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The Teachers Service Commission, TSC, has released the latest revised and final Promotions interviews marking scheme. According to the Commission’s latest scoring areas, knowledge on Institutional Financial Management has been awarded the highest score at 18 marks.

Read also: TSC promotion interview questions and their answers

INTERVIEW AREA	MAXIMUM MARKS
Education Qualifications	4 Marks
Management Courses	3 Marks
Computer Literacy	2 Marks
Honours/ Awards	3 Marks
Professional Knowledge	3 Marks
Online Teaching Modes/ ICT Integration	6 Marks
TPAD	6 Marks
Knowledge on TSC Operations	15 Marks
Knowledge on emerging issues and Reforms	10 Marks
Institutional Financial Management	18 Marks
Compliance with Legal Documents.	10 Marks
Performance at KCPE/ KCSE Exams (Mean Scores).	10 Marks
Current issues.	3 Marks
Presentation and Communication Skills	2 Marks

 

NAME:………………… INDEX NO:………….…….

 CANDIDATE’S SIGNATURE: ……  DATE: ………

101/2

ENGLISH

PAPER 2

(Comprehension, Literary Appreciation

and Grammar)

2 ½ HOURS

TRIAL ONE EVALUATION TEST

Kenya Certificate of Secondary Education (KCSE)

INSTRUCTIONS TO CANDIDATES

• Write your name and index in the spaces provided above. Sign and write the date of examination in the spaces provided above.
• Answer ALL the questions in this question paper.
• ALL your answers must be written in the spaces provided in the question paper.

For Examiner’s Use Only

Question	Maximum	Score
1	20
2	25
3	20
4	15
Total score

 

This paper consists of 12 printed pages. Candidates should check the question paper to ensure that all the pages are printed as indicates and no questions are missing.

Question 1

Read the passage below and answer the questions that follow:

A growing number of people are being afflicted with a condition christened “ Cyberchondria” – the techies’ term for patients who use the internet for self-diagnosis, and then present  “misinformation”  to their doctor.

Many of those who result to internet for solutions to their health problems say they want information on a medically diagnosed disease or illness, or to find alternative treatments.

Others seek support from fellow sufferers. There are those who turn to online diagnosis because of the state of government health services in the country. Our health service is itself not entirely healthy, and it barely survives with shoe-string budgets, low staffing levels and long queues of patients praying for help.

Internet statistics show that sites offering information on issues related to skin disorders, HIV and Aids, diabetes, cancer and sexually transmitted infections register more hits than other health information searches on the internet.

Self-diagnosis using medical websites is tricky, especially because these websites cover specific medical conditions. For example, if you type the symptom ‘stomach pain into the search engine of a general health and medical website, it may offer gastritis, appendicitis, irritable bowel syndrome or gastoententis for further information.

It would be a mistake to consider these suggestions as possible diagnoses, because your stomach pain could be caused by a build-up of gas following the baked beans and eggs you had for lunch! The same goes for physical pains, such as pain in the lower back, or neurological symptoms. You can’t rely on a website for self-diagnosis because it is going to spit out the most common reason for your symptoms, which  may not be accurate.

It is better to present your aching stomach or back to your physician or health care professional for a more informed diagnosis. Otherwise, you may be wasting time, prayers, money and emotional energy taking care of a disease that you don’t actually have. Whereas internet information can be valuable, the online health consumer should be wary because medical misinformation or ‘ cyberquackery’ is rife on the internet. It is important to know how  to recognize a reputable site.

You also can’t use the Internet as the gospel truth to determine health problems because not all diseases and disorders are going to be catalogued on websites.

Even a seemingly thorough site like WebMD.com is going to miss something, and you don’t want to rely on incomplete information.

Furthermore, you have no way of guaranteeing that the health information found on the Internet is accurate, I’m a public health technician, with bias in monitoring and evaluation, but I can start a website that says just about anything notwithstanding. I don’t have to have a medical degree to give faulty advice.

Online diagnosis may breed panic and apprehension. Many people look up their symptoms on the internet and discover that they could have any number of terminal or serious illnesses when they really have nothing to worry about at all.

Why put those types of thought in your head? Generally speaking, doctors advise that one should ask a few questions about the information collected from an internet site before deciding whether to trust it.

You need to find out who is responsible for the content, and whether the website is owned or sponsored by reputable organization. Every affiliation should be clearly shown on the website. Look for credentials and qualifications of the professionals presenting the information. Be especially cautions if they are anonymous. Can you contact the owners of the website via email, telephone and regular mail? Be vigilant, if the owners don’t offer any means of communication.

(Adapted from the Daily Nation March 2008)

 

Questions

1. (a) According to the passage, what is cyberchondria?                     (1mk)

………………………………………………………………………………………………………………………………………………………………………………………………………………………….……………

…………………………………………………………………………………………………

(b)  Why do patients resort to internet for solution to their health problems?(2mks)

……………………………………………………………………………………………………

……………………………………………………………………………………………………

……………………………………………………………………………………………………

……………………………………………………………………………………………………

 

(c) What picture does the writer paint about government health services?          (2mks)

……………………………………………………………………………………………………

……………………………………………………………………………………………………

 

(d) Describe how one can access a medical solution from the internet.              (2mks)

………………………………………………………………………………………………………………………………………………………………………………………………………………………………..………

……………………………………………………………………………………………………

……………………………………………………………………………………………

(e) What caution does the writer give in paragraph six?                                     (2mks)

………………………………………………………………………………………………………

…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

 

(f) Why does the writer prefer health care professionals to the website?            (2mks)

……………………………………………………………………………………………………

……………………………………………………………………………………………………

……………………………………………………………………………………………………

(g) You also can’t use the internet as the gospel truth to determine health problems.

(Paraphrase the above statement)                                                                       (1mk)

……………………………………………………………………………………………………

……………………………………………………………………………………………………

……………………………………………………………………………………………………

 

(h) What is the attitude of the writer towards looking for medical solutions from the

Internet, illustrate.                                                                                             (2mks)

……………………………………………………………………………………………………

…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

• Online diagnosis may breed panic and apprehension

(Rewrite the sentence in the passive)                                                                    (1mk)

…………………………………………………………………………………………………………………………………………………………………………………………………………………………..…

• Make notes on the precautions to be taken if one wishes to get credible

diagnosis from the internet.                                                                      (3mks)

………………………………………………………………………………………………………………………………………………………………………………………………………………………………….……

………………………………………………………………………………………………………………………………………………………………………………………………………………………………..…

………………………………………………………………………………………………….

……………………………………………………………………………………………………

……………………………………………………………………………………………………

• Explain the meaning of the following expressions as used in the passage:

Shoe string budget

……………………………………………………………………………………………………

……………………………………………………………………………………………………

Hits

…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………..

2. Read the excerpt below and then answer the questions that follow:

 

Nora:                                      I didn’t find it dull.

Helmer:                                  (smilling )But there was precious little result, Nora.

Nora:                                      Oh, you shouldn’t tease me about that again. How could I help the cat’s going in and tearing everything to pieces?

Helmer:                                  Of course you couldn’t, poor little girl. You had the best of intentions to please us all, and that’s the main thing. But it is a good thing that our hard times are over.

Nora:                                      Yes, it is really wonderful.

Helmer:                                  This time I needn’t sit here and be dull all alone, and you needn’t ruin your dear eyes and your pretty little hands-

Nora:                                      (clapping her hands) No, Tovald, I needn’t any longer, need I! It’s wonderfully lovely to hear you say so! (Taking his arm) Now I will tell you how I have been thinking we ought to arrange things, Torvald. As soon as Christmas is over-(A bell rings in the hall.) There’s the bell. (She tidies the room a little.) There’s someone at the door. What a nuisance!

Helmer:                                  If it is a caller, remember I am not at home.

Maid:                                      (in the doorway) A lady to see you, ma’am,–a stranger.

Nora:                                      Ask her to come in.

Maid:                                      (to Helmer)The doctor came at the same time, sir.

Helmer:                                  Did he go straight into my room?

Maid:                                      Yes, sir.

Questions

1. What does Nora refer to in her opening words in this extract? (2 marks)

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

2. What has happened that has made the couple happy? (2 marks)

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

3. Discuss two themes evident in this extract. (4 marks)

…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………….

4. “There’s someone at the door.” Add a question tag (1 mark)

……………………………………………………………………………………………………………………………………………..

5. A lady has come to see Nora as reported by the maid. Who is this lady and how does her coming affect the Helmers from the rest of the play. Write your answer in note form. (6 marks)

………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………..

6. Discuss one aspect of style in this extract. (2 marks)

………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………….

7. Briefly explain what happens after this extract. (4 marks)

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………..

8 Explain the meaning of the following expressions as used in the excerpt.                       (4 marks)

1. i) Tease………………………………………………………………………………………………………………………………..
2. ii) Dull………………………………………………………………………………………………………………………………….

iii) Nuisance………………………………………………………………………………………………………………………….

iv)A caller……………………………………………………………………………………………………………………………

Q3. Read the oral Narrative below and answer the questions that follow. (20 marks)

THE WARRIOR WHO HAD EIGHT LOVERS

A long time ago there was a warrior whose bravery and handsome looks made the girls of the village fall in love with him. Eight girls, at least, were known to want to marry the young warrior, for they had composed many songs in his praise.

 

Now, this warrior was one time getting ready to go on alone raid in faraway country. Before he set off for the raid, he called the youngest of his eight lovers and told her to put fresh milk in a small guard. He also instructed her to keep checking on the colour of the milk every day.“Should the milk turn red,” the warrior told the girl, “it would mean that I have been killed or I am seriously wounded.”

 

The young girl was so touched by the departure of her lover that she composed the following song for him.

My warrior whom I love

                        For whom I open the sweet curdled

                        Milk of my father’s herd,

                        And to whom I give fat rams

                        Of my father to slaughter,

                        To whom I give my slender

                        Thighs to lie on,

                        With whom are you going on a raid next?

 

It happened that many days after the departure of the warrior, the girl noticed that the milk was turning red. She wept bitterly, for she knew that her lover was either dead or dying in a faraway country. Without telling anyone, the girl set off to look for her dying lover.

 

For many days she traveled, and as she walked through plains and forests she sang the song she had composed for her warrior. She travelled on and as she travelled she checked the colour of the milk in the gourd. Each day that she saw the milk turn a little more red, she traveled faster. And each day she hoped that she would find her lover alive.

 

On the ninth day the girl sang louder and louder as she travelled. Each time she sang she would listen to hear if there was any reply. And as she listened at one time, she heard a faint voice. There was no mistake about it. It was her lover’s voice. She ran and ran and after a while she found her lover. He was extremely weak and badly wounded. When the dying warrior saw her, he told her, “When I am finished, you take my attire and weapons home. When you get a son give them to him and with that the warrior seemed to be dying.

 

But the girl did not listen to him she quickly looked for water and washed his wounds. And after that she began to look for food for him. It did not take long before she saw a deer passing by. With her lover’s spear she killed it, and wasted the meat for her lover. For many months the young women washed the wounds of her lover and fed him until he was well again.

 

Back at home everybody thought that the young woman and her lover were dead, and they insisted that their death rites be performed. However, the father of the warrior kept postponing the death rites. But at last the old man agreed to perform the rites because his youngest son was to be circumcised, and could not be circumcised before the rites were performed.

 

So preparations for the death rites for the lost warrior were made. But on the morning of the day that the rites were to be performed, and as people were gathering, one of the people in the gathering heard a war song coming from the other side of the valley. He asked other people to listen. The father of the warrior could not mistake his son’s voice. He was almost crying as he gazed on the other side of the valley. The singing voice became clearer and before long the warrior and his lover emerged, driving a large herd of cattle. The bells that were tied around the necks of the oxen played to the tune of the war song.

 

There was great rejoicing as people ran to meet the lost warrior and his young lover. On their arrival back home a big bull was slaughtered and there was a great feast. People ate and drank. And the warrior and his lover were married. The two became man and wife and lived happily. And my story ends.

 

From Oral Literature. A Junior Course

                                    By A. Bukenya and M. Gachanja,

                                    Longhorn Kenya.

 

Questions

1. With appropriate illustrations, classify the oral narrative above (2mks)                                           ………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………….

 

2. Identify two instances of repetition in the passage                                  (2mks)

………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………….

3. Identify and illustrate one character traits of:

(i) The warrior   (2mks)

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

 

(ii) The youngest lover (2mks)

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

 

4. Give two functions of the song.                               (2mks)

…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………..

5. Identify instances of irony in the passage                                  (2mks)

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

 

6. Name and explain one economic activity of this community.                                  (2mks)

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

1. Which devices have been used to start and end this story? What are their functions? (4mks)

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

 

1. Using a proverb, summarizethe moral lesson in the story.                             (2mks)

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………….

 

 

4. GRAMMAR

(a)     Rewrite the following sentences as instructed.(5mks)

 

(i) The young man did not join a public university. He did not pass the exam.

(Rewrite as one sentence beginning: If …)

……………………………………………………………………………………………………

……………………………………………………………………………………………………

 

• I waited until it was my turn to see the doctor. I was sitting on the couch.( Rewrite as one sentence beginning: Sitting

 

……………………………………………………………………………………………………..

……………………………………………………………………………………………………..

 

• I never thought I would ever be a minister. (Begin: Not once………)

 

……………………………………………………………………………………………………

 

1. iv) Come with me. (Add a question tag)

……………………………………………………………………………………………………

 

1. v) Most students benefit when they read in the morning. (Rewrite the sentence beginning with a gerund)

………………………………………………………………………………………………………………………………………………………………………………………………………………………………

 

1. b) Replace the underlined words with suitable phrasal verbs. (3mks)

i)The girl had to raise her siblings after the mother passed on.

……………………………………………………………………………………………

 

1. ii) Telcom Kenya sacked most of its workers.

…………………………………………………………………………………………

 

iii)The government has abolished certain taxes.

 

…………………………………………………………………………………………………

 

c)Use the correct form of the word given in brackets to fill in the gap in each sentence.                                                                                (4mks)

 

• (i) The police said it was a strange………………………….. ( occur)
• (ii) The …………………of his speech was appreciated by many.(clear)
• The baby had………….. in the cot for hours when the mother returned. (lie)
• Your ……………………………..of words should be clear for people to understand you. (pronounce)

 

 

1. d) Fill the blank spaces with correct prepositions.(3mks)

 

1. He also came home to congratulate me ……………. my graduation.
2. As a businessman, Ole Kaelo deals………….. agricultural products.

• The thief was oblivious ……………. the trap.

THE  END

---

---

---

TRIAL ONE EVALUATION TEST 

MARKING SCHEME

ENGLISH

Paper 2

101/2

Comprehension

(Instructionsfor marking: remove ½ mark for faulty expression)

1. (a) cyberchondria – a term for patients who use the internet for self diagnosis

and  then present ‘misinformation’ to their doctor.              b        (1mk)

(b)       Patients say they want information on medically diagnosed disease /illness b

or to find alternative treatment.b                                                     (1mk)

(c )      Our health service is not entirely healthy due to:

– it barely survives on shoe-string budgets    b                                (1mk)

– low staffing levels                                                                           (1mk)

– long queues of patients praying for help  (1mk)

( Any two, each one mark)

• One goes to the search engine (1mk) band types the symptom (1mk)b
• You can’t rely on a website for self-diagnosis because it can spit out the most common reason for your symptoms (1mk) bwhich may not be accurate b(1mk)
• Professional health providers give informed diagnosisb (1mk) whereas internet can give medical misinformation (1mk)b (cyberquakery)
• The information on the internet is not the ultimate truthb (1mk)
• He dislikes/mistrusts/doubtful b                                            (1mk)

Illustrations:

• You can’t rely on a website for self diagnosisb(1mk)
• It is better to present your aching stomach or back to your physician or health care profession
• You may be wasting time, prayers money and emotional energy….
• On line health consumer should be wary because of medical misinformation or cyberquackery
• You can’t use the internet as the gospel truth to determine health problems

( identification: 1mk, Illustration: 1mk)

( Accept any relevant illustration. Any 1 illustration.1 mark. No mark for illustration without identification)

(i)        Panic and apprehension may be bred by online diagnosis.b                      (1mk)

(j)        Find out who is responsible for the content                                      (1mk)

– Find out whether the website is owned or sponsored by a reputable      organization    b                                                                                        (1mk)

• Look for the credentials and qualifications of the professionals presenting the information.b       (1mk)

( Answer must be in note form. This is use of dashes (-) dots (.)

numbering (1, 2, 3,) Numerals (I, ii, iii) if not deduct 50% of the final mark)

• Shoe string budget –b limited /inadequate resources. (1mk)

Hits – access the website       b                                                        (1mk)

• Opening a website
• Visit the website

1. 2 :Answers to Sample Extracts

Sample extract 1

1.She refers to the previous Christmas when she spent every evening for three full weeks making ornaments for the Christmas tree. Her husband says it was the dullest three weeks he had spent.(2 marks)

2.The couple is happy that Mr. Helmer has been to the level of a bank manager. They are happy that the position will ease their financial burden and they will have a lot of money to spend.(2 marks)

3.Parental responsibility/the role of women in the society. Nora is committed to please her family. Krogstad says, “You had the best of intentions to please us all”

Falsehood-Mr Helmer tells Nora to say he is not at home if it’s a visitor. (4 marks)

4. ,isn’t there? (1 mark)
5. -The lady is Mrs. Linde.

-She is Nora’s school days times.

-She asks Nora talk to her husband to give her job because of his title.

-She gets the job which is to make Krogstad lose his job.

– Krogstad writes a letter exposing Nora’s forgery to prevent Helmer from sacking him.

-Mrs Linde speaks to Krogstad on her behalf and Krogstad withdraws his revenge plans.

-The Helmers reputation is protected although their marriage finally breaks. (6 marks)

6.Flashback. “How could I help the cat’s going in and tearing everything to pieces?” This refers to the previous Christmas                                                                                            (2 marks)

7. Helmer goes into his room while the maid ushers in Mrs. Linde. Mrslinde greets Nora in a dejected voice but Nora does not seem to recognize her. Later she remembers her as christen and observes that Christine has changed greatly. She agrees she has changed because they last met nine or ten years back.                         (4 marks)

8i) Provoke

1. ii) Boring/uninteresting

iii) Bother, an inconvinience

1. iv) Visitor (4 marks)

 

3. Answers – The Warrior who had Eight lovers
4. Dilemma 1mk (Identification)

Warrior’s choices going to bring cows and risk being killed or stay and marry wonderful lovers and remain poor.                1mk ( Explanation)

• No mark if identification is missing.

