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Grade 7 Business Studies Schemes of Work Free Editable Word, PDF Downloads
Grade 7 Business Studies Schemes of Work Free Editable Word, PDF Downloads
GRADE 7 BUSINESS STUDIES SCHEME OF WORK TERM 3
SCHOOL………………………….TEACHER’S NAME…………………. TERM THREE YEAR 2023
| Week | Lesson | Strand | Sub-strand | Specific-Learning outcomes | Learning Experience | Key Inquiry Question(S) | Learning
Resources |
Assessment Methods | Reflection |
| 1 | 1 | Government and global influence in business | Government and business | By the end of the lesson, the learner should be able to:
a) Define the term investor. b) Identify the need for government involvement in business in Kenya. c) Justify the need for government involvement in business activities in Kenya. d) Appreciate the importance of government in conducting businesses. |
Learners are guided to define the term investor.
Learners to identify the need for government involvement in business in Kenya.
In groups or in pairs, learners to justify the need for government involvement in business activities in Kenya. |
Who is an investor? | Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 133-135
Photographs Pictures Video clips Charts Computing devices |
Oral questions Oral Report Observation
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| 2 | Government and global influence in business | Government and business | By the end of the lesson, the learner should be able to:
a) Explain ways in which the shopkeeper in their local community may exploit his or her customer. b) Debate on the need for government involvement in business in Kenya. c) Debate on the motion: “There is need for the government of Kenya to be involved in Business” d) Have fun and enjoy the debate. |
Learners to explain ways in which the shopkeeper in their local community may exploit his or her customer.
In groups or in pairs, learners are guided to debate on the need for government involvement in business in Kenya, on the motion: “There is need for the government of Kenya to be involved in Business”
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What is exploitation?
What is foreign dominance? |
Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 135-137
Photographs Pictures Video clips Charts Realia Computing devices |
Oral questions Oral Report Observation
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| 3 | Government and global influence in business | Legal requirement for starting and operating a simple business in Kenya | By the end of the lesson, the learner should be able to:
a) Identify the legal requirement for starting and operating a simple business in Kenya. b) Give examples of businesses in Kenya that require a Health Certificate. c) Draw the certificate of registration in learner’s book. d) Appreciate the importance of the legal requirements for starting and operating a simple business in Kenya. |
Learners to identify the legal requirement for starting and operating a simple business in Kenya.
Learners to give examples of businesses in Kenya that require a Health Certificate.
Individually, learners to draw the certificate of registration in learner’s book 7 pg. 139 |
What is considered a simple business in Kenya?
What are the examples of simple business in Kenya? |
Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 137-141
Photographs Pictures Video clips Charts Computing devices
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Oral questions Oral Report Observation
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| 2 | 1 | Government and global influence in business | Extended activity | By the end of the lesson, the learner should be able to:
a) Organize to visit county or sub-county office of the Ministry of Industrialization, Trade and Enterprise Development or their local county offices. b) Write a report of their findings. c) Have fun and enjoy the visit to the county or sub-county offices. |
As a class, learners to organize visit county or sub-county office of the Ministry of Industrialization, Trade and Enterprise Development or their local county offices.
Learners to write a report of their findings.
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Which office did you visit? | Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 141
Photographs Pictures Video clips Charts Realia Computing devices
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Oral questions Oral Report Observation
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| 2 | Government and global influence in business | Role of government in business in Kenya | By the end of the lesson, the learner should be able to:
a) Outline the role of government in business in Kenya. b) List any malpractices by traders against which consumers may need protection from the government. c) Recognise the role of government in business in Kenya. d) Appreciate the role of government in business in Kenya.
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Learners to outline the role of government in business in Kenya.
Learners to list any malpractices by traders against which consumers may need protection from the government.
Learners are guided to recognise the role of government in business in Kenya. |
Why does the government of Kenya train and advise people in business? | Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 141-144
Photographs Pictures
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Oral questions Oral Report Observation
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| 3 | Government and global influence in business | Taxation in Kenya | By the end of the lesson, the learner should be able to:
a) Define the term tax and taxation. b) Read the case study in learner’s book. c) Have a desire to learn more about tax.
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Learners to define the term tax and taxation.
In groups, learners to read the case study in learner’s book 7 pg. 145 |
What is the meaning to tax and taxation? | Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 144-146
Pictures Video clips Computing devices
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Oral questions Oral Report Observation
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| 3 | 1 | Government and global influence in business | Taxation in Kenya | By the end of the lesson, the learner should be able to:
a) Identify values that patriotic Kenyan citizens who pays taxes possess are hidden. b) Play the game; unscramble the letters to find the hidden words. c) Have fun and enjoy playing the game.
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Learners to identify values that patriotic Kenyan citizens who pays taxes possess are hidden.
In groups or in pairs, learners to play the game; unscramble the letters to find the hidden words.
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Who is a patriotic Kenyan citizen? | Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 146
Assessment books. Computing devices
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Oral questions Oral Report Observation
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| 2 | Government and global influence in business | Importance of paying taxes in Kenya | By the end of the lesson, the learner should be able to:
a) Examine the importance of paying taxes. b) Role play the conversation in learner’s book. c) Have fun and enjoy the role playing. |
Learners to examine the importance of paying taxes.
In groups or in pairs, learners to role play the conversation in learner’s book 7 pg. 149 |
What is the importance of paying taxes? | Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 148-150
Photographs Pictures Video clips Computing devices
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Oral questions Oral Report Observation
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| 3 | Government and global influence in business | Importance of paying taxes in Kenya | By the end of the lesson, the learner should be able to:
a) State the importance of paying taxes. b) Make a poster with messages on the importance of paying taxes. c) Appreciate the importance of paying taxes.
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In groups or in pairs, learners to state the importance of paying taxes.
In groups or in pairs, learners to make a poster with messages on the importance of paying taxes
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What other ways can the government utilize taxes to promote development in the community? | Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 150-152
Pictures Video clips Computing devices
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Oral questions Oral Report Observation
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| 4 | 1 | Government and global influence in business | The need for paying taxes in Kenya | By the end of the lesson, the learner should be able to:
a) Identify the need for paying taxes in Kenya. b) Investigate the need for paying taxes in Kenya. c) Appreciate the importance of paying taxes in Kenya. |
Learners to identify the need for paying taxes in Kenya.
In groups or in pairs, learners to investigate the need for paying taxes in Kenya. |
What services provided by the government do they think their tax money should be used for? | Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 152-154
Photographs Pictures Computing devices
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Oral questions Oral Report Observation
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| 2 | Government and global influence in business | Promoting regional unity | By the end of the lesson, the learner should be able to:
a) Explain the meaning of good governance. b) Read the story in learner’s book and answer the questions that follow. c) Develop a desire to pay taxes as a Kenyan citizen.
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Learners are guided to explain the meaning of good governance.
In pairs, learners are guided to read the story in learner’s book 7 pg. 155 and answer the questions that follow |
What is good governance? | Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 154-156
Photographs Pictures Video clips Computing devices
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Oral questions Oral Report Observation
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| 3 | Financial records in business | Business transactions | By the end of the lesson, the learner should be able to:
a) Define the terms business transaction and financial documents. b) Differentiate business and non-business transactions. c) Have a desire to do business transactions. |
Learners to define the terms business transaction and financial documents.
Learners to differentiate business and non-business transactions.
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What is business transaction?
What are financial documents? |
Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 157-159
Photographs Pictures Video clips Computing devices
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Oral questions Oral Report Observation
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| 5 | 1 | Financial records in business | Extended activity | By the end of the lesson, the learner should be able to:
a) Identify and visit two or three local business people in their neighbourhood. b) Talk to them about the activities they carry out in their businesses daily. c) Have fun and enjoy visiting local business people.
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As a class, learners are guided to identify and visit two or three local business people in their neighbourhood.
As a class, learners are guided to talk to them about the activities they carry out in their businesses daily. |
What are donation?
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Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 159-160
Photographs Pictures Computing devices |
Oral questions Oral Report Observation
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| 2 | Financial records in business | Cash and credit transactions in business | By the end of the lesson, the learner should be able to:
a) Explain the meaning of cash and credit transactions in business. b) Match each business transaction with the correct description of how it occurs. c) Have a desire to learn more about cash and credit transactions in business.
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Learners to explain the meaning of cash and credit transactions in business.
In pairs, learners are guided to match each business transaction with the correct description of how it occurs.
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What is cash and credit transactions?
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Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 160-161
Photographs Pictures Video clips Computing devices
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Oral questions Oral Report Observation
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| 3 | Financial records in business | Extended activity | By the end of the lesson, the learner should be able to:
a) Role-play the scenarios of various business transactions (people buying and selling) b) Identify whether they are cash transactions or credit transactions. c) Take turns until everyone has demonstrated business transactions and other members have identified their types. d) Enjoy demonstrating to their class members.
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In groups or in pairs, learners to role-play the scenarios of various business transactions (people buying and selling)
Learners to identify whether they are cash transactions or credit transactions.
In groups or in pairs, learners to Take turns until everyone has demonstrated business transactions and other members have identified their types. |
What have you learnt about cash and credit transactions?
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Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 161-164
Photographs Pictures Video clips Charts Realia Computing devices
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Oral questions Oral Report Observation
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| 6 | 1 | Financial records in business | Financial documents used in buying and selling | By the end of the lesson, the learner should be able to:
a) Brainstorm on and list financial documents used in buying and selling. b) Share experiences on financial documents in buying and selling in everyday life. c) Appreciate the documents used in buying and selling.
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In groups or in pairs, learners to brainstorm on and list financial documents used in buying and selling.
In groups or in pairs, learners to share experiences on financial documents in buying and selling in everyday life. |
Which other documents are used in buying and selling? | Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 164
Photographs Pictures Video clips Charts Computing devices |
Oral questions Oral Report Observation
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| 2 | Financial records in business | Financial documents used in buying and selling | By the end of the lesson, the learner should be able to:
a) Identify financial documents used in buying and selling. b) Act out the conversation in learner’s book. c) Have fun and enjoy acting out the conversation.
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Learners to identify financial documents used in buying and selling.
In groups or in pairs, learners to act out the conversation in learner’s book 7 pg. 165 |
What is the conversation about? | Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 165-166
Photographs Pictures Video clips |
Oral questions Oral Report Observation
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| 3 | Financial records in business | Extended activity | By the end of the lesson, the learner should be able to:
a) Search for samples of financial documents used in buying and selling. b) Draw the samples in learner’s book and answer the questions that follow.] c) Appreciate the financial documents used in buying and selling. |
In groups or in pairs, learners to search for samples of financial documents used in buying and selling.
In pairs or individually, learners to draw the samples in learner’s book 7 pg. 167 and answer the questions that follow |
Who sends the financial document and to whom is it sent? | Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 167-168
Pictures Realia Computing devices
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Oral questions Oral Report Observation
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| 7 | 1 | Financial records in business | Extended activity | By the end of the lesson, the learner should be able to:
a) Visit a financial expert to discuss financial documents used in a business transaction. b) Identify each sample financial document that he or she will provide. c) Have fun and enjoy the visit to a financial expert.
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As a class, learners are guided to visit a financial expert to discuss financial documents used in a business transaction.
As a class, learners to identify each sample financial document that he or she will provide.
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What information is found on the financial document? | Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 168-169
Photographs Pictures Video clips Charts Computing devices |
Oral questions Oral Report Observation
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| 2 | Financial records in business | Extended activity | By the end of the lesson, the learner should be able to:
a) Organise and visit business people in the local market or trading centre neighbouring the school. b) Identify the gaps in keeping financial records. c) Enjoy visiting business men and women.
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As a class, learners are guided to organise and visit business people in the local market or trading centre neighbouring the school
As a class, learners are guided to identify the gaps in keeping financial records |
What is the financial document used for? | Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 169-179
Photographs Pictures Video clips Charts Realia Computing devices |
Oral questions Oral Report Observation
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| 3 | Financial records in business | Methods used in making payments for goods and services | By the end of the lesson, the learner should be able to:
a) Identify the methods used in making payments for goods and services. b) Analyse methods of payment for goods and services. c) Discuss how each method of payment is used. d) Appreciate the methods used in making payments for goods and services.
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Learners are guided to identify the methods used in making payments for goods and services.
In groups or in pairs, learners to analyse methods of payment for goods and services.
In groups or in pairs, learners to discuss how each method of payment is used. |
What are the methods used in making payments for goods and services?
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Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 173-174
Photographs Pictures
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Oral questions Oral Report Observation
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| 8 | 1 | Financial records in business | Methods used in making payments for goods and services | By the end of the lesson, the learner should be able to:
a) State the advantages and disadvantages of each method of payment for goods and services. b) Match methods of payments for goods and services. c) Appreciate the advantages each method of payment for goods and services. |
Learners are guided to state the advantages and disadvantages of each method of payment for goods and services.
Learners are guided to match methods of payments for goods and services. |
What are the advantages and disadvantages of each method of payment for goods and services?
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Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 175-176
Photographs Video clips Charts Computing devices
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Oral questions Oral Report Observation
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| 2 | Financial records in business | Extended activity | By the end of the lesson, the learner should be able to:
a) Interview the school bursar or business people in the local community. b) Discuss with them the method of payment which they preferred and why. c) Appreciate the role of financial documents in record keeping business. |
As a class or in groups, learner to interview the school bursar or business people in the local community.
As a class or in groups, learner to discuss with them the method of payment which they preferred and why. |
What is the difference between mobile money transfer and electronic funds transfer? | Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 176-178
Video clips Computing devices
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Oral questions Oral Report Observation
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| 3 | REVISION | ||||||||
| 9 | ASSESSEMENT |
Grade 7 CBC Visual Arts Complete Schemes of Work Free
GRADE 7 SCHEMES OF WORK
VISUAL ARTS ACTIVITIES
Grade 7 Visual curriculum Design
Top Scholar Visual Arts Grade 7 Teachers Guide/Learners Book
NAME OF THE TEACHER SCHOOL YEAR TERM III
| Week | LSN | strand | Sub-strand | Specific Learning Outcomes | Key Inquiry Question(s) | Learning Experiences | Learning Resources | Assessment Methods | Refl |
| 1 | 1 | INDIGENOUS CRAFTS | Making a mask using papier mache | By the end of the lesson the learner should be able:
Knowledge a. Discuss samples of relief sculptures. Skill b. Prepare papier Mache for making a mask Attitude c. Appreciate own and others sculptural work. |
1. How does one create models using clay?
2. What is papier Mache? 3. Why is an armature important in creating clay models? |
Learner is guided to:
• prepare the pulp with emphasis on; – selection of paper (recycled paper) – shredding – soaking – pounding – addition of adhesive |
Samples of relief sculptures, pictures, samples of mask, assortment of papers, glue, containers, a strong stick for pounding.
TOP scholar Visual Arts P.b pg. 88 |
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| 2 | By the end of the lesson the learner should be able:
Knowledge a. Discuss samples of relief sculptures. Skill b. Prepare papier Mache for making a mask Attitude c. Appreciate own and others sculptural work. |
1. How does one create models using clay?
2. What is papier Mache? Why is an armature important in creating clay models? |
Learner is guided to:
• prepare the pulp with emphasis on; – selection of paper (recycled paper) – shredding – soaking – pounding – addition of adhesive |
Samples of relief sculptures, pictures, samples of mask, assortment of papers, glue, containers, a strong stick for pounding.
TOP scholar Visual Arts P.b pg. 88 |
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| 3 | By the end of the lesson the learner should be able to:
Knowledge a. Discuss samples of relief sculptures. Skill b. Create a mask in low relief using papier Mache. Attitude c. Appreciate own and others sculptural work. |
1. How does one create models using clay?
2. What is papier Mache? Why is an armature important in creating clay models? |
Learner is guided to:
Create a low relief mask with the emphasis on; – balance – drying in a dry, cool and well aerated place – finishing (decorate by painting) |
Samples of relief sculptures, pictures, samples of mask, assortment of papers, glue, containers, a strong stick for pounding
TOP scholar Visual Arts P.b pg. 89-90. |
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| 2 | 1 | By the end of the lesson the learner should be able to:
Knowledge a. Discuss samples of relief sculptures. Skill b. Create a mask in low relief using papier Mache. Attitude c. Appreciate own and others sculptural work. |
1. How does one create models using clay?
2. What is papier Mache? Why is an armature important in creating clay models? |
Learner is guided to:
Create a low relief mask with the emphasis on; – balance – drying in a dry, cool and well aerated place – finishing (decorate by painting) |
Samples of relief sculptures, pictures, samples of mask, assortment of papers, glue, containers, a strong stick for pounding.
