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CHEMISTRY NOTES FORM 3 PDF

October 5, 2025 Maverick John Leave a comment

UNIT 2: NITROGEN AND ITS COMPOUNDS.

Unit checklist.

Introduction
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

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Action of heat on nitrates.
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

Nitric (V) acid

Laboratory preparation
Industrial manufacture of nitric (V) acid: The Otswalds process.
Reactions of dilute nitric acid
Reactions of concentrated nitric acid
Uses of nitric acid

Test for nitrates.
Pollution effects of nitrogen and its compounds
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) + CO_2(g) K₂CO_3(aq) + H₂O_(l)

Then

K₂CO_3(aq) + H₂O_(l) + CO_2(g) 2KHCO_3(aq)

Thus;

KOH_(aq) + CO_2(g) KHCO_3(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) + O_2(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:

NH₄NO_3(s) ^Heat N_2(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) + CO_2(g) K₂CO_3(aq) + H₂O_(l)

Then:

K₂CO_3(aq) + H₂O_(l) + CO_2(g) 2KHCO_3(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 -25^oC.

– 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 -196^oC (77K) and that of oxygen is -183^oC (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 (-183^oC) 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 -25^oC) 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.

NaNO_2(aq) + NH₄Cl_(aq) ^heat NaCl_(aq) + N_2(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) + 2NH_3(g) 3Cu_(s) + N_2(g) + 3H₂O_(l)

(Black) (Brown) (Colourless)

Zinc (II) oxide

3ZnO_(s) + 2NH_3(g) 3Zn_(s) + N_2(g) + 3H₂O_(l)

(Yellow-hot) (Grey) (Colourless)

(White-cold)

Lead (II) oxide

3PbO_(s) + 2NH_3(g) 3Cu_(s) + N_2(g) + 3H₂O_(l)

(Red-hot) (Grey) (Colourless)

(Yellow-cold)

Properties of nitrogen.

(a). Physical properties.

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

(b). Chemical properties.

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

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:

N_2(g) + 3H_2(g) 2NH_3(g)

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

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) + N_2(g) ^Heat Mg₃N_2(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

Mg₃N_2(s) + 6H₂O_(l) 2NH_3(g) + 3Mg(OH)_2(aq)

In sodium hydroxide:

Mg₃N_2(s) + NaOH_(aq)

Reaction with oxygen.

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

Equation:

N_2(g) + O_2(g) 2NO_(g)

Note:

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

Uses of nitrogen

Used in the Haber process in the manufacture of ammonia.
Due to its inert nature, it is mixed with argon to fill electric bulbs (to avoid soot formation).
In liquid state it is used as an inert refrigerant e.g. storage of semen for artificial insemination.
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.
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

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:

NH₄NO_3(s) ^Heat NO_2(g) + 2H₂O_(l)

Properties:

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

Equations:

N₂O_(g) + Mg_(s) ^Heat MgO_(s) + N_2(g)

2N₂O_(g) + S_(s) ^Heat SO_2(g) + 2N_2(g)

2N₂O_(g) + C_(s) ^Heat CO_2(g) + 2N_2(g)

5N₂O_(g) + 2P_(s) ^Heat P₂O_5(g) + 5N_2(g)

Magnesium decomposes the gas and continues to burn in it.

Equation:

N₂O_(g) + Mg_(s) ^Heat MgO_(s) + N_2(g)

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

Equation:

N₂O_(g) + Cu_(s) ^Heat CuO_(s) + N_2(g)

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.

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) + 8HNO_3(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) + O_2(g) 2NO_2(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.

It is a colourless, insoluble and neutral to litmus. It is also slightly denser than air.
Readily combines with oxygen in air and forms brown fumes of nitrogen (IV) oxide.
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) + N_2(g)

4P_(s) + 10NO_(g) 2P₂O_5(s) + 5N_2(g)

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

Equation:

2Cu_(s) + 2NO_(g) 2CuO_(s) + N_2(g)

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:

FeSO_4(aq) + NO_(g) FeSO4.NO_(aq)

Green solution Dark brown

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

It is also a reducing agent.

Equation:

Cl_2(g) + 2NO_(g) 2ClNO_(l)

Chloro nitrogen (II) oxide.

Reaction with hydrogen.

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

Equation:

2H_2(g) + 2NO_(g) 2H₂O_(l) + N_2(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.

It is an intermediate material in the manufacture of nitric acid

Nitrogen (IV) oxide.

Preparation:

(i). Apparatus.

(ii). Procedure:

– Action of conc. Nitric acid on copper metal.

Equation:

Cu_(s) + 4HNO_3(l) Cu(NO₃)_2(aq) + 2NO_2(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) + 4NO_2(g) + O_2(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:
2NO_2(g) N₂O_4(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.

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

Equation:

2NO_2(g) + H₂O_(l) HNO_3(aq) + HNO_2(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:

2NHO_2(aq) + O_2(g) 2NHO_3(aq)

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

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) + 2NO_2(g) 4MgO_(s) + 2N_2(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) + 2NO_2(g) 4CuO_(s) + N_2(g)

(ii). Phosphorus:

8P_(s) + 10NO_2(g) 4P₂O_5(s) + 5N_2(g)

(iii). Sulphur:

2S_(s) + 2NO_2(g) 2SO_2(g) + N_2(g)

Note:

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

Equation:

2NO_2(g) ^Heat 2NO_(g) + O_2(g)

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

– The resultant oxygen supports the burning.

Effects of heat:

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

Equation:

2NO_2(g) ^Heat 2NO_(g) + O_2(g)

– 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

N₂O_4(l) 2NO_2(g) 2NO_(g) + O_2(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 150^oC when the gas is entirely NO₂ and is almost black.

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

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) + 2NO_2(g) 2NaNO_3(g) + NaNO_2(aq) + H₂O_(l)

Ionically:

2OH^–_(aq) + 2NO_2(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.

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.

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

Examples:

2NaNO_3(s) ^Heat 2NaNO_2(s) + O_2(g)

2KNO_3(s) ^Heat 2KNO_2(s) + O_2(g)

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) + 4NO_2(g) + O_2(g)

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

Further examples:

2Ca(NO₃)_2(s) ^Heat 2CaO_(s) + 4NO_2(g) + O_2(g)

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

2Mg(NO₃)_2(s) ^Heat 2MgO_(s) + 4NO_2(g) + O_2(g)

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

2Zn(NO₃)_2(s) ^Heat 2ZnO_(s) + 4NO_2(g) + O_2(g)

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

2Cu(NO₃)_2(s) ^Heat 2CuO_(s) + 4NO_2(g) + O_2(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) + 4NO_2(g) + O_2(g)

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

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) + 2NO_2(g) + O_2(g)

(White solid) Brown Fumes Colourless gas

2AgNO_3(s) ^Heat 2Ag_(s) + 2NO_2(g) + O_2(g)

(White solid) Brown Fumes Colourless gas

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

Example:

NH₄NO_3(s) ^Heat N₂O_(g) + O_2(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 NH_3(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) CaCl_2(aq) + 2NH_3(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.

