number of
Protons Neutrons Electrons
32S
34S

 

(ii). Define relative atomic mass.                                                                                                (2 marks)

(iii). Calculate the relative atomic mass of the volcanic sulphur.                                              (2 marks)

  1. Iridium, atomic number 77, is a very dense metal. Scientists believe that meteorites have deposited virtually all the iridium present on earth. A fragment of a meteorite was analysed using a mass spectrometer and a section of the mass spectrum showing the isotopes present in iridium is shown below.

 

(a). Explain the term isotopes.                                                                                                       (1mark)

(b). Use the mass spectrum to help you complete the table below for each iridium isotope in the meteorite.

 

Isotope Percentage composition Number of
Protons Neutrons
6Ir
7Ir

 

(c) (i). Define the term relative atomic mass.                                                                                           (1mark)

 

(ii). Calculate the relative atomic mass of the iridium in this meteorite.                                  (3 marks)

 

 

Ion formation.

  • Introduction:

– Atoms whose outermost energy levels contain the maximum possible number of electrons are said to be stable.

– Thus atoms with energy levels 2, 2.8 and 2.8.8 are said to be stable.

– Electron configuration 2 is said to have a stable duplet state while electron configuration 2.8 and 2.8.8 is said to have a stable octet state.

– These electron configurations resemble those of noble gases and as such they are stable and do not react.

– Atoms without this stability acquire it by either electron gain or electron loss.

– Whether an atom loses or gains electro(s) depend on the number of electrons in the outermost energy level.

 

– Take the case of sodium.

– Atomic number is 11 with an electron configuration of 2.8.1.

– Thus sodium has two options in to become stable:

  • to lose the single electron and acquire a stable electron configuration of 2.8.
  • to gain 7 electrons in its outermost energy level and acquire a stable electron configuration of 2.8.8

– Gaining a single electrons and losing a single electrons requires equal amounts of energy.

– Thus it is cheaper and faster in terms of energy for sodium to lose the single electron in the outermost energy level than to gain 7 electrons into its outermost energy level.

– Thus sodium acquires a stable electron configuration 2.8 by losing the single electron in its outermost energy level.

 

Diagram

 

 

 

 

 

 

 

Sodium atom                                                                          Sodium ion.

 

Equation:

Na → Na+ + e

 

Further examples:

Element Electron arrangement Options for stability Best (cheapest) option
Chlorine 2.8.7 2.8 or 2.8.8 2.8.8
Potassium 2.8.8.1 2.8.8 or 2.8.8.8 2.8.8
Aluminium 2.8.3 2.8 or 2.8.8 2.8
Magnesium 2.8.2 2.8 or 2.8.8 2.8
Carbon 2.4 2 or 2.8 2. or 2.8
Oxygen 2.6 2. or 2.8 2.8

 

  • Ion:

Definition: an ion is a charged particle of an element.

– Are formed when an atom of an element either loses or gains electrons.

 

  • Illustration:

– For a neutral atom the number of electrons in the energy levels (negative charges) is equal and thus

completely balances the number of protons in the nucleus (positive charges).

– Thus the net charge in a neutral atom is zero (0).

– When an atom gains electron(s), the number of electrons becomes higher than the number of protons

resulting to a net negative charge hence an ion.

– Oppositely when an atom loses electron(s) the number of protons becomes higher than the number of

electrons resulting into a net positive charge hence an ion.

– The charge on the ion is usually indicated as a superscript to the right of the chemical symbol.

– Thus ions are of two types:

  • Cations
  • Anions

 

  • Cations:

– Are positively charged ions.

– Are formed when atoms lose electrons resulting into the number of protons being higher than the number of electrons.

– Are mostly ions of metallic elements since most metals react by electron loss.

 

Examples:

(i). Magnesium:

– It has atomic number 12, with electron arrangement 2.8.2.

– It has 12 protons and 12 electrons hence a net charge of “0” hence the atom is written simply as Mg.

– It will form its ions by losing the two electrons from the outermost energy level.

– Thus the number of electrons decreases to 10 while the number of protons remains 12.

– This leads to a net charge of +2, giving the ion with the formula Mg2+.

 

  • Diagrammatic illustration:

 

 

 

 

 

 

 

 

 

 

(ii). Phosphorus

– It has atomic number 13, with electron arrangement 2.8.3.

– It has 13 protons and 13 electrons hence a net charge of “0” hence the atom is written simply as Al.

– It will form its ions by losing three (3) electrons out of the outermost energy level.

– Thus the number of electrons decreases by three to 10 while the number of protons remains 13.

– This leads to a net charge of +3, giving the ion with the formula Al+3.

 

  • Diagrammatic illustration:

 

 

 

 

 

 

 

  • Anions:

– Are negatively charged ions.

– Are formed when atoms gain electrons resulting into the number of electrons being higher than the number of protons.

– Are mostly ions of non-metallic elements since most non-metals ionize (react) by electron gain.

 

Examples.

(i). Chlorine:

– It has atomic number 17, with electron arrangement 2.8.7.

– It has 17 protons and 17 electrons hence a net charge of “0” hence the atom is written simply as Cl.

– It will form its ions by gaining a single electron into the outermost energy level.

– Thus the number of electrons increases to 18 while the number of protons remains 17.

– This leads to a net charge of -1, giving the ion with the formula Cl.

 

  • Diagrammatic illustration:

 

 

 

 

 

 

 

Note: the electros gained must be represented by a different notation from the initial electrons in the atom. E.g. if the initial electrons are represented with crosses (x) then the gained electrons should be represented by dots (.) and vise versa.

 

(ii). Phosphorus

– It has atomic number 15, with electron arrangement 2.8.5.

– It has 15 protons and 15 electrons hence a net charge of “0” hence the atom is written simply as P.

– It will form its ions by gaining three (3) electrons into the outermost energy level.

– Thus the umber of electrons increases by three to 18 while the number of protons remains 15.

– This leads to a net charge of -2, giving the ion with the formula P-3.

 

  • Diagrammatic illustration:

 

 

 

 

 

 

 

Electron transfer during chemical reactions.

– Atoms react either by electron gain or electron loss.

– Generally metals react by electron gain while non-metals react by electron loss.

 

  • Illustration:

– Consider the reaction between sodium and chlorine.

– Sodium attains stability // reacts by losing the single electron form its outermost energy level.

– Chlorine attains stability // reacts by gaining a single electron into its outermost energy level.

– Thus during the reaction between the two elements the single electron lost by the sodium atom to

form the sodium ion is the same one gained by the chlorine atom to form the chloride ion.

 

Some basic concepts.

  • Valence electrons:

– Refers to the number electrons in the outermost energy level.

 

Examples:

– Calcium, with electron arrangement 2.8.8.2 has 2 valence electrons.

– Oxygen with electron arrangement 2.6 has 6 valence electrons.

– Phosphorus with electron arrangement 2.8.5 has 5 valence electrons.

 

  • Valency:

– Refers to the number of electrons an atom loses or gains during a chemical reaction.

– Valency is also known as the combining power of an element.

 

Examples:

– Calcium, with electron arrangement 2.8.8.2 loses 2 electrons during chemical reactions and hence has

a valency of 2.

– Oxygen with electron arrangement 2.6 gains 2 electrons during chemical reactions and thus has a

valency of 2.

– Phosphorus with electron arrangement 2.8.5 has gains 3 electrons during chemical reactions and

hence has a valency of 3.

– Aluminium, with electron arrangement 2.8.3 loses 3 electrons during chemical reactions and hence

has a valency of 2.

 

Note: Some elements have variable valencies and are usually termed the transitional elements (metals)

 

Examples:

– Iron can have valency 2 or 3;

– Copper can have valency 1 or 2

– Lead can have valency 2 or 4.

 

 

 

 

 

 

 

 

 

Summary on valencies of common elements.

 

  Valency 1 Valency 2 Valency 3
Metals Sodium

Potassium

Calcium

Barium

Magnesium

Zinc

Iron

Lead

Copper

Aluminium

Iron

 

Non-metals Nitrogen

Chlorine

Fluorine

Hydrogen

Nitrogen

Oxygen

Sulphur

Nitrogen

Phosphorus

 

Radicals.

– Are groups of atoms with a net charge that exist and react as a unit during chemical reactions.

– Radicals also have a valency, which is equivalent to the value of its charge.

 

Summary on valencies of some common radicals.

 

  Valency 1 Valency 2 Valency 3
Radicals Ammonium (NH4+)

Hydroxide (OH)

Nitrate (NO3)

Hydrogen carbonate (HCO3)

Hydrogen sulphate (HSO4)

Carbonate (CO32-)

Sulphate (SO42-)

Sulphite (SO32-)

Phosphate (PO43-)

 

Oxidation number.

– Refers to the number of electrons an atom loses or gains during a chemical reaction.

– In writing the oxidation number the sign (+ or -) to show gain or loss is written followed by the

number of electrons lost or gained respectively.

 

Illustration.

– Atoms are electrically neutral and are thus assigned an oxidation state of 0 since the number of

protons in the nucleus is equal to the number of electrons in the energy levels.

– However when atoms react they either lose or gain electrons and thus acquire a new state.

– This new state is a new oxidation state and the atom thus acquires a new oxidation number

 

Examples:

Atom E. arrangement Ion formula Valency Oxidation number
Sodium 2.8.1 Na+ 1 +1
Magnesium 2.8.2 Mg2+ 2 +2
Aluminium 2.8.3 Al3+ 3 +3
Nitrogen 2.5 N3- 3 -3
Sulphur 2.8.6 S2- 2 -2
Chlorine 2.8.7 Cl 1 -1

 

Further examples:

Particle Oxidation number
Copper metal, Cu 0
Lead (II) ion, Pb2+ +2
Bromide ion, Br -1
Aluminium ion, Al2+ +2
Sulphide ion, S2- -2
Magnesium metal, Mg 0

 

  • Note: oxidation number (state) and charge of an element.

Oxidation state is written with the positive or the negative sign coming before the element.

Examples: -2, 3, +1, -1 etc.

Charge on an element is write as a superscript of the element with the number coming before the

positive r the negative sign

Examples: Mg2+, Al3+, Na+, Cl etc.

 

Chemical formulae.

– Refers to a representation of a chemical substance using chemical symbols.

– In a single atom it is equivalent to the chemical symbol of the element.

– In a compound it shows the constituent elements and the proportions in which they are combined.

 

  • Deriving the chemical formula of compounds.

– In order to write the correct formula of a compound the following must be known:

  • The symbols of the constituent elements or radicals.
  • The valencies of the elements or radicals

– The chemical formula should start with the element which is more likely to lose electron (s) followed

by the element that is more likely to gain.

 

Worked examples.

 

  1. Deriving the formula of sodium chloride.
Elements Sodium Chlorine
Formula Na Cl
valencies 1 1
balancing x1 x1

 

Balancing ratios as subscripts: Na1Cl1

Formula: NaCl

 

Explanation:

– For sodium to combine with chlorine to form sodium chloride, sodium loses an electron while

chlorine gains an electron.

– Thus every sodium atom needs only a single chlorine atom for both t be fully stable

 

Note:

– When the balancing ratio // subscript is 1, it is usually not written since the symbol of the

element itself represents a single atom.

  1. Deriving the formula of magnesium chloride.
Elements Magnesium Chlorine
Formula Mg Cl
valencies 2 1
balancing X1 X2

 

Balancing ratios as subscripts: Mg1Cl2

Formula: MgCl2

 

  1. Deriving the formula of magnesium oxide.
Elements Magnesium Oxygen
Formula Mg O
valencies 2 2
balancing X1 X1

 

Balancing ratios as subscripts: Mg1O1

Formula: MgO

 

 

Further worked examples.

  1. Derive the chemical formula of each of the following compounds.
  • Calcium fluoride
  • Carbon (II) oxide
  • Carbon (IV) oxide
  • Aluminium nitrate
  • Calcium hydrogen carbonate

 

  1. Complete the table below for elements A, B and C
Element Valency Chemical formula of various compounds
hydroxides Sulphates carbonates nitrates phosphates Hydrogen carbonates
A 1
B 2
B 3

 

 

 

 

 

 

 

 

 

 

 

 

 

Chemical equations.

– Refers to representations of a chemical reaction by means of chemical symbols and formula.

 

  • Key features of a chemical equation.

– The correct formulae of the reactants are on the left of the equation.

– The correct formulae of the products are on the right of the equation.

– The reactants and products are separated by an arrow pointing to the right.

– The state symbols of the reactants and products must be stated as subscripts to the right of the symbols

– The number of each atom on the reactants side must be equal to the number of the same atom on the products side.

 

Example.

Reaction between hot copper metals and oxygen gas.

  • Word equation: Copper + oxygen → Copper (II) oxide.

 

  • Chemical equation: 2Cu(s) + O2(g) → 2CuO(s).

 

Balancing chemical equations.

– A chemical equation is only valid if it is balanced.

– A chemical equation is said to be balanced if the number of each atom on the reactants side is equal to that on the products side.

– This is because atoms are neither created nor destroyed during a chemical reaction.

 

Rules and guidelines in balancing chemical equations.

  • Step 1: Write the chemical equation in words.

Example: Copper metal + oxygen gas.

 

  • Step II: Write the correct formulae of both reactants and products

Example: Cu + O2 → CuO

 

  • Step III: Check whether the number of atoms of each element on the reactants side is equal to that on the products side.

– If equal proceed to step (V);

– If not equal proceed to step (IV).

 

Example: Cu + O2 → CuO

– In this case there are two oxygen atoms on the reactants side yet there is only one oxygen atom on the products side. Thus we proceed to step IV

 

  • Step IV: Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple.

Example: Cu + O2 → CuO

– In this case the chemical formula with the unbalanced atom is CuO on the products side. We thus multiply it by 2.

– The new equation now reads: Cu + O2 → 2CuO

 

 

 

  • Step V: check again to ensure that all atoms are balanced.

– If all atoms are balanced proceed to step VI.

– If not then repeat step IV until all atoms are balanced.

Example: Cu + O2 → 2CuO

– In this case multiplying CuO by 2 offsets the balancing of Cu; which is now unbalanced!

– We therefore repeat step IV in order to balance Cu.

– There is only 1 Cu atom on the reactants side yet there are 2 Cu atoms on the products side.

– We thus multiply the formula with the unbalanced atom (s) by the lowest common multiple, in this case 2.

– The new equation at this step thus becomes: 2Cu + O2 → 2CuO
– We then repeat step V; in this case all atoms are now balanced.

 

  • Step VI: The physical states of the reactants and the products are then indicated.

– If this is not done the chemical equation is considered incorrect.

 

  • Types of state symbols.

– There four main state symbols.

  • Solid; denoted as (s)
  • Liquid; denoted as (l)
  • Aqueous (in solution in water); denoted as (aq)
  • Gaseous; denoted as (g)

– In a chemical equation the state symbols are written with their denotations as subscripts to the right of the chemical formulae.