TOP scholar Visual Arts P.b pg. 89-90. |
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| 2 | Making an armature for modelling a sculpture in the round | By the end of the lesson the learner should be able to:
Knowledge a. Discuss virtual and actual samples of clay models in the round. Skill b. Create an animal form sculpture in the round using clay. Attitude c. Appreciate own and others sculptural work. |
1. How does one create models using clay?
2. What is armature? 3. Why is an armature important in creating clay models? |
Learner is guided to:
• make a sketch of an animal form and construct an armature with emphasis on; – size – materials (paper and strings) – balance |
Samples of clay models of sculpture in the round , pictures ,assortment of papers, strings.
TOP scholar Visual Arts P.b pg. 90-95. |
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| 3 | By the end of the lesson the learner should be able to:
Knowledge a. Discuss virtual and actual samples of clay models in the round. Skill b. Create an animal form sculpture in the round using clay. Attitude c. Appreciate own and others sculptural work. |
1. How does one create models using clay?
2. What is armature? Why is an armature important in creating clay models? |
Learner is guided to:
• make a sketch of an animal form and construct an armature with emphasis on; – size – materials (paper and strings) – balance |
Samples of clay models of sculpture in the round , pictures ,assortment of papers, strings.
TOP scholar Visual Arts P.b pg. 90-95. |
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| 3 | 1 | Model a sculpture in the round | By the end of the lesson the learner should be able to:
Knowledge a. Discuss virtual and actual samples of clay models in the round. Skill b. Create an animal form sculpture in the round using clay. Attitude c. Appreciate own and others sculptural work. |
1. How is stability achieved in sculpture?
2. Why is an armature important in creating clay models? |
Learner is guided to:
• model a sculpture in the round based on animal forms, using clay with emphasis on: – balance – solids – voids – planes – finishing (texturing) – drying in a dry, cool and well aerated place • display and critique own and others’ work |
Samples of clay models of sculpture in the round, pictures of animals, armature, clay.
TOP scholar Visual Arts P.b pg. 90-95. |
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| 2 | By the end of the lesson the learner should be able to:
Knowledge a. Discuss virtual and actual samples of clay models in the round. Skill b. Create an animal form sculpture in the round using clay. Attitude c. Appreciate own and others sculptural work. |
1. How is stability achieved in sculpture?
2. Why is an armature important in creating clay models? |
Learner is guided to:
• model a sculpture in the round based on animal forms, using clay with emphasis on: – balance – solids – voids – planes – finishing (texturing) – drying in a dry, cool and well aerated place • display and critique own and others’ work |
Samples of clay models of sculpture in the round, pictures of animals, armature, clay
TOP scholar Visual Arts P.b pg. 90-95. |
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| 3 | Designing armlets | By the end of the lesson the learner should be able to:
Knowledge a. Distinguish between ornaments and jewelry from samples. Skill b. Analyze materials and tools used in making wire armlets. Attitude c. Appreciate own and others armlets designs. |
1. What distinguishes jeweler from ornaments?
2. Why is it important to consider the choice of materials and tools when making jewelry? |
The learner is guided to:
• distinguish between ornaments and jewellery from virtual and actual sources with focus on; – function – materials – Tools |
Samples of ornaments and jewelry, pliable wires, pliers, round nose pliers, flat nose pliers, hammers, tin snip, wire cutters, scissors, ICT devices, internet connectivity, papers and pencils.
TOP scholar Visual Arts P.b pg. 96-104 |
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| 4 | 1 | By the end of the lesson the learner should be able to:
Knowledge a. Distinguish between ornaments and jewelry from samples. Skill b. Create an armlet using wires for self-expression. Attitude c. Appreciate own and others armlets designs. |
1. What distinguishes jewelry from ornaments?
2. Why is it important to consider the choice of materials and tools when making jewelry?
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The learner is guided to:
• distinguish between ornaments and jewellery from virtual and actual sources with focus on; – function – materials – Tools |
Samples of ornaments and jewelry, pliable wires, pliers, round nose pliers, flat nose pliers, hammers, tin snip, wire cutters, scissors, ICT devices, internet connectivity, papers and pencils.
TOP scholar Visual Arts P.b pg. 96-104 |
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| 2 | Making armlets | By the end of the lesson the learner should be able to:
Knowledge a. Name the materials and tools used in making wire armlets. Skill b. Create an armlet using wires for self-expression. Attitude c. Appreciate own and others armlets. |
1. What safety precautions should be observed when handling materials and tools used in making armlets?
2. How can one ensure that the armlets made from wire is safe for use? |
The learner is guided to:
• analyze materials and tools used in making a wire armlet with emphasis on; – suitable wire (pliable) – appropriate tools scissors, tin snips, hammer, pliers-, flat nose pliers, round nose pliers and wire cutters, pencil • design and create an armlet using wires with focus on; – design – size of the armlet – making the armlet – appropriate finishing (filing of the edges) |
Samples of ornaments and jewelry, pliable wires, pliers, round nose pliers, flat nose pliers, hammer, tin snip, wire cutters, scissors, ICT devices, internet connectivity, papers and pencils.
TOP scholar Visual Arts P.b pg. 96-104 |
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| 3 | By the end of the lesson the learner should be able to:
Knowledge a. Name the materials and tools used in making wire armlets. Skill b. Create an armlet using wires for self-expression. Attitude c. Appreciate own and others armlets. |
What safety precautions should be observed when handling materials and tools used in making armlets?
How can one ensure that the armlets made from wire is safe for use? |
The learner is guided to:
• analyze materials and tools used in making a wire armlet with emphasis on; – suitable wire (pliable) – appropriate tools scissors, tin snips, hammer, pliers-, flat nose pliers, round nose pliers and wire cutters, pencil • design and create an armlet using wires with focus on; – design – size of the armlet – making the armlet – appropriate finishing (filing of the edges) |
Samples of ornaments and jewelry, pliable wires, pliers, round nose pliers, flat nose pliers, hammer, tin snip, wire cutters, scissors, ICT devices, internet connectivity, papers and pencils.
TOP scholar Visual Arts P.b pg. 96-104 |
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| 5 | 1 | Designing earrings | By the end of the lesson the learner should be able to:
Knowledge a. describe earrings made using natural materials for inspiration to create own. Skill b. observe samples of earrings. Attitude c. Appreciate own and others erring’s. |
What materials can be used to make earrings?
Why is it important to consider the finish and design while making earrings? |
The learner is guided to:
• study and analyze samples of earrings from virtual or actual sources made using natural materials with focus on; -design -materials used (inedible seeds, wood, horn, shells and animal hide) -finishing techniques |
Samples of erring’s, pliable wires, natural materials for making the earring such as inedible seeds, shells, wood, seed pods, horns and stick, oil paints, brushes,
TOP scholar Visual Arts P.b pg. 105-114 |
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| 2 | By the end of the lesson the learner should be able to:
Knowledge a. describe earrings made using natural materials for inspiration to create own. Skill b. collect materials needed for the making of an earring. Attitude c. Appreciate own and others erring’s. |
What materials can be used to make earrings?
Why is it important to consider the finish and design while making earrings? |
The learner is guided to:
• study and analyze samples of earrings from virtual or actual sources made using natural materials with focus on; -design -materials used (inedible seeds, wood, horn, shells and animal hide) -finishing techniques
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Samples of erring’s, pliable wires, natural materials for making the earring such as inedible seeds, shells, wood, seed pods, horns and stick, oil paints, brushes,
TOP scholar Visual Arts P.b pg. 105-114 |
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| 3 | Making earring | By the end of the lesson the learner should be able to:
Knowledge a. describe earrings made using natural materials for inspiration to create own. Skill b. Make earrings using natural materials. Attitude c. Appreciate own and others erring’s. |
What safety precautions should be observed when handling materials and tools used in making earrings?
How would one effectively display earrings for self-expression? |
The learner is guided to:
• design and create earrings with emphasis on: – design – preparation of the materials (cleaning, cutting/drilling holes twisting, folding, sanding) – decoration (painting, incising, burning, varnishing, staining) – stringing the materials – finishing (clasps, hooks,stoppers, spacers,) |
Samples of erring’s, pliable wires, natural materials for making the earring such as inedible seeds, shells, wood, seed pods, horns and stick, oil paints, brushes,
TOP scholar Visual Arts P.b pg. 105-114 |
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| 6 | 1 | By the end of the lesson the learner should be able to:
Knowledge a. describe earrings made using natural materials for inspiration to create own. Skill b. Make earrings using natural materials. Attitude c. Appreciate own and others erring’s. |
What safety precautions should be observed when handling materials and tools used in making earrings?
How would one effectively display earrings for self-expression? |
The learner is guided to:
• design and create earrings with emphasis on: – design – preparation of the materials (cleaning, cutting/drilling holes twisting, folding, sanding) – decoration (painting, incising, burning, varnishing, staining) – stringing the materials – finishing (clasps, hooks,stoppers, spacers,) |
Samples of erring’s, pliable wires, natural materials for making the earring such as inedible seeds, shells, wood, seed pods, horns and stick, oil paints, brushes,
TOP scholar Visual Arts P.b pg. 105-114 |
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| 2 | Frame loom | By the end of the lesson the learner should be able to:
Knowledge a. Analyze samples of 2/2 twill weave structure. Skill b. Make a frame loom using mitre joints. Attitude c. Appreciate own and others work. |
What articles can be made using the weaving techniques?
How is a twill weave formed?
Why is a loom necessary for weaving?
|
The learner is guided to:
• analyze virtual or actual 2/2 twill weave samples with focus on: – materials and tools used – weave structure – weaving device (frame loom) • make a frame loom using mitre joints to be used in weaving with emphasis on: – alignment joints – nail spacing – sturdiness of the frame |
Papers, pencils, digital devices, samples of woven items, samples of a frame loom and sample of weaving
TOP scholar Visual Arts P.b pg. 115-119 |
||||
| 3 | By the end of the lesson the learner should be able to:
Knowledge a. Analyze samples of 2/2 twill weave structure. Skill b. Make a frame loom using mitre joints. Attitude c. Appreciate own and others work. |
What articles can be made using the weaving techniques?
How is a twill weave formed?
Why is a loom necessary for weaving?
|
The learner is guided to:
• analyze virtual or actual 2/2 twill weave samples with focus on: – materials and tools used – weave structure – weaving device (frame loom) • make a frame loom using mitre joints to be used in weaving with emphasis on: – alignment joints – nail spacing – sturdiness of the frame |
Papers, pencils, digital devices, samples of woven items, samples of a frame loom and sample of weaving
TOP scholar Visual Arts P.b pg. 115-119 |
|||||
| 7 | 1 | By the end of the lesson the learner should be able to:
Knowledge a. Analyze samples of 2/2 twill weave structure. Skill b. Make a frame loom using mitre joints. Attitude c. Appreciate own and others work. |
What articles can be made using the weaving techniques?
How is a twill weave formed?
Why is a loom necessary for weaving?
|
The learner is guided to:
• analyze virtual or actual 2/2 twill weave samples with focus on: – materials and tools used – weave structure – weaving device (frame loom) • make a frame loom using mitre joints to be used in weaving with emphasis on: – alignment joints – nail spacing – sturdiness of the frame |
Papers, pencils, digital devices, samples of woven items, samples of a frame loom and sample of weaving
TOP scholar Visual Arts P.b pg. 115-119 |
||||
| 2 | Preparing weaving materials | By the end of the lesson the learner should be able to:
Knowledge a. State the meaning of a frame loom. Skill b. Prepare recyclable materials for weaving. Attitude c. Appreciate own and others prepared weaving materials. |
How is a frame loom made?
Why should a frame loom be firm and sturdy?
Why are the nails on a frame loom fixed with part of the nails sticking out? |
The learner is guided to:
• select and reuse flexible weaving materials available in the locality, namely; – woolen yarn – Plain coloured cloth |
Papers, pencils, digital devices, samples of woven items, samples of a frame loom and sample of weaving
TOP scholar Visual Arts P.b pg. 120 |
||||
| 3 | Weaving using a frame loom | By the end of the lesson the learner should be able to:
Knowledge a. State the meaning of a frame loom. Skill b. Create a scarf in 2/2 twill weave using two colors. Attitude c. Appreciate own and others woven artwork. |
How is a frame loom made?
Why should a frame loom be firm and sturdy?
Why are the nails on a frame loom fixed with part of the nails sticking out? |
The learner is guided to:
• warp the loom in pairs, with emphasis on: – equal tension of warps • weave (2/2) a scarf in two colours on a frame loom, in pairs, taking into account the following weaving defects: – Waisting – Sagging
|
Papers, pencils, digital devices, samples of woven items, samples of a frame loom and sample of weaving
TOP scholar Visual Arts P.b pg. 122-128 |
||||
| 8 | 1 | By the end of the lesson the learner should be able to:
Knowledge a. State the meaning of a frame loom. Skill b. Create a scarf in 2/2 twill weave using two colors. Attitude c. Appreciate own and others woven artwork. |
How can you ensure uniform tension of the warp on a frame loom?
How does waisting differ from sagging?
How can you prevent the wasting and sagging defects on a woven item? |
The learner is guided to:
• warp the loom in pairs, with emphasis on: – equal tension of warps • weave (2/2) a scarf in two colours on a frame loom, in pairs, taking into account the following weaving defects: – Waisting – Sagging
|
Papers, pencils, digital devices, samples of woven items, samples of a frame loom and sample of weaving
TOP scholar Visual Arts P.b pg. 122-128 |
||||
| 2 | By the end of the lesson the learner should be able to:
Knowledge a. State the meaning of a frame loom. Skill b. Create a scarf in 2/2 twill weave using two colors. Attitude c. Appreciate own and others woven artwork. |
How can you ensure uniform tension of the warp on a frame loom?
How does waisting differ from sagging?
How can you prevent the waisting and sagging defects on a woven item? |
The learner is guided to:
• warp the loom in pairs, with emphasis on: – equal tension of warps • weave (2/2) a scarf in two colours on a frame loom, in pairs, taking into account the following weaving defects: – Waisting – Sagging
|
Papers, pencils, digital devices, samples of woven items, samples of a frame loom and sample of weaving
TOP scholar Visual Arts P.b pg. 122-128 |
|||||
| 3 | Finishing techniques | By the end of the lesson the learner should be able to:
Knowledge a. describe a 2/2 twill scarf weaved using two colors. Skill b. Apply finishing techniques on the woven scarf. Attitude c. Appreciate own and others woven artworks. |
Why should one finish a woven scarf?
Which techniques can be used to finish a woven scarf? |
The learner is guided to:
• explore and apply finishing techniques on the woven scarf with emphasis on; – knotting – tasseling (trimming /alignment, uniformity, colour alternation) • display and critique own and others’ work. |
Papers, pencils, digital devices, samples of woven items, samples of a frame loom and sample of weaving
TOP scholar Visual Arts P.b pg. 122-128 |
||||
| 9 | END YEAR ASSESSMENT/CLOSING | ||||||||
CHEMISTRY FORM 1, 2, 3 & 4 FREE PDF NOTES
UNIT 1: ATOMIC STRUCTURE AND THE PERIODIC TABLE.
Checklist
- Structure of the atom
- The subatomic particles;
- Protons
- Neutrons
- Electrons;
GET THE FREE CHEMISTRY NOTES IN PDF BELOW;
Chemistry High School notes for form 1-4 (Free updated pdf downloads)
CHEMISTRY FORM FOUR NOTES: NEW
- Atomic number and mass number
- Isotopes;
- Energy levels and electron arrangements.