2NH_3(g) + H₂SO_4(l) (NH₄)₂SO_4(aq)

With fused calcium chloride:

CaCl_2(aq) + 4NH_3(g) CaCl₂.4NH_3(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) + NH_3(g)

Ionically;

Na⁺_(aq) + OH^–_(aq) + NH₄Cl_(s) Na⁺_(aq) + Cl^–_(aq) + H₂O_(l) + NH_3(g)

Hence; NH₄Cl_(s) + OH^–_(aq) Cl^–_(aq) + H₂O_(l) + NH_3(g)

(ii). With caustic potash:

KOH_(aq) + NH₄Cl_(s) KCl_(aq) + H₂O_(l) + NH_3(g)

Ionically;

K⁺_(aq) + OH^–_(aq) + NH₄Cl_(s) K⁺_(aq) + Cl^–_(aq) + H₂O_(l) + NH_3(g)

Hence; NH₄Cl_(s) + OH^–_(aq) Cl^–_(aq) + H₂O_(l) + NH_3(g)

Note:

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

Equations:

(i). With caustic soda:

2NaOH_(aq) + (NH₄)₂SO_4(s) Na₂SO_4(aq) + 2H₂O_(l) + 2NH_3(g)

Ionically;

2Na⁺_(aq) + 2OH^–_(aq) + (NH₄)₂SO_4(s) 2Na⁺_(aq) + SO₄^2-_(aq) + H₂O_(l) + NH_3(g)

Hence; (NH₄)₂SO_4(s) + 2OH^–_(aq) SO₄^2-_(aq) + 2H₂O_(l) + 2NH_3(g)

(ii). With caustic potash:

2KOH_(aq) + (NH₄)₂SO_4(s) K₂SO_4(aq) + 2H₂O_(l) + 2NH_3(g)

Ionically;

2K⁺_(aq) + 2OH^–_(aq) + (NH₄)₂SO_4(s) 2K⁺_(aq) + SO₄^2-_(aq) + H₂O_(l) + NH_3(g)

Hence; (NH₄)₂SO_4(s) + 2OH^–_(aq) SO₄^2-_(aq) + 2H₂O_(l) + 2NH_3(g)

(iii). With calcium hydroxide:

Ca(OH)_2(aq) + (NH₄)₂SO_4(s) CaSO_4(aq) + 2H₂O_(l) + 2NH_3(g)

Ionically;

Ca²⁺_(aq) + 2OH^–_(aq) + (NH₄)₂SO_4(s) Ca²⁺_(aq) + SO₄^2-_(aq) + H₂O_(l) + NH_3(g)

Hence; (NH₄)₂SO_4(s) + 2OH^–_(aq) SO₄^2-_(aq) + 2H₂O_(l) + 2NH_3(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.

NH_3(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.

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

Equation:

NH_3(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.

Smell: has a characteristic pungent smell.
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:

NH_3(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.

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) + H₂SO_4(aq) (NH₄)₂SO_4(aq) + H₂O_(l)

Ionically:

2NH₄⁺_(aq) + 2OH^–_(aq) + 2H⁺_(aq) + SO₄^2-_(aq) 2NH₄⁺_(aq) + SO₄^2-_(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)₂SO_4(s) + 2NaOH_(aq) Na₂SO_4(aq) + 2NH_3(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)

HNO_3(aq) + NH₄OH_(aq) NH₄NO_3(aq)+ H₂O_(l)

Ionic equation:

NH_3(g) + H⁺_(aq) NH₄⁺_(aq)

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:

4NH_3(g) + 3O_2(g) 2N_2(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.

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) + 2NH_3(g) 3Cu_(s) + 3H₂O_(l) + N_2(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.

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:

2NH_3(g) + 3Cl_2(g) 6HCl_(g) + N_2(g)

Then;

6HCl_(g) + 6NH_3(g) 6NH₄Cl_(s)

Overall equation:

8NH_3(g)+ 3Cl_2(g) 6NH₄Cl_(s) + N_2(g)

Ammonia solution as an alkali.

– Solution of ammonia in water contains hydroxyl ions.

Equation:

NH_3(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.

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:

4NH_3(g) + 5O_2(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) + O_2(g) 2NO_2(g)

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
Metal ions in solution	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:

Mg²⁺_(aq) from MgCl₂;

MgCl_2(aq) + 2NH₄OH_(aq) Mg(OH)_2(s) + 2NH₄Cl_(aq)

Ionically:

Mg²⁺_(aq) + 2OH^–_(aq) Mg(OH)_2(s)

(White ppt)

Fe²⁺ from Fe(NO₃)₂;

Fe(NO₃)_2(aq) + 2NH₄OH_(aq) Fe(OH)_2(s) + 2NH₄NO_3(aq)

Ionically:

Fe²⁺_(aq) + 2OH^–_(aq) Fe(OH)_2(s)

(Pale green ppt)

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.

Zinc ions and ammonia solution.

With little ammonia:

ZnCl_2(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) + 4NH_3(aq) [Zn(NH₃)₄]²⁺_(aq) + 2OH^–_(aq)

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

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) + 4NH_3(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:

Ammonia gas is used in the manufacture of nitric acid and nylon.
Ammonia gas is important in the preparation of ammonium salts used as fertilizers.
It liquefies fairly easily (B.P is -33^oC) and the liquid is used as a refrigerant in large cold storages and ice cream factories.
Liquid ammonia is injected directly into the soil as a high nitrogen content fertilizer.
Ammonia solution is used in laundry work as a water softener and a cleansing agent (stain remover)
Ammonia is used in the manufacture of sodium carbonate in the Solvay process.
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
Metal ions in solution	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.

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)

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

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)	IDENTITY
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)
2. Yellow	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^–;	CO₃^2-; O^2-; SO₄^2-; OH^–;	White precipitate insoluble in excess	White precipitate insoluble in excess, on standing;
Al³⁺	Cl^–; NO₃^–; SO₄^2-	O^2-; 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^–; SO₄^2-	CO₃^2-; O^2-; SO₄^2-; OH^–;	White precipitate soluble in excess	White precipitate soluble in excess
Fe²⁺	Cl^–; NO₃^–; SO₄^2-	CO₃^2-; O^2-; OH^–;	(Dark) pale green precipitate insoluble in excess	(Dark) pale green precipitate insoluble in excess
Fe³⁺	Cl^–; NO₃^–; SO₄^2-	CO₃^2-; O^2-; OH^–;	(Red) brown precipitate insoluble in excess	(Red) brown precipitate insoluble in excess
Cu²⁺	Cl^–; NO₃^–; SO₄^2-	CO₃^2-; O^2-; 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.