Example: 2Cu(s) + O2(g) → 2CuO(s)

 

Thus the balanced chemical equation for the reaction between copper metal ad oxygen is:

  • 2Cu(s) + O2(g) → 2CuO(s)

 

Worked examples:

  1. Balance equations for each of the following reactions.
  • Sodium hydroxide and dilute hydrochloric acid

 

  • Zinc oxide and dilute sulphuric (VI) acid

 

  • Zinc metal and dilute nitric (V) acid

 

  • Calcium carbonate and dilute sulphuric (VI) acid

 

  • Sodium and water

 

  1. Balance each of the following equations.
  • Mg(s) + HCl(aq) →MgCl2(aq) + H2(g)

 

  • Na(s) + H2O(l) → NaOH(aq) + H2(g)

 

  • NaOH(aq) + H2SO4(aq) →Na2SO4(aq) + H2O(l)

 

  • CuCO3(s) + HNO3(aq) → Cu(NO3)2(aq) + CO2(g) + H2O(l)

 

  • H2S(g) + O2(g) → O2(g) + H2O(l)

 

  • C2H6(g) + O2(g) → CO2(g) + H2O(l)

 

  • Pb(NO3)2(s) → PbO(s) + NO2(g) + O2(g)

 

  • Fe(s) + Cl2(g) → FeCl3(s)

 

  • Al(s) + H2SO4(aq) → Al2(SO4)3(aq) + H2(g).

 

 

UNIT 2: CHEMICAL FAMILIES; PATTERNS AND PROPERTIES.

Checklist.

– Meaning of chemical families.

– Main chemical families;

  • Alkali metals
    • Meaning and members
    • Trends down the group
      • Atomic radius
      • Ionic radius
      • Energy levels
    • Physical properties
      • Appearance
      • Ease of cutting
      • Melting and boiling [points
      • Electrical conductivity
      • 1st ionization energy
    • Chemical properties
      • Burning in air
      • Exposure to air
      • Reaction with water
      • Reaction with chlorine
    • Similarity of ions and formula of compounds of alkali metals
    • Uses of alkali metals.
  • Alkaline earth metals
    • Meaning and members
    • Trends down the group
      • Atomic radius
      • Ionic radius
      • Energy levels
    • Physical properties
      • Melting and boiling [points
      • Electrical conductivity
      • 1st ionization energy
      • 2nd ionization energy
    • Chemical properties
      • Reaction with air and water
      • Reaction with steam
      • Reaction with chlorine
      • Reaction with dilute acids
    • Similarity of ions and formula of compounds of alkaline earth metals
    • Uses of alkaline earth metals.
  • The Halogens
    • Meaning and members
    • Trends down the group
      • Atomic radius
      • Ionic radius
      • Energy levels
    • Physical properties
      • Preparation and properties of chlorine
      • Appearance and physical states of halogens at room temperature
      • Melting and boiling points
      • Electrical conductivity
    • Chemical properties
      • Ion formation
      • Electron affinity
      • Reaction with metals
      • Reaction with water
      • Reactivity trend of halogens.
    • Similarity of ions and formula of compounds of alkali metals
    • Uses of alkali metals.
  • The noble gases
    • Meaning and members
    • Trends down the group
      • Atomic radius
      • Ionic radius
      • Energy levels
    • Physical properties
      • Melting and boiling [points
      • Electrical conductivity
      • 1st ionization energy
    • Chemical properties
    • Uses of the noble gases

 

Introduction:

– Elements are classified and hence positioned in the periodic table based on the number of valence electrons and the number of energy levels.

– The number of valence electrons is equal to the group to which the element belongs; while the number of energy levels is equal to the period to which the element belongs.

– Elements in the same group are said to belong to the same chemical family.

– Thus a chemical family refers to a group of elements in the same number of valence electrons ad hence in the same group of the periodic table.

 

 

 

 

 

 

Characteristics of a chemical family:

– have same number of valence electrons;

– Show a uniform gradation in physical properties;

– have similar chemical properties;

 

Main chemical families.

– Four main chemical families will be studies in this section.

  • The Alkali metals
  • The Alkaline earth metals
  • The halogens
  • The noble gases

 

  1. The Alkali metals.

– Are the elements with one valence electron and hence in group I of the periodic table.

– All are metallic in nature.

– The members of the family in order down the group is as follows:

  • Lithium
  • Sodium
  • Potassium
  • Rubidium
  • Caesium
  • Francium

 

Electron arrangement of the first three alkali metals.

Elements Electron arrangement
Lithium 2.1
Sodium 2.8.1
Potassium 2.8.81

 

Diagram: Part of periodic table showing the alkali metals

 

 

 

 

 

 

 

 

Gradation in properties of alkali metals.

  1. Atomic and ionic radius.
  • Atomic radius:

– Refers to the distance between the centre of the nucleus of an atom and the outermost energy level occupied by electron(s)

  • Ionic radius:

– Refers to the distance between the centre of the nucleus of an ion and the outermost energy level occupied by electron(s)

 

Trend:

– The ionic radius and the atomic radius of alkali metals increase down the group.

Reason:

– There is an increase in the number of energy levels down the group from lithium to Francium.

 

Illustration:

– Lithium (2.1) has only 2 energy levels; sodium (2.8.1) has 3 energy levels while potassium (2.8.8.1) has 4 energy levels.

– Thus the outermost electron in potassium is further from the nucleus than the outermost electron in sodium and lithium.

 

  1. Atomic and ionic radius of the same element.

– For the same alkali metals the atomic radius is larger than the ionic radius.

Reason:

– Alkali metals form ions by losing the valence electron, lading to the loss of an entire outermost energy level. Thus the atoms have more energy levels than the corresponding ion hence a larger radius in the atom than I the ion.

 

Illustration

– Potassium atom has electron arrangement of 2.8.8.1 hence 4 energy levels.

– During ion formation potassium reacts by losing the single valence electron to acquire a new electron arrangement of 2.8.8 hence 3 energy levels.

– Thus the ion has a smaller radius than the atom.

 

Diagrammatically: Potassium atom and potassium ion

 

 

 

 

 

 

 

 

 

Summary: changes in atomic and ionic radius among alkali metals.

Element Symbol Atomic number Atomic radius (nm) Ionic radius (nm)
Lithium Li 3 0.133 0.060
Sodium Na 11 0.157 0.095
Potassium K 19 0.203 0.133

Physical properties of alkali metals.

  1. Appearance.

– Alkali metals have metallic luster when freshly cut. This refers to a shiny appearance on th cut surface.

– This surface however tarnishes due to reaction on exposure to air.

 

  1. Ease of cutting.

– They are soft and easy to cut.

– The softness and ease of cutting increase down the group.

Reason:

– Alkali metals have giant metallic structures held together by metallic bonds.

– Metallic bond is due to attraction between the positively charged nucleus of one atom and the electrons in the outermost energy level of the next atom.

– Thus the force of attraction is stronger is smaller atoms than in larger atoms.

– The increase in atomic radius down the group implies that the strength of metallic bonds also decrease down the group (hence ease of cutting and softness).

 

  1. They have relatively low melting and boiling points (in comparison to other metals).

Reason: they have relatively weaker metallic bonds.

– The melting and boiling points decrease down the group.

Reason: – The size of the atoms increase down the group due to increasing number of energy levels hence decrease in the strength of the metallic bonds (down the group).

 

  1. Electrical conductivity.

– Alkali metals are good conductors of heat ad electricity.

Reason: they have delocalized electrons in the outermost energy level.

– The electrical conductivity is similar for all alkali metals.

Reason: all alkali metals have the same number of delocalized electron (a single electron) in the outermost energy level.

Note:

– In metals the electrons in the outermost energy level do not remain in one fixed position. They move randomly throughout the metallic structure and are thus said to be delocalsised.

 

  1. Ionization energy.
Ionization energy is the minimum energy required to remove an electron from the outermost energy level of an atom in its gaseous state.

 

 

 

 

 

– The number of ionization energies an element may have is equivalent to the number of valence electrons.

– The first ionization energy is the minimum energy required to remove the first electron from the outermost energy level of an atom in its gaseous state.

– The first ionization among alkali metals decreases down the group.

Reason:

– The effective force of attraction of on the outermost electron by the positive nucleus decreases with increasing atomic size and distance from the nucleus.

– Note that the atomic radius increases down the group due to increase in the number of energy levels.

 

 

 

Summary on physical properties of Alkali metals.

Element Appearance Ease of cutting Melting point (oC) Boiling point (oC) Electrical conductivity Atomic radius (nm) Ionic radius (nm)
Lithium Silver white Slightly hard 180 1330 Good 0.133 520
Sodium Shiny white Easy 98 890 Good 0.157 496
Potassium Shiny grey Easy 64 774 Good 0.203 419

 

Chemical properties of alkali metals.

  1. Reaction with air.

– When exposed to air alkali metals react with atmospheric moisture to form the corresponding metal hydroxide and hydrogen gas.

 

  • General equation:

Metal + Water → Metal hydroxide + hydrogen gas.

 

– The metal hydroxide further reacts with atmospheric carbon (IV) oxide to form hydrated metal carbonate.

 

  • General equation:

Metal hydroxide + carbon (IV) oxide → Hydrated metal carbonate.

 

Examples:

  • Lithium

With moisture: 2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g);

Then with carbon (IV) oxide:2LiOH(aq) + CO2(g) → Li2CO3.H2O(s);

 

  • Sodium

With moisture: 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g);

Then with carbon (IV) oxide: 2NaOH(aq) + CO2(g) → Na2CO3.H2O(s);

 

  • Potassium

With moisture: 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g);

Then with carbon (IV) oxide: 2KOH(aq) + CO2(g) → K2CO3.H2O(s);

 

  1. Burning in air.

– Alkali metals burn in air with characteristic flame colours to form corresponding metal oxides.

 

Examples:

– Burns in air to form lithium oxide as the only product.

 

Equation:

4Li(s) + O2(g) → 2Li2O(s)

 

– Burns in air to with a yellow flame to form sodium oxide as the only product.

Equation:

4Na(s) + O2(g) → 2Na2O(s)

 

Note:

– When burned in air enriched with oxygen or pure oxygen sodium burns with a yellow flame to fom sodium peroxide (instead of sodium oxide).

Equation: 2Na(s) + O2(g) → Na2O2(s)

 

– Burns in air with a lilac flame to form potassium oxide as the only product.

 

Equation:

4K(s) + O2(g) → 2K2O(s)

 

  1. Reaction with water.

– Alkali metals react with water to form the corresponding hydroxides and hydrogen gas.

 

Examples:

Procedure:
– A small piece of potassium metal is cut and dropped into a trough containing water;

– The resultant solution is tested with litmus paper;

 

Diagram of apparatus:

 

 

 

 

 

 

 

 

 

Observations and explanations:

– The metal floats on the water surface; because it is less dense than water;

– A hissing sound is produced; due to production of hydrogen gas;

– It explosively melts into a silvery ball then disappears because reaction between water and potassium is exothermic (produces heat). The resultant heat melts the potassium due to its low melting point.

– It darts on the surface; due to propulsion by hydrogen;

– The metal bursts into a lilac flame; because hydrogen explodes into a flame which then burns the small quantities potassium vapour produced during the reaction;

– The resultant solution turns blue; because potassium hydroxide solution formed is a strong base;

 

(b). Reaction equations.

Equation I

2K(s) + 2H2O(l) → 2KOH(aq) + H2(g);

 

Equation II

4K(s) + O2 (g) → 2K2O(s);

Equation III:

K2O(s) + H2O(l) → 2KOH(aq)

 

Effect of resultant solution on litmus paper;

– Litmus paper turns blue; sodium hydroxide formed is highly soluble in water; releasing a large number of hydroxyl ions which result into alkaline conditions // high pH;

 

Procedure:
– A small piece of sodium metal is cut and dropped into a trough containing water;

– The resultant solution is tested with litmus paper;

 

Diagram of apparatus:

 

 

 

 

 

 

 

Observations and explanations:

– The metal floats on the water surface; because it is less dense than water;

– A hissing sound is produced; due to production of hydrogen gas;

– It vigorously melts into a silvery ball then disappears because reaction between water and sodium is exothermic (produces heat). The resultant heat melts the sodium due to its low melting point.

– It darts on the surface; due to propulsion by hydrogen;

– The metal may burst into a golden yellow flame; because hydrogen may explode into a flame which then burns the sodium;

– The resultant solution turns blue; because sodium hydroxide solution formed is a strong base;

 

(b). Reaction equations.

Equation I

2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g);

 

Equation II

4Na(s) + O2 (g) → 2Na2O(s);

 

Equation III:

Na2O(s) + H2O(l) → 2NaOH(aq)

 

Effect of resultant solution on litmus paper;

– Litmus paper turns blue; sodium hydroxide formed is highly soluble in water; releasing a large number of hydroxyl ions which result into alkaline conditions // high pH;

 

 

 

 

 

 

Procedure:
– A small piece of lithium metal is cut and dropped into a trough containing water;

– The resultant solution is tested with litmus paper;

 

Diagram of apparatus:

 

 

 

 

 

 

Observations and explanations:

– The metal floats on the water surface; because it is less dense than water;

– A hissing sound is produced; due to production of hydrogen gas;

– It reacts less vigorously than sodium and des not melt since the melting point of lithium is relatively higher.

– It darts on the surface; due to propulsion by hydrogen;

– The gas does not ignite spontaneously;

– The resultant solution turns blue; because lithium hydroxide solution formed is a strong base;

 

(b). Reaction equations.

Equation I

2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g);

 

Equation II

4Li(s) + O2 (g) → 2Li2O(s);

 

Equation III:

Li2O(s) + H2O(l) → 2LiOH(aq)

 

Effect of resultant solution on litmus paper;

– Litmus paper turns blue; sodium hydroxide formed is highly soluble in water; releasing a large number of hydroxyl ions which result into alkaline conditions // high pH;

 

Summary on reaction rate.

Metal Reaction rate
Lithium Vigorous
Sodium More vigorous
Potassium Explosive

 

  1. Reaction with chlorine

– All alkali metals react with chlorine to form corresponding metal chlorides.

 

  • General procedure:

– A small piece of the alkali metal is cut and placed in a deflagrating spoon;

– It is then warmed and quickly lowered into a gas jar containing chlorine.

 

– The experiment should be done in a working fume chamber or in the open

Reason: Chlorine gas is poisonous.

– This experiment should not be attempted in the laboratory with potassium

Reason: the reaction is too explosive and very dangerous.

 

  • Observations:

Sodium:

– The metal bursts into a yellow flame.

– White fumes of sodium chloride are formed.

Equation: 2Na(s) + Cl2(g) → 2NaCl(s)

 

Lithium:

– The metal reacts less vigorously than sodium without bursting into a flame.

– White fumes of lithium chloride are formed.

Equation: 2Li(s) + Cl2(g) → 2LiCl(s)

 

Potassium:

– The metal bursts into a lilac flame.

– White fumes of potassium chloride are formed.

Equation: 2K(s) + Cl2(g) → 2KCl(s)

 

The general trend in reactivity of alkali metals.

– The reactivity of alkali metals increase down the group.

Reason:

– Alkali metals react by losing the single valence electron. The ease of loss of the valence electron increases down the group due to decrease in attraction of the valence electron towards the positive nucleus. This in turn is due to the increase in atomic radius down the group as a result of increase in number of energy levels.

 

Similarity of ions and formulae of some compounds of Alkali metals.

– Alkali metals have a similar charge on their ions since they all have a single valence electron.

– Thus they have the same general formula in their ions and compounds.

 

Examples.

Alkali metal ion Hydroxide (OH) Oxides (O2-) Chloride (Cl) Sulphates (SO42-)
Li+
Na+
K+

 

Note:

– Alkali metals are usually not found as free elements; but rather in their combined states in the earth’s crust.

Reason:

– They have high reactivities.

 

 

 

Uses of Alkali metals and their compounds.

  1. Sodium is used in the manufacture of sodium cyanide for gold extraction.
  2. Lithium is used in the manufacture of special high strength glasses ad ceramics.
  3. Lithium compounds are used in the manufacture of dry cells for use in mobile phones, laptops, stopwatches, and zero emission electric vehicles.
  4. A molten mixture of sodium and potassium is used as a coolant in nuclear reactors.
  5. Sodium vapour is used to produce the yellow glow in street ad advertisement lights.
  6. Molten sodium is used as a reducing agent in the extraction of titanium.