 

2. Sang louder and louder (1mk)

– She ran and ran                                                                                (1mk)

 

3. (ii) The warrior

– Brave (1mk) went for a lone raid                                                   (1mk)

(iii) the youngest lover

– loving (1mk) risked her life going to look for her lover               (1mk)

(Any other correct trait and illustration)                               Total 4 mks

No mark if trait is missing.

4. (i) Expresses the girl’s love for the warrior (1mk)

(ii) Helped the girl trace her warrior lover                                        (1mk)              Total 2 mks

5. (i) The warrior was ready to die but instead he recovered (1mk)

(ii) Everybody thought the young woman and her lover were dead and insisted

their death rites be performed.           (1mk)                          Any other relevant irony

 

 

6. (i) Pastoralists (½ ) – bull, cows (½ )

(ii) Hunting ½  – she killed a deer with lover’s spear ½ mks

 

7. (i) Opening formula

• to call for attention
• alert audience for start of narration
• Removes the audience from the world of reality and takes them into the world of fiction

 

(ii) Closing formula

• to mark the end of the story.
• To bring the audience back to the world of reality from the world of fiction

8. A friend in need is a friend indeed

• the girl was willing to risk her life to save her lover

4. Grammar ( Marking instructions: Deduct ½ mark for faulty expression)

(a)       (i) If the young man had passed the exam, he would have joined a public

University

(Marking instructions: The comma is mandatory, if missing deduct ½ marks)

• Not once did I ever think that I would ever be a minister
• Sitting on the couch, I waited until it was my turn to see the doctor.
• Come with me, will you?
• Reading in the morning benefits most students

1. b) i) look after
2. ii) laid off

iii) done away with

(Marking instructions: The phrasal verbs must be in the correct tense)

(c)

( i)       occurrence

(ii)       clarity

• Lain
• Pronunciation

(Marking instructions: If wrong spelling = 0 mk

(d)

(i) on

(ii) in

(iii) of

The CUK Nairobi CBD Training Institute Course List, Fees, Requirements, How to Apply

Academic Programmes.

MAIN ELECTRICITY

1. What current will a 500W resistor connected to a source of 240V draw?
2. Name a device used to change light energy directly into electrical energy.
3. When a current of 2.0A flows in a resistor for 10 minutes, 15000 Joules of electrical energy is dissipated. Determine the voltage across the resistor.
4. An electric bulb rated 40W is operating on 240v mains. Determine the resistance of its filament.
5. An electric heater rated 240V, 3000V is to be connected to a 240V mains supply, through a 10A fuse. Determine whether the fuse is suitable or not.
6. A 60W bulb is used continuously for 36 hours. Determine the energy consumed, giving your answer in kilowatt hour (Kwh)
7. How many 100W electric irons could be safely be connected to a 240V mains circuit fitted with a 13A fuse?
8. Find the maximum number of 75W bulbs that can be connected to a 13A fuse on a mains supply of 240V.
9. Determine the cost of using an electrical iron box rated 1500W, for a total of 30 hours given that the cost of electricity per Kwh is Kshs. 8.
10. State Ohm’s law.
11. Electrical energy costs Kshs. 1 per Kwh unit. Find the cost of using an electric heater of power 1.5 KW for a day.
12. The figure below represents part of the main circuit.

 

 

 

 

 

 

 

 

 

 

 

 

1. i) Explain why it is not advisable to fix a fuse on neutral line.
2. ii) Explain why there are fuses of different rating in the distribution box.
3. Calculate the power of a devise which has a p.d of 250V applied across it when a current of 0.5A passes through it.
4. An electric iron box is rated 2500W and uses a voltage of 240V. Given that electricity costs Kshs. 1.10 per Kwh, what is the cost of using it for 6 hours?
5. Four 40 w bulbs and seven 100W are switched on for 3 hours a day for domestic monthly bill given the cost of electricity is sh. 3.50 per unit (Take 1 month = 30 days)
6. a) Give one advantage of transmitting mains electricity as a.c and not d.c.
7. b) The figure below shows part of a wiring circuit for a house.

 

Identify two faults in the wiring.

1. c) The figure shows a connection to the pin plug.

 

1. i) State the colour of the leads labelled A, B and C.

A ……………………………

B ……………………………

C ………………………………

1. ii) Why is the earth pin normally longer than the two pins?

iii) What is the function of the fuse?

1. d) The power company supplied electrical energy and charges the consumption to ordinary domestic wiring as follows;

– A monthly fixed charge of Ksh. 75

– Ksh. 1.55 per unit for the first 50 units consumed.

– Ksh. 6.65 per unit for the next 51- 300 units

1 unit = 1kilowatt- hour (kwh)

A consumer uses 1.98 x 10⁵ kilojoules of electrical energy in a given month. Determine the total month bill.

17. The figure below shows a connection to the three pin plug.
18. i) Identify the mistakes in this wiring.
19. ii) State two reasons why the earth pin is normally longer than the other two pins.
20. (a) Figure 10 shows a section of a house wiring syste

(i).       Name:

The circuit labelled P

The terminals labelled X and Y

X……………………………………

Y……………………………………

(ii)      State the purpose of R in the circuit

(iii)     Give the reason why R is connected to Y but not to X

(iv)     Why is the earthing necessary in such a circuit?

(b)      Determine the cost of using an electric iron rated 1500W, for a total of 30 hours given that the cost of electricity per kWh is Ksh. 8

19. (a)A transformer has 800 turns in the primary windings. The alternating e.m.f connected to the primary is 240V and the current flowing is 0.2A. Find the power in the secondary coil if the transformer is 90% efficient

(b) The figure below shows a domestic wiring system.

 

 

 

(i) Point out two faults in the circuit.

(ii)     What is the fuse made of and why is it necessary in the circuit?

(c) State 2 ways of minimizing power loss in transmission lines.

20. A transformer has 800 turn in the primary and 40 turns in the secondary winding. The alternating e.m.f connected to the primary is 240v and the current is 0.5A.

Determine

1. The secondary e.m.f
2. The power in the secondary if the transformer is 95% efficient.
3. c) Explain how energy losses in a transformer are reduced by having
4. A soft-iron core
5. A laminated core. (
6. A transformer with 2000 turns in the primary circuit and 150 turns in the secondary circuit has its primary circuit connected to a 800V a.c source. It is found that when a heater is connected to the secondary circuit, it produced heat at the rate of 1000W. Assuming 100% efficiency, determine the;
7. Voltage in the secondary circuit.
8. Current in the primary circuit.

• Current in the secondary circuit.

1. State the type of transformer represented above.
2. State the reason why long distance power transmission is done at a very high voltage and using thick cables.
3. Calculate the cost of using the following appliances in one month (30 days ) if the company rate is ksh. 9.50 per unit.

I        A 2000W water heater for 2 hours per day.

II      A 75W bulb for 10 hour per day.

III     A 1500W electric iron for 1 hour per day.

22. Figure 6 below shows a part of housing wiring system

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1. Name (i) The  circuit labeled p
2. The terminal labeled X and Y

• State the purpose of R in the circuit

1. Give a reason why R is connected to Y but not to x
2. Why is earthing necessary in such a circuit.

Determine the cost of using an electric iron rated at 1.5Kw for a total of 40 hours if he cost of  electricity is Ksh 8 per unit.

23. (a) Long distance transmission of power is done at a very high voltage.

• Figure 10 below shows a section of a house wiring system.

 

           

 

 

 

 

 

 

 

 

 

 

• Name:

I           The circuit labelled F.

II          The terminals A and B.

• Give a reason why R is connected to B and not A.
• What is R?
• Why is earthing necessary in such a circuit?

• 54kg of water in a metallic cylinder of heat capacity 9000JKˉ¹ is heated from 10°C to 80°C using an immersion heater rated 1800W, 240V. Assuming that no heat is lost to the surrounding and the immersion heater works at its correct voltage.
• The current flowing through the water.
• The cost of heating the same mass of water from 10°C to 80°C, every day for 30 days, if electricity cost sh.6.70 per unit.

24. (i) State the reason why long distance power transmission is done at a very high voltage and using thick cables.

(ii)  Calculate the cost of using the following appliances in one month (30 days) if the company rate is Ksh. 9.50 per unit.

1. A 2000W water heater for 2 hours per day.
2. A 75W bulb for 10 hours per day.

• A 1500W electric iron for 1 hour per day.

Find the total monthly bill for the above household if in addition to the energy consumed, the power company charges each consumer a standing charge of Ksh. 200 and fuel cost at 70 cents per unit.

25. The following appliances are operated in a home for the time indicated everyday.

Applicance                              Time

Three 40W bulbs                    3hrs

One 2000W television            5 hrs

Four 60W bulbs                      30 min

One 500W freezer                  10 hrs

Calculate:

1. a) The total power used by the applicance everyday.
2. b) The total electrical power consumed in 30 days in Kwh.
3. c) If the cost per unit is ksh. 6.20, calculate the cost of using the applicances in 30 days.
4. (a) Appliances which draw current from a ring’s main circuit have a third cable connected to the earth.  Give a reason why?
   • In a lighting circuit the wires used are relatively thinner than those of a cooker circuit. Give an explanation for this.
   • A transformer with 6000 turns in the primary circuit and 300 turns in the secondary circuit has its primary circuit connected to a 400V a.c. source. A heater connected to the secondary circuit produces heat at the rate of 600W.  Assuming that the transformer is 100% efficient determine:-
     • The voltage in the secondary circuit.
     • The current in the primary circuit.
     • The current in the secondary circuit.

 

27. State why alternating current (a.c.) is used for transmitting electricity over long distances.
28. State the reason why electrical power is transmitted over long distances at very high voltages.
29. A house has one 100W bulb. Two 60W bulbs and one 30W bulb. Determine the cost of having all the bulbs switched on for 70 hours, given that the cost of electricity is 40cents per kilowatt hour.
30. The figure below illustrates a system for transmitting electricity around a country, from the PowerStation to the consumers

• At different points in the systems, the voltage is 220v, 11000v or 132v. At the three places marked, write the appropriate values of the voltage.
• State one advantage of using high voltages for transmission of electricity

31. Figure below shows the arrangement of elements in a circuit of an electric cooker. The two elements P and Q are each of resistance 40 ohms. One side of the mains is connected to terminals A. The other side of the mains is connected to a selector with three positions.

 

 

 

 

 

 

 

 

 

1. i) In the first position of the switch, the mains are connected to terminal which terminals?
2. ii) Determine the magnitude of the current through the element

iii) In the second position of the switch the mains are connected to terminal B and to terminal A. Calculate the total resistance in the circuit.

1. iv) Determine the magnitude of current in the second connection.
2. v) In the third connection of the switch, terminal B is connected to terminal A and then to one side of the mains while the other side of the mains is connected to terminal C.

(i) . Determine the total resistance in the circuit

(ii) . Determine the magnitude of the power consumed by this connection

32. a) Give a reason why a fluorescent tube is preferred to a filament bulb for domestic lighting

(b)       What property does a fuse wire have that makes it suitable for controlling excessive current in circuit?

1. c) State two factors which affect heating in electric circuits.
2. d) An electric heater takes a current of 12.5A from 240V power supply.
3. i) Calculate
4. a) Its power rating.
5. b) Its resistance
6. e) State Faraday’s law of Electromagnetic induction
7. (a) Electrical energy is distributed as alternating current (a.c) at high voltage.

Explain why:

(i)        A very high voltage is necessary.

(ii)       Thick aluminium transmission lines are recommended to carry the current.

(b)       The figure shows the electric wiring of an electric cooker X, Y and Z are main wires.

 

 

 

 

 

 

 

 

 

 

 

Identify Y and X giving reasons.

(c)       A student has a large number of 240V, 60W coloured bulbs he wishes to use for decorations so that the bulbs operate normally.

(i)        How many bulbs can be connected to a 240V supply through a 5A fuse?

(ii)       If electric energy cost Kshs. 3.00 per unit, what will be the cost of running the above circuit for 5 hours a night for 20 nights?

34. A hydro-electric power station produces 500KW at a voltage of 10KV. The voltage is then stepped up to 150KV and the power is transmitted through cables of resistance 200Ώ to a step down transformer in a sub-station. Assuming that both transformers are 100% efficient calculate;

(i) The current produced by the generator

(ii) The current that flows through the transmission cables

(iii) The voltage drop across the transmission cables

(iv) The power loss during transmission

(v) The power that reaches the sub-station

35. a)         State one danger of high voltage transmission
36. b) In domestic ring there are various circuits which may be used. Name three
37. c) Figure 5 shows part of a house wiring system

 

1. i) I using X and Y identify the live and the neutral cables respectively
2. ii) Insert in the circuit, S_L, S_S and S_C for switches of the lamps, sockets and

cooker respectively

iii)       Insert the earth connection in the circuit diagram

1. iv) Explain why the lamps are connected in parallel
2. An electric iron rated 240V, 750w is to be connected to a 240v mains supply through a 3A fuse. Determine whether the fuse is suitable or not.
3. (a) The earth pin of the three-pin plug is longer to open the blind for neutral and live pins. State another reason for its being longer.

(b) An electrician installed electric wiring in a house and connected the bulbs and the switches as shown in figure below.

 

(i)        State what happens when switch:

I           S₁ only is closed

II          S₂ only is closed

III        S₃ only is closed

IV        All the switches are closed.

(ii)      Redraw a diagram showing the best position the bulbs should be installed.

I:          Explain why you consider this your arrangement to be the best.

(c)      A 75W lamp and a 1.5kW water heater are connected in turn to a 240V supply.

(i)        Calculate the current in each appliance.

(ii)      Explain the difference you would expect to find between the wiring used to connect the bulb and the water heater to the mains supply.

38. a) State one source of mains electricity
39. b) How are the high voltages required during power transmission achieved?
40. c) i) Explain why three pin plugs are preferred for two pin plugs in the use with iron box
41. ii) Explain the letters L, N and E usually marked on the three pin plug
42. (a) Long distance transmission of power is done at a very high voltage.
43. b) Figure 10 below shows a section of a house wiring system.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

• Name:

I     The circuit labelled F.

II    The terminals A and B.

(ii) Give a reason why R is connected to B and not A.

(iii)      What is R?

(iv)      Why is earthing necessary in such a circuit?

• 54kg of water in a metallic cylinder of heat capacity 9000JKˉ¹ is heated from 10°C to 80°C using an immersion heater rated 1800W, 240V. Assuming that no heat is lost to the surrounding and the immersion heater works at its correct voltage.

(i)  The current flowing through the water.

(ii) The cost of heating the same mass of water from 10°C to 80°C, every day for 30 days, if electricity cost sh.6.70 per unit.

40. Define power as used in electric circuits.
41. a) A hoist motor powered by a 240V mains supply requires a current of 30A to lift a load of mass 3 tonnes at the rate of 5m per minute.

Calculate:

1. i) The power input
2. ii) The power output

iii)       Overall efficiency E

41. The following appliances are operated in a home for the time indicated everyday.

Appliance                               Time

Three 40W bulbs                    3hrs

One 2000W television                        5 hrs

Four 60W bulbs                      30 min

One 500W freezer                  10 hrs

Calculate:

1. a) The total power used by the appliance everyday.
2. b) The total electrical power consumed in 30 days in Kwh.
3. If the cost per unit is ksh. 6.20 calculate the cost of using the appliances in 30 days.   The voltmeter in figure 7(a) and the ammeter in figure 7(b) below read 2.1v and 0.35A respectively.

 

Fig. 7(a)

Fig 7(b)

 

 

 

Determine the reading in the voltmeter and ammeter in figure 7 (c ) and 7 (d) respectively. (The lamps are identical)

 

 

Fig. 7(d)

Fig. 7(c)

 

 

 

43. Give any two causes of energy loss in a transformer
44. a) A part from the amount of current drawn by an electric kettle, state the other factors which affects its heating effects.

(b) If the current passing through the kettle is doubled, the power dissipated increases by 3000W.

Determine:

1. i) The power dissipated before doubling the current.
2. ii) The current flowing if the internal resistance of the kettle element is 10W
3. (a) What is the purpose of a fuse?

(b) The diagram in figure 9 below shows a ring – main circuit used by an electrician in a certain house.

 

• Identify two faults in the installation.
• Explain why the 3 – pin plug fuse has a longer earth pin than the live and neutral pins.
• Identify the type of transformer T used in the diagram in Fig. 9

(c) A cooker rated 2.0kW was operated for 40minutes each for 30days.

If the cost of each kilo – watt – hour unit is Shs. 15.50, calculate the cost of electricity used.

46. (a) A house has five rooms each with 240V, 60W bulbs. If the bulbs are switched on from 7:00pm to 10:30 pm
    • Calculate the power consumed in the month of April in Kilowatt-hours
    • Find the cost per month for lighting these rooms at Ksh6.70 per unit.
      • A student designed a transformer to provide power to an electric bell marked 24W, 6V from a mains supply 240V. He wound coils, 50 turns and N turns on an iron ring core. When he connected the coil of 50 turns to the bell and N turns coil to the a.c source, he found out that the transformer was only 80% efficient. Find;

(i) The value of N

(ii) The current in the primary coil.

• The figure below shows a connection to the three- pin plug.
  • Name the cables A, B and C and state their colours.
  • Why is the fuse connected to cable C.?
  • State one reason why in domestic wiring system appliances are connected in parallel.

47. (a) The figure below shows the inner parts of a three-pin plug.
    • Identify the pins A and B. (2 marks)
    • State the reason why the pin B is normally longer than the other two pins A and C.

Kisii University Course List, Fees, Requirements, How to Apply

• Bachelors
• Certificate
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UNIT 1: ATOMIC STRUCTURE AND THE PERIODIC TABLE.

Checklist

1. Structure of the atom
2. The subatomic particles;

• Protons
• Neutrons
• Electrons;

3. Atomic number and mass number
4. Isotopes;
5. Energy levels and electron arrangements.
6. The periodic table;

• Groups
• Periods;

7. Relative atomic mass and isotopes;
8. Ion formation;

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

9. Valency and oxidation numbers;
10. Chemical formula;
11. Chemical equations
12. Balancing chemical equations;

 

The atom:

– Refers to the smallest particle of an element hat can take part in a chemical reaction;

– It has an average diameter of 10^-8 cm with a nucleus of about 10^-13 cm;

 

Parts of an atom

– The atom is made of two main parts:

• The nucleus
• The energy levels;

 

1. The nucleus:

– Is the positively charged part of an atom;

– The nucleus contains two subatomic particles; neutrons and protons;

– The positive charge is due to presence of protons;

– The nuclei of all atoms contain neutrons except the hydrogen atom;

– The protons and the neutrons are together referred to as the nucleons;

 

2. The energy levels.

– They contain the electrons;

– Electrons are so small and move so fast that their path cannot be traced directly;

– Thus the energy level simple represents the region where the electrons are most likely to be found;

 

Structure of the atom.