- The periodic table;
- Groups
- Periods;
- Relative atomic mass and isotopes;
- Ion formation;
- Ion;
- Cations;
- Anions;
- Ionization energy;
- Electron affinity;
- Valency and oxidation numbers;
- Chemical formula;
- Chemical equations
- Balancing chemical equations;
The atom:
– Refers to the smallest particle of an element hat can take part in a chemical reaction;
– It has an average diameter of 10-8 cm with a nucleus of about 10-13 cm;
Parts of an atom
– The atom is made of two main parts:
- The nucleus
- The energy levels;
- The nucleus:
– Is the positively charged part of an atom;
– The nucleus contains two subatomic particles; neutrons and protons;
– The positive charge is due to presence of protons;
– The nuclei of all atoms contain neutrons except the hydrogen atom;
– The protons and the neutrons are together referred to as the nucleons;
- The energy levels.
– They contain the electrons;
– Electrons are so small and move so fast that their path cannot be traced directly;
– Thus the energy level simple represents the region where the electrons are most likely to be found;
Structure of the atom.
Note:
– The atom can still however be split into smaller particles termed the sub-atomic particles;
The sub-atomic particles.
– Are generally three:
- Protons;
- Neutrons;
- Protons;
- Protons.
– Are the positively charged sub-atomic particles;
– Are found in the nucleus and thus form part of the nucleons;
– The number of protons in the nucleus is equal to the number of electrons in the energy levels;
- Neutrons.
– Are neutrally charged sub-atomic particles found in the nucleus of the atom;
– They are thought to probably prevent the positively charged protons from getting too close to each other;
- Electrons.
– Are negatively charged sub-atomic particles found in the energy levels;
– The number of electrons in the energy levels is equal to the number of protons in the nucleus;
– This makes the atom to be electrically neutral;
Atomic number and mass number.
Atomic number.
– Refers to the number of protons in the nucleus of an atom;
Examples.
- Sodium has 11 protons I the nucleus and thus said to have atomic number 11;
- Chlorine has 17 protons in the nucleus and thus said to have atomic number 17;
Mass number;
– Refers to the sum of the number of protons and neutrons in an atom of an element;
Examples:
- Sodium has 2 neutrons and 11 protons hence a mass number of 23;
- Chlorine has 18 neutrons and 17 protons hence a mass number of 35.
Notation of atomic number and mass number;
– Both atomic number and mass number of an element can be written along with the symbol of an element;
Mass number;
– Is conventionally represented as a superscript to the left of the symbol;
Examples:
Sodium; 23Na;
Magnesium 24Mg;
Atomic number;
– Is conventionally represented as a superscript to the left of the symbol;
Examples:
Sodium; 11Na;
Magnesium 12Mg;
Thus the elements can be conventionally represented as:
Sodium 23Na
Magnesium 24Mg
Atomic properties of the first 20 elements.
| Element | Symbol | Number
of electrons |
Number
of protons |
Number
Of neutrons |
Atomic number | Mass number |
| Hydrogen | H | |||||
| Helium | He | |||||
| Lithium | Li | |||||
| Beryllium | Be | |||||
| Boron | B | |||||
| Carbon | C | |||||
| Nitrogen | N | |||||
| Oxygen | O | |||||
| Fluorine | F | |||||
| Neon | Ne | |||||
| Sodium | Na | |||||
| Magnesium | Mg | |||||
| Aluminium | Al | |||||
| Silicon | Si | |||||
| Phosphorus | P | |||||
| Sulphur | S | |||||
| Chlorine | Cl | |||||
| Argon | Ar | |||||
| Potassium | K | |||||
| Calcium | Ca |
Isotopes.
– Are atoms of the same element with same atomic number but different mass number due to different number of neutrons.
Examples of isotopes.
| Element | Isotope | Atomic No. | Number of protons | Number of neutrons | Mass number | Isotopic representation |
| Hydrogen | ||||||
| Carbon | ||||||
| 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 Meneleevs periodic law which states that:
The properties of elements are a periodic functions of their relative atomic masses
– The modern periodic law itself states that:
The properties of elements are a periodic functions of their atomic numbers
Design of the modern periodic table.
It has vertical columns called groups and horizontal rows called periods.
- Groups:
– Are the vertical columns of a periodic table.
– Are eight in number; ad numbered in capital Roman numerals I all through to VIII.
Note: group VIII is also called group zero because the elements have little tendency to gain o lose electrons during chemical reactions
– Between group 2 and group 3 is a group of elements called the transition metals;
- Periods:
– Are the horizontal rows in a periodic table.
– They are 8 in number in a modern periodic table.
- Transition metals.
– Are elements that form a shallow rectangle between group II and group III.
– These elements are generally metallic and hence the name transition metals”
– They are not fitted in any group because they have variable valencies.
– Are sometimes called the d-block elements.
– They are much less reactive than the elements in groups I and II
– They have some unique characteristics that make them not fit in the 8 groups of the periodic table
– These include:
- Have variable valencies; hence show different oxidation states in their compounds;
- They form coloured compounds as solids and in aqueous solutions;
- Have very high melting and boiling points (than metals in groups I and II).
- They do not react with water;
- Have very high densities (compared to metals in groups I ad II)
- Lanthanides and Actinides.
– They form a block of elements within the transition metals.
– Are sometimes called the inner transition metals.
– The lanthanides consist of 14 elements from Cerium (Ce) to Lutetium (Lu).
– The Actinides are the 14 elements from Thorium (Tho) to Lawrencium (Lr).
Placing an elements in the periodic table
– The position of an elements in the periodic table is governed by the atomic number and hence the
electron arrangement.
- The period:
– The period to which an element belongs is determined by the number of energy levels.
– The number of energy levels is equal to the period to which an elements belongs.
Examples:
| Elements | Symbol | Atomic number | Electron arrangement | Number of energy levels | Period |
| Lithium | Li | 3 | 2.1 | 2 | 2 |
| Sodium | Na | 11 | 2.8.1 | 3 | 3 |
| Calcium | Ca | 20 | 2.8.8.2 | 4 | 4 |
| Nitrogen | N | 7 | 2.5 | 2 | 2 |
| Helium | He | 2 | 2 | 1 | 1 |
- Group:
– The group to which an element belongs to is governed by the number of electrons in the
outermost energy level.
– The number of electrons in the outermost energy level is equal to the group to which the element
belongs.
Examples:
| Elements | Symbol | Atomic number | Electron arrangement | Outermost electrons | Group |
| Potassium | K | 19 | 2.8.8.1 | 1 | 1 |
| Aluminium | Al | 13 | 2.8.3 | 3 | 3 |
| Silicon | Si | 14 | 2.8.4 | 4 | 4 |
| Oxygen | O | 8 | 2.6 | 6 | 6 |
| Chlorine | Cl | 17 | 2.8.7 | 7 | 7 |
Diagram of a full periodic table.
Note: In the modern periodic table atomic masses are used instead of mass numbers. The atomic
masses are preferable because they take care of elements with isotopes unlike mass numbers.
Diagram: part of the periodic table showing the first 20 elements
Relative Atomic Mass and Isotopes.
- Introduction:
– The masses of individual atoms of elements are very negligible and thus quite difficult to weigh.
– On average the mass of an atom is approximately 10-22g which cannot be determined by an
ordinary laboratory balance.
– For this reason the mass of atom has been expressed relative to that of a chosen standard element
hence the term relative atomic mass.
– The initial reference element was hydrogen which was later replaced with oxygen.
– Later the oxygen scale was found unsuitable;
Reason:
Oxygen exists I several isotopes and thus led to problems when deciding the mass of an oxygen atom.
– For this reason oxygen was replaced with carbon as the reference atom and to date relative atomic
masses of elements are based on an atom of carbon-12 (note that carbon is isotopic and exists as
Carbon -12 or carbon-14).
- Definition:
Relative atomic mass (R.A.M) of an element refers to the average mass of an atom of the element
compared with a twelfth (1/12) of an atom of carbon-12.
|
- Measurement of Relative atomic mass
– RAM of elements is determined by an instrument called Mass Spectrometer.
– The instrument can also be used to determine the relative abundance of isotopes.
– The use of a mass spectrometer in determining the RAM of elements is called mass spectrometry.
- How mass spectrometry works.
– In the mass spectrometer atoms and molecules are converted into ions.
– The ions are then separated as a result of the deflection which occurs in a magnetic field.
– Each ion (from an atom, isotope or molecule) gives a deflection which is amplified into a trace.
– The height of each peak measures the relative abundance of the ion which gives rise to that peak.
Note:
– Generally the relative atomic mass of an element is closest in value to the mass of the most
abundant isotope of the element.
Example: Diagram of a spectrometer trace for Lithium
- Explanations:
– The trace has two peaks indicating that there are two isotopes for lithium.
– The fist peak occurs at a relative isotopic mass of 6 and the second at 7; these are the RAM of the two
isotopes respectively.
– The percentage abundance of the isotope with RAM of 6 (6Li) is 9 while the RAM of the isotope with
RAM 7 (7Li) is 91.
Calculating relative atomic masses of isotopic elements.
– Information form a spectrometer trace is usually extracted and used in calculation the relative atomic
mass of elements.
Worked examples.
- The mass spectrum below shows the isotopes present in a sample of lithium.
(i). Use this mass spectrum to help you complete the table below for each lithium isotope in the sample. (3 marks)
| Isotope | Percentage composition | Number of | |
| Protons | Neutrons | ||
| 6Li | |||
| 7Li | |||
(ii). Calculate the relative atomic mass of this lithium sample. Your answer should be given to three significant figures. (3 marks)
- Element X with atomic number 16 has two isotopes. ⅔ of 33X and ⅓ of 30X. What is the relative atomic mass of element X? (2 marks)
- Calculate the relative atomic mass of an element whose isotopic masses and relative abundances are shown below. (2 marks)
- A neutral atom of silicon contains 14 electrons, 92% of silicon – 28, 5% silicon 29 and 3% silicon 30
(i). What is the atomic number of silicon? (1mark)
(ii). Calculate the relative atomic mass of silicon. (1mark)
- Oxygen exists naturally as isotopes of mass numbers 16, 17 and 18 in the ratio 96:2:2 respectively. Calculate its R.A.M (2 marks)
- Calculate the relative atomic mass of potassium from the isotopic composition given below.
Isotope Relative abundance
39K 93.1
40K 0.01
41K 6.89
- Sulphur and sulphur compounds are common in the environment.
(a). A sample of sulphur form a volcano contained 88% by mass of 32S and 12% by mass of 34S.
(i). Complete the table below to show the atomic structure of each isotope of sulphur.
| Isotope | Number of | ||
| Protons | Neutrons | Electrons | |
| 32S | |||
| 34S | |||
(ii). Define relative atomic mass. (2 marks)
(iii). Calculate the relative atomic mass of the volcanic sulphur. (2 marks)
- Iridium, atomic number 77, is a very dense metal. Scientists believe that meteorites have deposited virtually all the iridium present on earth. A fragment of a meteorite was analysed using a mass spectrometer and a section of the mass spectrum showing the isotopes present in iridium is shown below.
(a). Explain the term isotopes. (1mark)
(b). Use the mass spectrum to help you complete the table below for each iridium isotope in the meteorite.
| Isotope | Percentage composition | Number of | |
| 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.
- Deriving the formula of sodium chloride.
| Elements | Sodium | Chlorine |
| Formula | Na | Cl |
| valencies | 1 | 1 |
| balancing | x1 | x1 |
Balancing ratios as subscripts: Na1Cl1
Formula: NaCl
Explanation:
– For sodium to combine with chlorine to form sodium chloride, sodium loses an electron while
chlorine gains an electron.
– Thus every sodium atom needs only a single chlorine atom for both t be fully stable
Note:
– When the balancing ratio // subscript is 1, it is usually not written since the symbol of the
element itself represents a single atom.
- Deriving the formula of magnesium chloride.
| Elements | Magnesium | Chlorine |
| Formula | Mg | Cl |
| valencies | 2 | 1 |
| balancing | X1 | X2 |
Balancing ratios as subscripts: Mg1Cl2
Formula: MgCl2
- Deriving the formula of magnesium oxide.
| Elements | Magnesium | Oxygen |
| Formula | Mg | O |
| valencies | 2 | 2 |
| balancing | X1 | X1 |
Balancing ratios as subscripts: Mg1O1
Formula: MgO
Further worked examples.
- Derive the chemical formula of each of the following compounds.
- Calcium fluoride
- Carbon (II) oxide
- Carbon (IV) oxide
- Aluminium nitrate
- Calcium hydrogen carbonate
- Complete the table below for elements A, B and C
| Element | Valency | Chemical formula of various compounds | |||||
| hydroxides | Sulphates | carbonates | nitrates | phosphates | Hydrogen carbonates | ||
| A | 1 | ||||||
| B | 2 | ||||||
| B | 3 | ||||||
Chemical equations.
– Refers to representations of a chemical reaction by means of chemical symbols and formula.
- Key features of a chemical equation.
– The correct formulae of the reactants are on the left of the equation.
– The correct formulae of the products are on the right of the equation.
– The reactants and products are separated by an arrow pointing to the right.
– The state symbols of the reactants and products must be stated as subscripts to the right of the symbols
– The number of each atom on the reactants side must be equal to the number of the same atom on the products side.
Example.
Reaction between hot copper metals and oxygen gas.
- Word equation: Copper + oxygen → Copper (II) oxide.
- Chemical equation: 2Cu(s) + O2(g) → 2CuO(s).
Balancing chemical equations.
– A chemical equation is only valid if it is balanced.
– A chemical equation is said to be balanced if the number of each atom on the reactants side is equal to that on the products side.
– This is because atoms are neither created nor destroyed during a chemical reaction.
Rules and guidelines in balancing chemical equations.
- Step 1: Write the chemical equation in words.
Example: Copper metal + oxygen gas.
- Step II: Write the correct formulae of both reactants and products
Example: Cu + O2 → CuO
- Step III: Check whether the number of atoms of each element on the reactants side is equal to that on the products side.
– If equal proceed to step (V);
– If not equal proceed to step (IV).
Example: Cu + O2 → CuO
– In this case there are two oxygen atoms on the reactants side yet there is only one oxygen atom on the products side. Thus we proceed to step IV
- Step IV: Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple.
Example: Cu + O2 → CuO
– In this case the chemical formula with the unbalanced atom is CuO on the products side. We thus multiply it by 2.
– The new equation now reads: Cu + O2 → 2CuO
- Step V: check again to ensure that all atoms are balanced.
– If all atoms are balanced proceed to step VI.
– If not then repeat step IV until all atoms are balanced.
Example: Cu + O2 → 2CuO
– In this case multiplying CuO by 2 offsets the balancing of Cu; which is now unbalanced!
– We therefore repeat step IV in order to balance Cu.
– There is only 1 Cu atom on the reactants side yet there are 2 Cu atoms on the products side.
– We thus multiply the formula with the unbalanced atom (s) by the lowest common multiple, in this case 2.
– The new equation at this step thus becomes: 2Cu + O2 → 2CuO
– We then repeat step V; in this case all atoms are now balanced.
- Step VI: The physical states of the reactants and the products are then indicated.
– If this is not done the chemical equation is considered incorrect.
- Types of state symbols.
– There four main state symbols.
- Solid; denoted as (s)
- Liquid; denoted as (l)
- Aqueous (in solution in water); denoted as (aq)
- Gaseous; denoted as (g)
– In a chemical equation the state symbols are written with their denotations as subscripts to the right of the chemical formulae.
Example: 2Cu(s) + O2(g) → 2CuO(s)
Thus the balanced chemical equation for the reaction between copper metal ad oxygen is:
- 2Cu(s) + O2(g) → 2CuO(s)
Worked examples:
- Balance equations for each of the following reactions.