Equation: Na(l) + TiCl4(g) → Ti(s) + 4NaCl(l)

  1. Sodium chloride is used as a food additive.
  2. A mixture of sodium hydroxide (caustic soda) and carbon disulphide is used in the manufacture of artificial silk called rayon.

 

The Alkaline Earth metals.

– Are the elements with two valence electrons and hence in group II of the periodic table.

– All are metallic in nature.

– The members of the family in order down the group is as follows:

  • Beryllium
  • Magnesium
  • Calcium
  • Strontium
  • Barium
  • Radium

 

Electron arrangement of the first three alkaline earth metals.

Elements Electron arrangement
Beryllium 2.2
Magnesium 2.8.2
Calcium 2.8.8.2

 

Diagram: Part of periodic table showing the alkaline earth metals

 

 

 

 

 

 

 

 

Dot and cross diagrams for the first three Alkaline Earth metals

 

 

 

 

 

 

 

 

Gradation in properties of alkali metals.

  1. Atomic and ionic radius.
  • Atomic radius:

– Refers to the distance between the centre of the nucleus of an atom and the outermost energy level occupied by electron(s)

  • Ionic radius:

– Refers to the distance between the centre of the nucleus of an ion and the outermost energy level occupied by electron(s)

 

Trend:

– The ionic radius and the atomic radius of alkaline earth metals increase down the group.

Reason:

– There is an increase in the number of energy levels down the group from Beryllium to radium.

 

Illustration:

– Beryllium (2.2) has only 2 energy levels; Magnesium (2.8.2) has 3 energy levels while calcium (2.8.8.2) has 4 energy levels.

– Thus the outermost electron in calcium is further from the nucleus than the outermost electron in magnesium and beryllium.

 

  1. Atomic and ionic radius of the same element.

– For the same alkaline earth metal the atomic radius is larger than the ionic radius.

Reason:

– Alkaline earth metals form ions by losing the valence electrons, leading to the loss of an entire outermost energy level. Thus the atoms have more energy levels than the corresponding ion hence a larger radius in the atom than I the ion.

 

Illustration

– Calcium atom has electron arrangement of 2.8.8.2 hence 4 energy levels.

– During ion formation it reacts by losing the 2 valence electrons to acquire a new electron arrangement of 2.8.8 hence 3 energy levels.

– Thus the ion has a smaller radius than the atom.

 

Diagrammatically: calcium atom and calcium ion

 

 

 

 

 

 

 

 

Summary: changes in atomic and ionic radius among alkali metals.

Element Symbol Atomic number Electron arrangement Atomic radius (nm) Ionic radius (nm)
Beryllium Be 4 2.2 0.089 0.031
Magnesium Mg 12 2.8.2 0.136 0.065
Calcium Ca 20 2.8.8.2 0.174 0.099

Physical properties of alkali metals.

  1. Appearance.

– Alkaline earth metals acquire a metallic luster when polished. This refers to a shiny appearance on the cut surface.

– They however lose this metallic luster when exposed to air due to oxidation.

  • Note:

– The purpose of polishing alkaline earth metals before using them in experiment is to remove the oxide coating that slows down and prevents them from reacting.

  1. Ease of cutting.

– Magnesium is hard to cut with a knife; but is however malleable and ductile.

– Calcium cannot also be cut with a knife because it is brittle.

Note:

  • Malleability:
    – Refers to the ability of a material to be hammered into sheets.

Example: Iron sheets are possible to be made because iron metal is malleable.

 

  • Ductility:

– The ability of a material to be rolled into wires.

Example: electric cables are made of aluminium because aluminium metal is ductile.

 

  • Brittle:

– Refers to a substance which is hard and likely to break.

 

  1. They have relatively high melting and boiling points in comparison to alkali metals.

Reason: they have relatively stronger metallic bonds (than alkali metals).

– The melting and boiling points decrease down the group.

Reason:

– The size of the atoms increase down the group due to increasing number of energy levels. As the atomic radius increase the force of attraction between the positive nucleus and the delocalized electrons decrease. This leads to a decrease in the strength of the metallic bonds (down the group).

 

  1. Electrical conductivity.

– Alkaline earth metals are good conductors of heat and electricity.

Reason: they have delocalized electrons in the outermost energy level.

– The electrical conductivity is similar for all the alkaline earth metals.

Reason: all alkaline earth metals have the same number of delocalized electron (two valence electrons) in the outermost energy level.

 

  1. Ionization energy.
Ionization energy is the minimum energy required to remove an electron from the outermost energy level of an atom in its gaseous state.

 

 

 

 

 

– The number of ionization energies an element may have is equivalent to the number of valence electrons.

– Thus alkaline earth metals have two ionization energies.

– The first ionization energy is the minimum energy required to remove the first electron from the outermost energy level of an atom in its gaseous state.

– The second ionization energy is the minimum energy required to remove the second electron from the outermost energy level of an atom in its gaseous state.

– The first and second ionization energies among alkaline earth metals decreases down the group.

Reason:

– The effective force of attraction of on the outermost electron by the positive nucleus decreases with increasing atomic size and distance from the nucleus.

– Note that the atomic radius increases down the group due to increase in the number of energy levels.

 

  • Variation between 1st and 2nd Ionization energies.

– The first ionization energy is always lower than the second ionization energy for the same element.

Reason:

– After losing the first electron from an atom, the overall positive charge holds the remaining electrons more firmly. Thus removing a second electron from the ion requires more energy than the first electron

Note:

– The third ionization energy will also be higher tha the second ionization energy for the same reason.

 

Example:
Magnesium.

  • First ionization energy: Mg(g) → Mg+(g) + e- (1stE = 736 kJ per mole)
  • Second ionization energy: Mg+(g) → Mg2+(g) + e- (2ndE = 1450 kJ per mole)

 

Summary on physical properties of alkaline earth metals.

Element Atomic number Melting point (oC) Boiling point (oC) Atomic radius (nm) 1st I.E

(kJmol-1)

2nd I.E (kJmol-1)
Beryllium 4 1280 2450 0.089 900 1800
Magnesium 12 650 1110 0.136 736 1450
Calcium 20 850 1140 0.174 590 1150
Strontium 38 789 1330 0.210 550 1060
Barium 56 725 1140 0.220 503 970

 

Chemical properties of Alkaline earth metals.

  1. Burning alkaline earth metals in air.

– Alkaline earth metals react burn in air to form corresponding oxides.

– More reactive alkaline earth metals may also react with atmospheric nitrogen to form corresponding nitrides.

 

Examples:

– Burns in air with a blinding brilliant flame forming a white solid.

– The white solid is a mixture of magnesium oxide ad magnesium nitride.

 

Equations:

Reacting with oxygen: 2Mg(s) + O2(g) → 2MgO(s)

Reacting with nitrogen: 3Mg(s) + N2(g) → Mg3N2(s)

 

– Burns in air with a faint orange flame forming a white solid.

– The white solid is a mixture of calcium oxide ad calcium nitride.

 

 

Equations:

Reacting with oxygen: 2Ca(s) + O2(g) → 2CaO(s)

Reacting with nitrogen: 3Ca(s) + N2(g) → Ca3N2(s)

 

Note:

– The trend in the reactivity of alkaline earth metals when burning in air is not clear; due to the oxide coating on the calcium that tends to slow down the reaction of calcium in air.

– For this reason it is important to polish the surfaces of alkaline earth metals before using them in experiments.

 

  1. Reaction of alkaline earth metals with cold water.

– Alkaline earth metals react slowly with cold water to form corresponding hydroxides and hydrogen gas.

 

Examples

  • Magnesium:

– Reacts slowly with water to form magnesium hydroxide and hydrogen gas.

– The reaction is very slow and the amount f hydrogen gas evolved is very low hence the hydrogen gas bubbles stick on the surface of the metal.

– The magnesium hydroxide formed dissolves slightly in water to form magnesium hydroxide.

– Thus the resultant solution is slightly alkaline.

 

Equation:

Mg(s) + 2H2O(l) → Mg(OH)2(aq) + H2(g).

 

  • Calcium:

Diagram of apparatus:

Observations and explanations:

– Calcium sinks to the bottom of the beaker; because it is denser than water;

– Slow effervescence of a colourless gas; due to slow evolution of hydrogen gas;

– Soapy solution formed; due to formation of alkaline calcium hydroxide;

– A white suspension is formed; because calcium hydroxide is slightly soluble in water;

 

Reaction equation:

Ca(s) + H2O (l) → Ca (OH) 2(aq) + H2 (g);

 

Effect of resultant solution on litmus paper;

– Litmus paper slowly turns blue; calcium hydroxide formed is slightly soluble in water; releasing a small number of hydroxyl ions which result into alkaline conditions // high pH;

 

  1. Reaction with steam.

– Alkaline earth metals react with steam to produce corresponding metal oxide and hydrogen gas.

– The reactivity with stem is faster ad more vigorous for each alkaline earth metals as compared to reaction with cold water.

 

Examples:

  • Magnesium

Procedure:
– A small amount of wet sand is put at the bottom of a boiling tube;

– A small piece of magnesium ribbon is cleaned and put in the middle of the combustion tube;

– The wet sand is heated gently first then the magnesium ribbon is heated strongly until it glows.

Reason:

– To generate steam that drives out the air that would otherwise react with the magnesium (preventing reaction with steam)

– The delivery tube is removed from the water before heating stops.

Reason:

To prevent sucking back (of the gas) as the apparatus cools

– The gas produced is tested using a burning splint.

 

Diagram of apparatus:

Observations and explanations.

– Magnesium burns with a white blinding flame;

– Grey solid (magnesium) forms a white solid; due to formation of magnesium oxide;

– Evolution of a colourless gas that produces a pop sound when exposed to a burning splint; confirming it is hydrogen;

 

Reaction equation.

Magnesium + Steam → Magnesium oxide + Hydrogen gas;

Mg(s) + H2O(g) → MgO(s) + H2(g);

 

  • Calcium

– Reaction between calcium and steam would produce calcium oxide and hydrogen gas.

– However the reaction is too explosive to be done under laboratory conditions.

 

  1. Reaction with chlorine.

– Alkaline earth metals react with chlorine to form corresponding chlorides as the only products.

Condition: presence of heat, hence the metal must be heated first.

Precaution: reaction should be done in a working fume chamber because chlorine gas is poisonous).

 

 

 

Examples:

Procedure:
– A piece of burning magnesium is lowered into a gas jar containing chlorine.

 

Observations:
– The metal continues to burn with a brilliant white flame.

– They grey solid forms a white powder.

 

Explanation.

– Reaction between magnesium and chlorine is exothermic.

– The heat produced keeps the metal burning; thus facilitates the reaction between magnesium and chlorine to form magnesium chloride, which is the white powder.

 

Equation:
Mg(s) + Cl2(g) → MgCl2(s)

 

Procedure:
– A piece of burning calcium is lowered into a gas jar containing chlorine.

 

Observations and explanations.

– The metal burns shortly with an orange flame but soon smolders off.

– There is no steady reaction between calcium and chlorine.

Reason:

– When calcium is heated a coating of the metal oxide is formed first which prevents furtheer reaction between the metal and chlorine.

– However under suitable conditions calcium reacts with chlorine to form a white powder of calcium chloride.

 

Equation:
Ca(s) + Cl2(g) → CaCl2(s)

 

  1. Reaction of alkaline earth metals and dilute acids

– Generally alkaline earth metals react with dilute acids to form salts and hydrogen gas.

 

Examples:

  • With hydrochloric acid.
    • Beryllium:

– When a piece of beryllium is dropped into a beaker containing hydrochloric acid, there is effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of beryllium chloride remains in the test tube // boiling tube.

 

Equation: Be(s) + 2HCl(aq) → BeCl2(aq) + H2(g)

 

 

 

  • Magnesium:

– When a piece of magnesium is dropped into a beaker containing hydrochloric acid, there is effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of magnesium chloride remains in the test tube // boiling tube.

 

Equation: Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

 

  • Calcium:

– When a piece of calcium is dropped into a beaker containing hydrochloric acid, there is effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of calcium chloride remains in the test tube // boiling tube.

 

Equation: Ca(s) + 2HCl(aq) → CaCl2(aq) + H2(g)

 

  • With sulphuric (VI) acid.
    • Beryllium:

– When a piece of beryllium is dropped into a beaker containing sulphuric (VI) acid, there is effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of beryllium sulphate remains in the test tube // boiling tube.

 

Equation: Be(s) + H2SO4(aq) → BeSO4(aq) + H2(g)

 

  • Magnesium:

– When a piece of magnesium is dropped into a beaker containing sulphuric (VI) acid, there is effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of magnesium sulphate remains in the test tube // boiling tube.

 

Equation: Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g)

 

  • Calcium:

– When a piece of calcium is dropped into a beaker containing sulphuric (VI) acid, there is slight effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– The reaction however quickly stops and there is formation of a white precipitate in the test tube or boiling tube.

Reason:
– When calcium reacts with dilute sulphuric (VI) cid there is formation of calcium sulphate which is insoluble hence the formation of a white precipitate. The insoluble calcium sulphate coats the surface of the (calcium) metal stopping further reaction.

 

Equation: Ca(s) + H2SO4(aq) → CaSO4(s) + H2(g)

 

  • With nitric (V) acid.
    • Beryllium:

– When a piece of beryllium is dropped into a beaker containing nitric (V) acid, there is effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of beryllium nitrate remains in the test tube // boiling tube.

 

Equation: Be(s) + 2HNO3(aq) → Be(NO3)2(aq) + H2(g)

 

  • Magnesium:

– When a piece of magnesium is dropped into a beaker containing nitric (V) acid, there is effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of magnesium nitrate remains in the test tube // boiling tube.

 

Equation: Mg(s) + 2HNO3(aq) → Mg(NO3)2(aq) + H2(g)

 

  • Calcium:

– When a piece of calcium is dropped into a beaker containing nitric (V) acid, there is slight effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of calcium nitrate remains in the test tube // boiling tube.

 

Equation: Ca(s) + 2HNO3(aq) → Ca(NO3)2(aq) + H2(g)

 

Trend in reactivity in alkaline earth metals.

– The reactivity of the alkaline earth metals increases down the group.

Reason:

– The atomic radius increases down the group due to increasing number of energy levels. The attraction of the two valence electrons towards the positive nucleus thus decreases down the group. Since alkaline earth metals react by losing electrons the ease with which the valence electrons are lost therefore decreases down the group

 

Similarity of ions and formulae of some compounds of Alkaline earth metals.

– Alkaline earth metals have the same valency (2) and a similar charge (2+) on their ions since they all have a single valence electron.

– Thus they have the same general formula in their ions and compounds.

 

Examples.

Alkali metal ion Hydroxide (OH) Oxides (O2-) Chloride (Cl) Sulphates (SO42-)
Be2+
Mg2+
Ca2+

Uses of some alkaline earth metals and their compounds

  1. Magnesium is used in the manufacture of magnesium hydroxide which is used as an anti-acid medicine.

Reason: magnesium hydroxide is a non-toxic base.