 

 

 

 

 

 

 

 

Note:

– The atom can still however be split into smaller particles termed the sub-atomic particles;

 

The sub-atomic particles.

– Are generally three:

• Protons;
• Neutrons;
• Protons;

 

1. Protons.

– Are the positively charged sub-atomic particles;

– Are found in the nucleus and thus form part of the nucleons;

– The number of protons in the nucleus is equal to the number of electrons in the energy levels;

 

2. 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;

 

3. Electrons.

– Are negatively charged sub-atomic particles found in the energy levels;

– The number of electrons in the energy levels is equal to the number of protons in the nucleus;

– This makes the atom to be electrically neutral;

 

Atomic number and mass number.

Atomic number.

– Refers to the number of protons in the nucleus of an atom;

Examples.

• Sodium has 11 protons I the nucleus and thus said to have atomic number 11;
• Chlorine has 17 protons in the nucleus and thus said to have atomic number 17;

 

Mass number;

– Refers to the sum of the number of protons and neutrons in an atom of an element;

Examples:

• Sodium has 2 neutrons and 11 protons hence a mass number of 23;
• Chlorine has 18 neutrons and 17 protons hence a mass number of 35.

 

Notation of atomic number and mass number;

– Both atomic number and mass number of an element can be written along with the symbol of an element;

 

Mass number;

– Is conventionally represented as a superscript to the left of the symbol;

Examples:

Sodium; ²³Na;

Magnesium ²⁴Mg;

 

Atomic number;

– Is conventionally represented as a superscript to the left of the symbol;

Examples:

Sodium; ₁₁Na;

Magnesium ₁₂Mg;

 

Thus the elements can be conventionally represented as:

Sodium ²³Na

Magnesium ²⁴Mg

 

Atomic properties of the first 20 elements.

Element	Symbol	Number of electrons	Number of  protons	Number Of  neutrons	Atomic number	Mass number
Hydrogen	H
Helium	He
Lithium	Li
Beryllium	Be
Boron	B
Carbon	C
Nitrogen	N
Oxygen	O
Fluorine	F
Neon	Ne
Sodium	Na
Magnesium	Mg
Aluminium	Al
Silicon	Si
Phosphorus	P
Sulphur	S
Chlorine	Cl
Argon	Ar
Potassium	K
Calcium	Ca

 

Isotopes.

– Are atoms of the same element with same atomic number but different mass number due to different number of neutrons.

 

Examples of isotopes.

Element	Isotope	Atomic No.	Number of protons	Number of neutrons	Mass number	Isotopic representation
Hydrogen
Carbon
Oxygen
Chlorine

 

Energy levels and electron arrangements.

• Energy levels:

– Are definite orbits in an atom that the electrons occupy.

– The energy levels are numbered 1, 2, 3……starting with the one closest to the nucleus.

– Electrons occupying the same energy level have approximately the same amount of energy.

– Each energy level can only accommodate a given maximum number of electrons.

 

 

Maximum number of electrons per energy level

Energy level	Maximum number of electrons
1st	2
2nd	8
3rd	8 (only for the first 20 elements)

 

Illustrations:

• Hydrogen

– It has only one electron and thus this electron occupies the first energy level.

– Since the first energy level is not yet full, hydrogen does not have the second energy level;

– The electron arrangement of hydrogen is thus 1.

 

• Helium:

– Helium is atomic number 2 and has only two electrons, which occupy the first energy level.

– The first energy level is thus completely full, but since there are no other r electrons lithium also has only one energy level;

– The electron arrangement is thus 2.

 

• Chlorine:

– Chlorine has atomic number 17 and thus has 17 electrons;

– The first two electrons occupy the fist energy level which is thus completely filled up;

– The remaining 15 electrons occupy the second energy level, which can however accommodate only 8 to be completely filled up;

– Thus the remaining 7 electrons move to the third energy level; which needs 8 to be completely filled up;

– Since the third energy level is not yet full chlorine does not have a fourth energy level;

– The electron arrangement is thus 2.8.7.

 

Electron arrangement.

– Refers to the distribution of electrons in the energy levels of an atom.

 

Example: electron arrangement for the first 20 elements.

Element	Symbol	Atomic number	No. of electrons	Electron arrangement
Hydrogen
Helium
Lithium
Beryllium
Boron
Carbon
Nitrogen
Oxygen
Fluorine
Neon
Sodium
Magnesium
Aluminium
Silicon
Phosphorus
Sulphur
Chlorine
Argon
Potassium
Calcium

 

Dot and cross diagrams.

– Is a diagrammatic representation of the electron arrangements in an atom in which the energy levels are represented by concentric lines while electrons are represented by dots or crosses.

– However all electrons are the same regardless of whether they are represented as dots or crosses.

 

Examples:

Lithium	Magnesium	Aluminium	Carbon
No. of electrons =	No. of electrons	No. of electrons	No. of electrons
No. of protons   =	No. of protons	No. of protons	No. of protons
No. of neutrons =	No. of neutrons	No. of neutrons	No. of neutrons

 

Beryllium	Nitrogen	Chlorine	Argon
No. of electrons =	No. of electrons =	No. of electrons =	No. of electrons =
No. of protons   =	No. of protons   =	No. of protons   =	No. of protons   =
No. of neutrons =	No. of neutrons =	No. of neutrons =	No. of neutrons =

 

Potassium	Boron	Silicon	Phosphorus
No. of electrons =	No. of electrons =	No. of electrons =	No. of electrons =
No. of protons   =	No. of protons   =	No. of protons   =	No. of protons   =
No. of neutrons =	No. of neutrons =	No. of neutrons =	No. of neutrons =

The Periodic table

– Is a table showing the arrangements of elements in order of their relative atomic masses.

– Is based on the ideas of Dmitri Ivanovich Medeleev.

– The modern periodic table is based on Meneleev’s periodic law which states that:

                        “The properties of elements are a periodic functions of their relative atomic masses”

– The modern periodic law itself states that:

                        “The properties of elements are a periodic functions of their atomic numbers”

 

Design of the modern periodic table.

It has vertical columns called groups and horizontal rows called periods.

• Groups:

– Are the vertical columns of a periodic table.

– Are eight in number; ad numbered in capital Roman numerals I all through to VIII.

Note: group VIII is also called group zero because the elements have little tendency to gain o lose electrons during chemical reactions

– Between group 2 and group 3 is a group of elements called the transition metals;

 

• Periods:

– Are the horizontal rows in a periodic table.

– They are 8 in number in a modern periodic table.

 

• Transition metals.

– Are elements that form a shallow rectangle between group II and group III.

– These elements are generally metallic and hence the name “transition metals”

– They are not fitted in any group because they have variable valencies.

– Are sometimes called the “d-block” elements.

– They are much less reactive than the elements in groups I and II

– They have some unique characteristics that make them not fit in the 8 groups of the periodic table

– These include:

• Have variable valencies; hence show different oxidation states in their compounds;
• They form coloured compounds as solids and in aqueous solutions;
• Have very high melting and boiling points (than metals in groups I and II).
• They do not react with water;
• Have very high densities (compared to metals in groups I ad II)

 

• Lanthanides and Actinides.

– They form a block of elements within the transition metals.

– Are sometimes called the inner transition metals.

– The lanthanides consist of 14 elements from Cerium (Ce) to Lutetium (Lu).

– The Actinides are the 14 elements from Thorium (Tho) to Lawrencium (Lr).

 

Placing an elements in the periodic table

– The position of an elements in the periodic table is governed by the atomic number and hence the

electron arrangement.

 

• The period:

– The period to which an element belongs is determined by the number of energy levels.

– The number of energy levels is equal to the period to which an elements belongs.

 

Examples:

Elements	Symbol	Atomic number	Electron arrangement	Number of energy levels	Period
Lithium	Li	3	2.1	2	2
Sodium	Na	11	2.8.1	3	3
Calcium	Ca	20	2.8.8.2	4	4
Nitrogen	N	7	2.5	2	2
Helium	He	2	2	1	1

 

• Group:

– The group to which an element belongs to is governed by the number of electrons in the

outermost energy level.

– The number of electrons in the outermost energy level is equal to the group to which the element

belongs.

 

Examples:

Elements	Symbol	Atomic number	Electron arrangement	Outermost electrons	Group
Potassium	K	19	2.8.8.1	1	1
Aluminium	Al	13	2.8.3	3	3
Silicon	Si	14	2.8.4	4	4
Oxygen	O	8	2.6	6	6
Chlorine	Cl	17	2.8.7	7	7

 

Diagram of a full periodic table.

Note: In the modern periodic table atomic masses are used instead of mass numbers. The atomic

masses are preferable because they take care of elements with isotopes unlike mass numbers.

Diagram: part of the periodic table showing the first 20 elements

 

 

 

 

 

 

 

 

 

 

 

Relative Atomic Mass and Isotopes.

• Introduction:

– The masses of individual atoms of elements are very negligible and thus quite difficult to weigh.

– On average the mass of an atom is approximately 10^-22g which cannot be determined by an

ordinary laboratory balance.

– For this reason the mass of atom has been expressed relative to that of a chosen standard element

hence the term relative atomic mass.

– The initial reference element was hydrogen which was later replaced with oxygen.

– Later the oxygen scale was found unsuitable;

Reason:

Oxygen exists I several isotopes and thus led to problems when deciding the mass of an oxygen atom.

– For this reason oxygen was replaced with carbon as the reference atom and to date relative atomic

masses of elements are based on an atom of carbon-12 (note that carbon is isotopic and exists as

Carbon -12 or carbon-14).

 

• Definition:

Relative atomic mass (R.A.M) of an element refers to the average mass of an atom of the element

compared with a twelfth (1/12) of an atom of carbon-12.

	RAM = Average mass of one atom of an element               1/12 mass of an atom of carbon-12

 

 

 

 

 

• Measurement of Relative atomic mass

– RAM of elements is determined by an instrument called Mass Spectrometer.

– The instrument can also be used to determine the relative abundance of isotopes.

– The use of a mass spectrometer in determining the RAM of elements is called mass spectrometry.

 

• How mass spectrometry works.

– In the mass spectrometer atoms and molecules are converted into ions.

– The ions are then separated as a result of the deflection which occurs in a magnetic field.

– Each ion (from an atom, isotope or molecule) gives a deflection which is amplified into a trace.

– The height of each peak measures the relative abundance of the ion which gives rise to that peak.

 

Note:

– Generally the relative atomic mass of an element is closest in value to the mass of the most

abundant isotope of the element.

 

Example: Diagram of a spectrometer trace for Lithium

• Explanations:

– The trace has two peaks indicating that there are two isotopes for lithium.

– The fist peak occurs at a relative isotopic mass of 6 and the second at 7; these are the RAM of the two

isotopes respectively.

– The percentage abundance of the isotope with RAM of 6 (⁶Li) is 9 while the RAM of the isotope with

RAM 7 (⁷Li) is 91.

 

Calculating relative atomic masses of isotopic elements.

– Information form a spectrometer trace is usually extracted and used in calculation the relative atomic

mass of elements.

 

Worked examples.

1. The mass spectrum below shows the isotopes present in a sample of lithium.

(i). Use this mass spectrum to help you complete the table below for each lithium isotope in the sample.                                                                                                                                   (3 marks)

 

Isotope	Percentage composition	Number of
		Protons	Neutrons
6Li
7Li

 

(ii). Calculate the relative atomic mass of this lithium sample. Your answer should be given to three significant figures.                                                                                                                      (3 marks)

 

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

 

 

4. 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)

 

5. 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)

 

6. 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

 

7. Sulphur and sulphur compounds are common in the environment.

(a). A sample of sulphur form a volcano contained 88% by mass of ³²S and 12% by mass of ³⁴S.

(i). Complete the table below to show the atomic structure of each isotope of sulphur.

 

Isotope	Number of
	Protons	Neutrons	Electrons
32S
34S

 

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

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

8. Iridium, atomic number 77, is a very dense metal. Scientists believe that meteorites have deposited virtually all the iridium present on earth. A fragment of a meteorite was analysed using a mass spectrometer and a section of the mass spectrum showing the isotopes present in iridium is shown below.

 

(a). Explain the term isotopes.                                                                                                       (1mark)

(b). Use the mass spectrum to help you complete the table below for each iridium isotope in the meteorite.

 

Isotope	Percentage composition	Number of
		Protons	Neutrons
6Ir
7Ir

 

(c) (i). Define the term relative atomic mass.                                                                                           (1mark)

 

(ii). Calculate the relative atomic mass of the iridium in this meteorite.                                  (3 marks)

 

 

Ion formation.

• Introduction:

– Atoms whose outermost energy levels contain the maximum possible number of electrons are said to be stable.

– Thus atoms with energy levels 2, 2.8 and 2.8.8 are said to be stable.

– Electron configuration 2 is said to have a stable duplet state while electron configuration 2.8 and 2.8.8 is said to have a stable octet state.

– These electron configurations resemble those of noble gases and as such they are stable and do not react.

– Atoms without this stability acquire it by either electron gain or electron loss.

– Whether an atom loses or gains electro(s) depend on the number of electrons in the outermost energy level.

 

– Take the case of sodium.

– Atomic number is 11 with an electron configuration of 2.8.1.

– Thus sodium has two options in to become stable:

• to lose the single electron and acquire a stable electron configuration of 2.8.
• to gain 7 electrons in its outermost energy level and acquire a stable electron configuration of 2.8.8

– Gaining a single electrons and losing a single electrons requires equal amounts of energy.

– Thus it is cheaper and faster in terms of energy for sodium to lose the single electron in the outermost energy level than to gain 7 electrons into its outermost energy level.

– Thus sodium acquires a stable electron configuration 2.8 by losing the single electron in its outermost energy level.

 

Diagram

 

 

 

 

 

 

 

Sodium atom                                                                          Sodium ion.

 

Equation:

Na → Na⁺ + e^–

 

Further examples:

Element	Electron arrangement	Options for stability	Best (cheapest) option
Chlorine	2.8.7	2.8 or 2.8.8	2.8.8
Potassium	2.8.8.1	2.8.8 or 2.8.8.8	2.8.8
Aluminium	2.8.3	2.8 or 2.8.8	2.8
Magnesium	2.8.2	2.8 or 2.8.8	2.8
Carbon	2.4	2 or 2.8	2. or 2.8
Oxygen	2.6	2. or 2.8	2.8

 

• Ion:

Definition: an ion is a charged particle of an element.

– Are formed when an atom of an element either loses or gains electrons.

 

• Illustration:

– For a neutral atom the number of electrons in the energy levels (negative charges) is equal and thus

completely balances the number of protons in the nucleus (positive charges).

– Thus the net charge in a neutral atom is zero (0).

– When an atom gains electron(s), the number of electrons becomes higher than the number of protons

resulting to a net negative charge hence an ion.

– Oppositely when an atom loses electron(s) the number of protons becomes higher than the number of

electrons resulting into a net positive charge hence an ion.

– The charge on the ion is usually indicated as a superscript to the right of the chemical symbol.

– Thus ions are of two types:

• Cations
• Anions

 

• Cations:

– Are positively charged ions.

– Are formed when atoms lose electrons resulting into the number of protons being higher than the number of electrons.

– Are mostly ions of metallic elements since most metals react by electron loss.

 

Examples:

(i). Magnesium:

– It has atomic number 12, with electron arrangement 2.8.2.

– It has 12 protons and 12 electrons hence a net charge of “0” hence the atom is written simply as Mg.

– It will form its ions by losing the two electrons from the outermost energy level.

– Thus the number of electrons decreases to 10 while the number of protons remains 12.

– This leads to a net charge of +2, giving the ion with the formula Mg²⁺.

 

• Diagrammatic illustration:

 

 

 

 

 

 

 

 

 

 

(ii). Phosphorus

– It has atomic number 13, with electron arrangement 2.8.3.

– It has 13 protons and 13 electrons hence a net charge of “0” hence the atom is written simply as Al.

– It will form its ions by losing three (3) electrons out of the outermost energy level.

– Thus the number of electrons decreases by three to 10 while the number of protons remains 13.

– This leads to a net charge of +3, giving the ion with the formula Al⁺³.

 

• Diagrammatic illustration:

 

 

 

 

 

 

 

• Anions:

– Are negatively charged ions.

– Are formed when atoms gain electrons resulting into the number of electrons being higher than the number of protons.

– Are mostly ions of non-metallic elements since most non-metals ionize (react) by electron gain.

 

Examples.

(i). Chlorine:

– It has atomic number 17, with electron arrangement 2.8.7.

– It has 17 protons and 17 electrons hence a net charge of “0” hence the atom is written simply as Cl.

– It will form its ions by gaining a single electron into the outermost energy level.

– Thus the number of electrons increases to 18 while the number of protons remains 17.

– This leads to a net charge of -1, giving the ion with the formula Cl^–.

 

• Diagrammatic illustration:

 

 

 

 

 

 

 

Note: the electros gained must be represented by a different notation from the initial electrons in the atom. E.g. if the initial electrons are represented with crosses (x) then the gained electrons should be represented by dots (.) and vise versa.

 

(ii). Phosphorus

– It has atomic number 15, with electron arrangement 2.8.5.

– It has 15 protons and 15 electrons hence a net charge of “0” hence the atom is written simply as P.

– It will form its ions by gaining three (3) electrons into the outermost energy level.

– Thus the umber of electrons increases by three to 18 while the number of protons remains 15.

– This leads to a net charge of -2, giving the ion with the formula P^-3.

 

• Diagrammatic illustration:

 

 

 

 

 

 

 

Electron transfer during chemical reactions.

– Atoms react either by electron gain or electron loss.

– Generally metals react by electron gain while non-metals react by electron loss.

 

• Illustration:

– Consider the reaction between sodium and chlorine.

– Sodium attains stability // reacts by losing the single electron form its outermost energy level.

– Chlorine attains stability // reacts by gaining a single electron into its outermost energy level.

– Thus during the reaction between the two elements the single electron lost by the sodium atom to

form the sodium ion is the same one gained by the chlorine atom to form the chloride ion.

 

Some basic concepts.

• Valence electrons:

– Refers to the number electrons in the outermost energy level.

 

Examples:

– Calcium, with electron arrangement 2.8.8.2 has 2 valence electrons.

– Oxygen with electron arrangement 2.6 has 6 valence electrons.

– Phosphorus with electron arrangement 2.8.5 has 5 valence electrons.

 

• Valency:

– Refers to the number of electrons an atom loses or gains during a chemical reaction.

– Valency is also known as the combining power of an element.