	SO₄^2-_(aq)	Cl^–_(aq)	NO₃^–_(aq)	CO₃^2-_(aq)
TEST	Add Ba²⁺_(aq) ions from Ba(NO₃)_2(aq); acidify with dilute HNO_3(aq)	Add Ag⁺_(aq) from AgNO_3(aq). Acidify with dilute HNO₃ Alternatively; Add Pb²⁺ from Pb(NO₃)₂ and warm	Add FeSO_4(aq); Tilt the tube and carefully add 1-2 cm³ of concentrated H₂SO_4(aq)	Add dilute HNO_3(aq); bubble gas through lime water;
OBSERVATION	The formation of a white precipitate shows presence of SO₄^2- 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 CO₃^2- 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 CO₃^2- or HCO₃^– will liberate carbon (IV) oxide with dilute acids

Checklist:

Why is it not possible to use dilute sulphuric acid in the test for SO₄^2- ions;
Why is it not possible to use dilute hydrochloric acid in the test for chloride ions?
Why is it best to use dilute nitric acid instead of the other two mineral acids in the test for CO₃^2-ions?
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	CO₃^2- of Zn²⁺; Pb²⁺; Fe²⁺; Fe³⁺; Cu²⁺;
4. Carbon (IV) oxide and water vapour (condensed)	HCO₃^–
5. Nitrogen (IV) oxide	NO₃^–of Cu²⁺; Al³⁺; Zn²⁺; Pb²⁺; Fe²⁺; Fe³⁺
6. Oxygen	NO₃^– or BaO₂; MnO₂; PbO₂;

Industrial manufacture of ammonia-The Haber process.

– Most of the worlds 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

N_2(g) + 3 H_2(g) 2NH_3(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:H₂respectively), 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

N_2(g) + 3H_2(g) 2NH_3(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:

Availability of raw materials (natural gas and crude oil)
Presence of cheap sources of energy.
Availability of transport and marketing.
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:

2NH_3(g) + H₂SO_4(aq) (NH₄)₂SO_4(aq)

Note: It is a cheap fertilizer.

(b). Ammonium nitrate

– Is prepared by neutralization nitric acid by ammonia.

Equation:

NH_3(g) + HNO_3(aq) NH₄NO_3(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:

3NH_3(g) + H₃PO_4(aq) (NH₄)₃ PO_4(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:

NH_3(g) +CO_2(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

KNO_3(g) + H₂SO_4(aq) KHSO_4(aq) +HNO_3(g)

– If Lead (II) nitrate was used;

Pb(NO₃)_2(s) + H₂SO_4(aq) PbSO_4(s) + 2HNO_3(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:

4HNO_3(aq) 4NO_2(g) + O_2(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 83^oC 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.

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

4HNO_3(l) 4NO_2(g) + 2H₂O_(l) + O_2(g)

(Brown fumes)

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

Equation:

2NO_2(g) + H₂O_(l) HNO_2(aq) + HNO_3(aq)

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

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) + HNO_3(l) NO_2(g) + CO_2(g) +H₂O_(g)

Saw dust

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

Equation:

C_(s) + 4HNO_3(l) 2H₂O_(l) + 4NO_2(g) + CO_2(g)

Reaction with sulphur

Procedure:

– 2 cm³of 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) + 6HNO_3(l) H₂SO_4(aq) + 6NO_2(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) + SO₄^2-_(aq) BaSO_4(s)

(White precipitate)

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, NO₂and water.

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

Equation:

C_u(s) + HNO_3(l) C_u(NO₃)_2(aq) + 2NO_2(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.

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:

6FeSO_4(s) + 3H₂SO_4(aq) +3HNO_3(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) + O_2(g) 2NO_2(g)

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) + 2HNO_3(l) 2H₂O_(l) + 2NO_2(g) +S_(s)

Reduction

Properties of dilute nitric (V) acid

Reaction with metals

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

Example:

3Mg_(s) + 8HNO_3(aq) 3Mg(NO₃)_2(aq) +2NO_(g) + 4H₂O_(l)

– In fact HNO₃is 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.

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) + 2HNO_3(aq) Cu (NO₃)_2(aq)+ H₂O_(l)

(Black) (Blue)

Zn(OH)_2(s) + 2HNO_3(aq) Zn (NO₃)_2(aq) + 2H₂O_(l)

(White ppt) (Colourless)

KOH_(aq) + HNO_3(aq) KNO_3(aq) + H₂O_(l)

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

Reaction with metal carbonates and hydrogen carbonates

– Dilute HNO₃reacts with metal carbonates and hydrogen carbonates to form a nitrate, CO₂and water.

Examples.

CuCO₃(s) + 2HNO_3(aq) Cu(NO₃)_2(aq) + CO_2(g) + H₂O_(l)

(Green) (Blue solution)

NaHCO_3(s) + HNO_3(aq) NaNO_3(aq) + CO_2(g) + H₂O_(l)

Test for nitrates/nitric acid

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^– NO_2(g) + H₂O_(l) (reduced)

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. H₂SO₄and the KNO₃reacted to produce HNO₃.

Equation:

KNO₃(aq) +H₂SO_4(aq) KHSO_4(aq) + HNO_3(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:

FeSO_4(aq) + NO_(g) FeSO₄. NO_(aq)

(Green) (Brown)

Ionically:

Fe²⁺_(aq) + 5H₂O_(l) + NO_(g) [Fe(H₂O)₅NO]²⁺_(aq)

(Green) (Brown)

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) + 4NO_2(g) + O_2(g)

(Brown, acidic)

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

Equation:

2Na NO_3(s) ^heat 2NaNO_2(s) + O_2(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 Otswalds 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 900^oC and then passed over a platinum-rhodium catalytic chamber.

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

Equation:

4NH_3(g) + 5O_2(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) + O_2(g) 2NO_2(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:

2NO_2(g) + H₂O_(l) HNO_3(aq) + HNO_2(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 HNO_2(aq) + O_2(g) 2HNO_3(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 otswalds process.

Ammonia

HEAT EXCHANGER

CATALYTIC CHAMBER

Air

Water Unreacted NO_(g)

NO + air;

Nitric (V) acid

Pollution effects of nitrogen compounds.

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.

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.

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.

Recycling unreacted gases in manufacture of nitric acid to prevent release into environment.
Treating sewage and industrial effluents to remove nitrogen compounds before releasing to rivers and lakes.
Fitting exhausts systems of vehicles with catalytic converters which convert nitrogen oxides into harmless nitrogen gas.
Adding lime to lakes and soils in surrounding regions to reduce acidity.
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:

Occurrence of sulphur
Extraction of sulphur

The Frasch pump
Extraction process

Properties of sulphur

Physical properties
Chemical properties

Uses of sulphur
Allotropes of sulphur

Rhombic sulphur
Monoclinic sulphur

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

Sulphur (VI) oxide

Laboratory preparation
Properties of sulphur (VI) oxide

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

Hydrogen sulphide gas

Laboratory preparation
Properties of hydrogen sulphide
Physical properties of hydrogen sulphide
Chemical properties of hydrogen sulphide

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 170^oC

Froth of molten sulphur

Cross section of the Frasch pump

Outermost pipe: brings superheated water at 170^oC

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 170^o 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 120^oC, but this is enough to melt sulphur whose M.P is 114^oC.