  1. A low-density alloy of magnesium and aluminium is used in aeroplanes construction.
  2. Hydrated calcium sulphate (plaster of Paris) is used in hospitals to set fractured bones.
  3. Cement is made by heating a mixture of calcium carbonate (limestone), clay and sand.
  4. Calcium carbonate is used in the extraction of iron.
  5. Calcium oxide (quicklime) is added tom acidic soils to raise pH for agricultural purposes.
  6. Calcium nitrate is used as a nitrogenous fertilizer.
  7. Magnesium oxide is used in the lining of furnaces.
  8. Barium sulphate is used in the diagnosis of ulcers.
  9. Barium nitrate is used to produce the green flame in fireworks.
  10. Calcium carbonate is mixed with oil to make putty.

 

The Halogens.

– Are the elements with seven valence electrons and hence in group VII of the periodic table.

– All are non-metallic in nature.

– The members of the family in order down the group is as follows:

  • Chlorine
  • Bromine
  • Iodine
  • Astatine

 

Electron arrangement of the first three halogens.

Elements Electron arrangement
Fluorine 2.7
Chlorine 2.8.7
Bromine 2.8.18.7

 

Note:

Halogen is a derivative of two Greek words: halo and gen.

 

Diagram: Part of periodic table showing the halogens

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Gradation in properties of halogens.

  1. Atomic and ionic radius.
  • Atomic radius:

– Refers to the distance between the centre of the nucleus of an atom and the outermost energy level occupied by electron(s)

  • Ionic radius:

– Refers to the distance between the centre of the nucleus of an ion and the outermost energy level occupied by electron(s)

 

Trend:

– The ionic radius and the atomic radius of halogens increase down the group.

Reason:

– There is an increase in the number of energy levels down the group from Fluorine to iodine.

Illustration:

– Fluorine (2.7) has only 2 energy levels; Chlorine (2.8.7) has 3 energy levels while bromine (2.8.18.7) has 4 energy levels.

– Thus the outermost electron in bromine is further from the nucleus than the outermost electron in chlorine and fluorine.

 

  1. Atomic and ionic radius of the same element.

– For the same halogen the ionic radius is larger than the atomic radius.

Reason:

– Halogens form ions by gaining (an) electron into the outermost energy level. This increases the electron-electron repulsion in the outermost energy level hence decreasing the nuclear attraction of the outermost electrons towards the nucleus.

 

Illustration

– Chlorine atom has electron arrangement of 2.8.7 hence 3 energy levels.

– During ion formation it reacts by gaining an electron into the third energy level to acquire a new electron arrangement of 2.8.8.

– In the atom 17 protons are attracting 17 electrons; while in the chloride ion there are 17 protons attracting 18 electrons in the outermost energy level.

– Thus in the ion the effect of the positive nucleus is lower.

– This is due to increased repulsive effect between the existing electrons in the outermost energy level and the incoming electron (electron-electro repulsion)

 

Diagrammatically: chlorine atom and chloride ion

 

 

 

 

 

 

 

 

 

 

 

Summary: changes in atomic and ionic radius among alkali metals.

Element Symbol Atomic number Electron arrangement Atomic radius (nm) Ionic radius (nm)
Fluorine F 9 2.7 0.064 0.136
Chlorine Cl 17 2.8.7 0.099 0.181
Bromine Br 35 2.8.18.7 0.114 0.195
Iodine I 53 2.8.18.18.7 0.133 0.216

 

 

Laboratory preparation of chlorine.

Note: It is usually prepared by oxidation of concentrated hydrochloric acid by removal of hydrogen.

 

Equation:

2HCl(aq) + [O]                Cl2(g) + H2O(l)

– The [O] is from a substance containing oxygen (an oxidizing agent).

– The main oxidizing agents normally used for preparation of chlorine are:

  • Potassium manganate (VII); KMnO4.
  • Manganese (IV) oxide; MnO2

 

(a). Preparation of chlorine from MnO2 and HCl.

(i). Apparatus:

 

 

 

 

 

 

 

 

 

 

 

(ii). Conditions:

– Heating;

– Presence of an oxidizing agent; in this case it is manganese (IV) oxide.

 

(iii). Procedure:

– Hydrochloric acid is reacted with manganese (IV) oxide (dropwise);

Equation:

MnO2(s) + 4HCl(aq)     Heat        MnCl2(aq) + 2H2O(l) + Cl2(g)

 

(iv). Explanation:

– Manganese (IV) oxide oxidizes hydrochloric acid by removing hydrogen resulting into chlorine.

– The manganese (IV) oxide is reduced to water and manganese chloride.

– The resultant chlorine gas is passed through a bottle containing water.

Reason:

– To remove hydrogen chloride fumes (gas) which is very soluble in water.

– Next it is passed through concentrated sulphuric acid or anhydrous calcium chloride; to dry the gas.

 

(v). Collection:

(a). Wet chlorine is collected over brine (saturated sodium chloride solution) or hot water.

Reason:

– It does not dissolve in brine and is less soluble in water

 

(b). Dry chlorine is collected by downward delivery (upward displacement of air)

Reason:

– It is denser than air (2.5 times).

Note:

– Chlorine may also be dried by adding calcium chloride to the jar of chlorine.

 

(c). The first bottle must contain water and the second concentrated sulphuric acid.

Reason:

– If the gas is first passed through concentrated sulphuric acid in the first bottle then to the water; it will be made wet again.

 

Physical properties of Halogens.

  1. Physical state and appearance at room temperature.

 

Halogen State and appearance
Fluorine Pale yellow gas
Chlorine Green-yellow gas
Bromine Volatile brown liquid
Iodine Shiny dark grey solid.

 

  1. Solubility

(a). In water

Experiment: To investigate solubility of halogens in water.

  • Procedure:

– A boiling tube containing chlorine gas is inverted into a beaker containing water.

– The experiment is repeated with fluorine, bromine and a few crystals of iodine.

 

  • Diagram:

 

 

 

 

 

 

 

 

  • Observations:
  • Fluorine:
    – The level of solution rises in the boiling tube.

– The pale yellow colour of fluorine disappears.

 

  • Fluorine:
    – The level of solution rises in the boiling tube.

– The green-yellow colour of fluorine disappears.

 

  • Bromine
    – The level of solution rises in the boiling tube.

– The brown colour of fluorine disappears.

 

  • Iodine:
    – The level of solution remains the same in the boiling tube.

– The shiny dark grey crystals remain in the beaker.

  • The rise in water level is higher in fluorine than in chlorine while the rise in chlorine is higher than in bromine.

 

Diagrams: observations after some time

 

 

 

 

 

 

 

 

 

 

  • Explanations:

– Fluorine, chlorine and bromine are all soluble in water, while iodine is insoluble in water.

– When a boiling tune containing the soluble halogens is inverted into a beaker containing water, the halogen dissolves in the water.

– This creates a partial vacuum and the water in the beaker thus rises to occupy the space left by the dissolved gas.

– The halogens dissolve in water to form acidic solutions.

– The more soluble the halogen, the higher the rise in water level in the boiling tube.

– Thus the solubility of halogens decreases down the group from fluorine to bromine.

Reason:

– All halogens have molecular structures with wreak van der waals forces between the molecules.

During the dissolution the Van der Waals must be broken. The strength of Van der Waals increase as the atomic size and hence the molecular size increases which occurs down the group.

 

(b). In tetrachloromethane.

– The same procedure (of dissolving halogens in water) is followed with tetrachloromethane.

 

  • Observations:

– All halogens are soluble in tetrachloromethane.

– The solubility of each halogen is higher in tetrachloromethane than in water.

Reason: Halogens are molecular thus non-polar and thus are more soluble in polar organic solvents like tetrachloromethane than in polar solvents like water.

 

  1. Effect of heat.

– Fluorine and chlorine are gases at room temperature and simply expand and increase in volume when heated.

– Bromine is a brown liquid at room temperature and evolves brown fumes when heated.

– Iodine exists as shiny dark grey solid at room temperature and sublimes to give brown fumes when heated.

 

  1. Electrical conductivity.

– All halogens are on conductors of heat and electricity.

Reason:

– Halogens are molecular and thus have neither delocalized electrons nor free mobile ions for electrical conductivity.

  1. Trend in melting and boiling points.

– The melting and boiling points of halogens increase down the group.

Reason:

– Halogens exist as diatomic molecules and thus have molecular structures;

– The molecules are held together by intermolecular forces called the Van der Waals which have to be broken during melting and boiling;

– The strength of the Van der Waals increases as the size of the atoms and hence molecules increases which occurs down the group.

 

Summary on some physical properties of halogens>

Halogen Formula Atomic number Appearance Melting point (oC) Boiling point (oC)
Fluorine F 9 Pale yellow gas -238 -188
Chlorine Cl 17 Green yellow gas -101 -35
Bromine Br 35 Brown liquid -7 59
Iodine I 53 Shiny dark grey solid 114 184

 

Chemical properties of Halogens.

Note: It is not easy for non-metals like halogens to form cations.

Reason: the ionization energy (amount of energy required to lose an electron(s) from the outermost energy level of a gaseous atom) is very large.

– Thus most non-metals react by forming anions (negatively charged ions) by electron loss.

 

  1. Ion formation.

– Halogens react by gaining a single electron into the outermost energy level to form a stable electron configuration and corresponding anions.

– During ion formation by electron loss energy is released, and the energy change for this process is called electron affinity

 

Note: Definition.

– Electron affinity is thus the energy released when an atom in gaseous state gains an electron to form a negative ion (anion)

 

 

 

 

 

  • Trend in electron affinity of halogens:

– Generally the electron affinity of halogens decreases down the group.

Reason:

– The size of the atoms increases (due to increasing number of energy levels) down the group and thus the force of attraction of the electrons in the outermost energy level towards the nucleus decreases.

– Thus down the group the ease with which electrons are gained decreases and the faster the ease of electron gain, the more the energy released hence the more the electron affinity.

 

Summary:

Element Ionization equation Electron affinity (kJ per mole
Fluorine F(g) + e → F(g) -322
Chlorine Cl(g) + e → Cl(g) -349
Bromine Br(g) + e → Br(g) -325
Iodine I(g) + e → I(g) -295
  1. Reaction with metals.

(a). Chlorine.

  • Reaction of chlorine with iron.

(i). Apparatus.

 

Note:

– The set up can also be modified by using sodium hydroxide to absorb excess chlorine gas as shown below.

 

 

 

 

 

 

 

 

 

(ii). Precaution.

– Experiment should be done in a functional fume cupboard or in the open.

Reason:

– Chlorine gas is poisonous and will thus be harmful to the human body.

 

(iii). Procedure:

– A stream of dry chlorine gas is passed over heated iron wool as per the diagram.

 

(iv). Conditions.

  • Chlorine gas has to be dry (done by the anhydrous calcium chloride in the U-tube)

Reason:

To prevent hydration hence oxidation of iron (which will then form Fe2O3.5H2O) hence preventing reaction between iron and chlorine.

 

  • Iron metal must be hot; and this is done by heating.

Reason:

To provide activation energy i.e. the minimum kinetic energy which the reactants must have to form products.

  • Anhydrous calcium chloride.

– In the U-tube; to dry the chlorine gas.

– In the thistle funnel; to prevent atmospheric water vapour (moisture) from getting into the apparatus and hence reacting with iron (III) chloride.

 

Note: In the guard tube, calcium oxide is preferable to anhydrous calcium chloride.

Reason:

– Other than preventing atmospheric water vapour from getting into the flask with iron wool; it also absorbs excess chlorine thus preventing environmental pollution

 

(v). Observations:

– Iron metal glows red-hot.

– Red brown fumes (FeCl3(g)) are formed in the combustion tube.

– A black solid (FeCl3(s)) is collected in the flask.

Note:

– Iron (III) chloride cannot be easily collected in the combustion tube.

Reason:

– It sublimes when heated and hence the hotter combustion tube causes it to sublime and its vapour is collected on the cooler parts of the flask.

 

(vi). Reaction equation.

2Fe(s) + 3Cl2(g) → 2FeCl3(g)

 

(vii). Conclusion.

– Iron (III) chloride sublimes on heating; the black solid changes to red-brown fumes on heating.

Equation:

FeCl3(s)                                  FeCl3(g)

(black)                                        (Red-brown)

 

 

  • Reaction with aluminium.

– Chlorine reacts with aluminium I a similar manner to iron to from aluminium chloride.

Equation:

2Al(s) + 3Cl2(g) → 2AlCl3(s)

2Al(s) + 3Cl2(g) → Al2Cl6(s)

 

Note:

– Aluminium chloride also sublimes on heating.

Equation:

AlCl3(s)                                  AlCl3(g)

(White)                                        (White)

 

  • Reaction with burning magnesium.

(i). Procedure:

– Burning magnesium is lowered into a gar jar of chlorine gas.

 

(ii). Observations:

– The magnesium continues to burn with a bright blinding flame;

– Formation of white fumes (MgCl2); which cools into a white powder.

 

(iii). Equation:

Mg(s) + Cl2(g) → MgCl2(s)

 

– Generally chlorine reacts with most metals when hot top form corresponding chlorides.

Note:

Where a metal forms two chlorides when it reacts with chlorine, the higher chloride is usually formed.

Reason:

The higher chloride is stable. This explains why reactions of chlorine with iron results into iron (III) chloride and not iron (II) chloride.

 

(b). Bromine.

  • Reaction of bromine with iron

Apparatus

 

 

 

 

 

 

 

 

 

 

(ii). Precaution.

– Experiment should be done in a functional fume cupboard or in the open.

Reason:

– Bromine gas is poisonous and will thus be harmful to the human body.

 

(iii). Procedure:

– Bromine liquid is heated to generate bromine vapour (fumes).

– The iron wool is then heated and a stream of the bromine gas is passed over heated iron wool as per the diagram.

 

(iv). Conditions.

  • Bromine must be heated to generate fumes before heating the iron.

Reason:

So that bromine vapour will drive out air from the boiling tube to prevent oxidation of iron with oxygen which would otherwise prevent reaction between iron and bromine.

 

  • Iron metal must be hot; and this is done by heating.

Reason:

To provide activation energy i.e. the minimum kinetic energy which the reactants must have to form products.

 

(v). Observations:

– Brown fumes of bromine are produced on heating bromine liquid.

– The iron wool glows red-hot upon heating

– The iron wool forms dark red crystals (of iron (III) bromide)

 

(vi). Reaction equation.

Word: Iron + bromine → Iron (III) bromide.

Chemical: 2Fe(s) + 3Br2(g) → 2FeBr3(s)

Dark red crystals.

 

  • Reaction of bromine with zinc

– Using the same set up bromine also reacts with zinc to form zinc bromide.

Equation: Zn(s) + Br2(g) → ZnBr2(s).

 

  • Reaction of bromine with magnesium

– Using the same set up bromine also reacts with zinc to form magnesium bromide.

Equation: Mg(s) + Br2(g) → MgBr2(s).

 

  • Reaction of bromine with sodium

– Using the same set up bromine also reacts with zinc to form sodium bromide.

Equation: 2Na(s) + Br2(g) → 2NaBr(s).

 

(c). Iodine.

  • Reaction of iodine with iron

Apparatus

 

 

 

 

 

 

 

 

 

(ii). Precaution.

– Experiment should be done in a functional fume cupboard or in the open.

Reason:

– Iodine is poisonous and will thus be harmful to the human body.

 

(iii). Procedure:

– Iodine crystals are heated to generate iodine vapour (fumes).

– The iron wool is then heated and a stream of the iodine gas is passed over heated iron wool as per the diagram.

 

(iv). Conditions.

  • Iodine must be heated to generate fumes before heating the iron.

Reason:

So that iodine vapour will drive out air from the boiling tube to prevent oxidation of iron with oxygen which would otherwise prevent reaction between iron and iodine.

 

  • Iron metal must be hot; and this is done by heating.