 

Examples:

– Calcium, with electron arrangement 2.8.8.2 loses 2 electrons during chemical reactions and hence has

a valency of 2.

– Oxygen with electron arrangement 2.6 gains 2 electrons during chemical reactions and thus has a

valency of 2.

– Phosphorus with electron arrangement 2.8.5 has gains 3 electrons during chemical reactions and

hence has a valency of 3.

– Aluminium, with electron arrangement 2.8.3 loses 3 electrons during chemical reactions and hence

has a valency of 2.

 

Note: Some elements have variable valencies and are usually termed the transitional elements (metals)

 

Examples:

– Iron can have valency 2 or 3;

– Copper can have valency 1 or 2

– Lead can have valency 2 or 4.

 

 

 

 

 

 

 

 

 

Summary on valencies of common elements.

 

	Valency 1	Valency 2	Valency 3
Metals	Sodium Potassium	Calcium Barium Magnesium Zinc Iron Lead Copper	Aluminium Iron
Non-metals	Nitrogen Chlorine Fluorine Hydrogen	Nitrogen Oxygen Sulphur	Nitrogen Phosphorus

 

Radicals.

– Are groups of atoms with a net charge that exist and react as a unit during chemical reactions.

– Radicals also have a valency, which is equivalent to the value of its charge.

 

Summary on valencies of some common radicals.

 

	Valency 1	Valency 2	Valency 3
Radicals	Ammonium (NH4+) Hydroxide (OH–) Nitrate (NO3–) Hydrogen carbonate (HCO3–) Hydrogen sulphate (HSO4–)	Carbonate (CO32-) Sulphate (SO42-) Sulphite (SO32-)	Phosphate (PO43-)

 

Oxidation number.

– Refers to the number of electrons an atom loses or gains during a chemical reaction.

– In writing the oxidation number the sign (+ or -) to show gain or loss is written followed by the

number of electrons lost or gained respectively.

 

Illustration.

– Atoms are electrically neutral and are thus assigned an oxidation state of 0 since the number of

protons in the nucleus is equal to the number of electrons in the energy levels.

– However when atoms react they either lose or gain electrons and thus acquire a new state.

– This new state is a new oxidation state and the atom thus acquires a new oxidation number

 

Examples:

Atom	E. arrangement	Ion formula	Valency	Oxidation number
Sodium	2.8.1	Na+	1	+1
Magnesium	2.8.2	Mg2+	2	+2
Aluminium	2.8.3	Al3+	3	+3
Nitrogen	2.5	N3-	3	-3
Sulphur	2.8.6	S2-	2	-2
Chlorine	2.8.7	Cl–	1	-1

 

Further examples:

Particle	Oxidation number
Copper metal, Cu	0
Lead (II) ion, Pb2+	+2
Bromide ion, Br–	-1
Aluminium ion, Al2+	+2
Sulphide ion, S2-	-2
Magnesium metal, Mg	0

 

• Note: oxidation number (state) and charge of an element.

– Oxidation state is written with the positive or the negative sign coming before the element.

Examples: -2, 3, +1, -1 etc.

– Charge on an element is write as a superscript of the element with the number coming before the

positive r the negative sign

Examples: Mg²⁺, Al³⁺, Na⁺, Cl^– etc.

 

Chemical formulae.

– Refers to a representation of a chemical substance using chemical symbols.

– In a single atom it is equivalent to the chemical symbol of the element.

– In a compound it shows the constituent elements and the proportions in which they are combined.

 

• Deriving the chemical formula of compounds.

– In order to write the correct formula of a compound the following must be known:

• The symbols of the constituent elements or radicals.
• The valencies of the elements or radicals

– The chemical formula should start with the element which is more likely to lose electron (s) followed

by the element that is more likely to gain.

 

Worked examples.

 

1. Deriving the formula of sodium chloride.

Elements	Sodium	Chlorine
Formula	Na	Cl
valencies	1	1
balancing	x1	x1

 

Balancing ratios as subscripts: Na₁Cl₁

Formula: NaCl

 

Explanation:

– For sodium to combine with chlorine to form sodium chloride, sodium loses an electron while

chlorine gains an electron.

– Thus every sodium atom needs only a single chlorine atom for both t be fully stable

 

Note:

– When the balancing ratio // subscript is 1, it is usually not written since the symbol of the

element itself represents a single atom.

2. 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

 

3. Deriving the formula of magnesium oxide.

Elements	Magnesium	Oxygen
Formula	Mg	O
valencies	2	2
balancing	X1	X1

 

Balancing ratios as subscripts: Mg₁O₁

Formula: MgO

 

 

Further worked examples.

1. Derive the chemical formula of each of the following compounds.

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

 

2. Complete the table below for elements A, B and C

Element	Valency	Chemical formula of various compounds
		hydroxides	Sulphates	carbonates	nitrates	phosphates	Hydrogen carbonates
A	1
B	2
B	3

 

 

 

 

 

 

 

 

 

 

 

 

 

Chemical equations.

– Refers to representations of a chemical reaction by means of chemical symbols and formula.

 

• Key features of a chemical equation.

– The correct formulae of the reactants are on the left of the equation.

– The correct formulae of the products are on the right of the equation.

– The reactants and products are separated by an arrow pointing to the right.

– The state symbols of the reactants and products must be stated as subscripts to the right of the symbols

– The number of each atom on the reactants side must be equal to the number of the same atom on the products side.

 

Example.

Reaction between hot copper metals and oxygen gas.

• Word equation: Copper + oxygen → Copper (II) oxide.

 

• Chemical equation: 2Cu_(s) + 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:

1. Balance equations for each of the following reactions.

• Sodium hydroxide and dilute hydrochloric acid

 

• Zinc oxide and dilute sulphuric (VI) acid

 

• Zinc metal and dilute nitric (V) acid

 

• Calcium carbonate and dilute sulphuric (VI) acid

 

• Sodium and water

 

2. 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

 

1. The Alkali metals.

– Are the elements with one valence electron and hence in group I of the periodic table.

– All are metallic in nature.

– The members of the family in order down the group is as follows:

• Lithium
• Sodium
• Potassium
• Rubidium
• Caesium
• Francium

 

Electron arrangement of the first three alkali metals.

Elements	Electron arrangement
Lithium	2.1
Sodium	2.8.1
Potassium	2.8.81

 

Diagram: Part of periodic table showing the alkali metals

 

 

 

 

 

 

 

 

Gradation in properties of alkali metals.

1. Atomic and ionic radius.

• Atomic radius:

– Refers to the distance between the centre of the nucleus of an atom and the outermost energy level occupied by electron(s)

• Ionic radius:

– Refers to the distance between the centre of the nucleus of an ion and the outermost energy level occupied by electron(s)

 

Trend:

– The ionic radius and the atomic radius of alkali metals increase down the group.

Reason:

– There is an increase in the number of energy levels down the group from lithium to Francium.

 

Illustration:

– Lithium (2.1) has only 2 energy levels; sodium (2.8.1) has 3 energy levels while potassium (2.8.8.1) has 4 energy levels.

– Thus the outermost electron in potassium is further from the nucleus than the outermost electron in sodium and lithium.

 

2. Atomic and ionic radius of the same element.

– For the same alkali metals the atomic radius is larger than the ionic radius.

Reason:

– Alkali metals form ions by losing the valence electron, lading to the loss of an entire outermost energy level. Thus the atoms have more energy levels than the corresponding ion hence a larger radius in the atom than I the ion.

 

Illustration

– Potassium atom has electron arrangement of 2.8.8.1 hence 4 energy levels.

– During ion formation potassium reacts by losing the single valence electron to acquire a new electron arrangement of 2.8.8 hence 3 energy levels.

– Thus the ion has a smaller radius than the atom.

 

Diagrammatically: Potassium atom and potassium ion

 

 

 

 

 

 

 

 

 

Summary: changes in atomic and ionic radius among alkali metals.

Element	Symbol	Atomic number	Atomic radius (nm)	Ionic radius (nm)
Lithium	Li	3	0.133	0.060
Sodium	Na	11	0.157	0.095
Potassium	K	19	0.203	0.133

Physical properties of alkali metals.

1. Appearance.

– Alkali metals have metallic luster when freshly cut. This refers to a shiny appearance on th cut surface.

– This surface however tarnishes due to reaction on exposure to air.

 

2. 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).

 

3. 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).

 

4. 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.

 

5. Ionization energy.

	Ionization energy is the minimum energy required to remove an electron from the outermost energy level of an atom in its gaseous state.

 

 

 

 

– The number of ionization energies an element may have is equivalent to the number of valence electrons.

– The first ionization energy is the minimum energy required to remove the first electron from the outermost energy level of an atom in its gaseous state.

– The first ionization among alkali metals decreases down the group.

Reason:

– The effective force of attraction of on the outermost electron by the positive nucleus decreases with increasing atomic size and distance from the nucleus.

– Note that the atomic radius increases down the group due to increase in the number of energy levels.

 

 

 

Summary on physical properties of Alkali metals.

Element	Appearance	Ease of cutting	Melting point (oC)	Boiling point (oC)	Electrical conductivity	Atomic radius (nm)	Ionic radius (nm)
Lithium	Silver white	Slightly hard	180	1330	Good	0.133	520
Sodium	Shiny white	Easy	98	890	Good	0.157	496
Potassium	Shiny grey	Easy	64	774	Good	0.203	419

 

Chemical properties of alkali metals.

1. Reaction with air.

– When exposed to air alkali metals react with atmospheric moisture to form the corresponding metal hydroxide and hydrogen gas.

 

• General equation:

Metal + Water → Metal hydroxide + hydrogen gas.

 

– The metal hydroxide further reacts with atmospheric carbon (IV) oxide to form hydrated metal carbonate.

 

• General equation:

Metal hydroxide + carbon (IV) oxide → Hydrated metal carbonate.

 

Examples:

• Lithium

With moisture: 2Li_(s) + 2H₂O_(l) → 2LiOH_(aq) + 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);

 

2. 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)

 

3. 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

 

4. 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 (O2-)	Chloride (Cl–)	Sulphates (SO42-)
Li+
Na+
K+

 

Note:

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

Reason:

– They have high reactivities.

 

 

 

Uses of Alkali metals and their compounds.

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

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

7. Sodium chloride is used as a food additive.
8. A mixture of sodium hydroxide (caustic soda) and carbon disulphide is used in the manufacture of artificial silk called rayon.

 

The Alkaline Earth metals.

– Are the elements with two valence electrons and hence in group II of the periodic table.

– All are metallic in nature.

– The members of the family in order down the group is as follows:

• Beryllium
• Magnesium
• Calcium
• Strontium
• Barium
• Radium

 

Electron arrangement of the first three alkaline earth metals.

Elements	Electron arrangement
Beryllium	2.2
Magnesium	2.8.2
Calcium	2.8.8.2

 

Diagram: Part of periodic table showing the alkaline earth metals

 

 

 

 

 

 

 

 

Dot and cross diagrams for the first three Alkaline Earth metals

 

 

 

 

 

 

 

 

Gradation in properties of alkali metals.

1. Atomic and ionic radius.

• Atomic radius:

– Refers to the distance between the centre of the nucleus of an atom and the outermost energy level occupied by electron(s)

• Ionic radius:

– Refers to the distance between the centre of the nucleus of an ion and the outermost energy level occupied by electron(s)

 

Trend:

– The ionic radius and the atomic radius of alkaline earth metals increase down the group.

Reason:

– There is an increase in the number of energy levels down the group from Beryllium to radium.

 

Illustration:

– Beryllium (2.2) has only 2 energy levels; Magnesium (2.8.2) has 3 energy levels while calcium (2.8.8.2) has 4 energy levels.

– Thus the outermost electron in calcium is further from the nucleus than the outermost electron in magnesium and beryllium.

 

2. Atomic and ionic radius of the same element.

– For the same alkaline earth metal the atomic radius is larger than the ionic radius.

Reason:

– Alkaline earth metals form ions by losing the valence electrons, leading to the loss of an entire outermost energy level. Thus the atoms have more energy levels than the corresponding ion hence a larger radius in the atom than I the ion.

 

Illustration

– Calcium atom has electron arrangement of 2.8.8.2 hence 4 energy levels.

– During ion formation it reacts by losing the 2 valence electrons to acquire a new electron arrangement of 2.8.8 hence 3 energy levels.

– Thus the ion has a smaller radius than the atom.

 

Diagrammatically: calcium atom and calcium ion

 

 

 

 

 

 

 

 

Summary: changes in atomic and ionic radius among alkali metals.

Element	Symbol	Atomic number	Electron arrangement	Atomic radius (nm)	Ionic radius (nm)
Beryllium	Be	4	2.2	0.089	0.031
Magnesium	Mg	12	2.8.2	0.136	0.065
Calcium	Ca	20	2.8.8.2	0.174	0.099

Physical properties of alkali metals.

1. Appearance.

– Alkaline earth metals acquire a metallic luster when polished. This refers to a shiny appearance on the cut surface.

– They however lose this metallic luster when exposed to air due to oxidation.

• Note:

– The purpose of polishing alkaline earth metals before using them in experiment is to remove the oxide coating that slows down and prevents them from reacting.

2. 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.

 

3. 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).

 

4. 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.

 

5. 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)	1st I.E (kJmol-1)	2nd I.E (kJmol-1)
Beryllium	4	1280	2450	0.089	900	1800
Magnesium	12	650	1110	0.136	736	1450
Calcium	20	850	1140	0.174	590	1150
Strontium	38	789	1330	0.210	550	1060
Barium	56	725	1140	0.220	503	970

 

Chemical properties of Alkaline earth metals.

1. Burning alkaline earth metals in air.

– Alkaline earth metals react burn in air to form corresponding oxides.

– More reactive alkaline earth metals may also react with atmospheric nitrogen to form corresponding nitrides.

 

Examples:

– Burns in air with a blinding brilliant flame forming a white solid.

– The white solid is a mixture of magnesium oxide ad magnesium nitride.

 

Equations:

Reacting with oxygen: 2Mg_(s) + 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.

 

2. 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;

 

3. 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.

 

4. 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)

 

5. Reaction of alkaline earth metals and dilute acids

– Generally alkaline earth metals react with dilute acids to form salts and hydrogen gas.

 

Examples:

• With hydrochloric acid.
  • Beryllium:

– When a piece of beryllium is dropped into a beaker containing hydrochloric acid, there is effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of beryllium chloride remains in the test tube // boiling tube.

 

Equation: Be_(s) + 2HCl_(aq) → BeCl_2(aq) + H_2(g)

 

 

 

• Magnesium:

– When a piece of magnesium is dropped into a beaker containing hydrochloric acid, there is effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of magnesium chloride remains in the test tube // boiling tube.

 

Equation: Mg_(s) + 2HCl_(aq) → MgCl_2(aq) + H_2(g)

 

• Calcium:

– When a piece of calcium is dropped into a beaker containing hydrochloric acid, there is effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of calcium chloride remains in the test tube // boiling tube.

 

Equation: Ca_(s) + 2HCl_(aq) → CaCl_2(aq) + H_2(g)

 

• With sulphuric (VI) acid.
  • Beryllium:

– When a piece of beryllium is dropped into a beaker containing sulphuric (VI) acid, there is effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of beryllium sulphate remains in the test tube // boiling tube.

 

Equation: Be_(s) + H₂SO_4(aq) → BeSO_4(aq) + H_2(g)

 

• Magnesium:

– When a piece of magnesium is dropped into a beaker containing sulphuric (VI) acid, there is effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of magnesium sulphate remains in the test tube // boiling tube.

 

Equation: Mg_(s) + H₂SO_4(aq) → MgSO_4(aq) + H_2(g)

 

• Calcium:

– When a piece of calcium is dropped into a beaker containing sulphuric (VI) acid, there is slight effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– The reaction however quickly stops and there is formation of a white precipitate in the test tube or boiling tube.

Reason:
– When calcium reacts with dilute sulphuric (VI) cid there is formation of calcium sulphate which is insoluble hence the formation of a white precipitate. The insoluble calcium sulphate coats the surface of the (calcium) metal stopping further reaction.

 

Equation: Ca_(s) + H₂SO_4(aq) → CaSO_4(s) + H_2(g)

 

• With nitric (V) acid.
  • Beryllium:

– When a piece of beryllium is dropped into a beaker containing nitric (V) acid, there is effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of beryllium nitrate remains in the test tube // boiling tube.

 

Equation: Be_(s) + 2HNO_3(aq) → Be(NO₃)_2(aq) + H_2(g)

 

• Magnesium:

– When a piece of magnesium is dropped into a beaker containing nitric (V) acid, there is effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of magnesium nitrate remains in the test tube // boiling tube.

 

Equation: Mg_(s) + 2HNO_3(aq) → Mg(NO₃)_2(aq) + H_2(g)

 

• Calcium:

– When a piece of calcium is dropped into a beaker containing nitric (V) acid, there is slight effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of calcium nitrate remains in the test tube // boiling tube.

 

Equation: Ca_(s) + 2HNO_3(aq) → Ca(NO₃)_2(aq) + H_2(g)

 

Trend in reactivity in alkaline earth metals.

– The reactivity of the alkaline earth metals increases down the group.

Reason:

– The atomic radius increases down the group due to increasing number of energy levels. The attraction of the two valence electrons towards the positive nucleus thus decreases down the group. Since alkaline earth metals react by losing electrons the ease with which the valence electrons are lost therefore decreases down the group

 

Similarity of ions and formulae of some compounds of Alkaline earth metals.

– Alkaline earth metals have the same valency (2) and a similar charge (2+) on their ions since they all have a single valence electron.

– Thus they have the same general formula in their ions and compounds.

 

Examples.

Alkali metal ion	Hydroxide (OH–)	Oxides (O2-)	Chloride (Cl–)	Sulphates (SO42-)
Be2+
Mg2+
Ca2+

Uses of some alkaline earth metals and their compounds

1. Magnesium is used in the manufacture of magnesium hydroxide which is used as an anti-acid medicine.

Reason: magnesium hydroxide is a non-toxic base.

2. A low-density alloy of magnesium and aluminium is used in aeroplanes construction.
3. Hydrated calcium sulphate (plaster of Paris) is used in hospitals to set fractured bones.
4. Cement is made by heating a mixture of calcium carbonate (limestone), clay and sand.
5. Calcium carbonate is used in the extraction of iron.
6. Calcium oxide (quicklime) is added tom acidic soils to raise pH for agricultural purposes.
7. Calcium nitrate is used as a nitrogenous fertilizer.
8. Magnesium oxide is used in the lining of furnaces.
9. Barium sulphate is used in the diagnosis of ulcers.
10. Barium nitrate is used to produce the green flame in fireworks.
11. Calcium carbonate is mixed with oil to make putty.