(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

– 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.

Solubility

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

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

Effects of heat

– When sulphur is heated out of contact with air, it melts at low temperatures of about 113^oC 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 160^oC, 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 (S_{100 000}).

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 444^oC (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 750^oC 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

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) + O_2(g) SO_2(g)

Note:

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

Equation:

2SO_2(s) + O_2(g) 2SO_3(g)

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) + 2H₂SO_4(l) 3SO_2(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) + 6HNO_3(l) H₂SO_4(aq) + 6NO_2(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 BaSO_4(s)

Ionically;

Ba²⁺_(aq) + SO₄^2-_(aq) BaSO_4(s)

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

H_2(g) + S_(s) H₂S_(g)

Fluorine

S_(s) + F_2(g) SF_2(g)

Chlorine

S_(s) + Cl_2(g) SCl_2(g)

Bromine

2S_(s) +Br_2(g) S₂Br_2(g)

Phosphorous

10S_(s) + 4P_(s) P₄S_10(s)

Note:

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

Uses of sulphur

Industrial manufacture of sulphuric (VI) acid in the contact process.
It is used as a fungicide for treatment of fungal skin diseases.
It is used for vulcanization (hardening) of rubber
Manufacture of calcium hydrogen sulphite (Ca(HSO₃)₂ used for bleaching in paper and textile industries.
Manufacture of matches and fireworks.
Manufacture of dyes e.g. sulphur blacks that gives paint smooth texture.
Manufacture of sulphur ointments and drugs e.g. sulphur-guanidine for dysentery.
Manufacture of hair oil.
Small amounts of sulphur are added to concrete to prevent corrosion by acids.
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 96^oC	– Stable above 96^oC
Colour	– Bright yellow crystalline solid	– Pale yellow crystalline solid
Melting point	– Melts at 113^oC;	– Melts at 119^oC;
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.

96^oC 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. H₂SO_4(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.

Na₂SO_3(aq) + 2HCl_(aq) H₂O_(l) + SO_2(g) + 2NaCl_(aq)

Ionically;

2H⁺_(aq) + SO₃^2-_(aq) H₂O_(l) + SO_2(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:

2HNO_3(aq) + SO_2(g) 2NO_2(g) + H₂SO_4(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.

C_u(s) +2H₂SO_4(l) CuSO_4(aq) + 2H₂O_(l) + SO_2(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) + H₂SO_4(l) CuO_(s) + H₂O_(l) + SO_2(g)

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

Equation:

CuO_(s) + H₂SO_4(l) CuSO_4(aq) + H₂O_(l)

Overall equation:

Cu_(s) + H₂SO_4(l) CuSO_4(aq) + 2H₂O_(l) + SO_2(g)

(c). Roasting sulphur in air

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

Equation:

S_(s) + O_2(g) SO_2(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) + 3O_2(g) 2FeO_(s) + 2SO_2(g)

2ZnS_(g) + 3O_2(g) 2ZnO_(s) + 2SO_2(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

It is a colourless gas with an irritating (pungent) characteristic smell.
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.
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, SO₃^2-, acts as a supplier of electrons; the overall reaction results into formation of sulphate ions.

Equations:

H₂SO_3(aq) 2H⁺_(aq) + SO₃^2-_(aq) then;

SO₃^2-_(aq) + H₂O_(l) SO₄^2-_(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 H₂SO₄was 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:

5SO_2(g) + 2KMnO_4(aq) + 2H₂O K₂SO_4(aq) + 2MnSO_4(aq)+ H₂SO_4(aq)

Ionically;

2MnO^–_4(aq) + 5SO₃^2-_(aq) + 6H⁺_(aq) 2Mn²⁺_(aq) + 5SO₄^2-_(aq) + 3H₂O_(l)

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

Procedure

– To 2 cm³of 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

K₂Cr₂O_7(aq) + 3SO_2(aq) + H₂SO_4(aq) K₂SO_4(aq) + H₂O_(l) + Cr₂(SO₄)_3(aq)

(Orange) (Green)

Ionically: Oxidation

Cr₂O₇^2-_(aq)+ 3SO₃^2-_(aq) + 8H⁺_(aq) 2Cr³⁺_(aq) + 3SO₄^2-_(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 FeCl_2(aq).

Ionically

2Fe³⁺_(aq) + SO₃^2-_(aq) + H₂O_(l) Fe²⁺_(aq) + SO₄^2-_(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 BaCl₂solution.

Equation

Br_2(aq) + SO_2(g) + 2H₂O_(l) 2HBr_(aq) + H₂SO_4(aq)

Ionically: ^Oxidation

Br_2(aq) + H₂O_(l) + SO₃^2-_(aq) 2HBr_(aq) + SO₄^2-_(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) + SO₄^2-_(aq) BaSO_4(s)

Note

– This test confirms presence of SO₄^2- since a white precipitate insoluble in dilute hydrochloric acid is formed.

– CO₃^2-_(aq) and SO₃^2- also forms a white precipitate with BaCl_2(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 cm³of 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 (H₂SO_3(aq)) is oxidized to SO₄^2-_(aq)

Equation

H₂O_2(l) +SO₃^2-_(aq) H₂O_(l) + SO₄^2-_(aq)

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

– This confirms presence of sulphate ions.

Equation:

Ba²⁺_(aq) + SO₄^2-_(aq) BaSO_4(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:

2HNO_3(l) + SO_2(g) 2NO_2(g) + H₂SO_4(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 (SO₃^2-) into sulphate (SO₄^2-)

Equation:

2SO₃^2-_(aq) + O_2(g) 2SO₄^2-_(aq)

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

Equation:

Ba²⁺_(aq) + SO₄^2-_(aq) BaSO4_(s)

(White ppt)

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) + SO_2(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) + SO_2(g) 2MgO_(s) + S_(s)

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:

SO_2(g) + H₂O_(l) H₂SO_3(aq)

H₂SO_3(aq) 2H⁺_(aq) + SO₃^2-_(aq)

Then;

SO₃^2-_(aq) + [O] SO₄^2-_(aq)

From dye

General equation

SO_2(g) + H₂O_(l) + [Dye + (O)] Dye + H₂SO_4(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.