Reason:

To provide activation energy i.e. the minimum kinetic energy which the reactants must have to form products.

(v). Observations:

– Purple vapour (fumes) of iodine is produced on heating iodine crystals.

– The iron wool glows red-hot upon heating

– The iron wool forms grey black crystals (of iron (II) iodide)

 

(vi). Reaction equation.

Word: Iron + iodine → Iron (II) iodide.

Chemical: Fe(s) + I2(g) → FeI2(s)

Grey-black crystals.

 

  • Reaction of Iodine with zinc

– Using the same set up bromine also reacts with zinc to form zinc iodide.

Equation: Zn(s) + I2(g) → ZnI2(s).

 

  • Reaction of Iodine with magnesium

– Using the same set up bromine also reacts with zinc to form magnesium iodide.

Equation: Mg(s) + I2(g) → MgI2(s).

 

  • Reaction of Iodine with sodium

– Using the same set up bromine also reacts with zinc to form sodium iodide.

Equation: 2Na(s) + I2(g) → 2NaI(s).

 

Note:

– The reactivity of chlorine with metals is more vigorous than that of bromine, which is more than that of iodine.

– Thus reactivity of halogens with metals decreases down the group.

Reason:

– Halogens react by gaining electrons; the ease of gaining electrons decrease down the group as the atomic size increases which leads to progressive decrease in the force of attraction of electrons in the outermost energy levels by the positively charged nucleus.

 

  1. Reaction with water.

– Halogens that dissolve in water form a mixture of two acids.

 

Reaction of chlorine with water.

– Chlorine dissolves in water to form chlorine water, which is a mixture of two acids: chloric (I) acid and hydrochloric acid.

 

Equation:
Cl2(g) + H2O(l) → HOCl(aq) + HCl(aq)

Chloric (I) acid   Hydrochloric acid

 

 

  • Effects of sunlight on chlorine water.

(i). Procedure:

– Chlorine water is made by dissolving the gas in water.

– A long tube filled with chlorine water is inverted over a beaker containing water.

– It is then exposed to sunlight (bright light) as shown below.

 

(ii). Apparatus:

 

 

 

 

 

 

 

 

 

(iii). Observations:

– After sometime a gas collects in the tube and on applying a glowing splint, the splint is rekindles showing that the gas collected is oxygen.

 

(iv). Explanation:

– Chlorine water has two components.

Equation:

Cl2(g) + H2O(l)   ═ HCl(aq) + HOCl(aq)

 

– The HOCl being unstable will dissolve on exposure to sunlight, giving out oxygen.

Equation:

2HOCl(aq) → 2HCl(aq) + O2(g) (slow reaction)

 

Overall reaction:

2H2O(l) + 2Cl2(g) → 4HCl(aq) + O2(g)

 

  • Effect of chlorine water on litmus papers

(i). Procedure:

– A strip of blue ad a strip of red litmus papers are dropped into chlorine water in a beaker.

 

(ii). Observations:
– The blue litmus paper turns red; then both litmus papers are decolourised.

 

(iii). Explanations.

– Chlorine water contains a mixture of chloric (I) acid ad hydrochloric acid.

Equation:

Cl2(g) + H2O(l) ═ HCl(aq) + HOCl(aq)

 

– The two acids cause the litmus paper to turn red (from blue) while the red litmus paper remained red.

– The litmus papers are then decolourised due to bleaching action of chlorine water, which is due to the activity of chloric (I) acid.

– The chloric (I) acid is unstable and thus decomposes to give hydrochloric acid and oxygen atom.

Equation:

HOCl(aq) → HCl(aq) + [O]

 

– The oxygen atom combines with the chemical of the natural dye in the litmus thus decolourising it.

Equation:

Dye + HOCl(aq) → HCl(aq) + (Dye + [O])

Coloured                                                              Decolourised

Note: solutions of bromine and fluorine in water will behave in a similar manner.

 

Examples:

  • Fluorine dissolving in water

F2(g) + H2O(l)    ═ HF(aq) + HOF(aq)

                                Hydrofluoric acid     Fluoric (I) acid

– The mixture is flourine water

 

  • Bromine dissolving in water

Br2(g) + H2O(l)  ═ HBr(aq) + HOBr(aq)

                                Hydrobromic acid     Bromic (I) acid

– The mixture is bromine water.

 

Some uses of halogens and their compounds.

  1. Fluorine is a raw material in the preparation of a synthetic fibre known as polytetrafluoroethane.
  2. Some compounds of fluorine are added to water and some tooth pastes in small quantities to reduce tooth decay.
  3. Fluorine is used in the manufacture of hydrogen fluoride used to engrave words or pictures on glass.
  4. Chlorine is used to make bleaches used in paper and textile industries.
  5. Chlorine is added to water to kill micro-organisms in water treatment works.
  6. Chlorine is used in the manufacture of a plastic known as polyvinylchloride (PVC).
  7. Chlorine is used in large scale manufacture of hydrochloric acid.
  8. Bromine is used in the manufacture of silver bromide which is used to make the light sensitive photographic paper and films.
  9. A solution of iodine in alcohol (tincture of iodine) is used as a disinfectant.

 

The Noble gases.

– Are the elements with the maximum possible number of valence electrons and hence in group VIII of the periodic table.

– All are gaseous in nature.

– The members of the family in order down the group is as follows:

  • Helium
  • Neon
  • Argon
  • Krypton
  • Xenon
  • radon

– They are found as free atoms in nature and form about 1% of air with argon being the most abundant

 

Electron arrangement of the first three noble gases.

Elements Electron arrangement
Helium 2.
Neon 2.8.
Argon 2.8.8.

 

Note: – Helium with only two electrons has one occupied energy level; which is completely full and hence is said to have a duplet electron configuration

– The rest of the noble gases have eight electrons in their outermost occupied energy level hence are said to have the octet electron configuration.

Diagram: Part of periodic table showing the noble gases

 

 

 

 

 

 

 

 

 

 

 

Dot and cross diagrams for the first three Alkaline Earth metals

 

 

 

 

 

 

 

 

 

 

Physical properties of noble gases

  1. Atomic and ionic radius.
  • Atomic radius:

– Refers to the distance between the centre of the nucleus of an atom and the outermost energy level occupied by electron(s)

  • Ionic radius:

– Refers to the distance between the centre of the nucleus of an ion and the outermost energy level occupied by electron(s)

 

Trend:

– The atomic radius of noble gases increase down the group.

Reason:

– There is an increase in the number of energy levels down the group from Helium to radon.

 

Illustration:

– Helium (2.) has only 1 energy level; Neon (2.8) has 2 energy levels while Argon (2.8.8) has 3 energy levels.

– Thus the outermost electron in Argon is further from the nucleus than the outermost electron in neon and helium.

 

Note:

– Under normal circumstance the noble gases neither lose nor gain electrons and are thus stable hence unreactive.

– Due to this they rarely form ions and hence comparative studies of ionic radii among noble gases are not feasible.

 

  1. Ionization energies.

– Noble gases have very high ionization energies.

Reason:

– Noble gases are colourless monoatomic gases with very stable electron arrangements, either dupltet (helium) or octet (the rest).

– The nuclear attraction force of electrons in the outermost energy level towards the nucleus is thus very strong.

Trend:

– The ionization energy decrease down the noble gases group.

Reason:

– As the size of the atoms increase down the group (due to increase in the number of energy levels), the force with which the electros in the outermost energy levels are attracted towards the nucleus decrease hence increase in the ease of electron loss (from the outermost energy level).

 

  1. Melting and boiling points.

– Generally noble gases have very low melting and boiling points.

Reason:

– They exist as monoatomic gases with simple molecular structures with weak van der waals forces that are easy to break.

Trend:

– Melting and boiling points among the noble gases increase down the group.

Reason:

– Noble gases exist as monoatomic molecules and thus have simple molecular structures;

– The molecules (atoms) are held together by intermolecular (inter-atomic) forces called the Van der Waals which have to be broken during melting and boiling;

– The strength of the Van der Waals increases as the size of the atoms and hence molecules increases which occurs down the group.

 

Summary: changes in atomic and ionic radius among alkali metals.

Element Symbol Atomic number Atomic radius (nm) 1st Ionization energy (kJmol-1) Melting point (oC) Boiling point (oC)
Helium He 2 0.128 2372 -270 -269
Neon Ne 10 0.160 2080 -249 -246
Argon Ar 18 0.192 1520 -189 -186
Krypton Kr 36 0.197 1350 -157 -152
Xenon Xe 54 0.217 1170 -112 -108

 

Chemical properties of Noble gases.

– Generally the noble gases neither lose nor gain electrons and are thus stable hence unreactive.

– Due to this they rarely form ions and hence have no feasible chemical reactions.

– However xenon and radon with very large atomic radii and smaller ionization energies take part in some reactions and thus display come chemical properties.

 

Uses of some noble gases.

Note: The application of noble gases is ironically centered on their inert nature.

  1. Argon is used in light bulbs to provide an inert environment to prevent oxidation.
  2. Argon is used as an insulator in arch-welding.
  3. Neon gas is used in streets and advertisement lights.
  4. Helium mixed with oxygen is used in deep sea diving and mountaineering.
  5. Helium mixed with oxygen also is used in hospitals for breathing in patients with respiratory problems and those undergoing certain forms of surgery.
  6. Helium can be used instead of hydrogen in balloons for meteorological research.

Reason:

– Hydrogen is explosive I presence of air and may explode causing serious accidents.

– Helium is less dense than hydrogen.

  1. Helium is used in thermometers for measurements f very low temperatures.
  2. Liquid helium is used to keep certain metal alloys at temperatures low enough for them to become superconductors.

 

 

UNIT 3: PROPERTIES AND TRENDS ACROSS A PERIOD.

Checklist.

  1. Introduction
  2. Trends in physical properties of elements in period 3
  • Electrical conductivity
  • Atomic and ionic radii
  • Melting and boiling points
  1. Trends in chemical properties of elements in period 3
  • Reaction with oxygen
  • Reaction with water
  • Reaction with acids

 

Introduction:

– A period is a vertical row of elements in the periodic table.

– Elements in the same period have same number of energy levels.

– There are 7 periods in the periodic table except for lanthanides and actinides which are not assigned periods.

– Periods 1 – 3 have fewer elements because they lack the d-block elements and have only the s-block elements and the p-block elements.

Note:

  • s-block elements: group 1 and 2
  • d-block elements: transitional elements
  • p-block elements: groups III to VIII.

 

– To understand trends and properties across a typical period of the periodic table, we shall use period 3 as the reference.

 

The Period three of the periodic table.

– Consists of elements with three energy levels.

– Consists of the 8 elements from sodium to argon.

– It is only made of s-block and p-block elements ad lacks any element in the d-block group of elements.

 

Part of periodic table showing period 3 of the periodic table.

 

 

 

 

 

 

 

 

 

 

 

Trends in physical properties of elements in period 3.

  1. Electrical conductivity.

– Sodium, magnesium and aluminium are good conductors of electricity.

Reason:

They all have giant metallic structures with delocalized electros which conduct electricity;

– The electrical conductivity increases from sodium to aluminium.

Reason:

Electrical conductivity increases with increase in the number of delocalized electrons; thus aluminium with the highest number of delocalized electrons (3) in each atom will have the highest electrical conductivity;

– In the metals the electrical conductivity decreases with increase in temperature.

Reason:

– Increase in temperature distorts the alignment of electrons thus preventing their easy flow and hence poor conductivity;

– Phosphorus, sulphur, chlorine and argon do not conduct electric current.

Reason:
They all have molecular structures and all the electrons in the atoms are used in bonding; thus they lack delocalized electron or free ions for electrical conductivity.

– Silicon conducts electric current, and its electrical conductivity increases with increase in temperatures.

Reason:

– It is a semi-conductor; making it a very unique element in this period.

Note:

– A semi-conductor is a crystalline material which only conducts electricity under certain conditions.

 

  1. The atomic and ionic radii.

– The atomic radii of period 3 elements decrease across the period.

Reason.

– For the same number of energy levels the number of protons in the nucleus increases across the period; this leads to the increase in nuclear charge while the shielding effect remains the same hence decrease in atomic radius across the period.

 

  1. Melting and boiling points

– Sodium, magnesium and aluminium have very high melting and boiling points.

  • Reason:

– They have giant metallic structures with strong metallic bonds which need a lot of energy to break.

– The boiling and melting points increase from sodium to aluminium.

  • Reason:

– As you move across the period from sodium to aluminium, the nuclear charge increases while the energy levels remain the same hence decrease in atomic radius; the smaller the atomic radius (for metals) the stronger the metallic bonds.

– Silicon, though a non-metal also has a very high melting and boiling points.

  • Reason:

– Silicon has a giant atomic structure with strong covalent bonds throughout the structure, which need a lot of heat energy to break.

– Phosphorus, Sulphur, Chlorine and argon have low melting and boiling points.

  • Reason:

– They all have molecular structures with strong covalent bonds between the atoms (except in argon) but weak van der waals forces between the molecules which are easy to break.

– Note that argon exist as atoms and thus a monoatomic molecule.

– The melting and boiling points decreases from phosphorus to argon.

– As we move across the period from phosphorus to argon, the size of the atoms decreases leading to smaller atoms and hence molecules, which lead to decrease in the strength of the van der Waals (across the period)

Note:

– Phosphorus and sulphur exists as solids at room temperature while chlorine and argon exists as gases at room temperature.

Reason: phosphorus and sulphur have giant molecular structures while chlorine and argon have simple molecular structures.


Summary: Some physical properties f elements in period 3.

 

Property Na Mg Al Si P (white) S (monoclinic) Cl Ar
Physical state and appearance Silver Silver solid Silver solid Black solid White solid Yellow solid Green yellow gas Colourless gas
Electron arrangement 2.8.1 2.8.2 2.8..3 2.8.4 2.8.5 2.8.6 2.8.7 2.8.8
Valency 1 2 3 4 5 or 5 6 7 8
Atomic radius 0.157 0.136 0.125 0.117 0.110 0.104 0.09 0.192
MP (oC) 98 650 660 1410 44 119 -101 -189
BP (oC) 890 1110 2470 2360 280 445 -35 -186
Structure Giant metallic Giant metallic Giant metallic Giant atomic Molecular Molecular Simple molecular Simple molecular /atomic
Bond type metallic metallic metallic Covalent Covalent Covalent Covalent Van

der waals

 

Trends in chemical properties of the elements in period 3.

Note: Trends in reactivity.

– The reactivity among the metallic elements decreases across the period from sodium to aluminium.

Reason:

– There is a continuous increase in nuclear charge from sodium to aluminium which leads to increase in ionization energies hence increasing difficulty in removing an electron from the outermost energy level.

– Among the non-metallic elements, the reactivity increases across the period from phosphorus to chlorine.

Reason:

– There is increase in nuclear charge from phosphorus to chlorine, hence an increase in ease of electron gain (electro affinity) since non-metals react by gaining electrons.

– Argo is unreactive and can only react under very special conditions.

Reason:

– It’s a noble gas with a stable octet configuration.

 

  1. Reaction of period 3 elements with oxygen.

– All period three elements react with (burn in) oxygen with the exception of argon.

 

Experiment: To investigate the reactions between period 3 elements and oxygen

Procedure:

– A small piece of the element is placed in a deflagrating spoon and warmed gently until it catches fire.

– It is then lowered into a gas jar full of oxygen.

– The flame colour and the colour of the product are noted.

– 10 cm3 of water containing universal or litmus indicator is added into the gas jar with the products.

 

  • Sodium:

– Burns vigorously in oxygen with a golden yellow flame; to produce white solid of sodium oxide.