 

The Halogens.

– Are the elements with seven valence electrons and hence in group VII of the periodic table.

– All are non-metallic in nature.

– The members of the family in order down the group is as follows:

• Chlorine
• Bromine
• Iodine
• Astatine

 

Electron arrangement of the first three halogens.

Elements	Electron arrangement
Fluorine	2.7
Chlorine	2.8.7
Bromine	2.8.18.7

 

Note:

– Halogen is a derivative of two Greek words: halo and gen.

 

Diagram: Part of periodic table showing the halogens

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Gradation in properties of halogens.

1. Atomic and ionic radius.

• Atomic radius:

– Refers to the distance between the centre of the nucleus of an atom and the outermost energy level occupied by electron(s)

• Ionic radius:

– Refers to the distance between the centre of the nucleus of an ion and the outermost energy level occupied by electron(s)

 

Trend:

– The ionic radius and the atomic radius of halogens increase down the group.

Reason:

– There is an increase in the number of energy levels down the group from Fluorine to iodine.

Illustration:

– Fluorine (2.7) has only 2 energy levels; Chlorine (2.8.7) has 3 energy levels while bromine (2.8.18.7) has 4 energy levels.

– Thus the outermost electron in bromine is further from the nucleus than the outermost electron in chlorine and fluorine.

 

2. Atomic and ionic radius of the same element.

– For the same halogen the ionic radius is larger than the atomic radius.

Reason:

– Halogens form ions by gaining (an) electron into the outermost energy level. This increases the electron-electron repulsion in the outermost energy level hence decreasing the nuclear attraction of the outermost electrons towards the nucleus.

 

Illustration

– Chlorine atom has electron arrangement of 2.8.7 hence 3 energy levels.

– During ion formation it reacts by gaining an electron into the third energy level to acquire a new electron arrangement of 2.8.8.

– In the atom 17 protons are attracting 17 electrons; while in the chloride ion there are 17 protons attracting 18 electrons in the outermost energy level.

– Thus in the ion the effect of the positive nucleus is lower.

– This is due to increased repulsive effect between the existing electrons in the outermost energy level and the incoming electron (electron-electro repulsion)

 

Diagrammatically: chlorine atom and chloride ion

 

 

 

 

 

 

 

 

 

 

 

Summary: changes in atomic and ionic radius among alkali metals.

Element	Symbol	Atomic number	Electron arrangement	Atomic radius (nm)	Ionic radius (nm)
Fluorine	F	9	2.7	0.064	0.136
Chlorine	Cl	17	2.8.7	0.099	0.181
Bromine	Br	35	2.8.18.7	0.114	0.195
Iodine	I	53	2.8.18.18.7	0.133	0.216

 

 

Laboratory preparation of chlorine.

Note: It is usually prepared by oxidation of concentrated hydrochloric acid by removal of hydrogen.

 

Equation:

2HCl_(aq) + [O]                Cl_2(g) + H₂O_(l)

– The [O] is from a substance containing oxygen (an oxidizing agent).

– The main oxidizing agents normally used for preparation of chlorine are:

• Potassium manganate (VII); KMnO4.
• Manganese (IV) oxide; MnO2

 

(a). Preparation of chlorine from MnO₂ and HCl.

(i). Apparatus:

 

 

 

 

 

 

 

 

 

 

 

(ii). Conditions:

– Heating;

– Presence of an oxidizing agent; in this case it is manganese (IV) oxide.

 

(iii). Procedure:

– Hydrochloric acid is reacted with manganese (IV) oxide (dropwise);

Equation:

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.

 

Physical properties of Halogens.

1. Physical state and appearance at room temperature.

 

Halogen	State and appearance
Fluorine	Pale yellow gas
Chlorine	Green-yellow gas
Bromine	Volatile brown liquid
Iodine	Shiny dark grey solid.

 

2. Solubility

(a). In water

Experiment: To investigate solubility of halogens in water.

• Procedure:

– A boiling tube containing chlorine gas is inverted into a beaker containing water.

– The experiment is repeated with fluorine, bromine and a few crystals of iodine.

 

• Diagram:

 

 

 

 

 

 

 

 

• Observations:
• Fluorine:
  – The level of solution rises in the boiling tube.

– The pale yellow colour of fluorine disappears.

 

• Fluorine:
  – The level of solution rises in the boiling tube.

– The green-yellow colour of fluorine disappears.

 

• Bromine
  – The level of solution rises in the boiling tube.

– The brown colour of fluorine disappears.

 

• Iodine:
  – The level of solution remains the same in the boiling tube.

– The shiny dark grey crystals remain in the beaker.

• The rise in water level is higher in fluorine than in chlorine while the rise in chlorine is higher than in bromine.

 

Diagrams: observations after some time

 

 

 

 

 

 

 

 

 

 

• Explanations:

– Fluorine, chlorine and bromine are all soluble in water, while iodine is insoluble in water.

– When a boiling tune containing the soluble halogens is inverted into a beaker containing water, the halogen dissolves in the water.

– This creates a partial vacuum and the water in the beaker thus rises to occupy the space left by the dissolved gas.

– The halogens dissolve in water to form acidic solutions.

– The more soluble the halogen, the higher the rise in water level in the boiling tube.

– Thus the solubility of halogens decreases down the group from fluorine to bromine.

Reason:

– All halogens have molecular structures with wreak van der waals forces between the molecules.

During the dissolution the Van der Waals must be broken. The strength of Van der Waals increase as the atomic size and hence the molecular size increases which occurs down the group.

 

(b). In tetrachloromethane.

– The same procedure (of dissolving halogens in water) is followed with tetrachloromethane.

 

• Observations:

– All halogens are soluble in tetrachloromethane.

– The solubility of each halogen is higher in tetrachloromethane than in water.

Reason: Halogens are molecular thus non-polar and thus are more soluble in polar organic solvents like tetrachloromethane than in polar solvents like water.

 

3. Effect of heat.

– Fluorine and chlorine are gases at room temperature and simply expand and increase in volume when heated.

– Bromine is a brown liquid at room temperature and evolves brown fumes when heated.

– Iodine exists as shiny dark grey solid at room temperature and sublimes to give brown fumes when heated.

 

4. Electrical conductivity.

– All halogens are on conductors of heat and electricity.

Reason:

– Halogens are molecular and thus have neither delocalized electrons nor free mobile ions for electrical conductivity.

5. Trend in melting and boiling points.

– The melting and boiling points of halogens increase down the group.

Reason:

– Halogens exist as diatomic molecules and thus have molecular structures;

– The molecules are held together by intermolecular forces called the Van der Waals which have to be broken during melting and boiling;

– The strength of the Van der Waals increases as the size of the atoms and hence molecules increases which occurs down the group.

 

Summary on some physical properties of halogens>

Halogen	Formula	Atomic number	Appearance	Melting point (oC)	Boiling point (oC)
Fluorine	F	9	Pale yellow gas	-238	-188
Chlorine	Cl	17	Green yellow gas	-101	-35
Bromine	Br	35	Brown liquid	-7	59
Iodine	I	53	Shiny dark grey solid	114	184

 

Chemical properties of Halogens.

Note: It is not easy for non-metals like halogens to form cations.

Reason: the ionization energy (amount of energy required to lose an electron(s) from the outermost energy level of a gaseous atom) is very large.

– Thus most non-metals react by forming anions (negatively charged ions) by electron loss.

 

1. Ion formation.

– Halogens react by gaining a single electron into the outermost energy level to form a stable electron configuration and corresponding anions.

– During ion formation by electron loss energy is released, and the energy change for this process is called electron affinity

 

Note: Definition.

– Electron affinity is thus the energy released when an atom in gaseous state gains an electron to form a negative ion (anion)

 

 

 

 

• Trend in electron affinity of halogens:

– Generally the electron affinity of halogens decreases down the group.

Reason:

– The size of the atoms increases (due to increasing number of energy levels) down the group and thus the force of attraction of the electrons in the outermost energy level towards the nucleus decreases.

– Thus down the group the ease with which electrons are gained decreases and the faster the ease of electron gain, the more the energy released hence the more the electron affinity.

 

Summary:

Element	Ionization equation	Electron affinity (kJ per mole
Fluorine	F(g) + e → F–(g)	-322
Chlorine	Cl(g) + e → Cl–(g)	-349
Bromine	Br(g) + e → Br–(g)	-325
Iodine	I(g) + e → I–(g)	-295

2. Reaction with metals.

(a). Chlorine.

• Reaction of chlorine with iron.

(i). Apparatus.

 

Note:

– The set up can also be modified by using sodium hydroxide to absorb excess chlorine gas as shown below.

 

 

 

 

 

 

 

 

 

(ii). Precaution.

– Experiment should be done in a functional fume cupboard or in the open.

Reason:

– Chlorine gas is poisonous and will thus be harmful to the human body.

 

(iii). Procedure:

– A stream of dry chlorine gas is passed over heated iron wool as per the diagram.

 

(iv). Conditions.

• Chlorine gas has to be dry (done by the anhydrous calcium chloride in the U-tube)

Reason:

To prevent hydration hence oxidation of iron (which will then form Fe₂O₃.5H₂O) hence preventing reaction between iron and chlorine.

 

• Iron metal must be hot; and this is done by heating.

Reason:

To provide activation energy i.e. the minimum kinetic energy which the reactants must have to form products.

• Anhydrous calcium chloride.

– In the U-tube; to dry the chlorine gas.

– In the thistle funnel; to prevent atmospheric water vapour (moisture) from getting into the apparatus and hence reacting with iron (III) chloride.

 

Note: In the guard tube, calcium oxide is preferable to anhydrous calcium chloride.

Reason:

– Other than preventing atmospheric water vapour from getting into the flask with iron wool; it also absorbs excess chlorine thus preventing environmental pollution

 

(v). Observations:

– Iron metal glows red-hot.

– Red brown fumes (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)

 

 

• Reaction with aluminium.

– Chlorine reacts with aluminium I a similar manner to iron to from aluminium chloride.

Equation:

2Al_(s) + 3Cl_2(g) → 2AlCl_3(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)

 

• 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.

 

(b). Bromine.

• Reaction of bromine with iron

Apparatus

 

 

 

 

 

 

 

 

 

 

(ii). Precaution.

– Experiment should be done in a functional fume cupboard or in the open.

Reason:

– Bromine gas is poisonous and will thus be harmful to the human body.

 

(iii). Procedure:

– Bromine liquid is heated to generate bromine vapour (fumes).

– The iron wool is then heated and a stream of the bromine gas is passed over heated iron wool as per the diagram.

 

(iv). Conditions.

• Bromine must be heated to generate fumes before heating the iron.

Reason:

So that bromine vapour will drive out air from the boiling tube to prevent oxidation of iron with oxygen which would otherwise prevent reaction between iron and bromine.

 

• Iron metal must be hot; and this is done by heating.

Reason:

To provide activation energy i.e. the minimum kinetic energy which the reactants must have to form products.

 

(v). Observations:

– Brown fumes of bromine are produced on heating bromine liquid.

– The iron wool glows red-hot upon heating

– The iron wool forms dark red crystals (of iron (III) bromide)

 

(vi). Reaction equation.

Word: Iron + bromine → Iron (III) bromide.

Chemical: 2Fe_(s) + 3Br_2(g) → 2FeBr_3(s)

Dark red crystals.

 

• Reaction of bromine with zinc

– Using the same set up bromine also reacts with zinc to form zinc bromide.

Equation: Zn_(s) + Br_2(g) → ZnBr_2(s).

 

• Reaction of bromine with magnesium

– Using the same set up bromine also reacts with zinc to form magnesium bromide.

Equation: Mg_(s) + Br_2(g) → MgBr_2(s).

 

• Reaction of bromine with sodium

– Using the same set up bromine also reacts with zinc to form sodium bromide.

Equation: 2Na_(s) + Br_2(g) → 2NaBr_(s).

 

(c). Iodine.

• Reaction of iodine with iron

Apparatus

 

 

 

 

 

 

 

 

 

(ii). Precaution.

– Experiment should be done in a functional fume cupboard or in the open.

Reason:

– Iodine is poisonous and will thus be harmful to the human body.

 

(iii). Procedure:

– Iodine crystals are heated to generate iodine vapour (fumes).

– The iron wool is then heated and a stream of the iodine gas is passed over heated iron wool as per the diagram.

 

(iv). Conditions.

• Iodine must be heated to generate fumes before heating the iron.

Reason:

So that iodine vapour will drive out air from the boiling tube to prevent oxidation of iron with oxygen which would otherwise prevent reaction between iron and iodine.

 

• Iron metal must be hot; and this is done by heating.

Reason:

To provide activation energy i.e. the minimum kinetic energy which the reactants must have to form products.

(v). Observations:

– Purple vapour (fumes) of iodine is produced on heating iodine crystals.

– The iron wool glows red-hot upon heating

– The iron wool forms grey black crystals (of iron (II) iodide)

 

(vi). Reaction equation.

Word: Iron + iodine → Iron (II) iodide.

Chemical: Fe_(s) + I_2(g) → FeI_2(s)

Grey-black crystals.

 

• Reaction of Iodine with zinc

– Using the same set up bromine also reacts with zinc to form zinc iodide.

Equation: Zn_(s) + I_2(g) → ZnI_2(s).

 

• Reaction of Iodine with magnesium

– Using the same set up bromine also reacts with zinc to form magnesium iodide.

Equation: Mg_(s) + I_2(g) → MgI_2(s).

 

• Reaction of Iodine with sodium

– Using the same set up bromine also reacts with zinc to form sodium iodide.

Equation: 2Na_(s) + I_2(g) → 2NaI_(s).

 

Note:

– The reactivity of chlorine with metals is more vigorous than that of bromine, which is more than that of iodine.

– Thus reactivity of halogens with metals decreases down the group.

Reason:

– Halogens react by gaining electrons; the ease of gaining electrons decrease down the group as the atomic size increases which leads to progressive decrease in the force of attraction of electrons in the outermost energy levels by the positively charged nucleus.

 

3. Reaction with water.

– Halogens that dissolve in water form a mixture of two acids.

 

Reaction of chlorine with water.

– Chlorine dissolves in water to form chlorine water, which is a mixture of two acids: chloric (I) acid and hydrochloric acid.

 

Equation:
Cl_2(g) + H₂O_(l) → HOCl_(aq) + HCl_(aq)

Chloric (I) acid   Hydrochloric acid

 

 

• Effects of sunlight on chlorine water.

(i). Procedure:

– Chlorine water is made by dissolving the gas in water.

– A long tube filled with chlorine water is inverted over a beaker containing water.

– It is then exposed to sunlight (bright light) as shown below.

 

(ii). Apparatus:

 

 

 

 

 

 

 

 

 

(iii). Observations:

– After sometime a gas collects in the tube and on applying a glowing splint, the splint is rekindles showing that the gas collected is oxygen.

 

(iv). Explanation:

– Chlorine water has two components.

Equation:

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)

 

• Effect of chlorine water on litmus papers

(i). Procedure:

– A strip of blue ad a strip of red litmus papers are dropped into chlorine water in a beaker.

 

(ii). Observations:
– The blue litmus paper turns red; then both litmus papers are decolourised.

 

(iii). Explanations.

– Chlorine water contains a mixture of chloric (I) acid ad hydrochloric acid.

Equation:

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

 

– The two acids cause the litmus paper to turn red (from blue) while the red litmus paper remained red.

– The litmus papers are then decolourised due to bleaching action of chlorine water, which is due to the activity of chloric (I) acid.

– The chloric (I) acid is unstable and thus decomposes to give hydrochloric acid and oxygen atom.

Equation:

HOCl(aq) → HCl_(aq) + [O]

 

– The oxygen atom combines with the chemical of the natural dye in the litmus thus decolourising it.

Equation:

Dye + HOCl_(aq) → HCl_(aq) + (Dye + [O])

Coloured                                                              Decolourised

Note: solutions of bromine and fluorine in water will behave in a similar manner.

 

Examples:

• Fluorine dissolving in water

F_2(g) + H₂O_(l)    ═ HF_(aq) + HOF_(aq)

                                Hydrofluoric acid     Fluoric (I) acid

– The mixture is flourine water

 

• Bromine dissolving in water

Br_2(g) + H₂O_(l)  ═ HBr_(aq) + HOBr_(aq)

                                Hydrobromic acid     Bromic (I) acid

– The mixture is bromine water.

 

Some uses of halogens and their compounds.

1. Fluorine is a raw material in the preparation of a synthetic fibre known as polytetrafluoroethane.
2. Some compounds of fluorine are added to water and some tooth pastes in small quantities to reduce tooth decay.
3. Fluorine is used in the manufacture of hydrogen fluoride used to engrave words or pictures on glass.
4. Chlorine is used to make bleaches used in paper and textile industries.
5. Chlorine is added to water to kill micro-organisms in water treatment works.
6. Chlorine is used in the manufacture of a plastic known as polyvinylchloride (PVC).
7. Chlorine is used in large scale manufacture of hydrochloric acid.
8. Bromine is used in the manufacture of silver bromide which is used to make the light sensitive photographic paper and films.
9. A solution of iodine in alcohol (tincture of iodine) is used as a disinfectant.

 

The Noble gases.

– Are the elements with the maximum possible number of valence electrons and hence in group VIII of the periodic table.

– All are gaseous in nature.

– The members of the family in order down the group is as follows:

• Helium
• Neon
• Argon
• Krypton
• Xenon
• radon

– They are found as free atoms in nature and form about 1% of air with argon being the most abundant

 

Electron arrangement of the first three noble gases.

Elements	Electron arrangement
Helium	2.
Neon	2.8.
Argon	2.8.8.

 

Note: – Helium with only two electrons has one occupied energy level; which is completely full and hence is said to have a duplet electron configuration

– The rest of the noble gases have eight electrons in their outermost occupied energy level hence are said to have the octet electron configuration.

Diagram: Part of periodic table showing the noble gases

 

 

 

 

 

 

 

 

 

 

 

Dot and cross diagrams for the first three Alkaline Earth metals

 

 

 

 

 

 

 

 

 

 

Physical properties of noble gases

1. Atomic and ionic radius.

• Atomic radius:

– Refers to the distance between the centre of the nucleus of an atom and the outermost energy level occupied by electron(s)

• Ionic radius:

– Refers to the distance between the centre of the nucleus of an ion and the outermost energy level occupied by electron(s)

 

Trend:

– The atomic radius of noble gases increase down the group.

Reason:

– There is an increase in the number of energy levels down the group from Helium to radon.

 

Illustration:

– Helium (2.) has only 1 energy level; Neon (2.8) has 2 energy levels while Argon (2.8.8) has 3 energy levels.