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) + SO_2(g) Na₂SO_3(aq) + H₂O_(l)

With excess sulphur (IV) oxide:

NaOH_(aq) + SO_2(g) NaHSO_3(aq)

Reaction with chlorine:

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

Equation:

SO_2(g) + SO_2(g) H₂O_(l) H₂SO_4(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

Characteristic pungent smell.
Bleaches flower petals.
Decolourises purple potassium manganate (VII)
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:

SO_2(g) + H₂O_(l) H₂SO_3(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)+ O_2(g) SO_2(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

2SO_2(g) + O_2(g) 2SO_3(g) + Heat

450^oC

– 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.

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 450^oC 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.

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.

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 450^oC.

– 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 SO₃gas 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:

SO_3(g) + H₂SO_4(l) H₂S₂O_7(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:

H₂S₂O_7(l) + H₂O_(l) 2H₂SO_4(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

2SO_2(g) + O_2(g) 2SO_3(g) + Heat

450^oC

– 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) + SO_2(g) CaSO_3(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

– Colourless, odourless, oily liquid.
– Very dense; with density 1.84 gcm^-3.
– Soluble in water and gives out considerable heat when a solution is formed.
– 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

– 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) CuSO_4(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

C₁₂H₂₂O_11(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

H₂C₂O_4(s) CO_(g) + CO_2(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) C₂H_4(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)

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) + 2H₂SO_4(l) CuSO_4(aq) + SO_2(g) + 2H₂O_(l)

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

Zinc

Zn_(s) + 2H₂SO_4(l) ZnSO_4(aq) + SO_2(g) + 2H₂O_(l)

(Hot acid)

Zn_(s) + H₂SO_4(l) ZnSO_4(aq) + H_2(g)

(Cold acid)

Lead

Pb_(s) + 2H₂SO_4(l) PbSO_4(aq) + SO_2(g) + 2H₂O_(l)

(Hot; conc.) (Insoluble)

Note:

– Dilute sulphuric (VI) acid doesnt 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) + 2H₂SO_4(l) CO_2(g) + 2SO_2(g) + 2H₂O_(l)

Ø With sulphur

S_(s) + 2H₂SO_4(l) 3SO_2(aq) + 2H₂O_(l)

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

Reactions of dilute sulphuric acid

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) + H₂SO_4(aq) MgSO_4(aq) + H_2(g)

With zinc:

Zn_(s) + H₂SO_4(aq) ZnSO_4(aq) + H_2(g)

Note:

– Copper is below hydrogen in reactivity series hence cant displace hydrogen form dilute sulphuric (VI) acid.

Reaction with carbonates and hydrogen carbonates

– Dilute H₂SO_4(aq) reacts with carbonates and hydrogen carbonates to produce a salt, carbon (IV) oxide and water.

Equations

With sodium carbonate:

Na₂CO_3(s) + H₂SO_4(aq) Na₂SO_4(aq) + CO_2(g) + H₂O_(l)

With calcium hydrogen carbonate:

CaHCO_3(s) + H₂SO_4(aq) CaSO_4(aq) + CO_2(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.

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) + H₂SO_4(aq) MgSO_4(aq) + H₂O_(g)

(White) (Colourless solution)

With copper (II) oxide:

CuO_(s) + H₂SO_4(aq) CuSO_4(aq) + H₂O_(g)

(Black) (Blue solution)

With sodium hydroxide:

NaOH_(s) + H₂SO_4(aq) Na₂SO_4(aq) + 2H₂O_(g)

(White) (Colourless solution)

With lead (II) oxide:

PbO_(s) + H₂SO_4(aq) PbSO_4(aq) + H₂O_(g)

(Red) (White ppt; reaction stops immediately)

Uses of sulphuric (VI) acid

Manufacture of fertilizers.
Processing of metal ores.
Manufacture of detergents.
Manufacture of plastics.
Manufacture of dyes and paints.
Manufacture of lead and accumulators.
Manufacture of polymers.
Manufacture of petroleum (petroleum refinery).
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

(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) + FeCl_2(aq)

Ionically:

S^2-_(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

Colourless and very poisonous gas (similar to hydrogen cyanide)
Has a repulsive smell (similar to that of rotten eggs or decaying cabbages)
Soluble in water giving a weak acid (only slightly ionized)

Equation:

H2_S(g) + H₂O_(l) H₂S_(aq)

Then:

H₂S_(aq) H⁺_(aq) + HS^–_(aq) 2H⁺_(aq) + S^2-_(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

Combustion

– Burns in a blue flame in a limited supply of oxygen (air) forming a yellow deposit of sulphur and steam.

Equation:

2H₂S_(g) + O_2(g) 2SO_2(s) + 2H₂O_(g)

– In plentiful supply (excess) of Oxygen (air) it burns with a blue flame forming SO₂ and steam.

Equation:

2H₂S_(g) +3O_2(g) 2S_(s) + 2H₂O_(g)

It is a reducing agent

– It supplies electrons which are accepted by the oxidizing agent and forms sulphur.

Ionically:

H₂S_(aq) + 2H⁺_(aq) + S^2-_(aq)

Then

S^2-_(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:

Cr₂O₇^2-_(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:

FeCl_3(aq) + H₂S_(g) 2FeCl_2(aq) + 2HCl_(aq) + S_(s)

Ionically:

Reduction:

Fe³⁺_(aq) + S^2-_(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 S^2-_(aq) are oxidized to yellow sulphur.

(iv). Action with Conc. HNO₃

Equation:

2HNO_3(aq) + H₂S_(g) 2H₂O_(aq) + 2NO_2(aq) + S_(s)+ Heat

Ionically:

Reduction:

2H⁺_(aq) + 2NO₃^–_(aq) + H⁺_(aq) + S^2-_(aq) 2H₂O_(l) + 2NO_2(g) + S_(s)+ Heat

(Colourless solution) (Brown) (Yellow)

Oxidation

Observation:

– Evolution of brown fumes; and deposits of a yellow solid;

– HNO_3(aq) is reduced to brown NO_2(g); while S^2-_(aq) are oxidized to yellow sulphur;

Note: The solution also contains H₂SO₄ produced by the reaction:

Reduction

2HNO_3(aq) + H₂S_(g) H₂SO_4(aq) + 8NO_2(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) + O_2(g) 2H₂O_(l) + 2S_(s)

(vi). Action with concentrated sulphuric (VI) acid.

Equation

Reduction

H₂SO_4(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

Br_2(aq) + H₂S_(g) 2HBr_(aq) + S_(s)

(Red-brown) (Colourless) (Yellow suspension)

Oxidation

(viii). Action with hydrogen peroxide.