Equation:

Na(s) + O2(s) → Na2O(s)

– The resultant sodium oxide dissolves in water to form sodium hydroxide.

Equation:

Na2O(s) + H2O(l) → 2NaOH(aq);

– The sodium hydroxide is alkaline in nature and thus turns litmus indicator blue;

 

  • Magnesium:

– Burns vigorously in oxygen with a bright white light; to produce white solid of magnesium oxide.

Equation:

2Mg(s) + O2(s) → 2MgO(s)

– The resultant magnesium oxide is slightly soluble in water to form magnesium hydroxide.

Equation:

MgO(s) + H2O(l) → Mg(OH)2(aq);

– The magnesium hydroxide is alkaline in nature and thus turns litmus indicator blue;

 

  • Aluminium:

Note:

– Aluminium (foil) is usually coated with a thin layer of aluminium oxide, Al2O3; which prevents reaction with the oxygen.

– When polished, it reacts slowly with oxygen to form a white solid of aluminium oxide.

Equation:

Al(s) + 3O2(s) → Al2O3(s)

– The resultant aluminium (III) oxide is insoluble in water.

 

  • Silicon:

– Silicon powder can only bur in oxygen at very high temperatures of about 450oC to form solid silicon (IV) oxide.

Equation:

Si(s) + O2(s) → SiO2(s)

– The resultant silicon (IV) oxide is insoluble in water.

 

– Burns readily in oxygen with a bright orange flame; to produce dense white fumes (solid) of phosphorus (V) oxide

Equation:

P4(s) + 5O2(s) → 2P2O5(s)

Note: sulphur exists and therefore reacts as molecules of P4.

– The resultant phosphorus (V) oxide readily dissolves in water to form phosphoric (V) acid.

Equation:

P2O5(s) + 3H2O(l) → 2H3PO4(aq);

– The phosphoric acid is acidic in nature and thus turns litmus indicator pink / red;

 

– Burns in oxygen with a blue flame; to form a colourless gas of sulphur (IV) oxide

Equation:

S(s) + O2(s) → SO2(s)

– The resultant sulphur (IV) oxide readily dissolves in water to form sulphuric (IV) acid.

Equation:

SO2(s) + H2O(l) → H2SO3(aq);

– The phosphoric acid is acidic in nature and thus turns litmus indicator pink / red;

 

– The sulphuric (IV) acid is unstable and thus easily gets oxidized by oxygen to the stable sulphuric (VI) acid.

Equation:

H2SO3(aq) + O2(g) → H2SO4(aq);

 

– Burns in oxygen only under certain conditions to form acidic oxides.

Equation:

2Cl2(s) + O2(s) → 2Cl2O(s)

 

– Argon is unreactive.

 

Conclusion:

– Metallic elements burn in oxygen to form basic oxides.

– Non-metallic oxides burn in oxygen to form acidic oxides.

 

  1. Reaction of period 3 elements with water.

Procedure:
– A small piece of sodium metal is cut and dropped into a trough containing water;

– The resultant solution is tested with litmus paper;

 

Diagram of apparatus:

 

 

 

 

 

 

 

 

Observations and explanations:

– The metal floats on the water surface; because it is less dense than water;

– A hissing sound is produced; due to production of hydrogen gas;

– It vigorously melts into a silvery ball then disappears because reaction between water and sodium is exothermic (produces heat). The resultant heat melts the sodium due to its low melting point.

– It darts on the surface; due to propulsion by hydrogen;

– The metal may burst into a golden yellow flame; because hydrogen may explode into a flame which then burns the sodium;

– The resultant solution turns blue; because sodium hydroxide solution formed is a strong base;

 

Reaction equations.

Equation I

2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g);

 

Equation II

4Na(s) + O2 (g) → 2Na2O(s);

 

Equation III:

Na2O(s) + H2O(l) → 2NaOH(aq)

 

Effect of resultant solution on litmus paper;

– Litmus paper turns blue; sodium hydroxide formed is highly soluble in water; releasing a large number of hydroxyl ions which result into alkaline conditions // high pH;

 

Procedure:
– A small piece of magnesium ribbon is cut and dropped into a trough containing water;

– The resultant solution is tested with litmus paper;

 

Diagram of apparatus:

 

 

 

 

 

 

 

 

Observations and explanations:

– The metal sinks into the water surface; because it is denser than water;

– It reacts slowly with water leading to slow evolution of hydrogen gas.

– The resultant solution turns blue; because sodium hydroxide solution formed is a strong base;

 

Reaction equation.

Mg(s) + 2H2O(l) → Mg(OH)2(aq) + H2(g);

 

Effect of resultant solution on litmus paper;

– Litmus paper turns blue; magnesium hydroxide formed is highly soluble in water; releasing a large number of hydroxyl ions which result into alkaline conditions // high pH;

 

– Aluminium does not normally react with cold water or steam.

Reason:

– Aluminium is usually coated with a thin coating of aluminium oxide which prevents further reaction.

– However at very high temperatures of about 700oCsteam can react with aluminium to form aluminium oxide and hydrogen gas.

Equation:

2Al(s) + 3H2O(g) → Al2O3(s) + 3H2(g);

 

Note:

– Due to its inability t react with water and air aluminium is preferable for making cooking utensils like sufurias and coking pans.

 

 

 

 

  • Silicon, phosphorus and sulphur.

– These non-metals do not displace hydrogen from water and thus do not react with water.

 

– Dissolves in water t form chlorine water, which is a mixture of chloric (I) acid ad hydrochloric acid.

 

Equation:
Cl2(g) + H2O(l) → HOCl(aq) + HCl(aq)

Chloric (I) acid   Hydrochloric acid

 

  1. Reaction of period 3 elements with acids

Procedure:

– A piece of the element is dropped into 5 cm3 of an acid in a test tube.

– Any gas produced is tested.

 

  • Sodium:

– Reacts explosively with acids to form salts and hydrogen and thus reactions of sodium with acids should not be tried in the laboratory.

 

  • Magnesium:

– Reacts with both dilute hydrochloric and dilute sulphuric acid to form magnbsium salts and hydrogen gas.

 

  • With hydrochloric acid:

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

 

  • With sulphuric acid:

Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g)

 

  • Aluminium:

Note:

– It does not readily react with dilute acids.

Reason:

– This is due to presence of a thin aluminium oxide coating that prevents contact hence reaction with the acids.

– When polished t remove the oxide coating it reacts with both dilute hydrochloric and dilute sulphuric acid to form aluminium salts and hydrogen gas.

 

  • With hydrochloric acid:

2Al(s) + 6HCl(aq) → 3AlCl3(aq) + 3H2(g)

 

  • With sulphuric acid:

2Al(s) + 3H2SO4(aq) → Al2(SO4)2(aq) + 3H2(g)

 

  • Silicon, phosphorus, sulphur and chlorine.

– They do not react with dilute acids.

 

 

 

Summary: Chemical properties of period 3 elements.

Element Na Mg Al Si P S Cl
Reaction with air or oxygen

 

Readily reacts with air.

– Burns brightly in oxygen with a golden yellow flame to form Na2O

Reacts slowly with air.

– Burns in oxygen with a bright white flame to form MgO

 

– Forms a protective coating of Al2O3 when it burns in oxygen. Si powder burns at temperatures above 950oC to form SiO2 White phosphorus smolders in air;

– P burns in air with a bright orange flame to form P2O3 and P2O5

Burns in air or oxygen with a blue flame to form SO2 gas N reaction with air or oxygen under normal conditions.
Reaction with water

 

Reacts vigorously to produce H2 and NaOH Slow reaction with cold water to form Mg(OH)2 and H2;

– Reacts faster with steam to form MgO and H2

No reaction No reaction No reaction No reaction Dissolves in water to form chlorine water
Reaction with dilute acids Violent reaction giving out H2 and a sodium salt Rapid evolution of H2 gas and a Mg salt is formed Reacts slowly to give H2 and an Al salt No reaction No reaction No reaction No reaction

 

 

UNIT4: STRUCTURE AND BONDING.

  1. Meaning of structure and bond.
  2. Nature of the chemical bond.
  3. Types of bonds.
  • Ionic bonds
    • Meaning
    • Formation
    • Dot ad cross diagrams.
    • Giant Ionic structures
      • Examples
      • Meaning
      • Properties of giant ionic structures
    • Covalent and dative bonds
      • Meaning
      • Formation
      • Dot ad cross diagrams.
      • Giant covalent (atomic) structures
        • Examples
        • Meaning
        • Properties of giant covalent structures
        • Details of Diamond, graphite and silicon (IV) oxide
      • Molecular (simple and giant) structures
        • Examples
        • Meaning
        • Hydrogen bond and Van der waals
        • Properties of molecular structures
      • Metallic bonds
        • Meaning
        • Formation
        • Dot ad cross diagrams.
        • Giant Metallic structures
          • Examples
          • Meaning
          • Properties of giant metallic structures
  1. Bond types across period 3
  • Oxides of period 3 elements
    • Nature
    • Melting and boiling points
    • General trends in bond types
  • Chlorides of period 3 elements
    • Nature
    • Melting and boiling points
    • General trends in bond types
  1. Summary on bond characteristics

 

 

 

 

 

  1. Meaning of structure and bond.
  • Bond:

– The mutual force of attraction that holds particles together when atoms (similar or different) combine during chemical reactions.

  • Structure:

– A regular pattern of particles in a substance held together by chemical bonds.

 

  1. Nature of the chemical bond.

– Atoms are made up of energy levels and nucleus.

– The energy levels contain electrons while the nucleus contains protons and neutrons.

– The electrons are negatively charged, protons are positively charged while neutrons are electrically neutral.

– The noble gases are chemically inert since their outermost energy levels are completely filled with the maximum possible number of electrons.

– Thus noble gases have either a stable duplet state (2) like in helium or a table octet configuration (2.8 or 2.8.8) as in neon and argon respectively.

– Other atoms are unstable because the outermost energy levels are not yet completely filled with the maximum possible number of electrons.

– To attain the stable duplet or octet noble gas configuration, such atoms lose, gain or share their valance electrons.

– It is the act of losing, gaining or sharing valence electrons that lead to the chemical bonds.

– When atoms gain or lose valence electron(s) they become charged forming anions and cations respectively.

– Particles of the same charge repel each other while particles of different charges attract one another.

 

  1. Types of bonds.

– There are three main types of chemical bonds:

  • Ionic bonds
  • Covalent bond
  • Metallic bond

 

(a). Ionic bonds / electrovalent bond.

  • Meaning

– Is a bond formed due to complete transfer of electrons from one atom to another resulting into two oppositely charged ions.

 

Formation of an ionic bond.

– Formed due to complete transfer of electrons from one atom to another; and mainly formed between a metal and a non-metal.

– This occurs in a bid for both atoms to acquire a stable noble gas configuration.

– One atom loses all its valence electrons thus forming a cation (positively charged ions).

– The other atom gains all the lost valence electrons forming an anion (negatively charged ion).

– The cation and the anion are oppositely charged and thus develop a mutual force of attraction between them which is the ionic / electrovalent bond.

 

 

 

 

 

Illustration: formation of ionic bond between sodium and chlorine to form sodium chloride.

– Sodium metal (atomic number 11) has electronic configuration 2.8.1 and thus unstable with 1 valence electron.

– Chlorine gas (atomic number 17) has electronic configuration 2.8.7 and thus unstable with 7 valence electrons.

– Sodium is more electropositive (the tendency to lose electrons to form cations) than chlorine while chlorine is more electronegative (the tendency to gain electrons to form anions).

– Sodium loses its single valance electron to form sodium ion with electronic configuration 2.8 and a net charge of +1 (Na+)

– Chlorine atom accepts the single electron lost by sodium to form a chloride ion with electronic configuration 2.8.8 with a net charge of -1 (Cl).

– The positively charged sodium ion and the negatively charged chloride ion attract each other.

– The electrostatic forces of attraction develop between the two oppositely charged ions and this constitutes the ionic bond.

– Compounds formed due to ionic bonding are thus called ionic compounds.

 

Diagrammatically:

 

 

 

 

 

 

 

 

 

Examples of ionic compounds.

  • Sodium chloride.
  • Potassium fluoride.
  • Magnesium oxide
  • Aluminium (III) oxide.

 

Dot and cross diagrams for ionic compounds

  1. Potassium fluoride.

 

  1. Magnesium oxide

 

  1. Sodium oxide.

 

  1. Magnesium chloride.

 

  1. Aluminium (III) oxide.

 

 

 

 

 

 

 

Giant Ionic structures

– Ionic bonding results into one type of structure, the giant ionic structure.

– This is a type of structure in which all ions are bonded with strong ionic bonds throughout the structure.

– Each ion in the giant ionic structure is surrounded by several others resulting into giant pattern of several ions, hence giant ionic structure.

– Most ionic substances with the giant ionic structure are crystalline in nature, made up crystals.

  • Note:

A crystal is a solid form of a substance in which the particles are arranged in a definite pattern repeated regularly in 3 dimensions.

 

Illustration of the giant ionic structure: sodium chloride structure.

– The NaCl structure consists of many Na+ and Cl arranged and packed in a regular pattern.

– Each Na+ is surrounded by six Cl that are equidistant from it.

– Similarly each Cl is surrounded by six Na+ that are equidistant from it.

– This pattern occurs repeatedly in all directions.

– The result is a giant of ions in all directions hence giant ionic structure.

 

Diagram of the cubic structure of sodium chloride.

Properties of giant ionic structures.

  1. They are hard and brittle.
  • Reason:

– Ionic solids are hard because each ion is held in the crystal by strong attractions from the oppositely charged ions around it.

– They are brittle and thus may be split cleanly (cleaved) using a sharp-edged razor.

  • Explanation:

– When a crystal is tapped sharply along a particular plane it is possible to displace one layer of ions relative to the next.

– Due to the displacement, ions of similar charge come together leading to repulsive forces between the portions of the crystals.

– This forces the two portions of the crystals to split apart.

 

Diagram: crystal cleavage in ionic compounds.

 

 

 

 

 

 

 

 

  1. They have high melting and boiling points.
  • Reason:

– They have strong electrostatic forces / ionic bonds / electrovalent bonds between the oppositely charged ions throughout the structure which require large amounts of energy to break.

 

  1. Solubility.

(i). They are soluble in polar solvents like water, ethanol and acetone (propanone)

  • Reason:

– Water contains highly polar molecules.

– The positive ends of the polar water molecules are attracted to the negative ions in the crystal, and the negative ends of the water molecules are attracted to the positive ions in the crystal.

– This results to the formation of ion-solvent bonds which leads to release of energy.

– This energy is sufficient to cause the detachment of ions from the crystal lattice hence dissolution.

Note:
– This detachment of ions is called solvation, and the energy required for this is called solvation energy.

– Where the solvent is water the ions are said to have been hydrated, and the energy involved in the process is called hydration energy.

 

Diagrams: hydrated positive and negative ions.

 

 

 

 

 

 

 

(ii). They are insoluble in non-polar organic solvents like tetrachloromethane, benzene and hexane.

  • Reason:

– Non-polar molecules are held together by weak intermolecular forces, the Van der Waals forces.

– The Van der Waals are much smaller in magnitude compared to the ionic bonds in the ionic solid crystal lattice.

– Thus the ion-ion interactions in the ionic solid are stronger than the solvent-solvent interactions in the solvent or the solvent-ion interactions between the solid and the solvent.

– Thus the non-polar solvent molecules cannot penetrate the ionic lattice to cause salvation.

 

  1. Electrical conductivity.

– Ionic substances do not conduct electric current in solid state.