– Thus the outermost electron in Argon is further from the nucleus than the outermost electron in neon and helium.

 

Note:

– Under normal circumstance the noble gases neither lose nor gain electrons and are thus stable hence unreactive.

– Due to this they rarely form ions and hence comparative studies of ionic radii among noble gases are not feasible.

 

2. Ionization energies.

– Noble gases have very high ionization energies.

Reason:

– Noble gases are colourless monoatomic gases with very stable electron arrangements, either dupltet (helium) or octet (the rest).

– The nuclear attraction force of electrons in the outermost energy level towards the nucleus is thus very strong.

Trend:

– The ionization energy decrease down the noble gases group.

Reason:

– As the size of the atoms increase down the group (due to increase in the number of energy levels), the force with which the electros in the outermost energy levels are attracted towards the nucleus decrease hence increase in the ease of electron loss (from the outermost energy level).

 

3. Melting and boiling points.

– Generally noble gases have very low melting and boiling points.

Reason:

– They exist as monoatomic gases with simple molecular structures with weak van der waals forces that are easy to break.

Trend:

– Melting and boiling points among the noble gases increase down the group.

Reason:

– Noble gases exist as monoatomic molecules and thus have simple molecular structures;

– The molecules (atoms) are held together by intermolecular (inter-atomic) forces called the Van der Waals which have to be broken during melting and boiling;

– The strength of the Van der Waals increases as the size of the atoms and hence molecules increases which occurs down the group.

 

Summary: changes in atomic and ionic radius among alkali metals.

Element	Symbol	Atomic number	Atomic radius (nm)	1st Ionization energy (kJmol-1)	Melting point (oC)	Boiling point (oC)
Helium	He	2	0.128	2372	-270	-269
Neon	Ne	10	0.160	2080	-249	-246
Argon	Ar	18	0.192	1520	-189	-186
Krypton	Kr	36	0.197	1350	-157	-152
Xenon	Xe	54	0.217	1170	-112	-108

 

Chemical properties of Noble gases.

– Generally the noble gases neither lose nor gain electrons and are thus stable hence unreactive.

– Due to this they rarely form ions and hence have no feasible chemical reactions.

– However xenon and radon with very large atomic radii and smaller ionization energies take part in some reactions and thus display come chemical properties.

 

Uses of some noble gases.

Note: The application of noble gases is ironically centered on their inert nature.

1. Argon is used in light bulbs to provide an inert environment to prevent oxidation.
2. Argon is used as an insulator in arch-welding.
3. Neon gas is used in streets and advertisement lights.
4. Helium mixed with oxygen is used in deep sea diving and mountaineering.
5. Helium mixed with oxygen also is used in hospitals for breathing in patients with respiratory problems and those undergoing certain forms of surgery.
6. Helium can be used instead of hydrogen in balloons for meteorological research.

Reason:

– Hydrogen is explosive I presence of air and may explode causing serious accidents.

– Helium is less dense than hydrogen.

7. Helium is used in thermometers for measurements f very low temperatures.
8. Liquid helium is used to keep certain metal alloys at temperatures low enough for them to become superconductors.

 

 

UNIT 3: PROPERTIES AND TRENDS ACROSS A PERIOD.

Checklist.

1. Introduction
2. Trends in physical properties of elements in period 3

• Electrical conductivity
• Atomic and ionic radii
• Melting and boiling points

3. Trends in chemical properties of elements in period 3

• Reaction with oxygen
• Reaction with water
• Reaction with acids

 

Introduction:

– A period is a vertical row of elements in the periodic table.

– Elements in the same period have same number of energy levels.

– There are 7 periods in the periodic table except for lanthanides and actinides which are not assigned periods.

– Periods 1 – 3 have fewer elements because they lack the d-block elements and have only the s-block elements and the p-block elements.

Note:

• s-block elements: group 1 and 2
• d-block elements: transitional elements
• p-block elements: groups III to VIII.

 

– To understand trends and properties across a typical period of the periodic table, we shall use period 3 as the reference.

 

The Period three of the periodic table.

– Consists of elements with three energy levels.

– Consists of the 8 elements from sodium to argon.

– It is only made of s-block and p-block elements ad lacks any element in the d-block group of elements.

 

Part of periodic table showing period 3 of the periodic table.

 

 

 

 

 

 

 

 

 

 

 

Trends in physical properties of elements in period 3.

1. Electrical conductivity.

– Sodium, magnesium and aluminium are good conductors of electricity.

Reason:

They all have giant metallic structures with delocalized electros which conduct electricity;

– The electrical conductivity increases from sodium to aluminium.

Reason:

Electrical conductivity increases with increase in the number of delocalized electrons; thus aluminium with the highest number of delocalized electrons (3) in each atom will have the highest electrical conductivity;

– In the metals the electrical conductivity decreases with increase in temperature.

Reason:

– Increase in temperature distorts the alignment of electrons thus preventing their easy flow and hence poor conductivity;

– Phosphorus, sulphur, chlorine and argon do not conduct electric current.

Reason:
They all have molecular structures and all the electrons in the atoms are used in bonding; thus they lack delocalized electron or free ions for electrical conductivity.

– Silicon conducts electric current, and its electrical conductivity increases with increase in temperatures.

Reason:

– It is a semi-conductor; making it a very unique element in this period.

Note:

– A semi-conductor is a crystalline material which only conducts electricity under certain conditions.

 

2. The atomic and ionic radii.

– The atomic radii of period 3 elements decrease across the period.

Reason.

– For the same number of energy levels the number of protons in the nucleus increases across the period; this leads to the increase in nuclear charge while the shielding effect remains the same hence decrease in atomic radius across the period.

 

3. Melting and boiling points

– Sodium, magnesium and aluminium have very high melting and boiling points.

• Reason:

– They have giant metallic structures with strong metallic bonds which need a lot of energy to break.

– The boiling and melting points increase from sodium to aluminium.

• Reason:

– As you move across the period from sodium to aluminium, the nuclear charge increases while the energy levels remain the same hence decrease in atomic radius; the smaller the atomic radius (for metals) the stronger the metallic bonds.

– Silicon, though a non-metal also has a very high melting and boiling points.

• Reason:

– Silicon has a giant atomic structure with strong covalent bonds throughout the structure, which need a lot of heat energy to break.

– Phosphorus, Sulphur, Chlorine and argon have low melting and boiling points.

• Reason:

– They all have molecular structures with strong covalent bonds between the atoms (except in argon) but weak van der waals forces between the molecules which are easy to break.

– Note that argon exist as atoms and thus a monoatomic molecule.

– The melting and boiling points decreases from phosphorus to argon.

– As we move across the period from phosphorus to argon, the size of the atoms decreases leading to smaller atoms and hence molecules, which lead to decrease in the strength of the van der Waals (across the period)

Note:

– Phosphorus and sulphur exists as solids at room temperature while chlorine and argon exists as gases at room temperature.

Reason: phosphorus and sulphur have giant molecular structures while chlorine and argon have simple molecular structures.

–
Summary: Some physical properties f elements in period 3.

 

Property	Na	Mg	Al	Si	P (white)	S (monoclinic)	Cl	Ar
Physical state and appearance	Silver	Silver solid	Silver solid	Black solid	White solid	Yellow solid	Green yellow gas	Colourless gas
Electron arrangement	2.8.1	2.8.2	2.8..3	2.8.4	2.8.5	2.8.6	2.8.7	2.8.8
Valency	1	2	3	4	5 or 5	6	7	8
Atomic radius	0.157	0.136	0.125	0.117	0.110	0.104	0.09	0.192
MP (oC)	98	650	660	1410	44	119	-101	-189
BP (oC)	890	1110	2470	2360	280	445	-35	-186
Structure	Giant metallic	Giant metallic	Giant metallic	Giant atomic	Molecular	Molecular	Simple molecular	Simple molecular /atomic
Bond type	metallic	metallic	metallic	Covalent	Covalent	Covalent	Covalent	Van der waals

 

Trends in chemical properties of the elements in period 3.

Note: Trends in reactivity.

– The reactivity among the metallic elements decreases across the period from sodium to aluminium.

Reason:

– There is a continuous increase in nuclear charge from sodium to aluminium which leads to increase in ionization energies hence increasing difficulty in removing an electron from the outermost energy level.

– Among the non-metallic elements, the reactivity increases across the period from phosphorus to chlorine.

Reason:

– There is increase in nuclear charge from phosphorus to chlorine, hence an increase in ease of electron gain (electro affinity) since non-metals react by gaining electrons.

– Argo is unreactive and can only react under very special conditions.

Reason:

– It’s a noble gas with a stable octet configuration.

 

1. Reaction of period 3 elements with oxygen.

– All period three elements react with (burn in) oxygen with the exception of argon.

 

Experiment: To investigate the reactions between period 3 elements and oxygen

Procedure:

– A small piece of the element is placed in a deflagrating spoon and warmed gently until it catches fire.

– It is then lowered into a gas jar full of oxygen.

– The flame colour and the colour of the product are noted.

– 10 cm³ of water containing universal or litmus indicator is added into the gas jar with the products.

 

• Sodium:

– Burns vigorously in oxygen with a golden yellow flame; to produce white solid of sodium oxide.

Equation:

Na_(s) + O_2(s) → Na₂O_(s)

– The resultant sodium oxide dissolves in water to form sodium hydroxide.

Equation:

Na₂O_(s) + H₂O_(l) → 2NaOH_(aq);

– The sodium hydroxide is alkaline in nature and thus turns litmus indicator blue;

 

• Magnesium:

– Burns vigorously in oxygen with a bright white light; to produce white solid of magnesium oxide.

Equation:

2Mg_(s) + O_2(s) → 2MgO_(s)

– The resultant magnesium oxide is slightly soluble in water to form magnesium hydroxide.

Equation:

MgO_(s) + H₂O_(l) → Mg(OH)_2(aq);

– The magnesium hydroxide is alkaline in nature and thus turns litmus indicator blue;

 

• Aluminium:

Note:

– Aluminium (foil) is usually coated with a thin layer of aluminium oxide, Al₂O₃; which prevents reaction with the oxygen.

– When polished, it reacts slowly with oxygen to form a white solid of aluminium oxide.

Equation:

Al_(s) + 3O_2(s) → Al₂O_3(s)

– The resultant aluminium (III) oxide is insoluble in water.

 

• Silicon:

– Silicon powder can only bur in oxygen at very high temperatures of about 450^oC to form solid silicon (IV) oxide.

Equation:

Si_(s) + O_2(s) → SiO_2(s)

– The resultant silicon (IV) oxide is insoluble in water.

 

– Burns readily in oxygen with a bright orange flame; to produce dense white fumes (solid) of phosphorus (V) oxide

Equation:

P_4(s) + 5O_2(s) → 2P₂O_5(s)

Note: sulphur exists and therefore reacts as molecules of P₄.

– The resultant phosphorus (V) oxide readily dissolves in water to form phosphoric (V) acid.

Equation:

P₂O_5(s) + 3H₂O_(l) → 2H₃PO_4(aq);

– The phosphoric acid is acidic in nature and thus turns litmus indicator pink / red;

 

– Burns in oxygen with a blue flame; to form a colourless gas of sulphur (IV) oxide

Equation:

S_(s) + O_2(s) → SO_2(s)

– The resultant sulphur (IV) oxide readily dissolves in water to form sulphuric (IV) acid.

Equation:

SO_2(s) + H₂O_(l) → H₂SO_3(aq);

– The phosphoric acid is acidic in nature and thus turns litmus indicator pink / red;

 

– The sulphuric (IV) acid is unstable and thus easily gets oxidized by oxygen to the stable sulphuric (VI) acid.

Equation:

H₂SO_3(aq) + O_2(g) → H₂SO_4(aq);

 

– Burns in oxygen only under certain conditions to form acidic oxides.

Equation:

2Cl_2(s) + O_2(s) → 2Cl₂O_(s)

 

– Argon is unreactive.

 

Conclusion:

– Metallic elements burn in oxygen to form basic oxides.

– Non-metallic oxides burn in oxygen to form acidic oxides.

 

2. Reaction of period 3 elements with water.

Procedure:
– A small piece of sodium metal is cut and dropped into a trough containing water;

– The resultant solution is tested with litmus paper;

 

Diagram of apparatus:

 

 

 

 

 

 

 

 

Observations and explanations:

– The metal floats on the water surface; because it is less dense than water;

– A hissing sound is produced; due to production of hydrogen gas;

– It vigorously melts into a silvery ball then disappears because reaction between water and sodium is exothermic (produces heat). The resultant heat melts the sodium due to its low melting point.

– It darts on the surface; due to propulsion by hydrogen;

– The metal may burst into a golden yellow flame; because hydrogen may explode into a flame which then burns the sodium;

– The resultant solution turns blue; because sodium hydroxide solution formed is a strong base;

 

Reaction equations.

Equation I

2Na_(s) + 2H₂O_(l) → 2NaOH_(aq) + 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 magnesium ribbon is cut and dropped into a trough containing water;

– The resultant solution is tested with litmus paper;

 

Diagram of apparatus:

 

 

 

 

 

 

 

 

Observations and explanations:

– The metal sinks into the water surface; because it is denser than water;

– It reacts slowly with water leading to slow evolution of hydrogen gas.

– The resultant solution turns blue; because sodium hydroxide solution formed is a strong base;

 

Reaction equation.

Mg_(s) + 2H₂O_(l) → Mg(OH)_2(aq) + H_2(g);

 

Effect of resultant solution on litmus paper;

– Litmus paper turns blue; magnesium hydroxide formed is highly soluble in water; releasing a large number of hydroxyl ions which result into alkaline conditions // high pH;

 

– Aluminium does not normally react with cold water or steam.

Reason:

– Aluminium is usually coated with a thin coating of aluminium oxide which prevents further reaction.

– However at very high temperatures of about 700^oCsteam can react with aluminium to form aluminium oxide and hydrogen gas.

Equation:

2Al_(s) + 3H₂O_(g) → Al₂O_3(s) + 3H_2(g);

 

Note:

– Due to its inability t react with water and air aluminium is preferable for making cooking utensils like sufurias and coking pans.

 

 

 

 

• Silicon, phosphorus and sulphur.

– These non-metals do not displace hydrogen from water and thus do not react with water.

 

– Dissolves in water t form chlorine water, which is a mixture of chloric (I) acid ad hydrochloric acid.

 

Equation:
Cl_2(g) + H₂O_(l) → HOCl_(aq) + HCl_(aq)

Chloric (I) acid   Hydrochloric acid

 

3. Reaction of period 3 elements with acids

Procedure:

– A piece of the element is dropped into 5 cm³ of an acid in a test tube.

– Any gas produced is tested.

 

• Sodium:

– Reacts explosively with acids to form salts and hydrogen and thus reactions of sodium with acids should not be tried in the laboratory.

 

• Magnesium:

– Reacts with both dilute hydrochloric and dilute sulphuric acid to form magnbsium salts and hydrogen gas.

 

• With hydrochloric acid:

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

 

• With sulphuric acid:

Mg_(s) + H₂SO_4(aq) → MgSO_4(aq) + H_2(g)

 

• Aluminium:

Note:

– It does not readily react with dilute acids.

Reason:

– This is due to presence of a thin aluminium oxide coating that prevents contact hence reaction with the acids.

– When polished t remove the oxide coating it reacts with both dilute hydrochloric and dilute sulphuric acid to form aluminium salts and hydrogen gas.

 

• With hydrochloric acid:

2Al_(s) + 6HCl_(aq) → 3AlCl_3(aq) + 3H_2(g)

 

• With sulphuric acid:

2Al_(s) + 3H₂SO_4(aq) → Al₂(SO₄)_2(aq) + 3H_2(g)

 

• Silicon, phosphorus, sulphur and chlorine.

– They do not react with dilute acids.

 

 

 

Summary: Chemical properties of period 3 elements.

Element	Na	Mg	Al	Si	P	S	Cl
Reaction with air or oxygen	Readily reacts with air. – Burns brightly in oxygen with a golden yellow flame to form Na2O	Reacts slowly with air. – Burns in oxygen with a bright white flame to form MgO	– Forms a protective coating of Al2O3 when it burns in oxygen.	Si powder burns at temperatures above 950oC to form SiO2	White phosphorus smolders in air; – P burns in air with a bright orange flame to form P2O3 and P2O5	Burns in air or oxygen with a blue flame to form SO2 gas	N reaction with air or oxygen under normal conditions.
Reaction with water	Reacts vigorously to produce H2 and NaOH	Slow reaction with cold water to form Mg(OH)2 and H2; – Reacts faster with steam to form MgO and H2	No reaction	No reaction	No reaction	No reaction	Dissolves in water to form chlorine water
Reaction with dilute acids	Violent reaction giving out H2 and a sodium salt	Rapid evolution of H2 gas and a Mg salt is formed	Reacts slowly to give H2 and an Al salt	No reaction	No reaction	No reaction	No reaction

 

 

UNIT4: STRUCTURE AND BONDING.

1. Meaning of structure and bond.
2. Nature of the chemical bond.
3. Types of bonds.

• Ionic bonds
  • Meaning
  • Formation
  • Dot ad cross diagrams.
  • Giant Ionic structures
    • Examples
    • Meaning
    • Properties of giant ionic structures
  • Covalent and dative bonds
    • Meaning
    • Formation
    • Dot ad cross diagrams.
    • Giant covalent (atomic) structures
      • Examples
      • Meaning
      • Properties of giant covalent structures
      • Details of Diamond, graphite and silicon (IV) oxide
    • Molecular (simple and giant) structures
      • Examples
      • Meaning
      • Hydrogen bond and Van der waals
      • Properties of molecular structures
    • Metallic bonds
      • Meaning
      • Formation
      • Dot ad cross diagrams.
      • Giant Metallic structures
        • Examples
        • Meaning
        • Properties of giant metallic structures

3. Bond types across period 3

• Oxides of period 3 elements
  • Nature
  • Melting and boiling points
  • General trends in bond types
• Chlorides of period 3 elements
  • Nature
  • Melting and boiling points
  • General trends in bond types

4. Summary on bond characteristics

 

 

 

 

 

1. Meaning of structure and bond.

• Bond:

– The mutual force of attraction that holds particles together when atoms (similar or different) combine during chemical reactions.

• Structure:

– A regular pattern of particles in a substance held together by chemical bonds.

 

2. Nature of the chemical bond.

– Atoms are made up of energy levels and nucleus.

– The energy levels contain electrons while the nucleus contains protons and neutrons.

– The electrons are negatively charged, protons are positively charged while neutrons are electrically neutral.

– The noble gases are chemically inert since their outermost energy levels are completely filled with the maximum possible number of electrons.