Equation:

Reduction

H₂O_2(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) + 2HNO_3(aq)

(Colourless) (Black)

Ionically:

Pb²⁺_(aq) + S^2-_(aq) PbS_(s)

Copper (II) ions:

CuSO_4(aq) + H₂S_(aq) CuS_(s) + H₂SO_4(aq)

(Blue) (Black)

Ionically:

Cu²⁺_(aq) + S^2-_(aq) CuS_(s)

Iron (II) ions:

FeSO_4(aq) + H₂S_(aq) FeS_(s) + H2SO_4(aq)

(Pal green) (Black)

Ionically:

Fe²⁺_(aq) + S^2-_(aq) FeS_(s)

Zinc ions:

Zn(NO₃)_2(aq) + H₂S_(aq) ZnS_(s) + 2HNO_3(aq)

(Colourless) (Black)

Ionically:

Zn²⁺_(aq) + S^2-_(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 (SO₃^2-) and a metallic or ammonium cation

Effects of heat

– They decompose on heating, forming SO₂;

Example:

CuSO_3(s) ^Heat CuO_(s) + SO_2(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 SO_2(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) + SO₃^2-_(aq) BaSO_3(s)

(White precipitate)

On addition of dilute HCl(aq):

BaSO_3(s) + 2HCl_(aq) BaCl_2(aq) + SO_2(g) + H₂O_(l)

(White precipitate) (Colourless)

Ionically:

BaSO_3(s) + 2H⁺_(aq) Ba²⁺_(aq) + SO_2(g) + H₂O_(l)

Sulphates

– Are compounds of the sulphate radical (SO₄^2-) and a metallic or ammonium cation.

Effects of heat.

– Decompose on heating and liberate SO₂ and SO₃or SO₃alone;

– However quite a number of sulphates do not decompose on heating; and thus require very strong heating in order to decompose.

Examples:

2FeSO_4(s) ^Heat Fe₂O_3(s) + SO_2(g) + SO_3(g)

(Pale green) (Brown) (Colourless gases)

CuSO_4(s) ^Heat CuO_(s) + SO_3(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) + SO₄^2-_(aq) BaSO_4(s)

(white precipitate)

On addition of dilute acid:

BaSO_4(s) + 2HCl_(aq) BaSO_4(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:

About chlorine.
Preparation of chlorine.
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.

Industrial manufacture of chlorine (The mercury cathode cell)
Uses of chlorine and its compounds
Hydrogen chloride gas

Preparation
Properties

Test for chlorides.
Hydrochloric acid

Large scale manufacture
Uses of hydrochloric acid

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] Cl_2(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:

MnO_2(s) + 4HCl_(aq) ^Heat MnCl_2(aq) + 2H₂O_(l) + Cl_2(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.

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.

Solubility in water.

– It is fairly soluble in water forming green-yellow chlorine water.

Equation:

Cl_2(g) + H₂O_(l) HCl_(aq) + HOCl_(aq)

– Chlorine water is composed of two acids; chloric (I) acid (hypochlorous acid) and hydrochloric acid.

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:

Cl_2(g) + H₂O_(l) HCl_(aq) + HOCl_(aq)

Acidic solution

Then:

Dye + HOCl_(aq) HCl_(aq) + {Dye + [O]}

Coloured Colourless

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:

Cl_2(g) + H₂O_(l) HCl_(aq) + HOCl_(aq)

Acidic solution

Then:

Dye + HOCl_(aq) HCl_(aq) + {Dye + [O]}

Coloured Colourless

Action on a burning splint.

– The gas put out a glowing splint. It does not burn.

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 (FeCl_3(g)) are formed in the combustion tube.

– A black solid (FeCl_3(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) + 3Cl_2(g) 2FeCl_3(g)

(vii). Conclusion.

– Iron (III) chloride sublimes on heating; the black solid changes to red-brown fumes on heating.

Equation:

FeCl_3(s) FeCl_3(g)

(black) (Red-brown)

(b). Aluminium chloride.

2Al_(s) + 3Cl_2(g) 2FeCl_2(s)

2Al_(s) + 3Cl_2(g) Al₂Cl_6(s)

Note:

– Aluminium chloride also sublimes on heating.

Equation:

AlCl_3(s) AlCl_3(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) + Cl_2(g) MgCl_2(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.

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:

P_4(s) + 6Cl_2(g) 4PCl_3(s)

(With limited chlorine)

P_4(s) + 10Cl_2(g) 4PCl_5(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:

Cl_2(g) + H_2(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) + NH_3(g) NH₄Cl_(g)

White fumes.

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

Cl_2(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:

Cl_2(g) + H₂S_(aq) 2HCl_(aq)+ S_(s)

Ionic:

Cl_2(g) + S^2-_(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:

1^st portion: White precipitate formed indicating presence of SO₄^2-;

Explanations:

– The white precipitate indicate presence of SO₄^2-; the precipitate is barium sulphate Ba(SO₄)₂;

– Chlorine oxidizes SO₃^2- in Na₂SO₃ to SO₄^2- while itself is reduced to chloride ions;

Equations:

H₂O_(l) + Cl_2(g) + Na₂SO_3(aq) Na₂SO_4(aq) + 2HCl_(aq)

Ionically:

Cl_2(g) + SO₃^2-_(aq) + H₂O_(l) SO₄^2-_(aq) +2H⁺_(aq) + 2Cl^–_(aq)

– On adding barium nitrate (Ba(NO₃)₂); the Ba²⁺ ions react with the SO₄^2-to form insoluble BaSO₄; the white precipitate.

Ionically;

Ba²⁺_(aq)+ SO₄^2-_(aq)BaSO_4(s)

(White precipitate)

Note:

– The solution is first acidified (with HNO₃) before addition of Ba(NO₃)₂ to prevent precipitation of BaSO_3(s)and BaCO_3(s).

2^nd portion:

Observation:

– Formation of a white precipitate on addition of Pb(NO₃)₂solution.

– On warming the white precipitate dissolves then recrystalizes back on cooling.

Explanations:

– The white precipitate shows presence of either Cl^–; SO₃^2-orSO₄^2-

– However the fact that it dissolves on warming confirms the presence of Cl^–_(aq) and not SO₃^2-_(aq) and SO₃^2-_(aq)

Equation:

Pb²⁺_(aq)+ Cl^–_(aq)PbCl_2(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

8NH_3(g) + 3Cl_2(g) 6NH₄Cl_(g)+ N_2(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) + Cl_2(g) 2KCl_(aq)+ Br_2(l)

^Oxidation^{Red brown}

Ionically;

2Br^–_(aq) + Cl_2(g) 2Cl^–_(aq) + Br_2(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) + Cl_2(g) 2KCl_(aq)+ I_{2(l) (black)}

^Oxidation

Ionically;

2I^–_(aq) + Cl_2(g) 2l^–_(aq) + Br_2(l)

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:

Cl_2(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:

3Cl_2(g)+ 6NaOH_(l) 5NaCl_(aq) + NaClO_3(aq)+ 3H₂O_(l)

(b). Reaction with potassium hydroxide

– Follows the trend of sodium.