  • Reason:

The ions are held in static positions in the solid crystal lattice and thus cannot move to conduct electric current.

– They conduct electric current in molten and solution (aqueous) states.

  • Reason:

– In molten and aqueous states the ions are free and mobile and thus move about conducting electric current.

 

 

 

Gradation in properties of some ionic compounds of sodium.

Property of compound Compound of sodium
Sodium fluoride Sodium chloride Sodium bromide Sodium iodide
Solubility in water Soluble Soluble Soluble Soluble
Melting point (oC) 993 801 747 661
Boiling point (oC) 1695 1413 1390 1304
Electrical conductivity Solid Does not Does not Does not Does not
Molten / solution Conducts Conducts Conducts Conducts

 

Note:

– Solubility of the compounds decrease from sodium fluoride to sodium iodide.

– Melting and boiling points decrease from sodium fluoride to sodium iodide.

 

  1. The covalent bond.
  • Meaning:

– Refers to a bond formed when two atoms of the same or of different elements share electrons to become stable.

– Formation of covalent bond between atoms (similar or dissimilar) result to the formation of a molecule.

– Covalent bonds are usually formed by the association of non-metals.

 

  • Note:

A molecule is a group of atoms (two or more) of the same or different elements that are held together by strong covalent bonds.

 

Formation of a covalent bond.

– Covalent bonding is brought about by the facts that the electro-positivity and the electro-negativity of the elements involved are very close.

– For that reason, none of the atoms can completely lose its valence electrons to the next atom.

– For this reason, both atoms donate electrons which are then shared between them.

– Both atoms thus attain a stable noble gas (duplet or octet) configuration.

 

Illustrations:

  • Formation of chlorine molecule.

– Each chloride atom has electronic configuration 2.8.7 and thus need to gain a single electron in the outermost energy level to attain a stable noble gas configuration.

– Since both chloride atoms have same electro-negativities, none will easily lose an electron to the other.
– For this reason, both donate the number of electrons required by the other atom (in this case 1), which they share between them.

– Thus chlorine molecule is formed by sharing 2 electrons between two chlorine atoms, hence a single covalent bond.

 

Diagram: formation of chlorine molecule.

 

 

 

 

 

Dot and cross diagram for chlorine molecule.

 

 

 

 

 

 

 

 

Formation of oxygen molecule.

– Each oxygen atom has electronic configuration 2.8.6 and thus need to gain 2 electrons into the outermost energy level to attain a stable noble gas configuration.

– Since both oxygen atoms have same electro-negativities, none will easily lose an electron to the other.
– For this reason, both donate the number of electrons required by the other atom (in this case 2), which they share between them.

– Thus oxygen molecule is formed by sharing 4 electrons (2 from each atom) between two chlorine atoms, hence a double covalent bond.

 

Diagram: the oxygen molecule.

 

 

 

 

 

 

 

 

 

Note:

– A single covalent bond is represented in dot (.) and cross (x) diagrams using two dots, two crosses, a dot and a cross or a single line (–) between the atoms involved in the bond.

– Thus a single covalent bond like in chlorine can be represented as Cl – Cl, a double covalent bond like in oxygen can be represented as O = O while a triple covalent bond like in nitrogen can be represented as N ≡ N.

 

Dot (.) and cross (x) diagrams for various covalent compounds.

  1. Hydrogen, H2

 

  1. Hydrogen chloride, HCl

 

  1. Nitrogen, N2

 

  1. Water

 

  1. Carbon (IV) oxide, CO2

 

  1. Ammonia gas, NH3

 

  1. Phosphene, PH3

 

  1. Methane, CH4

 

  1. Ethane. CH2CH2

 

  1. Ethyne, C2H2

 

  1. Ethanol, C2H5OH

 

  1. Bromoethane, C2H4Br.

 

The coordinate bond.

– Refers to a type of covalent bond in which the shared pair of electrons forming the bond is contributed by only one of the atoms forming the bond.

– It is also called the dative bond.

 

Examples:

  • Formation of ammonium ion.

– Occurs when an ammonia gas molecule combines with a hydrogen ion (proton).

– All the atoms in the ammonia molecule have a stable noble gas configuration and thus the molecule is stable.

– However the nitrogen in the ammonia molecule has a lone pair of electrons (electrons that have not yet been used in bond formation)

– The hydrogen ion has lost its outermost single valence electron to form the hydrogen ion.

– Thus the hydrogen ion has no electron(s) in its outermost energy level.

– To be stable the hydrogen needs two electrons in its outermost energy level.

– The hydrogen ion thus accepts bonds with the lone pair (2) of electrons in the nitrogen of the ammonia molecule forming a dative bond.

– The total number of electrons in the ammonium ion is 11 while the total number of electrons is 10 leading to a net positive charge of +1

 

Note:

– In a dot (.) and cross (x) diagram where the covalent bond is represented by horizontal lines (–), the dative // coordinate bond is represented by an arrow (→) pointing the atom that “accepts” the electrons.

 

Diagram: formation of ammonium ion.

 

 

 

 

 

 

 

 

 

 

 

Formation of hydroxonium ion (H3O+)

– Occurs when a water molecule combines with a hydrogen ion (proton).

– All the atoms in the water molecule have a stable noble gas configuration and thus the molecule is stable.

– However the oxygen in the ammonia molecule has 2 lone pairs of electrons // four electrons (electrons that have not yet been used in bond formation)

– The hydrogen ion has lost its outermost single valence electron to form the hydrogen ion.

– Thus the hydrogen ion has no electron(s) in its outermost energy level.

– To be stable the hydrogen needs two electrons in its outermost energy level.

– The hydrogen ion thus accepts and bonds with two of the four electrons in the oxygen of the water molecule forming a dative bond // coordinate bond.

– The total number of electrons in the hydroxonium ion is 11 while the total number of electrons is 10 leading to a net positive charge of +1

 

Diagram: formation of the hydroxonium ion.

 

 

 

 

 

 

 

 

 

 

Note:

– The hydroxonium ion (H3O+) can still further react with another hydrogen ion to form another ion of the formula H4O2+.

– This is due to the presence of a single lone pair of electrons in the structure of the hydroxonium ion (H3O+).

 

Diagram: formation of H4O2+ from H3O+ and H+

 

 

 

 

 

 

 

 

 

 

 

Note:

– The H4O2+ cannot however react further since all the valence electrons in all its atoms have been used in bonding leaving no lone pairs.

 

 

 

Further examples of dative covalent bonds in other compounds.

  1. Carbon (IV) oxide

 

  1. Formation of PH4+

 

  1. Aluminium chlorine dimer (Al2Cl6).

 

  1. Ammonia-aluminium chloride vapours complex, AlCl3.NH3

 

Note:

– Substances with covalent bonds form two main types of structures:

  • Molecular structures
  • Giant atomic // Giant covalent structures.

 

  1. The Molecular structures.

– Refers to a structure in which covalent bonds holds atoms together to form molecules and the resultant molecules are held together by intermolecular forces.

– Substance with molecular structures are usually gases or liquids at room temperature.

– Few solids such as sulphur, iodine, fats, sugar, naphthalene and paraffin wax also have molecular structures.

– The intermolecular forces in molecular structures are of two types:

  • Van der Waals
  • Hydrogen bonds

 

(a). The Van der Waals.

– Are the weakest form of intermolecular forces due to induced dipole- induced dipole attractions between molecules.

– As the size of the molecule increases, the number of constituent electrons increases leading to increase in strength of the induced dipole – induced dipole interactions.

– The strength of the Van der Waals thus increases as the molecular size increases.

 

Note:

This explains the trend in boiling points for the halogens, which increase from fluorine to iodine.

  • Reason:

– From fluorine to iodine the size of the atoms and hence the molecules increases leading to increase in molecular masses that lead to stronger induced dipole – induced dipole interactions hence increasing strength of the Van der Waals (from F2 to I2)

– This also applies for the increase in boiling points for the homologous series of alkanes.

 

Diagram: Illustration of Van der Waals forces

(i). Iodine

(ii). Graphite.

 

(b). The hydrogen bonds.

– Is an intermolecular force in which the electropositive hydrogen atom of one molecule is attracted to an electronegative atom of another molecule.

– The essential requirements for the formation of a hydrogen bond (H – bond) are:

  • A hydrogen atom attached to a highly electronegative atom.
  • An unshared pair of electrons on the electronegative atom.

Note:

This explains why hydrogen bonds are common in molecules in which hydrogen are bonded to highly electronegative atoms like nitrogen, oxygen and fluorine.

 

The formation of hydrogen bonds

– Occurs when hydrogen atom is bonded to a highly electronegative atom like nitrogen, oxygen and fluorine.

– The electrons in the covalent bond (between hydrogen and the more electronegative atom) are drawn towards the electronegative atom.

– Hydrogen atom has no electrons other than the one it contributes to the covalent bond, which is also being pulled away from it (by the more electronegative atom).

– Hydrogen atom has no outer energy level of electrons making the single proton in the nucleus unusually “bare”.

– The proton is thus readily available for any form of dipole-dipole attractions.

– The “bare” proton of the hydrogen atom thus attracts the more electronegative atom (e.g. N, O and F) on either side.

– It thus exerts an attractive force on the more electronegative atom hence bonding them together.

– The two (more electronegative) larger atoms are drawn closer with a hydrogen atom effectively buried in their electron clouds.

– This constitutes the Hydrogen bond.

 

Illustration: formation of hydrogen bonds in water.

– In water two hydrogen atoms are bonded to an oxygen atom which highly electronegative.

– The electrons in the covalent bond (between each hydrogen and oxygen atom) are drawn towards the more electronegative oxygen atom.

– The hydrogen atoms have no electrons other than their share of those in the covalent bond, which are also being pulled away from them by the oxygen atom.

– Hydrogen atoms have no outer energy level of electrons making the single proton in their nucleus unusually “bare”.

– The proton is thus readily available for any form of dipole-dipole attractions (with oxygen in this case).

– The “bare” proton of each of the hydrogen atoms in one molecule thus attracts the more electronegative oxygen atom of the neighboring molecule.

– Each hydrogen atom thus exerts an attractive force on the oxygen atom of the next molecule hence bonding them together.

– The electronegative oxygen atom of one molecule is drawn to the electropositive hydrogen atom of the next molecule with the hydrogen atoms effectively buried in the electron clouds of oxygen.

– This constitutes the Hydrogen bond in the water molecule.

 

Diagram: hydrogen bonds in water molecules.

Note:

– Other compounds with hydrogen bonds include ethanol, ammonia, hydrogen fluoride etc.

– Hydrogen bonds are much stronger than the weak Van der Waals forces but still weaker than the covalent bonds.

 

Effect of hydrogen bonding on properties of molecular substances.

– Hydrogen bonding tends to disrupt the gradation in physical properties of molecular substances in relation to molecular weights.

– The effect on molecular masses on the melting and boiling points only apply when the intermolecular force is the same.

 

Examples.

  1. Both ethanol (C2H5OH) and dimethyl ether (C2H6O) have the same relative molecular mass of 46. However the boiling point of ethanol is higher at 78.5oC than that of dimethyl ether at only -24oC.

Reason:

– Even though both have molecular structures with covalent bonds between the atoms, the intermolecular forces in ethanol are hydrogen bonds which are much stronger than the intermolecular forces in dimethyl ether which are weak Van der Waals forces.

 

  1. Ethanol (C2H5OH) has a molecular mass of 46 while butane (C4H10) has a molecular mass of 58, yet the boiling point of ethanol is higher than that of butane.

Reason:

– Both have molecular structures with covalent bonds between the atoms. However, the intermolecular forces in ethanol are hydrogen bonds which are much stronger (and require more energy to break) than the intermolecular forces in dimethyl ether which are weak Van der Waals forces.

 

– Molecular substances are generally insoluble in polar solvents like water. However those with hydrogen bonding as the intermolecular forces are soluble in water since the hydrogen bonding confers them some polarity.

Examples:

– Sugar, ethanol, ethanoic acid etc

 

Properties of molecular structures.

  1. Melting and boiling points.

– The generally have low melting and boiling points.

Reason:

– Although the atoms forming the molecules are held together by strong covalent bonds, the intermolecular forces are weak Van der Waals forces which require low amounts of energy to break.

 

  1. Heat and electrical conductivity

– They are poor conductors of heat and electricity at any state.

Reason:

– They have neither delocalized electrons nor free mobile ions for electrical conductivity.

 

  1. Solubility

– Molecular substances are generally insoluble in polar solvents like water but soluble in non-polar organic solvents like benzene.

Reason:

– In polar solvents like water there are strong water – water attractions which are considerable stronger than the intermolecular forces (Van der Waals) attractions or molecule – water (solvent) attractions, making the molecules unable to penetrate the water (solvent) structure for dissolution to occur.

– In non-polar solvents like benzene, the benzene-benzene attractions are similar in strength to the intermolecular forces or the molecule – benzene (solvent) attractions, enabling the molecules to penetrate the solvent thus allowing dissolution.

 

Note:

– Molecular substances with hydrogen bonds as the intermolecular forces are soluble in polar solvents like water.

Reason:

– The hydrogen bonds in the molecules are equal in strength to the water – water interactions which are also hydrogen bonds, thus the molecules are able to penetrate the structure of water leading to salvation // dissolution // hydration i.e. they are polar like the water molecules.

 

 

 

 

 

 

Summary: Properties of some molecular substances.

Property of compound Molecular substance
Sugar (Sucrose) Naphthalene Iodine Rhombic sulphur Water Hydrogen sulphide
Solubility in water Soluble Insoluble Insoluble Insoluble Slightly soluble
Molecular mass 183 128 186 256 18 34
Melting point (oC) 186 82 113 114 0 -85
Boiling point (oC) 218 183 444 100 -60
Electrical conductivity Solid Does not Does not Does not Does not Does not Does not
Molten / solution Does not Does not Does not Does not Does not Does not

 

  1. Giant covalent structures // Giant atomic structures.

– Are molecular substances in which atoms are linked throughout the whole structure by very strong covalent bonds from one atom to the next.

– The result is an indefinite number of atoms which are all covalently bonded together.

– This pattern occurs repeatedly throughout the structure leading to a “giant” of atoms all covalently bonded.

 

Examples:

(a). Diamond.

– Is an allotrope of carbon.

  • Note: allotropes are different crystalline forms of the same element in the same physical state.

– In diamond, each carbon atom is bonded to four other carbon atoms by strong covalent bonds.

– The carbon atoms in diamond are covalently bonded into an octahedral pattern, which repeats itself in all directions resulting into a giant atomic structure.

– Since each carbon atom is bonded to four others, all the four valence electrons in each carbon are used in bonding hence no delocalized electrons in the structure of diamond.

– Diamond is the hardest substance known due to the fact that all the atoms are covalently bonded together and are closely packed together.

 

Diagram: Structure of diamond:

Properties of diamond.

  1. Have high melting and boiling points.

Reason:

– It has a giant atomic structure with strong covalent bonds throughout the structure which require large amounts of energy to break.

 

  1. It is insoluble in water.

Reason:

– It is non-polar and thus cannot dissolve in polar water molecules since there are no intermolecular interactions which would facilitate penetration into the water molecules for dissolution to occur.

iii. Does not conduct heat and electricity.

Reason:

– Each carbon atom in the structure of diamond is bonded to four others hence uses all its four valence electrons in bonding and thus lacks any delocalized electrons for electrical conductivity.

 

  1. It is the hardest substance known.

Reason:

– All carbon atoms are compactly bonded in a continuous octahedral pattern with strong covalent bonds throughout the structure which are very difficult to break.