- Sodium hydroxide and dilute hydrochloric acid
- Zinc oxide and dilute sulphuric (VI) acid
- Zinc metal and dilute nitric (V) acid
- Calcium carbonate and dilute sulphuric (VI) acid
- Sodium and water
- Balance each of the following equations.
- Mg(s) + HCl(aq) →MgCl2(aq) + H2(g)
- Na(s) + H2O(l) → NaOH(aq) + H2(g)
- NaOH(aq) + H2SO4(aq) →Na2SO4(aq) + H2O(l)
- CuCO3(s) + HNO3(aq) → Cu(NO3)2(aq) + CO2(g) + H2O(l)
- H2S(g) + O2(g) → O2(g) + H2O(l)
- C2H6(g) + O2(g) → CO2(g) + H2O(l)
- Pb(NO3)2(s) → PbO(s) + NO2(g) + O2(g)
- Fe(s) + Cl2(g) → FeCl3(s)
- Al(s) + H2SO4(aq) → Al2(SO4)3(aq) + H2(g).
UNIT 2: CHEMICAL FAMILIES; PATTERNS AND PROPERTIES.
Checklist.
– Meaning of chemical families.
– Main chemical families;
- Alkali metals
- Meaning and members
- Trends down the group
- Atomic radius
- Ionic radius
- Energy levels
- Physical properties
- Appearance
- Ease of cutting
- Melting and boiling [points
- Electrical conductivity
- 1st ionization energy
- Chemical properties
- Burning in air
- Exposure to air
- Reaction with water
- Reaction with chlorine
- Similarity of ions and formula of compounds of alkali metals
- Uses of alkali metals.
- Alkaline earth metals
- Meaning and members
- Trends down the group
- Atomic radius
- Ionic radius
- Energy levels
- Physical properties
- Melting and boiling [points
- Electrical conductivity
- 1st ionization energy
- 2nd ionization energy
- Chemical properties
- Reaction with air and water
- Reaction with steam
- Reaction with chlorine
- Reaction with dilute acids
- Similarity of ions and formula of compounds of alkaline earth metals
- Uses of alkaline earth metals.
- The Halogens
- Meaning and members
- Trends down the group
- Atomic radius
- Ionic radius
- Energy levels
- Physical properties
- Preparation and properties of chlorine
- Appearance and physical states of halogens at room temperature
- Melting and boiling points
- Electrical conductivity
- Chemical properties
- Ion formation
- Electron affinity
- Reaction with metals
- Reaction with water
- Reactivity trend of halogens.
- Similarity of ions and formula of compounds of alkali metals
- Uses of alkali metals.
- The noble gases
- Meaning and members
- Trends down the group
- Atomic radius
- Ionic radius
- Energy levels
- Physical properties
- Melting and boiling [points
- Electrical conductivity
- 1st ionization energy
- Chemical properties
- Uses of the noble gases
Introduction:
– Elements are classified and hence positioned in the periodic table based on the number of valence electrons and the number of energy levels.
– The number of valence electrons is equal to the group to which the element belongs; while the number of energy levels is equal to the period to which the element belongs.
– Elements in the same group are said to belong to the same chemical family.
|
Characteristics of a chemical family:
– have same number of valence electrons;
– Show a uniform gradation in physical properties;
– have similar chemical properties;
Main chemical families.
– Four main chemical families will be studies in this section.
- The Alkali metals
- The Alkaline earth metals
- The halogens
- The noble gases
- The Alkali metals.
– Are the elements with one valence electron and hence in group I of the periodic table.
– All are metallic in nature.
– The members of the family in order down the group is as follows:
- Lithium
- Sodium
- Potassium
- Rubidium
- Caesium
- Francium
Electron arrangement of the first three alkali metals.
| Elements | Electron arrangement |
| Lithium | 2.1 |
| Sodium | 2.8.1 |
| Potassium | 2.8.81 |
Diagram: Part of periodic table showing the alkali metals
Gradation in properties of alkali metals.
- Atomic and ionic radius.
- Atomic radius:
– Refers to the distance between the centre of the nucleus of an atom and the outermost energy level occupied by electron(s)
- Ionic radius:
– Refers to the distance between the centre of the nucleus of an ion and the outermost energy level occupied by electron(s)
Trend:
– The ionic radius and the atomic radius of alkali metals increase down the group.
Reason:
– There is an increase in the number of energy levels down the group from lithium to Francium.
Illustration:
– Lithium (2.1) has only 2 energy levels; sodium (2.8.1) has 3 energy levels while potassium (2.8.8.1) has 4 energy levels.
– Thus the outermost electron in potassium is further from the nucleus than the outermost electron in sodium and lithium.
- Atomic and ionic radius of the same element.
– For the same alkali metals the atomic radius is larger than the ionic radius.
Reason:
– Alkali metals form ions by losing the valence electron, lading to the loss of an entire outermost energy level. Thus the atoms have more energy levels than the corresponding ion hence a larger radius in the atom than I the ion.
Illustration
– Potassium atom has electron arrangement of 2.8.8.1 hence 4 energy levels.
– During ion formation potassium reacts by losing the single valence electron to acquire a new electron arrangement of 2.8.8 hence 3 energy levels.
– Thus the ion has a smaller radius than the atom.
Diagrammatically: Potassium atom and potassium ion
Summary: changes in atomic and ionic radius among alkali metals.
| Element | Symbol | Atomic number | Atomic radius (nm) | Ionic radius (nm) |
| Lithium | Li | 3 | 0.133 | 0.060 |
| Sodium | Na | 11 | 0.157 | 0.095 |
| Potassium | K | 19 | 0.203 | 0.133 |
Physical properties of alkali metals.
- Appearance.
– Alkali metals have metallic luster when freshly cut. This refers to a shiny appearance on th cut surface.
– This surface however tarnishes due to reaction on exposure to air.
- Ease of cutting.
– They are soft and easy to cut.
– The softness and ease of cutting increase down the group.
Reason:
– Alkali metals have giant metallic structures held together by metallic bonds.
– Metallic bond is due to attraction between the positively charged nucleus of one atom and the electrons in the outermost energy level of the next atom.
– Thus the force of attraction is stronger is smaller atoms than in larger atoms.
– The increase in atomic radius down the group implies that the strength of metallic bonds also decrease down the group (hence ease of cutting and softness).
- They have relatively low melting and boiling points (in comparison to other metals).
Reason: they have relatively weaker metallic bonds.
– The melting and boiling points decrease down the group.
Reason: – The size of the atoms increase down the group due to increasing number of energy levels hence decrease in the strength of the metallic bonds (down the group).
- Electrical conductivity.
– Alkali metals are good conductors of heat ad electricity.
Reason: they have delocalized electrons in the outermost energy level.
– The electrical conductivity is similar for all alkali metals.
Reason: all alkali metals have the same number of delocalized electron (a single electron) in the outermost energy level.
Note:
– In metals the electrons in the outermost energy level do not remain in one fixed position. They move randomly throughout the metallic structure and are thus said to be delocalsised.
- Ionization energy.
|
– The number of ionization energies an element may have is equivalent to the number of valence electrons.
– The first ionization energy is the minimum energy required to remove the first electron from the outermost energy level of an atom in its gaseous state.
– The first ionization among alkali metals decreases down the group.
Reason:
– The effective force of attraction of on the outermost electron by the positive nucleus decreases with increasing atomic size and distance from the nucleus.
– Note that the atomic radius increases down the group due to increase in the number of energy levels.
Summary on physical properties of Alkali metals.
| Element | Appearance | Ease of cutting | Melting point (oC) | Boiling point (oC) | Electrical conductivity | Atomic radius (nm) | Ionic radius (nm) |
| Lithium | Silver white | Slightly hard | 180 | 1330 | Good | 0.133 | 520 |
| Sodium | Shiny white | Easy | 98 | 890 | Good | 0.157 | 496 |
| Potassium | Shiny grey | Easy | 64 | 774 | Good | 0.203 | 419 |
Chemical properties of alkali metals.
- Reaction with air.
– When exposed to air alkali metals react with atmospheric moisture to form the corresponding metal hydroxide and hydrogen gas.
- General equation:
Metal + Water → Metal hydroxide + hydrogen gas.
– The metal hydroxide further reacts with atmospheric carbon (IV) oxide to form hydrated metal carbonate.
- General equation:
Metal hydroxide + carbon (IV) oxide → Hydrated metal carbonate.
Examples:
- Lithium
With moisture: 2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g);
Then with carbon (IV) oxide:2LiOH(aq) + CO2(g) → Li2CO3.H2O(s);
- Sodium
With moisture: 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g);
Then with carbon (IV) oxide: 2NaOH(aq) + CO2(g) → Na2CO3.H2O(s);
- Potassium
With moisture: 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g);
Then with carbon (IV) oxide: 2KOH(aq) + CO2(g) → K2CO3.H2O(s);
- Burning in air.
– Alkali metals burn in air with characteristic flame colours to form corresponding metal oxides.
Examples:
– Burns in air to form lithium oxide as the only product.
Equation:
4Li(s) + O2(g) → 2Li2O(s)
– Burns in air to with a yellow flame to form sodium oxide as the only product.
Equation:
4Na(s) + O2(g) → 2Na2O(s)
Note:
– When burned in air enriched with oxygen or pure oxygen sodium burns with a yellow flame to fom sodium peroxide (instead of sodium oxide).
Equation: 2Na(s) + O2(g) → Na2O2(s)
– Burns in air with a lilac flame to form potassium oxide as the only product.
Equation:
4K(s) + O2(g) → 2K2O(s)
- Reaction with water.
– Alkali metals react with water to form the corresponding hydroxides and hydrogen gas.
Examples:
Procedure:
– A small piece of potassium metal is cut and dropped into a trough containing water;
– The resultant solution is tested with litmus paper;
Diagram of apparatus:
Observations and explanations:
– The metal floats on the water surface; because it is less dense than water;
– A hissing sound is produced; due to production of hydrogen gas;
– It explosively melts into a silvery ball then disappears because reaction between water and potassium is exothermic (produces heat). The resultant heat melts the potassium due to its low melting point.
– It darts on the surface; due to propulsion by hydrogen;
– The metal bursts into a lilac flame; because hydrogen explodes into a flame which then burns the small quantities potassium vapour produced during the reaction;
– The resultant solution turns blue; because potassium hydroxide solution formed is a strong base;
(b). Reaction equations.
Equation I
2K(s) + 2H2O(l) → 2KOH(aq) + H2(g);
Equation II
4K(s) + O2 (g) → 2K2O(s);
Equation III:
K2O(s) + H2O(l) → 2KOH(aq)
Effect of resultant solution on litmus paper;
– Litmus paper turns blue; sodium hydroxide formed is highly soluble in water; releasing a large number of hydroxyl ions which result into alkaline conditions // high pH;
Procedure:
– A small piece of sodium metal is cut and dropped into a trough containing water;
– The resultant solution is tested with litmus paper;
Diagram of apparatus:
Observations and explanations:
– The metal floats on the water surface; because it is less dense than water;
– A hissing sound is produced; due to production of hydrogen gas;
– It vigorously melts into a silvery ball then disappears because reaction between water and sodium is exothermic (produces heat). The resultant heat melts the sodium due to its low melting point.
– It darts on the surface; due to propulsion by hydrogen;
– The metal may burst into a golden yellow flame; because hydrogen may explode into a flame which then burns the sodium;
– The resultant solution turns blue; because sodium hydroxide solution formed is a strong base;
(b). Reaction equations.
Equation I
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g);
Equation II
4Na(s) + O2 (g) → 2Na2O(s);
Equation III:
Na2O(s) + H2O(l) → 2NaOH(aq)
Effect of resultant solution on litmus paper;
– Litmus paper turns blue; sodium hydroxide formed is highly soluble in water; releasing a large number of hydroxyl ions which result into alkaline conditions // high pH;
Procedure:
– A small piece of lithium metal is cut and dropped into a trough containing water;
– The resultant solution is tested with litmus paper;
Diagram of apparatus:
Observations and explanations:
– The metal floats on the water surface; because it is less dense than water;
– A hissing sound is produced; due to production of hydrogen gas;
– It reacts less vigorously than sodium and des not melt since the melting point of lithium is relatively higher.
– It darts on the surface; due to propulsion by hydrogen;
– The gas does not ignite spontaneously;
– The resultant solution turns blue; because lithium hydroxide solution formed is a strong base;
(b). Reaction equations.
Equation I
2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g);
Equation II
4Li(s) + O2 (g) → 2Li2O(s);
Equation III:
Li2O(s) + H2O(l) → 2LiOH(aq)
Effect of resultant solution on litmus paper;
– Litmus paper turns blue; sodium hydroxide formed is highly soluble in water; releasing a large number of hydroxyl ions which result into alkaline conditions // high pH;
Summary on reaction rate.
| Metal | Reaction rate |
| Lithium | Vigorous |
| Sodium | More vigorous |
| Potassium | Explosive |
- Reaction with chlorine
– All alkali metals react with chlorine to form corresponding metal chlorides.
- General procedure:
– A small piece of the alkali metal is cut and placed in a deflagrating spoon;
– It is then warmed and quickly lowered into a gas jar containing chlorine.
– The experiment should be done in a working fume chamber or in the open
Reason: Chlorine gas is poisonous.
– This experiment should not be attempted in the laboratory with potassium
Reason: the reaction is too explosive and very dangerous.
- Observations:
Sodium:
– The metal bursts into a yellow flame.
– White fumes of sodium chloride are formed.
Equation: 2Na(s) + Cl2(g) → 2NaCl(s)
Lithium:
– The metal reacts less vigorously than sodium without bursting into a flame.
– White fumes of lithium chloride are formed.
Equation: 2Li(s) + Cl2(g) → 2LiCl(s)
Potassium:
– The metal bursts into a lilac flame.
– White fumes of potassium chloride are formed.
Equation: 2K(s) + Cl2(g) → 2KCl(s)
The general trend in reactivity of alkali metals.
– The reactivity of alkali metals increase down the group.
Reason:
– Alkali metals react by losing the single valence electron. The ease of loss of the valence electron increases down the group due to decrease in attraction of the valence electron towards the positive nucleus. This in turn is due to the increase in atomic radius down the group as a result of increase in number of energy levels.
Similarity of ions and formulae of some compounds of Alkali metals.
– Alkali metals have a similar charge on their ions since they all have a single valence electron.
– Thus they have the same general formula in their ions and compounds.
Examples.
| Alkali metal ion | Hydroxide (OH–) | Oxides (O2-) | Chloride (Cl–) | Sulphates (SO42-) |
| Li+ | ||||
| Na+ | ||||
| K+ |
Note:
– Alkali metals are usually not found as free elements; but rather in their combined states in the earths crust.
Reason:
– They have high reactivities.
Uses of Alkali metals and their compounds.
- Sodium is used in the manufacture of sodium cyanide for gold extraction.
- Lithium is used in the manufacture of special high strength glasses ad ceramics.
- Lithium compounds are used in the manufacture of dry cells for use in mobile phones, laptops, stopwatches, and zero emission electric vehicles.
- A molten mixture of sodium and potassium is used as a coolant in nuclear reactors.
- Sodium vapour is used to produce the yellow glow in street ad advertisement lights.
- Molten sodium is used as a reducing agent in the extraction of titanium.
Equation: Na(l) + TiCl4(g) → Ti(s) + 4NaCl(l)
- Sodium chloride is used as a food additive.
- A mixture of sodium hydroxide (caustic soda) and carbon disulphide is used in the manufacture of artificial silk called rayon.
The Alkaline Earth metals.
– Are the elements with two valence electrons and hence in group II of the periodic table.
– All are metallic in nature.
– The members of the family in order down the group is as follows:
- Beryllium
- Magnesium
- Calcium
- Strontium
- Barium
- Radium
Electron arrangement of the first three alkaline earth metals.
| Elements | Electron arrangement |
| Beryllium | 2.2 |
| Magnesium | 2.8.2 |
| Calcium | 2.8.8.2 |
Diagram: Part of periodic table showing the alkaline earth metals
Dot and cross diagrams for the first three Alkaline Earth metals
Gradation in properties of alkali metals.