– Thus noble gases have either a stable duplet state (2) like in helium or a table octet configuration (2.8 or 2.8.8) as in neon and argon respectively.

– Other atoms are unstable because the outermost energy levels are not yet completely filled with the maximum possible number of electrons.

– To attain the stable duplet or octet noble gas configuration, such atoms lose, gain or share their valance electrons.

– It is the act of losing, gaining or sharing valence electrons that lead to the chemical bonds.

– When atoms gain or lose valence electron(s) they become charged forming anions and cations respectively.

– Particles of the same charge repel each other while particles of different charges attract one another.

 

3. Types of bonds.

– There are three main types of chemical bonds:

• Ionic bonds
• Covalent bond
• Metallic bond

 

(a). Ionic bonds / electrovalent bond.

• Meaning

– Is a bond formed due to complete transfer of electrons from one atom to another resulting into two oppositely charged ions.

 

Formation of an ionic bond.

– Formed due to complete transfer of electrons from one atom to another; and mainly formed between a metal and a non-metal.

– This occurs in a bid for both atoms to acquire a stable noble gas configuration.

– One atom loses all its valence electrons thus forming a cation (positively charged ions).

– The other atom gains all the lost valence electrons forming an anion (negatively charged ion).

– The cation and the anion are oppositely charged and thus develop a mutual force of attraction between them which is the ionic / electrovalent bond.

 

 

 

 

 

Illustration: formation of ionic bond between sodium and chlorine to form sodium chloride.

– Sodium metal (atomic number 11) has electronic configuration 2.8.1 and thus unstable with 1 valence electron.

– Chlorine gas (atomic number 17) has electronic configuration 2.8.7 and thus unstable with 7 valence electrons.

– Sodium is more electropositive (the tendency to lose electrons to form cations) than chlorine while chlorine is more electronegative (the tendency to gain electrons to form anions).

– Sodium loses its single valance electron to form sodium ion with electronic configuration 2.8 and a net charge of +1 (Na⁺)

– Chlorine atom accepts the single electron lost by sodium to form a chloride ion with electronic configuration 2.8.8 with a net charge of -1 (Cl^–).

– The positively charged sodium ion and the negatively charged chloride ion attract each other.

– The electrostatic forces of attraction develop between the two oppositely charged ions and this constitutes the ionic bond.

– Compounds formed due to ionic bonding are thus called ionic compounds.

 

Diagrammatically:

 

 

 

 

 

 

 

 

 

Examples of ionic compounds.

• Sodium chloride.
• Potassium fluoride.
• Magnesium oxide
• Aluminium (III) oxide.

 

Dot and cross diagrams for ionic compounds

1. Potassium fluoride.

 

2. Magnesium oxide

 

3. Sodium oxide.

 

4. Magnesium chloride.

 

5. Aluminium (III) oxide.

 

 

 

 

 

 

 

Giant Ionic structures

– Ionic bonding results into one type of structure, the giant ionic structure.

– This is a type of structure in which all ions are bonded with strong ionic bonds throughout the structure.

– Each ion in the giant ionic structure is surrounded by several others resulting into giant pattern of several ions, hence giant ionic structure.

– Most ionic substances with the giant ionic structure are crystalline in nature, made up crystals.

• Note:

A crystal is a solid form of a substance in which the particles are arranged in a definite pattern repeated regularly in 3 dimensions.

 

Illustration of the giant ionic structure: sodium chloride structure.

– The NaCl structure consists of many Na⁺ and Cl^– arranged and packed in a regular pattern.

– Each Na⁺ is surrounded by six Cl^– that are equidistant from it.

– Similarly each Cl^– is surrounded by six Na⁺ that are equidistant from it.

– This pattern occurs repeatedly in all directions.

– The result is a giant of ions in all directions hence giant ionic structure.

 

Diagram of the cubic structure of sodium chloride.

Properties of giant ionic structures.

1. They are hard and brittle.

• Reason:

– Ionic solids are hard because each ion is held in the crystal by strong attractions from the oppositely charged ions around it.

– They are brittle and thus may be split cleanly (cleaved) using a sharp-edged razor.

• Explanation:

– When a crystal is tapped sharply along a particular plane it is possible to displace one layer of ions relative to the next.

– Due to the displacement, ions of similar charge come together leading to repulsive forces between the portions of the crystals.

– This forces the two portions of the crystals to split apart.

 

Diagram: crystal cleavage in ionic compounds.

 

 

 

 

 

 

 

 

2. They have high melting and boiling points.

• Reason:

– They have strong electrostatic forces / ionic bonds / electrovalent bonds between the oppositely charged ions throughout the structure which require large amounts of energy to break.

 

3. Solubility.

(i). They are soluble in polar solvents like water, ethanol and acetone (propanone)

• Reason:

– Water contains highly polar molecules.

– The positive ends of the polar water molecules are attracted to the negative ions in the crystal, and the negative ends of the water molecules are attracted to the positive ions in the crystal.

– This results to the formation of ion-solvent bonds which leads to release of energy.

– This energy is sufficient to cause the detachment of ions from the crystal lattice hence dissolution.

Note:
– This detachment of ions is called solvation, and the energy required for this is called solvation energy.

– Where the solvent is water the ions are said to have been hydrated, and the energy involved in the process is called hydration energy.

 

Diagrams: hydrated positive and negative ions.

 

 

 

 

 

 

 

(ii). They are insoluble in non-polar organic solvents like tetrachloromethane, benzene and hexane.

• Reason:

– Non-polar molecules are held together by weak intermolecular forces, the Van der Waals forces.

– The Van der Waals are much smaller in magnitude compared to the ionic bonds in the ionic solid crystal lattice.

– Thus the ion-ion interactions in the ionic solid are stronger than the solvent-solvent interactions in the solvent or the solvent-ion interactions between the solid and the solvent.

– Thus the non-polar solvent molecules cannot penetrate the ionic lattice to cause salvation.

 

4. Electrical conductivity.

– Ionic substances do not conduct electric current in solid state.

• Reason:

The ions are held in static positions in the solid crystal lattice and thus cannot move to conduct electric current.

– They conduct electric current in molten and solution (aqueous) states.

• Reason:

– In molten and aqueous states the ions are free and mobile and thus move about conducting electric current.

 

 

 

Gradation in properties of some ionic compounds of sodium.

Property of compound		Compound of sodium
		Sodium fluoride	Sodium chloride	Sodium bromide	Sodium iodide
Solubility in water		Soluble	Soluble	Soluble	Soluble
Melting point (oC)		993	801	747	661
Boiling point (oC)		1695	1413	1390	1304
Electrical conductivity	Solid	Does not	Does not	Does not	Does not
	Molten / solution	Conducts	Conducts	Conducts	Conducts

 

Note:

– Solubility of the compounds decrease from sodium fluoride to sodium iodide.

– Melting and boiling points decrease from sodium fluoride to sodium iodide.

 

2. The covalent bond.

• Meaning:

– Refers to a bond formed when two atoms of the same or of different elements share electrons to become stable.

– Formation of covalent bond between atoms (similar or dissimilar) result to the formation of a molecule.

– Covalent bonds are usually formed by the association of non-metals.

 

• Note:

– A molecule is a group of atoms (two or more) of the same or different elements that are held together by strong covalent bonds.

 

Formation of a covalent bond.

– Covalent bonding is brought about by the facts that the electro-positivity and the electro-negativity of the elements involved are very close.

– For that reason, none of the atoms can completely lose its valence electrons to the next atom.

– For this reason, both atoms donate electrons which are then shared between them.

– Both atoms thus attain a stable noble gas (duplet or octet) configuration.

 

Illustrations:

• Formation of chlorine molecule.

– Each chloride atom has electronic configuration 2.8.7 and thus need to gain a single electron in the outermost energy level to attain a stable noble gas configuration.

– Since both chloride atoms have same electro-negativities, none will easily lose an electron to the other.
– For this reason, both donate the number of electrons required by the other atom (in this case 1), which they share between them.

– Thus chlorine molecule is formed by sharing 2 electrons between two chlorine atoms, hence a single covalent bond.

 

Diagram: formation of chlorine molecule.

 

 

 

 

 

Dot and cross diagram for chlorine molecule.

 

 

 

 

 

 

 

 

Formation of oxygen molecule.

– Each oxygen atom has electronic configuration 2.8.6 and thus need to gain 2 electrons into the outermost energy level to attain a stable noble gas configuration.

– Since both oxygen atoms have same electro-negativities, none will easily lose an electron to the other.
– For this reason, both donate the number of electrons required by the other atom (in this case 2), which they share between them.

– Thus oxygen molecule is formed by sharing 4 electrons (2 from each atom) between two chlorine atoms, hence a double covalent bond.

 

Diagram: the oxygen molecule.

 

 

 

 

 

 

 

 

 

Note:

– A single covalent bond is represented in dot (.) and cross (x) diagrams using two dots, two crosses, a dot and a cross or a single line (–) between the atoms involved in the bond.

– Thus a single covalent bond like in chlorine can be represented as Cl – Cl, a double covalent bond like in oxygen can be represented as O = O while a triple covalent bond like in nitrogen can be represented as N ≡ N.

 

Dot (.) and cross (x) diagrams for various covalent compounds.

1. Hydrogen, H₂

 

2. Hydrogen chloride, HCl

 

3. Nitrogen, N₂

 

4. Water

 

5. Carbon (IV) oxide, CO₂

 

6. Ammonia gas, NH₃

 

7. Phosphene, PH₃

 

8. Methane, CH₄

 

9. Ethane. CH₂CH₂

 

10. Ethyne, C₂H₂

 

10. Ethanol, C₂H₅OH

 

12. Bromoethane, C₂H₄Br.

 

The coordinate bond.

– Refers to a type of covalent bond in which the shared pair of electrons forming the bond is contributed by only one of the atoms forming the bond.

– It is also called the dative bond.

 

Examples:

• Formation of ammonium ion.

– Occurs when an ammonia gas molecule combines with a hydrogen ion (proton).

– All the atoms in the ammonia molecule have a stable noble gas configuration and thus the molecule is stable.

– However the nitrogen in the ammonia molecule has a lone pair of electrons (electrons that have not yet been used in bond formation)

– The hydrogen ion has lost its outermost single valence electron to form the hydrogen ion.

– Thus the hydrogen ion has no electron(s) in its outermost energy level.

– To be stable the hydrogen needs two electrons in its outermost energy level.

– The hydrogen ion thus accepts bonds with the lone pair (2) of electrons in the nitrogen of the ammonia molecule forming a dative bond.

– The total number of electrons in the ammonium ion is 11 while the total number of electrons is 10 leading to a net positive charge of +1

 

Note:

– In a dot (.) and cross (x) diagram where the covalent bond is represented by horizontal lines (–), the dative // coordinate bond is represented by an arrow (→) pointing the atom that “accepts” the electrons.

 

Diagram: formation of ammonium ion.

 

 

 

 

 

 

 

 

 

 

 

Formation of hydroxonium ion (H₃O⁺)

– Occurs when a water molecule combines with a hydrogen ion (proton).

– All the atoms in the water molecule have a stable noble gas configuration and thus the molecule is stable.

– However the oxygen in the ammonia molecule has 2 lone pairs of electrons // four electrons (electrons that have not yet been used in bond formation)

– The hydrogen ion has lost its outermost single valence electron to form the hydrogen ion.

– Thus the hydrogen ion has no electron(s) in its outermost energy level.

– To be stable the hydrogen needs two electrons in its outermost energy level.

– The hydrogen ion thus accepts and bonds with two of the four electrons in the oxygen of the water molecule forming a dative bond // coordinate bond.

– The total number of electrons in the hydroxonium ion is 11 while the total number of electrons is 10 leading to a net positive charge of +1

 

Diagram: formation of the hydroxonium ion.

 

 

 

 

 

 

 

 

 

 

Note:

– The hydroxonium ion (H₃O⁺) can still further react with another hydrogen ion to form another ion of the formula H₄O²⁺.

– This is due to the presence of a single lone pair of electrons in the structure of the hydroxonium ion (H₃O⁺).

 

Diagram: formation of H₄O²⁺ from H₃O⁺ and H⁺

 

 

 

 

 

 

 

 

 

 

 

Note:

– The H₄O²⁺ cannot however react further since all the valence electrons in all its atoms have been used in bonding leaving no lone pairs.

 

 

 

Further examples of dative covalent bonds in other compounds.

1. Carbon (IV) oxide

 

2. Formation of PH₄⁺

 

3. Aluminium chlorine dimer (Al₂Cl₆).

 

4. Ammonia-aluminium chloride vapours complex, AlCl₃.NH₃

 

Note:

– Substances with covalent bonds form two main types of structures:

• Molecular structures
• Giant atomic // Giant covalent structures.

 

1. The Molecular structures.

– Refers to a structure in which covalent bonds holds atoms together to form molecules and the resultant molecules are held together by intermolecular forces.

– Substance with molecular structures are usually gases or liquids at room temperature.

– Few solids such as sulphur, iodine, fats, sugar, naphthalene and paraffin wax also have molecular structures.

– The intermolecular forces in molecular structures are of two types:

• Van der Waals
• Hydrogen bonds

 

(a). The Van der Waals.

– Are the weakest form of intermolecular forces due to induced dipole- induced dipole attractions between molecules.

– As the size of the molecule increases, the number of constituent electrons increases leading to increase in strength of the induced dipole – induced dipole interactions.

– The strength of the Van der Waals thus increases as the molecular size increases.

 

Note:

This explains the trend in boiling points for the halogens, which increase from fluorine to iodine.

• Reason:

– From fluorine to iodine the size of the atoms and hence the molecules increases leading to increase in molecular masses that lead to stronger induced dipole – induced dipole interactions hence increasing strength of the Van der Waals (from F₂ to I₂)

– This also applies for the increase in boiling points for the homologous series of alkanes.

 

Diagram: Illustration of Van der Waals forces

(i). Iodine

(ii). Graphite.

 

(b). The hydrogen bonds.

– Is an intermolecular force in which the electropositive hydrogen atom of one molecule is attracted to an electronegative atom of another molecule.

– The essential requirements for the formation of a hydrogen bond (H – bond) are:

• A hydrogen atom attached to a highly electronegative atom.
• An unshared pair of electrons on the electronegative atom.

Note:

This explains why hydrogen bonds are common in molecules in which hydrogen are bonded to highly electronegative atoms like nitrogen, oxygen and fluorine.

 

The formation of hydrogen bonds

– Occurs when hydrogen atom is bonded to a highly electronegative atom like nitrogen, oxygen and fluorine.

– The electrons in the covalent bond (between hydrogen and the more electronegative atom) are drawn towards the electronegative atom.

– Hydrogen atom has no electrons other than the one it contributes to the covalent bond, which is also being pulled away from it (by the more electronegative atom).

– Hydrogen atom has no outer energy level of electrons making the single proton in the nucleus unusually “bare”.

– The proton is thus readily available for any form of dipole-dipole attractions.

– The “bare” proton of the hydrogen atom thus attracts the more electronegative atom (e.g. N, O and F) on either side.

– It thus exerts an attractive force on the more electronegative atom hence bonding them together.

– The two (more electronegative) larger atoms are drawn closer with a hydrogen atom effectively buried in their electron clouds.

– This constitutes the Hydrogen bond.

 

Illustration: formation of hydrogen bonds in water.

– In water two hydrogen atoms are bonded to an oxygen atom which highly electronegative.

– The electrons in the covalent bond (between each hydrogen and oxygen atom) are drawn towards the more electronegative oxygen atom.

– The hydrogen atoms have no electrons other than their share of those in the covalent bond, which are also being pulled away from them by the oxygen atom.

– Hydrogen atoms have no outer energy level of electrons making the single proton in their nucleus unusually “bare”.

– The proton is thus readily available for any form of dipole-dipole attractions (with oxygen in this case).

– The “bare” proton of each of the hydrogen atoms in one molecule thus attracts the more electronegative oxygen atom of the neighboring molecule.

– Each hydrogen atom thus exerts an attractive force on the oxygen atom of the next molecule hence bonding them together.

– The electronegative oxygen atom of one molecule is drawn to the electropositive hydrogen atom of the next molecule with the hydrogen atoms effectively buried in the electron clouds of oxygen.

– This constitutes the Hydrogen bond in the water molecule.

 

Diagram: hydrogen bonds in water molecules.

Note:

– Other compounds with hydrogen bonds include ethanol, ammonia, hydrogen fluoride etc.

– Hydrogen bonds are much stronger than the weak Van der Waals forces but still weaker than the covalent bonds.

 

Effect of hydrogen bonding on properties of molecular substances.

– Hydrogen bonding tends to disrupt the gradation in physical properties of molecular substances in relation to molecular weights.

– The effect on molecular masses on the melting and boiling points only apply when the intermolecular force is the same.

 

Examples.

1. Both ethanol (C₂H₅OH) and dimethyl ether (C₂H₆O) have the same relative molecular mass of 46. However the boiling point of ethanol is higher at 78.5oC than that of dimethyl ether at only -24oC.

Reason:

– Even though both have molecular structures with covalent bonds between the atoms, the intermolecular forces in ethanol are hydrogen bonds which are much stronger than the intermolecular forces in dimethyl ether which are weak Van der Waals forces.

 

2. Ethanol (C₂H₅OH) has a molecular mass of 46 while butane (C₄H₁₀) has a molecular mass of 58, yet the boiling point of ethanol is higher than that of butane.

Reason:

– Both have molecular structures with covalent bonds between the atoms. However, the intermolecular forces in ethanol are hydrogen bonds which are much stronger (and require more energy to break) than the intermolecular forces in dimethyl ether which are weak Van der Waals forces.

 

– Molecular substances are generally insoluble in polar solvents like water. However those with hydrogen bonding as the intermolecular forces are soluble in water since the hydrogen bonding confers them some polarity.

Examples:

– Sugar, ethanol, ethanoic acid etc

 

Properties of molecular structures.

1. Melting and boiling points.

– The generally have low melting and boiling points.

Reason:

– Although the atoms forming the molecules are held together by strong covalent bonds, the intermolecular forces are weak Van der Waals forces which require low amounts of energy to break.

 

2. Heat and electrical conductivity

– They are poor conductors of heat and electricity at any state.

Reason:

– They have neither delocalized electrons nor free mobile ions for electrical conductivity.

 

3. Solubility

– Molecular substances are generally insoluble in polar solvents like water but soluble in non-polar organic solvents like benzene.

Reason:

– In polar solvents like water there are strong water – water attractions which are considerable stronger than the intermolecular forces (Van der Waals) attractions or molecule – water (solvent) attractions, making the molecules unable to penetrate the water (solvent) structure for dissolution to occur.