(c). Reaction with slaked lime {Ca(OH)_2(s)}

Equation:

Cl_2(g)+ Ca(OH)_2(l) CaOCl_2(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:

CaOCl_2(s) + CO_2(g) CaCO_3(s) + Cl_2(g)

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:

C₁₀H_16(l) + 8Cl_2(g) 16HCl_(g) + 10C_(s)

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:

Cl_2(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) + O_2(g) (slow reaction)

Overall reaction:

2H₂O_(l) + 2Cl_2(g) 4HCl_(aq) + O_2(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) Cl_2(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) + H_2(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.

Manufacture of hydrochloric acid.
Used in form of bleaching powder in textile and paper industries.
For sterilization of water for both domestic and industrial use and in swimming pools.
Used in sewage treatment e.g. NaOClO₃ solution used in latrines.
Manufacture of plastics (polyvinyl chloride; PVC)
Manufacture of germicides, pesticides and fungicides e.g. DDT and some CFCs.
CFCs are used to manufacture aerosol propellants.
Manufacture of solvents such as trichloromethane and some chlorofluorocarbons (CFCs).
CFCs are commonly freons are used as refrigerants in fridges and air condition units due to their low boiling points.
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:

H₂SO_4(l) + NaCl_(aq) NaHSO_4(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.

Colourless gas with a strong irritating pungent smell.
Slightly denser than air (1¼ times). This makes it possible to collect the gas by downward delivery.
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) ZnCl_2(aq) + H_2(g)

Mg_(s) + 2HCl_(aq) MgCl_2(aq) + H_2(g)

Fe_(s) + 2HCl_(aq) FeCl_2(aq) + H_2(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:

CaCO_3(s) + 2HCl_(aq) CaCl_2(aq) + H₂O_(l) + CO_2(g)

ZnCO_3(s) + 2HCl_(aq) ZnCl_2(aq) + H₂O_(l) + CO_2(g)

NaHCO_3(s) + HCl_(aq) NaCl_(aq) + H₂O_(l) + CO_2(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

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) + H_2(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.

Hydrogen chloride gas forms white fumes of ammonium chloride when reacted with ammonia gas;

Equation:

NH_3(g) + HCl_(g) NH₄Cl_(s)

Note: This is the chemical test for hydrogen chloride gas.

Hydrogen chloride is decomposed by oxidizing agents, giving off chlorine.

Examples:

PbO_2(s) + 4HCl_(g) PbCl_2(s) + 2H₂O_(l) + Cl_2(g)

MnO_2(s) + 4HCl_(g) MnCl_2(s) + 2H₂O_(l) + Cl_2(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) + AgNO_3(aq) NaNO_3(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) 2NaNO_3(aq) + PbCl_2(s)

White ppt.

Ionically;

Pb²⁺_(aq) + Cl^–_(aq) PbCl_2(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:

H_2(g) + Cl_2(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.

Sewage treatment.
Treatment of water (chlorination) at the waterworks.
Removing rust from metal e.g. descaling iron before it is galvanized or and other metals before they are electroplated.
Making dyes, drugs and photographic materials like silver chloride on photographic films.

Environmental pollution by chlorine and its compounds.

Chlorine may dissolve in rain and fall as acid rain, which has adverse effects on plants and animals, buildings and soil nutrients.
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.
PVCs are non-biodegradable.
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
COMPOUND	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 – C – H

│

Alkanes

Are the simplest hydrocarbons with the general formula; C_nH_{2n + 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 C_nH_2n+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, CH₄differs 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 78.4^oC

– 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 dont 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, – C₂H₅i.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:-

The largest continuous chain of carbon atoms in the molecule is used to deduce the parent name of the compound.
The carbon atoms of this chain are numbered such that the branching // substituents are attached to the carbon atom bearing the lowest number.
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;

2-methylpropane;

Further examples.

CH₃CH₂CH₂CH₃

│

CH₃

3-methylpentane;

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:

Butane.

Pentane;

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 120^oC and is collected, the of 40^oC intervals thereafter until the temperatures reach 350^oC.

(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

Methane

Ethane

Propane

Butane

Pentane

Hexane

Heptane

Octane;

Nonane

Decane

CH₄

C₂H₆

C₃H₈

C₄H₁₀

C₅H₁₂

C₆H₁₄

C₇H₁₆

C₈H₁₀

C₉H₂₀

C₁₀H₂₂

↑

Gaseous

↓

↑

Liquid

↓

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:

C_nH_{2n + 1}COONa + NaOH_(aq) → C_nH_{2n +2} + Na₂CO_3(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) → CH_4(g) + Na₂CO_3(aq)

Sodium ethanoate sodalime Methane Sodium carbonate

Physical properties of methane

It is a non-poisonous, colourless gas.
It is slightly soluble in water, but quite soluble in organic solvents such as ethanol and ether.
II is less denser than air and when cooled under pressure, it liquefies.

Chemical properties

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:

CH_4(g) + 2O_2(g) → CO_2(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.

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.

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 CH_4(g) + Br_2(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 400-700^oC

Equation

Example: Cracking of propane

Alkenes

– Are hydrocarbons with at least one carbon-carbon double bond, and have the general formula C_nH_2n.

– 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

The parent molecule is the longest carbon chain; and its prefix is followed by the suffix ene.
The carbon atoms in the chain are numbered such that the carbon atoms joined by the double bonds get the lowest possible numbers.
The position of the substituent groups is indicated by showing the position of the carbon atom to which they are attached.
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

Hex- l ene
Prop-l-ene

Hex-2-ene

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:-

Branching isomerism
Positional isomerism

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

i) Apparatus

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.

Equation

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

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.

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.

With oxidizing agents
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.

Reaction with acidified potassium chromate (VI) (K2Cr2O7)

Halogenations is the addition of halogen atoms across a double bond.
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.

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

Addition of hydrogen halides

With hydrobromic acid; HBr (aq)

With sulphuric acid

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.

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.

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

i) Conditions

High temperatures of about 200⁰C
High/elevated pressures of approximately 1000 atmospheres
A trace of oxygen catalyst.

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.

Equation:

Generally

Uses of polythene

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.

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

Manufacture of plastics, through polymerization.
Manufacture of ethanol; through hydrolysis reactions
Ripening of fruits.
Manufacture of ethan 1, 2-diol(glyco) which is used as a coolant.

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

Draw the structures of the following hydrocarbons
2,2 dimethyl-but-2-yne
propyne

4,4 diethyl-hex-2-yne.

Isomerism in alkynes

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

i) Isomers of Butyne

Branching isomerism occurs when alkyl group is present in the molecule.

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

–

0.695

0.716

Preparation and chemical properties of Ethyne.

Preparation
i) Apparatus

ii) Procedure:

Water is dripped over calcium carbide and is collected over water.

Reasons for over-water collection:-

Its insoluble in water
Unreactive and lighter than water.

Conditions
Room temperature

Equation

Properties of Ethyne
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.