- Atomic and ionic radius.
- Atomic radius:
– Refers to the distance between the centre of the nucleus of an atom and the outermost energy level occupied by electron(s)
- Ionic radius:
– Refers to the distance between the centre of the nucleus of an ion and the outermost energy level occupied by electron(s)
Trend:
– The ionic radius and the atomic radius of alkaline earth metals increase down the group.
Reason:
– There is an increase in the number of energy levels down the group from Beryllium to radium.
Illustration:
– Beryllium (2.2) has only 2 energy levels; Magnesium (2.8.2) has 3 energy levels while calcium (2.8.8.2) has 4 energy levels.
– Thus the outermost electron in calcium is further from the nucleus than the outermost electron in magnesium and beryllium.
- Atomic and ionic radius of the same element.
– For the same alkaline earth metal the atomic radius is larger than the ionic radius.
Reason:
– Alkaline earth metals form ions by losing the valence electrons, leading to the loss of an entire outermost energy level. Thus the atoms have more energy levels than the corresponding ion hence a larger radius in the atom than I the ion.
Illustration
– Calcium atom has electron arrangement of 2.8.8.2 hence 4 energy levels.
– During ion formation it reacts by losing the 2 valence electrons to acquire a new electron arrangement of 2.8.8 hence 3 energy levels.
– Thus the ion has a smaller radius than the atom.
Diagrammatically: calcium atom and calcium ion
Summary: changes in atomic and ionic radius among alkali metals.
| Element | Symbol | Atomic number | Electron arrangement | Atomic radius (nm) | Ionic radius (nm) |
| Beryllium | Be | 4 | 2.2 | 0.089 | 0.031 |
| Magnesium | Mg | 12 | 2.8.2 | 0.136 | 0.065 |
| Calcium | Ca | 20 | 2.8.8.2 | 0.174 | 0.099 |
Physical properties of alkali metals.
- Appearance.
– Alkaline earth metals acquire a metallic luster when polished. This refers to a shiny appearance on the cut surface.
– They however lose this metallic luster when exposed to air due to oxidation.
- Note:
– The purpose of polishing alkaline earth metals before using them in experiment is to remove the oxide coating that slows down and prevents them from reacting.
- Ease of cutting.
– Magnesium is hard to cut with a knife; but is however malleable and ductile.
– Calcium cannot also be cut with a knife because it is brittle.
Note:
- Malleability:
– Refers to the ability of a material to be hammered into sheets.
Example: Iron sheets are possible to be made because iron metal is malleable.
- Ductility:
– The ability of a material to be rolled into wires.
Example: electric cables are made of aluminium because aluminium metal is ductile.
- Brittle:
– Refers to a substance which is hard and likely to break.
- They have relatively high melting and boiling points in comparison to alkali metals.
Reason: they have relatively stronger metallic bonds (than alkali metals).
– The melting and boiling points decrease down the group.
Reason: –
– The size of the atoms increase down the group due to increasing number of energy levels. As the atomic radius increase the force of attraction between the positive nucleus and the delocalized electrons decrease. This leads to a decrease in the strength of the metallic bonds (down the group).
- Electrical conductivity.
– Alkaline earth metals are good conductors of heat and electricity.
Reason: they have delocalized electrons in the outermost energy level.
– The electrical conductivity is similar for all the alkaline earth metals.
Reason: all alkaline earth metals have the same number of delocalized electron (two valence electrons) in the outermost energy level.
- Ionization energy.
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– The number of ionization energies an element may have is equivalent to the number of valence electrons.
– Thus alkaline earth metals have two ionization energies.
– The first ionization energy is the minimum energy required to remove the first electron from the outermost energy level of an atom in its gaseous state.
– The second ionization energy is the minimum energy required to remove the second electron from the outermost energy level of an atom in its gaseous state.
– The first and second ionization energies among alkaline earth metals decreases down the group.
Reason:
– The effective force of attraction of on the outermost electron by the positive nucleus decreases with increasing atomic size and distance from the nucleus.
– Note that the atomic radius increases down the group due to increase in the number of energy levels.
- Variation between 1st and 2nd Ionization energies.
– The first ionization energy is always lower than the second ionization energy for the same element.
Reason:
– After losing the first electron from an atom, the overall positive charge holds the remaining electrons more firmly. Thus removing a second electron from the ion requires more energy than the first electron
Note:
– The third ionization energy will also be higher tha the second ionization energy for the same reason.
Example:
Magnesium.
- First ionization energy: Mg(g) → Mg+(g) + e- (1stE = 736 kJ per mole)
- Second ionization energy: Mg+(g) → Mg2+(g) + e- (2ndE = 1450 kJ per mole)
Summary on physical properties of alkaline earth metals.
| Element | Atomic number | Melting point (oC) | Boiling point (oC) | Atomic radius (nm) | 1st I.E
(kJmol-1) |
2nd I.E (kJmol-1) |
| Beryllium | 4 | 1280 | 2450 | 0.089 | 900 | 1800 |
| Magnesium | 12 | 650 | 1110 | 0.136 | 736 | 1450 |
| Calcium | 20 | 850 | 1140 | 0.174 | 590 | 1150 |
| Strontium | 38 | 789 | 1330 | 0.210 | 550 | 1060 |
| Barium | 56 | 725 | 1140 | 0.220 | 503 | 970 |
Chemical properties of Alkaline earth metals.
- Burning alkaline earth metals in air.
– Alkaline earth metals react burn in air to form corresponding oxides.
– More reactive alkaline earth metals may also react with atmospheric nitrogen to form corresponding nitrides.
Examples:
– Burns in air with a blinding brilliant flame forming a white solid.
– The white solid is a mixture of magnesium oxide ad magnesium nitride.
Equations:
Reacting with oxygen: 2Mg(s) + O2(g) → 2MgO(s)
Reacting with nitrogen: 3Mg(s) + N2(g) → Mg3N2(s)
– Burns in air with a faint orange flame forming a white solid.
– The white solid is a mixture of calcium oxide ad calcium nitride.
Equations:
Reacting with oxygen: 2Ca(s) + O2(g) → 2CaO(s)
Reacting with nitrogen: 3Ca(s) + N2(g) → Ca3N2(s)
Note:
– The trend in the reactivity of alkaline earth metals when burning in air is not clear; due to the oxide coating on the calcium that tends to slow down the reaction of calcium in air.
– For this reason it is important to polish the surfaces of alkaline earth metals before using them in experiments.
- Reaction of alkaline earth metals with cold water.
– Alkaline earth metals react slowly with cold water to form corresponding hydroxides and hydrogen gas.
Examples
- Magnesium:
– Reacts slowly with water to form magnesium hydroxide and hydrogen gas.
– The reaction is very slow and the amount f hydrogen gas evolved is very low hence the hydrogen gas bubbles stick on the surface of the metal.
– The magnesium hydroxide formed dissolves slightly in water to form magnesium hydroxide.
– Thus the resultant solution is slightly alkaline.
Equation:
Mg(s) + 2H2O(l) → Mg(OH)2(aq) + H2(g).
- Calcium:
Diagram of apparatus:
Observations and explanations:
– Calcium sinks to the bottom of the beaker; because it is denser than water;
– Slow effervescence of a colourless gas; due to slow evolution of hydrogen gas;
– Soapy solution formed; due to formation of alkaline calcium hydroxide;
– A white suspension is formed; because calcium hydroxide is slightly soluble in water;
Reaction equation:
Ca(s) + H2O (l) → Ca (OH) 2(aq) + H2 (g);
Effect of resultant solution on litmus paper;
– Litmus paper slowly turns blue; calcium hydroxide formed is slightly soluble in water; releasing a small number of hydroxyl ions which result into alkaline conditions // high pH;
- Reaction with steam.
– Alkaline earth metals react with steam to produce corresponding metal oxide and hydrogen gas.
– The reactivity with stem is faster ad more vigorous for each alkaline earth metals as compared to reaction with cold water.
Examples:
- Magnesium
Procedure:
– A small amount of wet sand is put at the bottom of a boiling tube;
– A small piece of magnesium ribbon is cleaned and put in the middle of the combustion tube;
– The wet sand is heated gently first then the magnesium ribbon is heated strongly until it glows.
Reason:
– To generate steam that drives out the air that would otherwise react with the magnesium (preventing reaction with steam)
– The delivery tube is removed from the water before heating stops.
Reason:
To prevent sucking back (of the gas) as the apparatus cools
– The gas produced is tested using a burning splint.
Diagram of apparatus:
Observations and explanations.
– Magnesium burns with a white blinding flame;
– Grey solid (magnesium) forms a white solid; due to formation of magnesium oxide;
– Evolution of a colourless gas that produces a pop sound when exposed to a burning splint; confirming it is hydrogen;
Reaction equation.
Magnesium + Steam → Magnesium oxide + Hydrogen gas;
Mg(s) + H2O(g) → MgO(s) + H2(g);
- Calcium
– Reaction between calcium and steam would produce calcium oxide and hydrogen gas.
– However the reaction is too explosive to be done under laboratory conditions.
- Reaction with chlorine.
– Alkaline earth metals react with chlorine to form corresponding chlorides as the only products.
Condition: presence of heat, hence the metal must be heated first.
Precaution: reaction should be done in a working fume chamber because chlorine gas is poisonous).
Examples:
Procedure:
– A piece of burning magnesium is lowered into a gas jar containing chlorine.
Observations:
– The metal continues to burn with a brilliant white flame.
– They grey solid forms a white powder.
Explanation.
– Reaction between magnesium and chlorine is exothermic.
– The heat produced keeps the metal burning; thus facilitates the reaction between magnesium and chlorine to form magnesium chloride, which is the white powder.
Equation:
Mg(s) + Cl2(g) → MgCl2(s)
Procedure:
– A piece of burning calcium is lowered into a gas jar containing chlorine.
Observations and explanations.
– The metal burns shortly with an orange flame but soon smolders off.
– There is no steady reaction between calcium and chlorine.
Reason:
– When calcium is heated a coating of the metal oxide is formed first which prevents furtheer reaction between the metal and chlorine.
– However under suitable conditions calcium reacts with chlorine to form a white powder of calcium chloride.
Equation:
Ca(s) + Cl2(g) → CaCl2(s)
- Reaction of alkaline earth metals and dilute acids
– Generally alkaline earth metals react with dilute acids to form salts and hydrogen gas.
Examples:
- With hydrochloric acid.
- Beryllium:
– When a piece of beryllium is dropped into a beaker containing hydrochloric acid, there is effervescence of a colourless gas.
– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.
– A colourless solution of beryllium chloride remains in the test tube // boiling tube.
Equation: Be(s) + 2HCl(aq) → BeCl2(aq) + H2(g)
- Magnesium:
– When a piece of magnesium is dropped into a beaker containing hydrochloric acid, there is effervescence of a colourless gas.
– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.
– A colourless solution of magnesium chloride remains in the test tube // boiling tube.
Equation: Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
- Calcium:
– When a piece of calcium is dropped into a beaker containing hydrochloric acid, there is effervescence of a colourless gas.
– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.
– A colourless solution of calcium chloride remains in the test tube // boiling tube.
Equation: Ca(s) + 2HCl(aq) → CaCl2(aq) + H2(g)
- With sulphuric (VI) acid.
- Beryllium:
– When a piece of beryllium is dropped into a beaker containing sulphuric (VI) acid, there is effervescence of a colourless gas.
– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.
– A colourless solution of beryllium sulphate remains in the test tube // boiling tube.
Equation: Be(s) + H2SO4(aq) → BeSO4(aq) + H2(g)
- Magnesium:
– When a piece of magnesium is dropped into a beaker containing sulphuric (VI) acid, there is effervescence of a colourless gas.
– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.
– A colourless solution of magnesium sulphate remains in the test tube // boiling tube.
Equation: Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g)
- Calcium:
– When a piece of calcium is dropped into a beaker containing sulphuric (VI) acid, there is slight effervescence of a colourless gas.
– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.
– The reaction however quickly stops and there is formation of a white precipitate in the test tube or boiling tube.
Reason:
– When calcium reacts with dilute sulphuric (VI) cid there is formation of calcium sulphate which is insoluble hence the formation of a white precipitate. The insoluble calcium sulphate coats the surface of the (calcium) metal stopping further reaction.
Equation: Ca(s) + H2SO4(aq) → CaSO4(s) + H2(g)
- With nitric (V) acid.
- Beryllium:
– When a piece of beryllium is dropped into a beaker containing nitric (V) acid, there is effervescence of a colourless gas.
– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.
– A colourless solution of beryllium nitrate remains in the test tube // boiling tube.
Equation: Be(s) + 2HNO3(aq) → Be(NO3)2(aq) + H2(g)
- Magnesium:
– When a piece of magnesium is dropped into a beaker containing nitric (V) acid, there is effervescence of a colourless gas.
– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.
– A colourless solution of magnesium nitrate remains in the test tube // boiling tube.
Equation: Mg(s) + 2HNO3(aq) → Mg(NO3)2(aq) + H2(g)
- Calcium:
– When a piece of calcium is dropped into a beaker containing nitric (V) acid, there is slight effervescence of a colourless gas.
– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.
– A colourless solution of calcium nitrate remains in the test tube // boiling tube.
Equation: Ca(s) + 2HNO3(aq) → Ca(NO3)2(aq) + H2(g)
Trend in reactivity in alkaline earth metals.
– The reactivity of the alkaline earth metals increases down the group.
Reason:
– The atomic radius increases down the group due to increasing number of energy levels. The attraction of the two valence electrons towards the positive nucleus thus decreases down the group. Since alkaline earth metals react by losing electrons the ease with which the valence electrons are lost therefore decreases down the group
Similarity of ions and formulae of some compounds of Alkaline earth metals.
– Alkaline earth metals have the same valency (2) and a similar charge (2+) on their ions since they all have a single valence electron.
– Thus they have the same general formula in their ions and compounds.
Examples.
| Alkali metal ion | Hydroxide (OH–) | Oxides (O2-) | Chloride (Cl–) | Sulphates (SO42-) |
| Be2+ | ||||
| Mg2+ | ||||
| Ca2+ |
Uses of some alkaline earth metals and their compounds
- 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.
- A low-density alloy of magnesium and aluminium is used in aeroplanes construction.
- Hydrated calcium sulphate (plaster of Paris) is used in hospitals to set fractured bones.
- Cement is made by heating a mixture of calcium carbonate (limestone), clay and sand.
- Calcium carbonate is used in the extraction of iron.
- Calcium oxide (quicklime) is added tom acidic soils to raise pH for agricultural purposes.
- Calcium nitrate is used as a nitrogenous fertilizer.
- Magnesium oxide is used in the lining of furnaces.
- Barium sulphate is used in the diagnosis of ulcers.
- Barium nitrate is used to produce the green flame in fireworks.
- 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.
- 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.
- Atomic and ionic radius of the same element.
– For the same halogen the ionic radius is larger than the atomic radius.
Reason:
– Halogens form ions by gaining (an) electron into the outermost energy level. This increases the electron-electron repulsion in the outermost energy level hence decreasing the nuclear attraction of the outermost electrons towards the nucleus.
Illustration
– Chlorine atom has electron arrangement of 2.8.7 hence 3 energy levels.
– During ion formation it reacts by gaining an electron into the third energy level to acquire a new electron arrangement of 2.8.8.
– In the atom 17 protons are attracting 17 electrons; while in the chloride ion there are 17 protons attracting 18 electrons in the outermost energy level.
– Thus in the ion the effect of the positive nucleus is lower.
– This is due to increased repulsive effect between the existing electrons in the outermost energy level and the incoming electron (electron-electro repulsion)
Diagrammatically: chlorine atom and chloride ion
Summary: changes in atomic and ionic radius among alkali metals.
| Element | Symbol | Atomic number | Electron arrangement | Atomic radius (nm) | Ionic radius (nm) |
| Fluorine | F | 9 | 2.7 | 0.064 | 0.136 |
| Chlorine | Cl | 17 | 2.8.7 | 0.099 | 0.181 |
| Bromine | Br | 35 | 2.8.18.7 | 0.114 | 0.195 |
| Iodine | I | 53 | 2.8.18.18.7 | 0.133 | 0.216 |
Laboratory preparation of chlorine.