– In non-polar solvents like benzene, the benzene-benzene attractions are similar in strength to the intermolecular forces or the molecule – benzene (solvent) attractions, enabling the molecules to penetrate the solvent thus allowing dissolution.

 

Note:

– Molecular substances with hydrogen bonds as the intermolecular forces are soluble in polar solvents like water.

Reason:

– The hydrogen bonds in the molecules are equal in strength to the water – water interactions which are also hydrogen bonds, thus the molecules are able to penetrate the structure of water leading to salvation // dissolution // hydration i.e. they are polar like the water molecules.

 

 

 

 

 

 

Summary: Properties of some molecular substances.

Property of compound		Molecular substance
		Sugar (Sucrose)	Naphthalene	Iodine	Rhombic sulphur	Water	Hydrogen sulphide
Solubility in water		Soluble	Insoluble	Insoluble	Insoluble	–	Slightly soluble
Molecular mass		183	128	186	256	18	34
Melting point (oC)		186	82	113	114	0	-85
Boiling point (oC)		–	218	183	444	100	-60
Electrical conductivity	Solid	Does not	Does not	Does not	Does not	Does not	Does not
	Molten / solution	Does not	Does not	Does not	Does not	Does not	Does not

 

2. Giant covalent structures // Giant atomic structures.

– Are molecular substances in which atoms are linked throughout the whole structure by very strong covalent bonds from one atom to the next.

– The result is an indefinite number of atoms which are all covalently bonded together.

– This pattern occurs repeatedly throughout the structure leading to a “giant” of atoms all covalently bonded.

 

Examples:

(a). Diamond.

– Is an allotrope of carbon.

• Note: allotropes are different crystalline forms of the same element in the same physical state.

– In diamond, each carbon atom is bonded to four other carbon atoms by strong covalent bonds.

– The carbon atoms in diamond are covalently bonded into an octahedral pattern, which repeats itself in all directions resulting into a giant atomic structure.

– Since each carbon atom is bonded to four others, all the four valence electrons in each carbon are used in bonding hence no delocalized electrons in the structure of diamond.

– Diamond is the hardest substance known due to the fact that all the atoms are covalently bonded together and are closely packed together.

 

Diagram: Structure of diamond:

Properties of diamond.

1. Have high melting and boiling points.

Reason:

– It has a giant atomic structure with strong covalent bonds throughout the structure which require large amounts of energy to break.

 

1. It is insoluble in water.

Reason:

– It is non-polar and thus cannot dissolve in polar water molecules since there are no intermolecular interactions which would facilitate penetration into the water molecules for dissolution to occur.

iii. Does not conduct heat and electricity.

Reason:

– Each carbon atom in the structure of diamond is bonded to four others hence uses all its four valence electrons in bonding and thus lacks any delocalized electrons for electrical conductivity.

 

1. It is the hardest substance known.

Reason:

– All carbon atoms are compactly bonded in a continuous octahedral pattern with strong covalent bonds throughout the structure which are very difficult to break.

 

(b). Graphite.

– Is also an allotrope of carbon.

– In graphite, each carbon atom is bonded to three other carbon atoms by strong covalent bonds.

– Since each carbon atom is bonded to only three others, only three of the four valence electrons in each carbon are used in bonding hence presence delocalized electrons in the structure of graphite.

– This explains the electrical conductivity of graphite.

– The carbon atoms in graphite are covalently bonded into hexagonal layers, which are joined to each other by weak Van der Waals forces.

– The presence of weak Van der Waals forces explains the slippery nature of graphite.

 

Diagram: Structure of graphite.

Properties of Graphite.

1. Have high melting and boiling points.

Reason:

– It has a giant atomic structure with strong covalent bonds throughout the hexagonal layers which require large amounts of energy to break. Even though there are Van der Waals between the layers the effect of the large number of covalent bonds still contribute to high melting and boiling points in graphite

 

1. Insoluble in water.

Reason:

– It is non-polar and thus cannot dissolve in polar water molecules since there are no intermolecular interactions which would facilitate penetration into the water molecules for dissolution to occur.

 

iii. It is a good conductor heat and electricity.

Reason:

– Each carbon atom in the structure of graphite is bonded to three others hence uses only three of its four valence electrons in bonding. This leads to presence of delocalized electrons in the structure of graphite which conducts heat and electricity.

 

1. It is soft and slippery.

Reason:

– The carbon atoms in graphite are covalently bonded into hexagonal layers which are joined to each other by weak Van der Waals forces.

– The weak Van der Waals forces easily slide over each other when pressed hence the soft and slippery feel.

 

(c). Silicon (IV) oxide.

– Is a covalent compound of silicon and oxygen.

– Silicon has four electrons in its outermost energy level while oxygen has six.

– Each silicon atom is bonded to four oxygen atoms by strong covalent bonds.

– Each oxygen atom is bonded to two silicon atoms by strong covalent bonds.

– This means there are no delocalized electrons in the structure of silicon (IV) oxide making it unable to conduct electric current.

– The silicon and oxygen atoms are all covalently bonded together (by strong covalent bonds) in a repeated manner leading to a giant of covalent bonds throughout the structure.

– The extra ordinarily strong covalent bonds in silicon (IV) oxide throuout the structure contribute to the very high melting (1728^oC) and boiling (2231^oC) points.

 

Diagram: structure of silicon (IV) oxide.

 

 

 

 

 

 

 

 

 

Properties of giant atomic // giant covalent structures.

1. – Have high melting and boiling points
2. – They are non-conductors of heat and electricity with the exception of graphite

iii. – They are insoluble in water

1. – Most are generally very hard, with the exception of graphite

 

• Note: for explanations of the above properties, refer to individual explanations of each compound.

 

3. The metallic bond.

– Is a bond formed due to electrostatic attraction between the positively charged nuclei and the negatively charged delocalized electrons that hold atoms together.

 

Formation of a metallic bond.

– In a metal there are usually many atoms surrounding any one atom.

– The valence electrons of any one atom are therefore mutually attracted to many nuclei.

– This leads to a situation in which the positive nuclei appear to be immersed in a sea of mobile electrons.

– The sea of mobile electrons are said to be delocalized which explains the ability of substances with metallic bonds to conduct electric current.

– This pattern of positive nuclei in a sea of electrons is repeated many times throughout the structure leading to a giant metallic structure, the only structure due to metallic bonds.

 

Diagram: model of the giant metallic structure.        

 

 

 

 

 

 

 

 

 

Properties of the giant metallic structure.

1. They have high melting and boiling points.

Reason:

– They have strong metallic bonds throughout the structure which need large amounts of energy to break.

– The strength of the metallic bond increases with decrease in atomic size, as well as with increase in the number of delocalized electrons.

– Thus metals with smaller atoms and more delocalized electrons tend to have stronger metallic bonds hence higher melting and boiling points.

– This explains why the melting and boiling points of metallic elements in period three increase from sodium to aluminium.

 

2. They are good conductors of heat and electricity.

Reason:

– The have delocalized electrons which heat and electric current.

Note:

– The electrical conductivity of metals increases with increase in the number of delocalized electrons in each atom in the structure.

– This explains why aluminium metal is a better conductor of heat and electricity than both magnesium and sodium.

 

3. They are insoluble in water.

Reason:

– There are no dipoles in the giant metallic structure and are thus non-polar, so cannot dissolve in polar water molecules.

Summary: some physical properties of metals

Metal	Valency	Melting point (oC)	Boiling point (oC)	Atomic radii (nm)	Electrical conductivity
Lithium	1	180	1330	0.133	Good
Sodium	1	98	890	0.155	Good
Potassium	1	64	774	0.203	Good
Magnesium	2	651	1110	0.136	Good
Aluminium	3	1083	2582	0.125	Good

 

Summary: Comparing various types of structures.

Attribute	Giant metallic	Giant atomic / giant  covalent	Giant ionic	Giant metallic
1. Structure i. Examples	Na, Fe, Cu.	Diamond, SiC, SiO2.	Ca2+O2-, (K+)2SO42-, Na+Cl–,	I2, S8, C10H8, HCl, CH4.
ii. Constituent particles	Atoms	Atoms	Ions	Molecules
iii. Type of substance compound	Metal element with low electronegativity	Non-metal element in group IV or its compound.	Metal/non-metal compound (a compound of elements with a large difference in electronegativity)	Non-metal element or non-metal/non-metal compound (elements with high electronegativity)
2. Bonding: In the solid	Attraction of outer mobile electrons for positive nuclei binds atoms together by strong metallic bonds	Atoms are linked through the whole structure by very strong covalent bonds from one atom to the next.	Attraction of positive ions for negative ions results in strong ionic bonds	Strong covalent bonds hold atoms together within the separate molecules; separate molecules are held together by weak intermolecular forces.
3. Properties. i. Volatility:     State at room temp.	Non-volatile. Very high melting and very high boiling points. Usually solid	Non-volatile. Very high melting and very high boiling points. Solid	Non-volatile. Very high melting and very high boiling points. Solid	Volatile. Low melting and low boiling points. Usually gases or volatile liquids
ii. Hardness // malleability	Hard, yet malleable.	Very hard and brittle	Hard and brittle	Soft
iii. Conductivity:	Good conductors when solid or molten.	Non-conductors in any state (except graphite)	Non-conductors when solid; good conductors when molten or in aqueous solution (electrolytes)	Non-conductors when solid, molten and in aqueous solution. (A few like HCl react with water to form electrolytes)
iv. Solubility:	Insoluble in polar and non-polar solvents but soluble in molten metals	Insoluble in all solvents	Soluble in polar solvents (e.g. H2O), insoluble in polar solvents like tetrachloromethane, CCl4.	Polar molecules e.g. HCl are soluble in polar solvents like water; but insoluble in non-polar solvents like CCl4 and vise-versa.

Types of bonds across a period.

– The number of valence electrons play an important role in determination of chemical bonding.

– Across a period in the periodic table, the nature of the bonds varies from metallic to covalent.

– The structure also thus varies from giant metallic to simple molecular.

– Thus similar compounds of the elements in period 3 will also exhibit variation in bond types, structures and properties.

 

Variation in bond types in oxides of period three elements.

Oxide	Na2O	MgO	Al2O3	SiO2	P2O5	SO2	Cl2O7
Physical state	Solid	Solid	Solid	Solid	Solid	Gas	Gas
M.P (oC)	1193	3075	2045	1728	563	-76	-60
B.P (oC)	1278	3601	2980	2231	301	-10	-9
Structure	Giant ionic	Giant ionic	Giant ionic	Giant atomic	Molecular	Molecular	Molecular
Bonding	Ionic	Ionic	Ionic	Covalent	Covalent with Van der Waals	Covalent with Van der Waals	Covalent with Van der Waals
Nature of oxide	Basic (alkaline)	Basic (weakly alkaline)	Amphoteric	Acidic	Acidic	Acidic	Acidic
Solubility in water	Dissolves to forms an alkaline solution	Dissolves to forms an alkaline solution	Insoluble	Insoluble	Dissolves in water to form acidic solution.	Dissolves in water to form acidic solution.	Dissolves in water to form acidic solution.
Reaction with acids	Reacts to form salt and water.	Reacts to form salt and water.	Reacts to form salt and water.	No reaction.	No reaction.	No reaction	No reaction

 

Note:

1. – The melting and boiling points of the magnesium oxide is higher than that of sodium oxide

Reason:

– Both have giant ionic structures. However the electrostatic forces of attraction between magnesium ions and oxide ions are stronger due to the fact that magnesium ion has a charge of +2 and is smaller in size than the sodium ion.

 

 

Effect of bond types of properties of chlorides of period 3 elements

– Most period 3 elements form stable chlorides.

– The trend in bond types, structures and properties of chlorides of period 3 elements show variation across the period.

 

Properties:

1. Reaction with water.

Procedure:

– A test tube is half filled with water and initial temperature of the water recorded.

– A spatula end full of sodium chloride is added and stirred until it dissolves.

– The hiughest temperature attained when all the solid dissolves is recored and the temperature change calculated.

– Two – three drops of universal indicator are added and the pH of the solution noted and recorded.

 

Observations:

Chloride	Observations
	Solubility	Temperature change (oC)	pH of solution
Sodium chloride	Dissolves	Drop in temperature	7
Magnesium chloride	Dissolves	Slight increase	6.5
Aluminium chloride	Hydrolyzed	Increases	3
Silicon (IV) oxide	Hydrolyzed	Increases	2
Phosphorus (III) chloride	Hydrolyzed	Increases	2
Phosphorus (V) chloride	Hydrolyzed	Increases	2

 

Explanations.

• Sodium and magnesium chlorides.

– Sodium chloride dissolves in water causing a slight drop in temperature.

– Magnesium dissolves readily with a small increase in temperature.

– Both chlorides are ionic and when added to water there is an immediate attraction of polar water molecules for ions in the chlorides.

– The solid thus readily dissolves forming aquated ions such as Na⁺_(aq) and Cl^–_(aq).

– These are separate metal and non-metal ions surrounded by polar water molecules.

– Since there is no production of either H⁺ or OH^– ions the solutions are neutral.

 

Equations:

NaCl_(s) + H₂O_(l) → Na⁺_(aq) + Cl^–_(aq) + H₂O_(l).

 

MgCl_2(s) + H₂O_(l) → Mg²⁺_(aq) + 2Cl^–_(aq) + H₂O_(l).

 

• Anhydrous aluminium chloride.

– It exits in molecular form as a dimeric molecule of Al₂Cl₆.

– The dimeric molecule is formed when aluminium chloride (AlCl₃) molecules vapour condense and combine forming larger molecules of Al₂Cl₆.

 

Diagram: Formation of a dimer in aluminium chloride.

– When added to water aluminium chloride is hydrolyzed to form an acidic solution.

– By so doing it behaves like a covalent chloride rather than an ionic chloride.

– The hydrolysis is an exothermic reaction accompanied by release of heat hence the increase in temperature.

 

Explanations.

– The hydrolysis is due to the very small but highly charged aluminium ion, Al³⁺.

– The Al³⁺ draws electrons away from its surrounding water molecules and causes them to give up H+ ions.

– This reaction usually involves aluminium ions combining with six water molecules to form hexa-aqua-aluminium (III) ions which dissociate to give H⁺.

 

Equation:

[Al(H₂O)₆]³⁺_(aq) → [Al(H₂O)₅(OH)]²⁺_(aq) + H⁺_(aq)

 

Note:

This reaction can be simplified as follows:

– Aluminium chloride reacting with water and hydrolyzing to give HCl_(aq) as one of the products.

• Equation: Al₂Cl_6(s) + 6H₂O_(l) → 2Al(OH)_3(aq) + 6HCl_(aq)

– Then the HCl dissociates to give H+ and Cl-

• Equation: 6HCl_(aq) → H⁺_(aq) + Cl^–_(aq)

– It is the resultant H⁺ that confer the solution its acidic properties.

 

• Silicon (IV) chloride

– Undergoes hydrolysis in water in an exothermic reaction producing a lot of heat.

– The products of the hydrolysis are silicon (IV) oxide solid and hydrogen chloride gas.

– The hydrogen chloride gas immediately dissolves in the water to form hydrochloric acid.

– The hydrochloric acid dissociated to liberate H⁺ which leads to acidic conditions.

 

Equations:

SiCl_4(s) + 2H₂O_(l) → SiO_2(s) + 4HCl_(aq)

Then: 4HCl_(aq) → 4H⁺_(aq) + 4Cl^–_(aq).

 

• Phosphorus (III) chloride and phosphorus (V) chloride

– Both undergo hydrolysis in water in an exothermic reaction producing a lot of heat.

– The products of the hydrolysis are phosphorus (III) acid and phosphoric (V) acid respectively, and hydrogen chloride gas.

– The hydrogen chloride gas immediately dissolves in the water to form hydrochloric acid.

– The hydrochloric acid dissociated to liberate H⁺ which leads to acidic conditions.

 

Equations:

• With phosphorus (III) chloride:

PCl_3(s) + 2H₂O_(l) → H₃PO_3(s) + 3HCl_(aq)

Then: 3HCl_(aq) → 3H⁺_(aq) + 3Cl^–_(aq).

 

• With phosphorus (V) chloride:

PCl_5(s) + 4H₂O_(l) → H₃PO_4(s) + 5HCl_(aq)

Then: 5HCl_(aq) → 5H⁺_(aq) + 5Cl^–_(aq).

 

Trends in bond types and properties of chlorides of elements in period 3

Property	Period 3 chloride
Formula	NaCl	MgCl2	Al2Cl6	SiCl4	PCl3 & PCl5	SCl2
Physical state at RT	Solid	Solid	Solid	Liquid	Liquid	Liquid
M.P (oC)	801	714	Sublimes at 180oC	-70	Sublimes at -94oC	-78
B.P (oC)	1467	1437	–	57	–	Decomposes at 57oC
Conductivity	Good	Good	V. poor	nil	nil	Nil
Structure	Giant ionic	Giant ionic	Molecular dimer	Molecular	Molecular	Molecular
Bond type	Ionic	Ionic	Ionic / covalent	Covalent	Covalent	Covalent
Effect on water	Readily dissolves	Readily dissolves	Hydrolyzed to give HCl fumes	Hydrolyzed to give HCl fumes	Hydrolyzed to give HCl fumes	Hydrolyzed to give HCl fumes
pH of solution	7	6.5	3	2	2	2

 

Summary: Characteristics of bonds

Property	Substances with
	Covalent bonds	Ionic bonds	Metallic bonds
Electrical conductivity	Non-conductors except graphite	– Solids do not conduct. – Aqueous solutions and molten state conduct	– Conducts
Thermal conductivity	– Non-conductors except graphite	– Do not conduct	– Conducts
Melting point (oC)	– Low for molecular substances – High for giant atomic structures	– Usually high.	– Generally high
Boiling point (oC)	– Low for molecular substances – High for giant atomic structures	– Usually high	– Generally high
Solubility	– Generally insoluble in water but soluble in organic solvents	– Generally soluble in water	– Some metals react with water

 

 

UNIT 5: SALTS.

1. Meaning.
2. Types of salts:

• Normal salts
• Acid salts
• Basic salts
• Double salts.

3. Solubility of salts in water

• Sulphates
• Chlorides
• Nitrates
• Carbonates
• Sodium, potassium and ammonium salts

4. Solubility of bases in water.
5. Obtaining crystals
6. Preparation of salts

• Insoluble salt
  • Double decomposition
• Soluble salts

A sodium, potassium and ammonium salts

 

A non-potassium, sodium and ammonium salt.

That reacts with water

Direct synthesis

That does not react with water

Acid and metal method

Acid + base method

Neutralization

7. Uses of salts
8. Action of heat on salts

• Carbonates
• Nitrates