 

(b). Graphite.

– Is also an allotrope of carbon.

– In graphite, each carbon atom is bonded to three other carbon atoms by strong covalent bonds.

– Since each carbon atom is bonded to only three others, only three of the four valence electrons in each carbon are used in bonding hence presence delocalized electrons in the structure of graphite.

– This explains the electrical conductivity of graphite.

– The carbon atoms in graphite are covalently bonded into hexagonal layers, which are joined to each other by weak Van der Waals forces.

– The presence of weak Van der Waals forces explains the slippery nature of graphite.

 

Diagram: Structure of graphite.

Properties of Graphite.

  1. Have high melting and boiling points.

Reason:

– It has a giant atomic structure with strong covalent bonds throughout the hexagonal layers which require large amounts of energy to break. Even though there are Van der Waals between the layers the effect of the large number of covalent bonds still contribute to high melting and boiling points in graphite

 

  1. Insoluble in water.

Reason:

– It is non-polar and thus cannot dissolve in polar water molecules since there are no intermolecular interactions which would facilitate penetration into the water molecules for dissolution to occur.

 

iii. It is a good conductor heat and electricity.

Reason:

– Each carbon atom in the structure of graphite is bonded to three others hence uses only three of its four valence electrons in bonding. This leads to presence of delocalized electrons in the structure of graphite which conducts heat and electricity.

 

  1. It is soft and slippery.

Reason:

– The carbon atoms in graphite are covalently bonded into hexagonal layers which are joined to each other by weak Van der Waals forces.

– The weak Van der Waals forces easily slide over each other when pressed hence the soft and slippery feel.

 

(c). Silicon (IV) oxide.

– Is a covalent compound of silicon and oxygen.

– Silicon has four electrons in its outermost energy level while oxygen has six.

– Each silicon atom is bonded to four oxygen atoms by strong covalent bonds.

– Each oxygen atom is bonded to two silicon atoms by strong covalent bonds.

– This means there are no delocalized electrons in the structure of silicon (IV) oxide making it unable to conduct electric current.

– The silicon and oxygen atoms are all covalently bonded together (by strong covalent bonds) in a repeated manner leading to a giant of covalent bonds throughout the structure.

– The extra ordinarily strong covalent bonds in silicon (IV) oxide throuout the structure contribute to the very high melting (1728oC) and boiling (2231oC) points.

 

Diagram: structure of silicon (IV) oxide.

 

 

 

 

 

 

 

 

 

Properties of giant atomic // giant covalent structures.

  1. – Have high melting and boiling points
  2. – They are non-conductors of heat and electricity with the exception of graphite

iii. – They are insoluble in water

  1. – Most are generally very hard, with the exception of graphite

 

  • Note: for explanations of the above properties, refer to individual explanations of each compound.

 

  1. The metallic bond.

– Is a bond formed due to electrostatic attraction between the positively charged nuclei and the negatively charged delocalized electrons that hold atoms together.

 

Formation of a metallic bond.

– In a metal there are usually many atoms surrounding any one atom.

– The valence electrons of any one atom are therefore mutually attracted to many nuclei.

– This leads to a situation in which the positive nuclei appear to be immersed in a sea of mobile electrons.

– The sea of mobile electrons are said to be delocalized which explains the ability of substances with metallic bonds to conduct electric current.

– This pattern of positive nuclei in a sea of electrons is repeated many times throughout the structure leading to a giant metallic structure, the only structure due to metallic bonds.

 

Diagram: model of the giant metallic structure.        

 

 

 

 

 

 

 

 

 

Properties of the giant metallic structure.

  1. They have high melting and boiling points.

Reason:

– They have strong metallic bonds throughout the structure which need large amounts of energy to break.

– The strength of the metallic bond increases with decrease in atomic size, as well as with increase in the number of delocalized electrons.

– Thus metals with smaller atoms and more delocalized electrons tend to have stronger metallic bonds hence higher melting and boiling points.

– This explains why the melting and boiling points of metallic elements in period three increase from sodium to aluminium.

 

  1. They are good conductors of heat and electricity.

Reason:

– The have delocalized electrons which heat and electric current.

Note:

– The electrical conductivity of metals increases with increase in the number of delocalized electrons in each atom in the structure.

– This explains why aluminium metal is a better conductor of heat and electricity than both magnesium and sodium.

 

  1. They are insoluble in water.

Reason:

– There are no dipoles in the giant metallic structure and are thus non-polar, so cannot dissolve in polar water molecules.

Summary: some physical properties of metals

Metal Valency Melting point (oC) Boiling point (oC) Atomic radii (nm) Electrical conductivity
Lithium 1 180 1330 0.133 Good
Sodium 1 98 890 0.155 Good
Potassium 1 64 774 0.203 Good
Magnesium 2 651 1110 0.136 Good
Aluminium 3 1083 2582 0.125 Good

 

Summary: Comparing various types of structures.

Attribute Giant metallic Giant atomic / giant  covalent Giant ionic Giant metallic
1. Structure

i. Examples

 

Na, Fe, Cu.

 

Diamond, SiC, SiO2.

 

Ca2+O2-, (K+)2SO42-,

Na+Cl,

 

I2, S8, C10H8, HCl, CH4.

ii. Constituent particles

 

Atoms Atoms Ions Molecules
iii. Type of substance compound Metal element with low electronegativity Non-metal element in group IV or its compound. Metal/non-metal compound (a compound of elements with a large difference in electronegativity)

 

Non-metal element or non-metal/non-metal compound (elements with high electronegativity)
2. Bonding:

In the solid

Attraction of outer mobile electrons for positive nuclei binds atoms together by strong metallic bonds Atoms are linked through the whole structure by very strong covalent bonds from one atom to the next.

 

Attraction of positive ions for negative ions results in strong ionic bonds Strong covalent bonds hold atoms together within the separate molecules; separate molecules are held together by weak intermolecular forces.

 

3. Properties.

i. Volatility:

 

 

State at room temp.

 

Non-volatile. Very high melting and very high boiling points.

Usually solid

 

 

Non-volatile. Very high melting and very high boiling points.

Solid

 

 

Non-volatile. Very high melting and very high boiling points.

Solid

 

Volatile. Low melting and low boiling points.

Usually gases or volatile liquids

 

ii. Hardness // malleability

 

Hard, yet malleable. Very hard and brittle Hard and brittle Soft
iii. Conductivity: Good conductors when solid or molten.

 

Non-conductors in any state (except graphite) Non-conductors when solid; good conductors when molten or in aqueous solution (electrolytes)

 

Non-conductors when solid, molten and in aqueous solution.

(A few like HCl react with water to form electrolytes)

 

iv. Solubility: Insoluble in polar and non-polar solvents but soluble in molten metals Insoluble in all solvents Soluble in polar solvents (e.g. H2O), insoluble in polar solvents like tetrachloromethane, CCl4.

 

Polar molecules e.g. HCl are soluble in polar solvents like water; but insoluble in non-polar solvents like CCl4 and vise-versa.

Types of bonds across a period.

– The number of valence electrons play an important role in determination of chemical bonding.

– Across a period in the periodic table, the nature of the bonds varies from metallic to covalent.

– The structure also thus varies from giant metallic to simple molecular.

– Thus similar compounds of the elements in period 3 will also exhibit variation in bond types, structures and properties.

 

Variation in bond types in oxides of period three elements.

Oxide Na2O MgO Al2O3 SiO2 P2O5 SO2 Cl2O7
Physical state Solid Solid Solid Solid

 

Solid Gas Gas
M.P (oC) 1193 3075 2045 1728

 

563 -76 -60
B.P (oC) 1278 3601 2980 2231

 

301 -10 -9
Structure Giant ionic Giant ionic Giant ionic Giant atomic

 

Molecular Molecular Molecular
Bonding Ionic Ionic Ionic Covalent Covalent with Van der Waals

 

Covalent with Van der Waals

 

Covalent with Van der Waals

 

Nature of oxide Basic (alkaline) Basic (weakly alkaline)

 

Amphoteric Acidic Acidic Acidic Acidic
Solubility in water Dissolves to forms an alkaline solution Dissolves to forms an alkaline solution

 

Insoluble Insoluble Dissolves in water to form acidic solution. Dissolves in water to form acidic solution. Dissolves in water to form acidic solution.
Reaction with acids Reacts to form salt and water. Reacts to form salt and water.

 

Reacts to form salt and water. No reaction. No reaction. No reaction No reaction

 

Note:

  1. – The melting and boiling points of the magnesium oxide is higher than that of sodium oxide

Reason:

– Both have giant ionic structures. However the electrostatic forces of attraction between magnesium ions and oxide ions are stronger due to the fact that magnesium ion has a charge of +2 and is smaller in size than the sodium ion.

 

 

Effect of bond types of properties of chlorides of period 3 elements

– Most period 3 elements form stable chlorides.

– The trend in bond types, structures and properties of chlorides of period 3 elements show variation across the period.

 

Properties:

  1. Reaction with water.

Procedure:

– A test tube is half filled with water and initial temperature of the water recorded.

– A spatula end full of sodium chloride is added and stirred until it dissolves.

– The hiughest temperature attained when all the solid dissolves is recored and the temperature change calculated.

– Two – three drops of universal indicator are added and the pH of the solution noted and recorded.

 

Observations:

Chloride Observations
Solubility Temperature change (oC) pH of solution
Sodium chloride Dissolves Drop in temperature 7
Magnesium chloride Dissolves Slight increase 6.5
Aluminium chloride Hydrolyzed Increases 3
Silicon (IV) oxide Hydrolyzed Increases 2
Phosphorus (III) chloride Hydrolyzed Increases 2
Phosphorus (V) chloride Hydrolyzed Increases 2

 

Explanations.

  • Sodium and magnesium chlorides.

– Sodium chloride dissolves in water causing a slight drop in temperature.

– Magnesium dissolves readily with a small increase in temperature.

– Both chlorides are ionic and when added to water there is an immediate attraction of polar water molecules for ions in the chlorides.

– The solid thus readily dissolves forming aquated ions such as Na+(aq) and Cl(aq).

– These are separate metal and non-metal ions surrounded by polar water molecules.

– Since there is no production of either H+ or OH ions the solutions are neutral.

 

Equations:

NaCl(s) + H2O(l) → Na+(aq) + Cl(aq) + H2O(l).

 

MgCl2(s) + H2O(l) → Mg2+(aq) + 2Cl(aq) + H2O(l).

 

  • Anhydrous aluminium chloride.

– It exits in molecular form as a dimeric molecule of Al2Cl6.

– The dimeric molecule is formed when aluminium chloride (AlCl3) molecules vapour condense and combine forming larger molecules of Al2Cl6.

 

Diagram: Formation of a dimer in aluminium chloride.

– When added to water aluminium chloride is hydrolyzed to form an acidic solution.

– By so doing it behaves like a covalent chloride rather than an ionic chloride.

– The hydrolysis is an exothermic reaction accompanied by release of heat hence the increase in temperature.

 

Explanations.

– The hydrolysis is due to the very small but highly charged aluminium ion, Al3+.

– The Al3+ draws electrons away from its surrounding water molecules and causes them to give up H+ ions.

– This reaction usually involves aluminium ions combining with six water molecules to form hexa-aqua-aluminium (III) ions which dissociate to give H+.

 

Equation:

[Al(H2O)6]3+(aq) → [Al(H2O)5(OH)]2+(aq) + H+(aq)

 

Note:

This reaction can be simplified as follows:

– Aluminium chloride reacting with water and hydrolyzing to give HCl(aq) as one of the products.

  • Equation: Al2Cl6(s) + 6H2O(l) → 2Al(OH)3(aq) + 6HCl(aq)

– Then the HCl dissociates to give H+ and Cl-

  • Equation: 6HCl(aq) → H+(aq) + Cl(aq)

– It is the resultant H+ that confer the solution its acidic properties.

 

  • Silicon (IV) chloride

– Undergoes hydrolysis in water in an exothermic reaction producing a lot of heat.

– The products of the hydrolysis are silicon (IV) oxide solid and hydrogen chloride gas.

– The hydrogen chloride gas immediately dissolves in the water to form hydrochloric acid.

– The hydrochloric acid dissociated to liberate H+ which leads to acidic conditions.

 

Equations:

SiCl4(s) + 2H2O(l) → SiO2(s) + 4HCl(aq)

Then: 4HCl(aq) → 4H+(aq) + 4Cl(aq).

 

  • Phosphorus (III) chloride and phosphorus (V) chloride

– Both undergo hydrolysis in water in an exothermic reaction producing a lot of heat.

– The products of the hydrolysis are phosphorus (III) acid and phosphoric (V) acid respectively, and hydrogen chloride gas.

– The hydrogen chloride gas immediately dissolves in the water to form hydrochloric acid.

– The hydrochloric acid dissociated to liberate H+ which leads to acidic conditions.

 

Equations:

  • With phosphorus (III) chloride:

PCl3(s) + 2H2O(l) → H3PO3(s) + 3HCl(aq)

Then: 3HCl(aq) → 3H+(aq) + 3Cl(aq).

 

  • With phosphorus (V) chloride:

PCl5(s) + 4H2O(l) → H3PO4(s) + 5HCl(aq)

Then: 5HCl(aq) → 5H+(aq) + 5Cl(aq).

 

Trends in bond types and properties of chlorides of elements in period 3

Property Period 3 chloride

 

Formula NaCl MgCl2 Al2Cl6 SiCl4 PCl3 & PCl5 SCl2

 

Physical state at RT Solid Solid Solid Liquid Liquid Liquid
M.P (oC) 801 714 Sublimes at 180oC -70 Sublimes at -94oC -78
B.P (oC) 1467 1437 57 Decomposes at 57oC
Conductivity Good Good V. poor nil nil Nil

 

Structure Giant ionic Giant ionic Molecular dimer Molecular Molecular Molecular
Bond type Ionic Ionic Ionic / covalent Covalent Covalent Covalent
Effect on water Readily dissolves Readily dissolves Hydrolyzed to give HCl fumes Hydrolyzed to give HCl fumes Hydrolyzed to give HCl fumes Hydrolyzed to give HCl fumes
pH of solution 7 6.5 3 2

 

2 2

 

Summary: Characteristics of bonds

Property Substances with
Covalent bonds Ionic bonds Metallic bonds
Electrical conductivity Non-conductors except graphite – Solids do not conduct.

– Aqueous solutions and molten state conduct

 

– Conducts
Thermal conductivity – Non-conductors except graphite

 

– Do not conduct – Conducts
Melting point (oC) – Low for molecular substances

– High for giant atomic structures

 

– Usually high. – Generally high
Boiling point (oC) – Low for molecular substances

– High for giant atomic structures

 

– Usually high – Generally high
Solubility  – Generally insoluble in water but soluble in organic solvents – Generally soluble in water – Some metals react with water

 

 

UNIT 5: SALTS.

  1. Meaning.
  2. Types of salts:
  • Normal salts
  • Acid salts
  • Basic salts
  • Double salts.
  1. Solubility of salts in water
  • Sulphates
  • Chlorides
  • Nitrates
  • Carbonates
  • Sodium, potassium and ammonium salts
  1. Solubility of bases in water.
  2. Obtaining crystals
  3. Preparation of salts
  • Insoluble salt
    • Double decomposition
  • Soluble salts

A sodium, potassium and ammonium salts

 

A non-potassium, sodium and ammonium salt.

That reacts with water

Direct synthesis

That does not react with water

Acid and metal method

Acid + base method

Neutralization

  1. Uses of salts
  2. Action of heat on salts
  • Carbonates
  • Nitrates