Note: It is usually prepared by oxidation of concentrated hydrochloric acid by removal of hydrogen.
Equation:
2HCl(aq) + [O] Cl2(g) + H2O(l)
– The [O] is from a substance containing oxygen (an oxidizing agent).
– The main oxidizing agents normally used for preparation of chlorine are:
- Potassium manganate (VII); KMnO4.
- Manganese (IV) oxide; MnO2
(a). Preparation of chlorine from MnO2 and HCl.
(i). Apparatus:
(ii). Conditions:
– Heating;
– Presence of an oxidizing agent; in this case it is manganese (IV) oxide.
(iii). Procedure:
– Hydrochloric acid is reacted with manganese (IV) oxide (dropwise);
Equation:
MnO2(s) + 4HCl(aq) Heat MnCl2(aq) + 2H2O(l) + Cl2(g)
(iv). Explanation:
– Manganese (IV) oxide oxidizes hydrochloric acid by removing hydrogen resulting into chlorine.
– The manganese (IV) oxide is reduced to water and manganese chloride.
– The resultant chlorine gas is passed through a bottle containing water.
Reason:
– To remove hydrogen chloride fumes (gas) which is very soluble in water.
– Next it is passed through concentrated sulphuric acid or anhydrous calcium chloride; to dry the gas.
(v). Collection:
(a). Wet chlorine is collected over brine (saturated sodium chloride solution) or hot water.
Reason:
– It does not dissolve in brine and is less soluble in water
(b). Dry chlorine is collected by downward delivery (upward displacement of air)
Reason:
– It is denser than air (2.5 times).
Note:
– Chlorine may also be dried by adding calcium chloride to the jar of chlorine.
(c). The first bottle must contain water and the second concentrated sulphuric acid.
Reason:
– If the gas is first passed through concentrated sulphuric acid in the first bottle then to the water; it will be made wet again.
Physical properties of Halogens.
- 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. |
- 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.
- 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.
- 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.
- 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.
- 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.
|
- 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 |
- Reaction with metals.
(a). Chlorine.
- Reaction of chlorine with iron.
(i). Apparatus.
Note:
– The set up can also be modified by using sodium hydroxide to absorb excess chlorine gas as shown below.
(ii). Precaution.
– Experiment should be done in a functional fume cupboard or in the open.
Reason:
– Chlorine gas is poisonous and will thus be harmful to the human body.
(iii). Procedure:
– A stream of dry chlorine gas is passed over heated iron wool as per the diagram.
(iv). Conditions.
- Chlorine gas has to be dry (done by the anhydrous calcium chloride in the U-tube)
Reason:
To prevent hydration hence oxidation of iron (which will then form Fe2O3.5H2O) hence preventing reaction between iron and chlorine.
- Iron metal must be hot; and this is done by heating.
Reason:
To provide activation energy i.e. the minimum kinetic energy which the reactants must have to form products.
- Anhydrous calcium chloride.
– In the U-tube; to dry the chlorine gas.
– In the thistle funnel; to prevent atmospheric water vapour (moisture) from getting into the apparatus and hence reacting with iron (III) chloride.
Note: In the guard tube, calcium oxide is preferable to anhydrous calcium chloride.
Reason:
– Other than preventing atmospheric water vapour from getting into the flask with iron wool; it also absorbs excess chlorine thus preventing environmental pollution
(v). Observations:
– Iron metal glows red-hot.
– Red brown fumes (FeCl3(g)) are formed in the combustion tube.
– A black solid (FeCl3(s)) is collected in the flask.
Note:
– Iron (III) chloride cannot be easily collected in the combustion tube.
Reason:
– It sublimes when heated and hence the hotter combustion tube causes it to sublime and its vapour is collected on the cooler parts of the flask.
(vi). Reaction equation.
2Fe(s) + 3Cl2(g) → 2FeCl3(g)
(vii). Conclusion.
– Iron (III) chloride sublimes on heating; the black solid changes to red-brown fumes on heating.
Equation:
FeCl3(s) FeCl3(g)
(black) (Red-brown)
- Reaction with aluminium.
– Chlorine reacts with aluminium I a similar manner to iron to from aluminium chloride.
Equation:
2Al(s) + 3Cl2(g) → 2AlCl3(s)
2Al(s) + 3Cl2(g) → Al2Cl6(s)
Note:
– Aluminium chloride also sublimes on heating.
Equation:
AlCl3(s) AlCl3(g)
(White) (White)
- Reaction with burning magnesium.
(i). Procedure:
– Burning magnesium is lowered into a gar jar of chlorine gas.
(ii). Observations:
– The magnesium continues to burn with a bright blinding flame;
– Formation of white fumes (MgCl2); which cools into a white powder.
(iii). Equation:
Mg(s) + Cl2(g) → MgCl2(s)
– Generally chlorine reacts with most metals when hot top form corresponding chlorides.
Note:
Where a metal forms two chlorides when it reacts with chlorine, the higher chloride is usually formed.
Reason:
The higher chloride is stable. This explains why reactions of chlorine with iron results into iron (III) chloride and not iron (II) chloride.
(b). Bromine.
- Reaction of bromine with iron
Apparatus
(ii). Precaution.
– Experiment should be done in a functional fume cupboard or in the open.
Reason:
– Bromine gas is poisonous and will thus be harmful to the human body.
(iii). Procedure:
– Bromine liquid is heated to generate bromine vapour (fumes).
– The iron wool is then heated and a stream of the bromine gas is passed over heated iron wool as per the diagram.
(iv). Conditions.
- Bromine must be heated to generate fumes before heating the iron.
Reason:
So that bromine vapour will drive out air from the boiling tube to prevent oxidation of iron with oxygen which would otherwise prevent reaction between iron and bromine.
- Iron metal must be hot; and this is done by heating.
Reason:
To provide activation energy i.e. the minimum kinetic energy which the reactants must have to form products.
(v). Observations:
– Brown fumes of bromine are produced on heating bromine liquid.
– The iron wool glows red-hot upon heating
– The iron wool forms dark red crystals (of iron (III) bromide)
(vi). Reaction equation.
Word: Iron + bromine → Iron (III) bromide.
Chemical: 2Fe(s) + 3Br2(g) → 2FeBr3(s)
Dark red crystals.
- Reaction of bromine with zinc
– Using the same set up bromine also reacts with zinc to form zinc bromide.
Equation: Zn(s) + Br2(g) → ZnBr2(s).
- Reaction of bromine with magnesium
– Using the same set up bromine also reacts with zinc to form magnesium bromide.
Equation: Mg(s) + Br2(g) → MgBr2(s).
- Reaction of bromine with sodium
– Using the same set up bromine also reacts with zinc to form sodium bromide.
Equation: 2Na(s) + Br2(g) → 2NaBr(s).
(c). Iodine.
- Reaction of iodine with iron
Apparatus
(ii). Precaution.
– Experiment should be done in a functional fume cupboard or in the open.
Reason:
– Iodine is poisonous and will thus be harmful to the human body.
(iii). Procedure:
– Iodine crystals are heated to generate iodine vapour (fumes).
– The iron wool is then heated and a stream of the iodine gas is passed over heated iron wool as per the diagram.
(iv). Conditions.
- Iodine must be heated to generate fumes before heating the iron.
Reason:
So that iodine vapour will drive out air from the boiling tube to prevent oxidation of iron with oxygen which would otherwise prevent reaction between iron and iodine.
- Iron metal must be hot; and this is done by heating.
Reason:
To provide activation energy i.e. the minimum kinetic energy which the reactants must have to form products.
(v). Observations:
– Purple vapour (fumes) of iodine is produced on heating iodine crystals.
– The iron wool glows red-hot upon heating
– The iron wool forms grey black crystals (of iron (II) iodide)
(vi). Reaction equation.
Word: Iron + iodine → Iron (II) iodide.
Chemical: Fe(s) + I2(g) → FeI2(s)
Grey-black crystals.
- Reaction of Iodine with zinc
– Using the same set up bromine also reacts with zinc to form zinc iodide.
Equation: Zn(s) + I2(g) → ZnI2(s).
- Reaction of Iodine with magnesium
– Using the same set up bromine also reacts with zinc to form magnesium iodide.
Equation: Mg(s) + I2(g) → MgI2(s).
- Reaction of Iodine with sodium
– Using the same set up bromine also reacts with zinc to form sodium iodide.
Equation: 2Na(s) + I2(g) → 2NaI(s).
Note:
– The reactivity of chlorine with metals is more vigorous than that of bromine, which is more than that of iodine.
– Thus reactivity of halogens with metals decreases down the group.
Reason:
– Halogens react by gaining electrons; the ease of gaining electrons decrease down the group as the atomic size increases which leads to progressive decrease in the force of attraction of electrons in the outermost energy levels by the positively charged nucleus.
- Reaction with water.
– Halogens that dissolve in water form a mixture of two acids.
Reaction of chlorine with water.
– Chlorine dissolves in water to form chlorine water, which is a mixture of two acids: chloric (I) acid and hydrochloric acid.
Equation:
Cl2(g) + H2O(l) → HOCl(aq) + HCl(aq)
Chloric (I) acid Hydrochloric acid
- Effects of sunlight on chlorine water.
(i). Procedure:
– Chlorine water is made by dissolving the gas in water.
– A long tube filled with chlorine water is inverted over a beaker containing water.
– It is then exposed to sunlight (bright light) as shown below.
(ii). Apparatus:
(iii). Observations:
– After sometime a gas collects in the tube and on applying a glowing splint, the splint is rekindles showing that the gas collected is oxygen.
(iv). Explanation:
– Chlorine water has two components.
Equation:
Cl2(g) + H2O(l) ═ HCl(aq) + HOCl(aq)
– The HOCl being unstable will dissolve on exposure to sunlight, giving out oxygen.
Equation:
2HOCl(aq) → 2HCl(aq) + O2(g) (slow reaction)
Overall reaction:
2H2O(l) + 2Cl2(g) → 4HCl(aq) + O2(g)
- Effect of chlorine water on litmus papers
(i). Procedure:
– A strip of blue ad a strip of red litmus papers are dropped into chlorine water in a beaker.
(ii). Observations:
– The blue litmus paper turns red; then both litmus papers are decolourised.
(iii). Explanations.
– Chlorine water contains a mixture of chloric (I) acid ad hydrochloric acid.
Equation:
Cl2(g) + H2O(l) ═ HCl(aq) + HOCl(aq)
– The two acids cause the litmus paper to turn red (from blue) while the red litmus paper remained red.
– The litmus papers are then decolourised due to bleaching action of chlorine water, which is due to the activity of chloric (I) acid.
– The chloric (I) acid is unstable and thus decomposes to give hydrochloric acid and oxygen atom.
Equation:
HOCl(aq) → HCl(aq) + [O]
– The oxygen atom combines with the chemical of the natural dye in the litmus thus decolourising it.
Equation:
Dye + HOCl(aq) → HCl(aq) + (Dye + [O])
Coloured Decolourised
Note: solutions of bromine and fluorine in water will behave in a similar manner.
Examples:
- Fluorine dissolving in water
F2(g) + H2O(l) ═ HF(aq) + HOF(aq)
Hydrofluoric acid Fluoric (I) acid
– The mixture is flourine water
- Bromine dissolving in water
Br2(g) + H2O(l) ═ HBr(aq) + HOBr(aq)
Hydrobromic acid Bromic (I) acid
– The mixture is bromine water.
Some uses of halogens and their compounds.
- Fluorine is a raw material in the preparation of a synthetic fibre known as polytetrafluoroethane.
- Some compounds of fluorine are added to water and some tooth pastes in small quantities to reduce tooth decay.
- Fluorine is used in the manufacture of hydrogen fluoride used to engrave words or pictures on glass.
- Chlorine is used to make bleaches used in paper and textile industries.
- Chlorine is added to water to kill micro-organisms in water treatment works.
- Chlorine is used in the manufacture of a plastic known as polyvinylchloride (PVC).
- Chlorine is used in large scale manufacture of hydrochloric acid.
- Bromine is used in the manufacture of silver bromide which is used to make the light sensitive photographic paper and films.
- 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
- 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.
- 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).
- 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.
- Argon is used in light bulbs to provide an inert environment to prevent oxidation.
- Argon is used as an insulator in arch-welding.
- Neon gas is used in streets and advertisement lights.
- Helium mixed with oxygen is used in deep sea diving and mountaineering.
- Helium mixed with oxygen also is used in hospitals for breathing in patients with respiratory problems and those undergoing certain forms of surgery.
- 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.
- Helium is used in thermometers for measurements f very low temperatures.
- 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.
- Introduction
- Trends in physical properties of elements in period 3
- Electrical conductivity
- Atomic and ionic radii
- Melting and boiling points
- 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.
- 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.
- 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.
- 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:
– Its a noble gas with a stable octet configuration.
- Reaction of period 3 elements with oxygen.
– All period three elements react with (burn in) oxygen with the exception of argon.
Experiment: To investigate the reactions between period 3 elements and oxygen
Procedure:
– A small piece of the element is placed in a deflagrating spoon and warmed gently until it catches fire.
– It is then lowered into a gas jar full of oxygen.
– The flame colour and the colour of the product are noted.
– 10 cm3 of water containing universal or litmus indicator is added into the gas jar with the products.
- Sodium:
– Burns vigorously in oxygen with a golden yellow flame; to produce white solid of sodium oxide.
Equation:
Na(s) + O2(s) → Na2O(s)
– The resultant sodium oxide dissolves in water to form sodium hydroxide.
Equation:
Na2O(s) + H2O(l) → 2NaOH(aq);
– The sodium hydroxide is alkaline in nature and thus turns litmus indicator blue;
- Magnesium:
– Burns vigorously in oxygen with a bright white light; to produce white solid of magnesium oxide.
Equation:
2Mg(s) + O2(s) → 2MgO(s)
– The resultant magnesium oxide is slightly soluble in water to form magnesium hydroxide.
Equation:
MgO(s) + H2O(l) → Mg(OH)2(aq);
– The magnesium hydroxide is alkaline in nature and thus turns litmus indicator blue;
- Aluminium:
Note:
– Aluminium (foil) is usually coated with a thin layer of aluminium oxide, Al2O3; which prevents reaction with the oxygen.
– When polished, it reacts slowly with oxygen to form a white solid of aluminium oxide.
Equation:
Al(s) + 3O2(s) → Al2O3(s)
– The resultant aluminium (III) oxide is insoluble in water.
- Silicon:
– Silicon powder can only bur in oxygen at very high temperatures of about 450oC to form solid silicon (IV) oxide.
Equation:
Si(s) + O2(s) → SiO2(s)
– The resultant silicon (IV) oxide is insoluble in water.
– Burns readily in oxygen with a bright orange flame; to produce dense white fumes (solid) of phosphorus (V) oxide
Equation:
P4(s) + 5O2(s) → 2P2O5(s)
Note: sulphur exists and therefore reacts as molecules of P4.
– The resultant phosphorus (V) oxide readily dissolves in water to form phosphoric (V) acid.
Equation:
P2O5(s) + 3H2O(l) → 2H3PO4(aq);
– The phosphoric acid is acidic in nature and thus turns litmus indicator pink / red;
– Burns in oxygen with a blue flame; to form a colourless gas of sulphur (IV) oxide
Equation:
S(s) + O2(s) → SO2(s)
– The resultant sulphur (IV) oxide readily dissolves in water to form sulphuric (IV) acid.
Equation:
SO2(s) + H2O(l) → H2SO3(aq);
– The phosphoric acid is acidic in nature and thus turns litmus indicator pink / red;
– The sulphuric (IV) acid is unstable and thus easily gets oxidized by oxygen to the stable sulphuric (VI) acid.