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;

Reaction with hydrogen (Hydrogenation)

Note: This reaction occurs under special conditions i.e. – Presence of a Nickel catalyst

Temperatures about 200⁰C

Reaction with halogens
i) Reaction with chlorine

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.

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)

Reaction with hydrogen halides

Uses of Ethyne

Industrial manufacture of compounds like adhesives and plastics
Its used in the oxy-acetylene flame which is used for welding and cutting metals.

Teachers' Resources

CHEMISTRY FORM 1, 2, 3 & 4 FREE PDF NOTES

October 5, 2025 Maverick John Leave a comment

UNIT 1: ATOMIC STRUCTURE AND THE PERIODIC TABLE.

Checklist

Structure of the atom
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

Atomic number and mass number
Isotopes;
Energy levels and electron arrangements.
The periodic table;

Groups
Periods;

Relative atomic mass and isotopes;
Ion formation;

Ion;
Cations;
Anions;
Ionization energy;
Electron affinity;

Valency and oxidation numbers;
Chemical formula;
Chemical equations
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;

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;

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;

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;

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;

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
Carbon
Oxygen


Chlorine
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, 3starting 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
1^st	2
2^nd	8
3^rd	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 Meneleevs 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.

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
Isotope	Percentage composition	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)

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)
Calculate the relative atomic mass of an element whose isotopic masses and relative abundances are shown below. (2 marks)

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)

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)

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

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
Isotope	Protons	Neutrons	Electrons
³²S
³⁴S

(ii). Define relative atomic mass. (2 marks)

(iii). Calculate the relative atomic mass of the volcanic sulphur. (2 marks)

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
Isotope	Percentage composition	Protons	Neutrons
⁶Ir
⁷Ir

(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 (CO₃^2-)

Sulphate (SO₄^2-)

Sulphite (SO₃^2-)

Phosphate (PO₄^3-)

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	N^3-	3	-3
Sulphur	2.8.6	S^2-	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, S^2-	-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.

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.

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

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.

Derive the chemical formula of each of the following compounds.

Calcium fluoride
Carbon (II) oxide
Carbon (IV) oxide
Aluminium nitrate
Calcium hydrogen carbonate

Complete the table below for elements A, B and C

Element	Valency	Chemical formula of various compounds
Element	Valency	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) + O_2(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) + O_2(g) → 2CuO_(s)

Thus the balanced chemical equation for the reaction between copper metal ad oxygen is:

2Cu_(s) + O_2(g) → 2CuO_(s)

Worked examples:

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

Balance each of the following equations.

Mg_(s) + HCl_(aq) →MgCl_2(aq) + H_2(g)

Na_(s) + H₂O_(l) → NaOH_(aq) + H_2(g)

NaOH_(aq) + H₂SO_4(aq) →Na₂SO_4(aq) + H₂O_(l)

CuCO_3(s) + HNO_3(aq) → Cu(NO₃)_2(aq) + CO_2(g) + H₂O_(l)

H₂S_(g) + O_2(g) → O_2(g) + H₂O_(l)

C₂H_6(g) + O_2(g) → CO_2(g) + H₂O_(l)

Pb(NO₃)_2(s) → PbO_(s) + NO_2(g) + O_2(g)

Fe_(s) + Cl_2(g) → FeCl_3(s)

Al_(s) + H₂SO_4(aq) → Al₂(SO₄)_3(aq) + H_2(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
  - 1^st 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
  - 1^st ionization energy
  - 2^nd 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
  - 1^st 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

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.

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.

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.

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.

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).

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).

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.

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.

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) + H_2(g);

Then with carbon (IV) oxide:2LiOH_(aq) + CO_2(g) → Li₂CO₃.H₂O_(s);

Sodium

With moisture: 2Na_(s) + 2H₂O_(l) → 2NaOH_(aq) + H_2(g);

Then with carbon (IV) oxide: 2NaOH_(aq) + CO_2(g) → Na₂CO₃.H₂O_(s);

Potassium

With moisture: 2K_(s) + 2H₂O_(l) → 2KOH_(aq) + H_2(g);

Then with carbon (IV) oxide: 2KOH_(aq) + CO_2(g) → K₂CO₃.H₂O_(s);

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) + O_2(g) → 2Li₂O_(s)

– Burns in air to with a yellow flame to form sodium oxide as the only product.

Equation:

4Na_(s) + O_2(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) + O_2(g) → Na₂O_2(s)

– Burns in air with a lilac flame to form potassium oxide as the only product.

Equation:

4K_(s) + O_2(g) → 2K₂O_(s)

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) + H_2(g);

Equation II

4K_(s) + O_{2 (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) + H_2(g);

Equation II

4Na_(s) + O_{2 (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) + H_2(g);

Equation II

4Li_(s) + O_{2 (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

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) + Cl_2(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) + Cl_2(g) → 2LiCl_(s)

Potassium:

– The metal bursts into a lilac flame.

– White fumes of potassium chloride are formed.

Equation: 2K_(s) + Cl_2(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 (O^2-)	Chloride (Cl^–)	Sulphates (SO₄^2-)
Li⁺
Na⁺
K⁺

Note:

– Alkali metals are usually not found as free elements; but rather in their combined states in the earths crust.

Reason:

– They have high reactivities.

Uses of Alkali metals and their compounds.

Sodium is used in the manufacture of sodium cyanide for gold extraction.
Lithium is used in the manufacture of special high strength glasses ad ceramics.
Lithium compounds are used in the manufacture of dry cells for use in mobile phones, laptops, stopwatches, and zero emission electric vehicles.
A molten mixture of sodium and potassium is used as a coolant in nuclear reactors.
Sodium vapour is used to produce the yellow glow in street ad advertisement lights.
Molten sodium is used as a reducing agent in the extraction of titanium.

Equation: Na_(l) + TiCl_4(g) → Ti_(s)+ 4NaCl_(l)

Sodium chloride is used as a food additive.
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.

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.

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.

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.

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.

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).

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.

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 1^st and 2^nd 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- (1^stE = 736 kJ per mole)
Second ionization energy: Mg⁺_(g)→ Mg²⁺_(g) + e- (2^ndE = 1450 kJ per mole)

Summary on physical properties of alkaline earth metals.

Element	Atomic number	Melting point (^oC)	Boiling point (^oC)	Atomic radius (nm)	1^st I.E (kJmol-1)	2^nd 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.

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) + O_2(g) → 2MgO_(s)

Reacting with nitrogen: 3Mg_(s) + N_2(g) → Mg₃N_2(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) + O_2(g) → 2CaO_(s)

Reacting with nitrogen: 3Ca_(s) + N_2(g) → Ca₃N_2(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.

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) + H_2(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) + H_{2 (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;

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) + H_2(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.

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) + Cl_2(g) → MgCl_2(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) + Cl_2(g) → CaCl_2(s)

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.