Equation:
H2SO3(aq) + O2(g) → H2SO4(aq);
– Burns in oxygen only under certain conditions to form acidic oxides.
Equation:
2Cl2(s) + O2(s) → 2Cl2O(s)
– Argon is unreactive.
Conclusion:
– Metallic elements burn in oxygen to form basic oxides.
– Non-metallic oxides burn in oxygen to form acidic oxides.
- Reaction of period 3 elements with water.
Procedure:
– A small piece of sodium metal is cut and dropped into a trough containing water;
– The resultant solution is tested with litmus paper;
Diagram of apparatus:
Observations and explanations:
– The metal floats on the water surface; because it is less dense than water;
– A hissing sound is produced; due to production of hydrogen gas;
– It vigorously melts into a silvery ball then disappears because reaction between water and sodium is exothermic (produces heat). The resultant heat melts the sodium due to its low melting point.
– It darts on the surface; due to propulsion by hydrogen;
– The metal may burst into a golden yellow flame; because hydrogen may explode into a flame which then burns the sodium;
– The resultant solution turns blue; because sodium hydroxide solution formed is a strong base;
Reaction equations.
Equation I
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g);
Equation II
4Na(s) + O2 (g) → 2Na2O(s);
Equation III:
Na2O(s) + H2O(l) → 2NaOH(aq)
Effect of resultant solution on litmus paper;
– Litmus paper turns blue; sodium hydroxide formed is highly soluble in water; releasing a large number of hydroxyl ions which result into alkaline conditions // high pH;
Procedure:
– A small piece of magnesium ribbon is cut and dropped into a trough containing water;
– The resultant solution is tested with litmus paper;
Diagram of apparatus:
Observations and explanations:
– The metal sinks into the water surface; because it is denser than water;
– It reacts slowly with water leading to slow evolution of hydrogen gas.
– The resultant solution turns blue; because sodium hydroxide solution formed is a strong base;
Reaction equation.
Mg(s) + 2H2O(l) → Mg(OH)2(aq) + H2(g);
Effect of resultant solution on litmus paper;
– Litmus paper turns blue; magnesium hydroxide formed is highly soluble in water; releasing a large number of hydroxyl ions which result into alkaline conditions // high pH;
– Aluminium does not normally react with cold water or steam.
Reason:
– Aluminium is usually coated with a thin coating of aluminium oxide which prevents further reaction.
– However at very high temperatures of about 700oCsteam can react with aluminium to form aluminium oxide and hydrogen gas.
Equation:
2Al(s) + 3H2O(g) → Al2O3(s) + 3H2(g);
Note:
– Due to its inability t react with water and air aluminium is preferable for making cooking utensils like sufurias and coking pans.
- Silicon, phosphorus and sulphur.
– These non-metals do not displace hydrogen from water and thus do not react with water.
– Dissolves in water t form chlorine water, which is a mixture of chloric (I) acid ad hydrochloric acid.
Equation:
Cl2(g) + H2O(l) → HOCl(aq) + HCl(aq)
Chloric (I) acid Hydrochloric acid
- Reaction of period 3 elements with acids
Procedure:
– A piece of the element is dropped into 5 cm3 of an acid in a test tube.
– Any gas produced is tested.
- Sodium:
– Reacts explosively with acids to form salts and hydrogen and thus reactions of sodium with acids should not be tried in the laboratory.
- Magnesium:
– Reacts with both dilute hydrochloric and dilute sulphuric acid to form magnbsium salts and hydrogen gas.
- With hydrochloric acid:
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
- With sulphuric acid:
Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g)
- Aluminium:
Note:
– It does not readily react with dilute acids.
Reason:
– This is due to presence of a thin aluminium oxide coating that prevents contact hence reaction with the acids.
– When polished t remove the oxide coating it reacts with both dilute hydrochloric and dilute sulphuric acid to form aluminium salts and hydrogen gas.
- With hydrochloric acid:
2Al(s) + 6HCl(aq) → 3AlCl3(aq) + 3H2(g)
- With sulphuric acid:
2Al(s) + 3H2SO4(aq) → Al2(SO4)2(aq) + 3H2(g)
- Silicon, phosphorus, sulphur and chlorine.
– They do not react with dilute acids.
Summary: Chemical properties of period 3 elements.
| Element | Na | Mg | Al | Si | P | S | Cl |
| Reaction with air or oxygen
|
Readily reacts with air.
– Burns brightly in oxygen with a golden yellow flame to form Na2O |
Reacts slowly with air.
– Burns in oxygen with a bright white flame to form MgO
|
– Forms a protective coating of Al2O3 when it burns in oxygen. | Si powder burns at temperatures above 950oC to form SiO2 | White phosphorus smolders in air;
– P burns in air with a bright orange flame to form P2O3 and P2O5 |
Burns in air or oxygen with a blue flame to form SO2 gas | N reaction with air or oxygen under normal conditions. |
| Reaction with water
|
Reacts vigorously to produce H2 and NaOH | Slow reaction with cold water to form Mg(OH)2 and H2;
– Reacts faster with steam to form MgO and H2 |
No reaction | No reaction | No reaction | No reaction | Dissolves in water to form chlorine water |
| Reaction with dilute acids | Violent reaction giving out H2 and a sodium salt | Rapid evolution of H2 gas and a Mg salt is formed | Reacts slowly to give H2 and an Al salt | No reaction | No reaction | No reaction | No reaction |
UNIT4: STRUCTURE AND BONDING.
- Meaning of structure and bond.
- Nature of the chemical bond.
- 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
- 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
- Summary on bond characteristics
- 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.
- 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.
- 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
- Potassium fluoride.
- Magnesium oxide
- Sodium oxide.
- Magnesium chloride.
- 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.
- 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.
- 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.
- 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.
- 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.
- 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.
- Hydrogen, H2
- Hydrogen chloride, HCl
- Nitrogen, N2
- Water
- Carbon (IV) oxide, CO2
- Ammonia gas, NH3
- Phosphene, PH3
- Methane, CH4
- Ethane. CH2CH2
- Ethyne, C2H2
- Ethanol, C2H5OH
- 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.
- Carbon (IV) oxide
- Formation of PH4+
- Aluminium chlorine dimer (Al2Cl6).
- 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.
- 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.
- 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.
- 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.
- 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.
- 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.
- 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 | |
- 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.
- 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.
- 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.
- 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.
- 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
- 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.
- It is soft and slippery.
Reason:
– The carbon atoms in graphite are covalently bonded into hexagonal layers which are joined to each other by weak Van der Waals forces.
– The weak Van der Waals forces easily slide over each other when pressed hence the soft and slippery feel.
(c). Silicon (IV) oxide.
– Is a covalent compound of silicon and oxygen.
– Silicon has four electrons in its outermost energy level while oxygen has six.
– Each silicon atom is bonded to four oxygen atoms by strong covalent bonds.
– Each oxygen atom is bonded to two silicon atoms by strong covalent bonds.
– This means there are no delocalized electrons in the structure of silicon (IV) oxide making it unable to conduct electric current.
– The silicon and oxygen atoms are all covalently bonded together (by strong covalent bonds) in a repeated manner leading to a giant of covalent bonds throughout the structure.
– The extra ordinarily strong covalent bonds in silicon (IV) oxide throuout the structure contribute to the very high melting (1728oC) and boiling (2231oC) points.
Diagram: structure of silicon (IV) oxide.
Properties of giant atomic // giant covalent structures.
- – Have high melting and boiling points
- – They are non-conductors of heat and electricity with the exception of graphite
iii. – They are insoluble in water
- – Most are generally very hard, with the exception of graphite
- Note: for explanations of the above properties, refer to individual explanations of each compound.
- 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.
- 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.
- 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.
- 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:
- – 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:
- Reaction with water.
Procedure:
– A test tube is half filled with water and initial temperature of the water recorded.
– A spatula end full of sodium chloride is added and stirred until it dissolves.
– The hiughest temperature attained when all the solid dissolves is recored and the temperature change calculated.
– Two three drops of universal indicator are added and the pH of the solution noted and recorded.
Observations:
| Chloride | Observations | ||
| Solubility | Temperature change (oC) | pH of solution | |
| Sodium chloride | Dissolves | Drop in temperature | 7 |
| Magnesium chloride | Dissolves | Slight increase | 6.5 |
| Aluminium chloride | Hydrolyzed | Increases | 3 |
| Silicon (IV) oxide | Hydrolyzed | Increases | 2 |
| Phosphorus (III) chloride | Hydrolyzed | Increases | 2 |
| Phosphorus (V) chloride | Hydrolyzed | Increases | 2 |
Explanations.
- Sodium and magnesium chlorides.
– Sodium chloride dissolves in water causing a slight drop in temperature.
– Magnesium dissolves readily with a small increase in temperature.
– Both chlorides are ionic and when added to water there is an immediate attraction of polar water molecules for ions in the chlorides.
– The solid thus readily dissolves forming aquated ions such as Na+(aq) and Cl–(aq).
– These are separate metal and non-metal ions surrounded by polar water molecules.
– Since there is no production of either H+ or OH– ions the solutions are neutral.
Equations:
NaCl(s) + H2O(l) → Na+(aq) + Cl–(aq) + H2O(l).
MgCl2(s) + H2O(l) → Mg2+(aq) + 2Cl–(aq) + H2O(l).
- Anhydrous aluminium chloride.
– It exits in molecular form as a dimeric molecule of Al2Cl6.
– The dimeric molecule is formed when aluminium chloride (AlCl3) molecules vapour condense and combine forming larger molecules of Al2Cl6.
Diagram: Formation of a dimer in aluminium chloride.
– When added to water aluminium chloride is hydrolyzed to form an acidic solution.
– By so doing it behaves like a covalent chloride rather than an ionic chloride.
– The hydrolysis is an exothermic reaction accompanied by release of heat hence the increase in temperature.
Explanations.
– The hydrolysis is due to the very small but highly charged aluminium ion, Al3+.
– The Al3+ draws electrons away from its surrounding water molecules and causes them to give up H+ ions.
– This reaction usually involves aluminium ions combining with six water molecules to form hexa-aqua-aluminium (III) ions which dissociate to give H+.
Equation:
[Al(H2O)6]3+(aq) → [Al(H2O)5(OH)]2+(aq) + H+(aq)
Note:
This reaction can be simplified as follows:
– Aluminium chloride reacting with water and hydrolyzing to give HCl(aq) as one of the products.
- Equation: Al2Cl6(s) + 6H2O(l) → 2Al(OH)3(aq) + 6HCl(aq)
– Then the HCl dissociates to give H+ and Cl-
- Equation: 6HCl(aq) → H+(aq) + Cl–(aq)
– It is the resultant H+ that confer the solution its acidic properties.
- Silicon (IV) chloride
– Undergoes hydrolysis in water in an exothermic reaction producing a lot of heat.
– The products of the hydrolysis are silicon (IV) oxide solid and hydrogen chloride gas.
– The hydrogen chloride gas immediately dissolves in the water to form hydrochloric acid.
– The hydrochloric acid dissociated to liberate H+ which leads to acidic conditions.
Equations:
SiCl4(s) + 2H2O(l) → SiO2(s) + 4HCl(aq)
Then: 4HCl(aq) → 4H+(aq) + 4Cl–(aq).
- Phosphorus (III) chloride and phosphorus (V) chloride
– Both undergo hydrolysis in water in an exothermic reaction producing a lot of heat.
– The products of the hydrolysis are phosphorus (III) acid and phosphoric (V) acid respectively, and hydrogen chloride gas.
– The hydrogen chloride gas immediately dissolves in the water to form hydrochloric acid.
– The hydrochloric acid dissociated to liberate H+ which leads to acidic conditions.
Equations:
- With phosphorus (III) chloride:
PCl3(s) + 2H2O(l) → H3PO3(s) + 3HCl(aq)
Then: 3HCl(aq) → 3H+(aq) + 3Cl–(aq).
- With phosphorus (V) chloride:
PCl5(s) + 4H2O(l) → H3PO4(s) + 5HCl(aq)
Then: 5HCl(aq) → 5H+(aq) + 5Cl–(aq).
Trends in bond types and properties of chlorides of elements in period 3
| Property | Period 3 chloride
|
|||||
| Formula | NaCl | MgCl2 | Al2Cl6 | SiCl4 | PCl3 & PCl5 | SCl2
|
| Physical state at RT | Solid | Solid | Solid | Liquid | Liquid | Liquid |
| M.P (oC) | 801 | 714 | Sublimes at 180oC | -70 | Sublimes at -94oC | -78 |
| B.P (oC) | 1467 | 1437 | – | 57 | – | Decomposes at 57oC |
| Conductivity | Good | Good | V. poor | nil | nil | Nil
|
| Structure | Giant ionic | Giant ionic | Molecular dimer | Molecular | Molecular | Molecular |
| Bond type | Ionic | Ionic | Ionic / covalent | Covalent | Covalent | Covalent |
| Effect on water | Readily dissolves | Readily dissolves | Hydrolyzed to give HCl fumes | Hydrolyzed to give HCl fumes | Hydrolyzed to give HCl fumes | Hydrolyzed to give HCl fumes |
| pH of solution | 7 | 6.5 | 3 | 2
|
2 | 2 |
Summary: Characteristics of bonds
| Property | Substances with | ||
| Covalent bonds | Ionic bonds | Metallic bonds | |
| Electrical conductivity | Non-conductors except graphite | – Solids do not conduct.
– Aqueous solutions and molten state conduct
|
– Conducts |
| Thermal conductivity | – Non-conductors except graphite
|
– Do not conduct | – Conducts |
| Melting point (oC) | – Low for molecular substances
– High for giant atomic structures
|
– Usually high. | – Generally high |
| Boiling point (oC) | – Low for molecular substances
– High for giant atomic structures
|
– Usually high | – Generally high |
| Solubility | – Generally insoluble in water but soluble in organic solvents | – Generally soluble in water | – Some metals react with water |
UNIT 5: SALTS.
- Meaning.
- Types of salts:
- Normal salts
- Acid salts
- Basic salts
- Double salts.
- Solubility of salts in water
- Sulphates
- Chlorides
- Nitrates
- Carbonates
- Sodium, potassium and ammonium salts
- Solubility of bases in water.
- Obtaining crystals
- 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
- Uses of salts
- Action of heat on salts
- Carbonates
- Nitrates
Download Chemistry lesson plans for all topics
Here are all the Chemistry secondary school lesson plans for all topics. You can also download the editable and pdf lesson plans below.
FREE CHEMISTRY LESSON PLANS (FORM 1-4)
FORM 3
TERM 1
SUBJECT: CHEMISTRY.
TOPIC 1: GAS LAWS.
TEACHER’S NAME: ………………………………………… TSC NO: …………….
SCHOOL/ INSTITUTION : ……………………………………………………………….
FORM: 3 TERM: 1 YEAR……………..
NUMBER OF STUDENTS……. SUBJECT: CHEMISTRY
TOPIC: GAS LAWS.
SUB-TOPIC: BOYLE’S LAW.
WEEK: …….. LESSON NUMBER: ……..
DATE: …….. TIME: ……….
OBJECTIVES:By the end of the lesson the learner should be able to;
-State Boyle’s law.
-Explain Boyle’s law using kinetic theory of matter.
-Represent Boyle’s law mathematically and graphically.
-Solve further problems involving Boyle’s law.
-Plot and interpret graphs involving pressure and volume of gases.
LESSON PRESENTATION
| TIME | CONTENT | LEARNING ACTIVITIES | RESOURCES | REFERENCE |
| 5 MINUTES | INTRODUCTION
State the Boyle’s law. |
Discussion.
Questions and answers.
|
-Chalk board/white board.
-A piece of chalk/ marker pen.
|
K.L.B. BK III
PP. 1-5
|
| 30 MINUTES | BODY DEVELOPMENT
-State Boyle’s law. -Explain Boyle’s law using kinetic theory of matter. -Represent Boyle’s law mathematically and graphically. -Solve further problems involving Boyle’s law. -Plot and interpret graphs involving pressure and volume of gases.
|
-State Boyle’s law.
-Explain Boyle’s law using kinetic theory of matter. -Represent Boyle’s law mathematically and graphically. -Solve further problems involving Boyle’s law. -Plot and interpret graphs involving pressure and volume of gases.
|
Chart
Volume-pressure relationship. Syringes. Calculators. Graph papers. |
K.L.B. BK III
PP. 1-5 Longhorn Book III PP 1 -8 |
| 5 MINUTES | CONCLUSION
Giving assignments. |
Questions and answers.
|
-Chalk board/white board.
-A piece of chalk/ marker pen.
|
K.L.B. BK III
PP. 1-5
|
SELF-EVALUATION:_______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
TEACHER’S NAME: ………………………………………… TSC NO: …………….
SCHOOL/ INSTITUTION : ……………………………………………………………….
FORM: 3 TERM: 1 YEAR……………..
NUMBER OF STUDENTS……. SUBJECT: CHEMISTRY
TOPIC: GAS LAWS.
SUB-TOPIC: CHARLE’S LAW.
WEEK: …….. LESSON NUMBER: ……..
DATE: …….. TIME: ……….
OBJECTIVES:By the end of the lesson the learner should be able to;
State Charles’ law.
Explain Charles’ law using kinetic theory of matter.
Convert temperature in degree Celsius to Kelvin and vice-versa.
Express Charles’ law with equations.
Give a graphical representation of Charles’ law.
Solve numerical problems based on Charles’ Law.
LESSON PRESENTATION
| TIME | CONTENT | LEARNING ACTIVITIES | RESOURCES | REFERENCE |
| 5 MINUTES | INTRODUCTION
-State Charles’ law.
|
Discussion.
Questions and answers.
|
-Chalk board/white board.
-A piece of chalk/ marker pen.
|
Longhorn Book III PP 9-14
|
| 30 MINUTES | BODY DEVELOPMENT
Teacher demonstration:- To show expansion of air when heated and contraction when pressure is constant. Explain increase in volume when temperature is raised. Q/A: – relation between volume and temperature, leading to Charles’ law. Teacher explains inter-conversion of the units. Students complete a table of temperature in the two units Derive equations from volume and temperature relationship.
Exposition: – Teacher exposes a volume-temperature graph and extrapolates it to obtain the absolute temperature. The definition of absolute temperature is exposed. Worked examples. Supervised exercise.
|
Teacher demonstration:- To show expansion of air when heated and contraction when pressure is constant.
Explain increase in volume when temperature is raised. Q/A: – relation between volume and temperature, leading to Charles’ law. Teacher explains inter-conversion of the units. Students complete a table of temperature in the two units Derive equations from volume and temperature relationship.
Exposition: – Teacher exposes a volume-temperature graph and extrapolates it to obtain the absolute temperature. The definition of absolute temperature is exposed. Worked examples. Supervised exercise.
|
Calculators.
Colored water, Glass tube, Warm water, Cork and Flask. |
K.L.B.
BK III P. 6-12
Longhorn Book III PP 9-14
|
| 5 MINUTES | CONCLUSION
Giving assignment on the sub-topic. |
Questions and answers.
|
-Chalk board/white board.
-A piece of chalk/ marker pen.
|
Longhorn Book III PP 9-14
|
SELF-EVALUATION:_______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
TEACHER’S NAME: ………………………………………… TSC NO: …………….
SCHOOL/ INSTITUTION : ……………………………………………………………….
FORM: 3 TERM: 1 YEAR……………..
NUMBER OF STUDENTS……. SUBJECT: CHEMISTRY
TOPIC: GAS LAWS.
SUB-TOPIC: COMBINED LAW.
WEEK: …….. LESSON NUMBER: ……..
DATE: …….. TIME: ……….
OBJECTIVES:By the end of the lesson the learner should be able to;
-Derive the Gas Law.
-Derive the combined gas law equation.
-Solve numerical problems using the equation.
LESSON PRESENTATION
| TIME | CONTENT | LEARNING ACTIVITIES | RESOURCES | REFERENCE |
| 5 MINUTES | INTRODUCTION
Explain and deduce a formulae for the combined gas equation. |
Discussion.
Questions and answers.
|
-Chalk board/white board.
-A piece of chalk/ marker pen.
|
K.L.B.
BK III P. 12
|
| 30 MINUTES | BODY DEVELOPMENT
Q/A: – Combining Boyle’s and Charles’ Laws. Worked examples. |
Q/A: – Combining Boyle’s and Charles’ Laws.
Worked examples. |
Calculators. | K.L.B.
BK III P. 12
Longhorn Book III PP 14-16
|
| 5 MINUTES | CONCLUSION
Oral evaluation on the sub-topic. |
Questions and answers.
|
-Chalk board/white board.
-A piece of chalk/ marker pen.
|
K.L.B.
BK III P. 12
|
SELF-EVALUATION:_______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
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ELECTROCHEMISTRY FORM 4 CHEMISTRY NOTES FREE
ELECTROCHEMISTRY
Objectives
By the end of the topic the learner should be able to:
(a) Explain reduction and oxidation in terms of gain and loss of electrons.
(b) Determine changes in oxidation number to keep track of the movement of electrons in redox reactions and write balanced redox equations.
(c) Describe the electrochemical cell and explain its working in terms of electron transfer process.
(d) Draw cell diagrams using correct cell notations.
(e) Use displacement reactions to compare reducing and oxidising power of ions.
(f) Calculate the electromotive force of a cell given standard electrode potentials.
(g) State the role of water during electrolysis and explain the factors affecting preferential discharge of ions.
(h) Relate the quantity of charge passed to the amount of substance liberated at the electrodes and explain some applications of electrolysis.
ELECTROCHEMISTRY
Electrochemistry is the study of how chemical reactions produce electrical energy and in turn how electrical energy causes chemical reactions.
These chemical reactions involve transfer of electrons.
Redox Reactions
A redox reaction is one in which reduction and oxidation processes occur simultaneously. Redox reactions involve electron gain and electron loss.
Gain of electrons is a reduction process. The species that gains electrons is an oxidising agent.
Loss of electrons is an oxidation process. The species that loses electrons is a reducing agent.
The reaction between iron fillings and copper(II) sulphate solution can be used to illustrate a redox reaction. When iron filings are added to a solution of copper(II) sulphate, a brown solid which is copper metal is formed. The colour of the solution changes from blue to light-green due to the formation of iron(II) ions in the solution.
Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s)
The reaction between iron and copper(II) ions involves transfer of electrons from the iron atoms to copper(II) ions. The reaction can be used to keep track of the transfer of electrons during the reaction as illustrated by the following ionic half equations
Fe (s) Fe2+ (aq) + 2e–
Cu2+(aq) + 2e– Cu(s)
In the first half equation, iron atoms lose electrons (getsoxidized) and formiron(II) ions. Iron atoms acts as a reducing agent.
In the second half equation copper(II) ions gain electrons (gets reduced) to form copper atoms. The copper (II)ions is an oxidising agent.
Oxidation Numbers
An oxidation number is the apparent charge that an element has in a compound or the charge on an ion.
Oxidation number is written with the plus (+) or minus (–) sign in front., ie +2 or – 3.
The knowledge of oxidation numbers helps one to keep track of electron movements in redox reaction and to understand the naming of inorganic compounds.
Rules of Assigning Oxidation Numbers
- The oxidation number of an uncombined element is zero (0), e.g., in molecules, O2, Cl2, H2, the oxidation number of all the atoms is zero.
- The charge on an ion containing one element is equal to the oxidation number of that element. For example,
| Ion | Na+ | S2- | Mg2+ | N3– | O2– | H+ |
| Oxidation number | +1 | -2 | +2 | –3 | –2 | +1 |
- Oxidation number of hydrogen in all compounds that contain it is + 1 except in metal hydrides where it is –1.
| Compound | HCl | H2O | HF | NaH | MgH2 |
| Oxidation number | +1 | +1 | +1 | –1 | –1 |
- Oxidation number of oxygen in all compounds that contain it is –2, except in peroxides where it is –1 and OF2 where it is +2.
| Compound | H2O | CuO | H2O2 | OF2 |
| Oxidation number | –2 | –2 | –1 | +2 |
- In compounds, the sum of the oxidation numbers of all constituent atoms is equal to zero, e.g.,
| Compound | H2O | Na2O2 | NaCl |
| Oxidation number | 2 (+1) –2=0 | 2 (+1) + 2(–1) = 0 | (+1) + (–1) = 0 |
- In ions containing more than one element, the overall charge is equal to the sum of the oxidation numbers of the constituent elements.
| NH4+ | OH– | SO42– |
| –3 + 4 (+1) = +1 | –2 + (+1) = –1 | + 6 + 4 (–2) = –2 |
When the oxidation number of an element in a compound or ion is not known, it is calculated from those of others by using rule 5 and 6 respectively.
Some elements have variable oxidation number. A good example is nitrogen.
| Species | Oxidation number | Species | Oxidation number |
| NO3– | + 5 | N2O | + 1 |
| NO2 | + 4 | N2 | 0 |
| NO | + 2 | NH3, NH4, Mg3N2 | – 3 |
Use of oxidation numbers
1. Naming compounds.
Oxidation numbers are used for assigning names to compounds of elements which have more than one oxidation number.
For example, copper has two oxidation numbers + 1 and + 2. Compounds containing copper with an oxidation number of +1 are referred to as copper(I) compounds. Those compounds containing copper with a oxidation number of +2 are called copper(II) compounds.
The oxidation state of an element in a compound is always denoted by a Roman number written in brackets, as shown in the table below.
| Substance | Oxidation number of: | IUPAC name |
| CuSO4 | Cu (+2) | Copper(II) sulphate |
| CuCl | Cu (+1) | Copper(I) chloride |
| FeS | Fe (+2) | Iron(II) sulphide |
| H2SO3 | S (+4) | Sulphuric(IV) acid |
| H2SO4 | S (+6) | Sulphuric(VI) acid |
| K2Cr2O7 | Cr (+6) | Potassium dichromate(VI) |
| KMnO4 | Mn (+7) | Potassium manganate(VII) |
| CO | C (+2) | Carbon(II) oxide |
| CO2 | C (+4) | Carbon(IV) oxide |
| SO2 | S (+4) | Sulphur(IV) oxide |
| SO3 | S (+6) | Sulphur(VI) oxide |
2. Keeping track of electron movement in redox reactions.
Knowledge of oxidation number helps in determining whether a reaction is a redox one or not. It also helps in determining which substance has been oxidised or reduced.
Oxidation involves increase in oxidation number while reduction involves decrease in oxidation number.
Consider the following reaction between acidified iron(II) sulphate and hydrogen peroxide.
2FeSO4(aq) + H2O2(aq) + H2SO4(aq) 2Fe2(SO4)3(aq) + 2H2O(l)
In the reaction above, iron(II) ions are converted to iron(III) ions, thus the oxidation number of iron increases from +2 to +3. Such a change is an oxidation.
On the other hand the oxygen from the hydrogen peroxide undergoes a reduction by having its oxidation number decrease from –1 to –2 on forming the water molecules.
The above is illustrated using an ionic equation thus,
Other Examples of Redox Reactions
- Reaction of a metal and water
Sodium undergoes oxidation because its oxidation number increases from 0 to + 1. Hydrogen undergoes reduction since its oxidation number decreases from + 1 to 0.
- Reaction of a metal and an acid
Magnesium undergoes oxidation and its oxidation number increases from 0 to + 2 while the hydrogen ion is reduced. The oxidation number of hydrogen decreases from +1 to 0.
Displacement Reactions
A displacement reaction takes place when a more reactive element takes the place of another element which is less reactive in a compound.
Reducing power of metals.
Metals higher in the reactivity series displace from solutions those metals which are lower in the series.
For example, when magnesium is reacted with copper(II) sulphate solution, a brown solid which is copper metal is formed. The colour of the solution changes from blue to colourless. This is because the blue copper (II) ions in the solution are displaced by magnesium ions which are colourless. The ionic half equations for the formation of magnesium ions and copper metal are:
Mg(s) Mg2+(aq) + 2e– (oxidation step)
Cu2+(aq) + 2e– Cu(s) (reduction step)
When the two ionic half equations are combined, the following overall ionic equation is obtained:
Mg(s) + Cu2+(aq) Mg2+(aq) + Cu(s)
The oxidation number of magnesium increases from 0 to +2 while that of copper decreases from +2 to 0. Magnesium is oxidised and Copper is reduced.
Displacement reactions are therefore redox reactions. The more reactive metal (Magnesium) is the reducing agent while copper(II) ions are the oxidising agent.
The more reactive elements such as sodium and calcium lose their electrons readily and are strong reducing agents.
The less reactive metals such as lead and copper lose electrons less readily and are weak reducing agents.
The order of reducing power is:
Potassium Strongest reducing agent
Sodium
Calcium
Magnesium
Aluminium Decreasing reducing power
Zinc
Iron
Lead
Copper
Silver Weakest reducing agent
Other Displacement Reactions
- Aluminium displaces copper from a solution of copper(II) ions.
Oxidation step; 2Al(s) 2Al3+(aq) + 6e–
Reduction step; 3Cu3+(aq) + 6e– 3Cu(s)
2Al(s) + 3Cu2+(aq) Al3+(aq) + 3Cu(s)
- Copper metal displaced from a solution by silver ions.
Oxidation step; Cu(s) Cu2+(aq) + 2e–
Reduction step; 2Ag+(aq) +2e– 2Ag(s)
Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s)
Oxidizing Power of Halogens.
Halogens have a tendency to accept electrons andare therefore strong oxidising agents.
Among the halogensfluorine is the strongest oxidising agent. However, in most reactions, chlorine is the most common halogen used as an oxidising agent.
The more reactive halogens oxidises the less reactive halogens and vice versa.
For example, Chlorine displaces bromine and iodine from theirsolutions.
Oxidation step 2Br–(aq) Br (l) + 2 e–
Reduction step Cl2 + 2e– 2Cl–(aq)
The oxidation number of bromine increases from –1 to 0 while that of chlorine decrease from 0 to –1. Bromine is oxidised while chlorine is reduced. Chlorine is the oxidising agent and the bromide ion is the reducing agent.
The ionic equation for the displacement of iodine by chlorine is:
Cl2(g) + 2I–(aq) 2Cl–(aq) + I2(s)
The oxidation number of iodine increases from –1 to 0 and that of chlorine decreases from 0 to –1. Iodine is oxidised while chlorine is reduced.
The above reactions show that chlorine has a greater tendency to accept electrons than both bromine and iodine. Chlorine takes electrons from the bromide and iodide ions forming bromine and iodine respectively.
Similarly, bromine is more reactive than iodine and takes electrons from iodide ions. Bromine has a higher tendency to accept electrons than iodine. The ionic equation for the reaction is:
Br2(l) + 2I–(aq) 2Br –(aq) + I2 (s)
The oxidation number of brominedecreases from 0 to –1 and that of iodine increases from –1 to 0.
The greater the tendency of an element to accept electrons, the higher is its oxidising power.
Among the halogens considered, chlorine is the strongest oxidising agent.
The order of oxidising power is;
Chlorine
Bromine Decreasing oxidising power
Iodine
The Electrochemical Cell
An electrochemical cell is a device that generates a potential difference between electrodes using chemical reactions.
Electrochemical Half-Cell
An electrochemical half-cell is formed by dipping a metal rod (electrode) in an aqueoussolution of its ions, where some of its surface atoms lose electrons and go into solution as ions. The lost electrons remain on the metal surface.
Metal atoms solution Metal ions in solution
M(s) Mn+(aq) + ne–
The metal rod develops a negative charge and this attracts the ions back again, some of the ions accept electrons from the rod and form atoms once more.
Mn+ (aq) + ne– M(s)
As the negative charge on the surface of the rod builds up, the rate at which the ions combine with electrons increases until eventually it is equal to the rate at which metal atoms lose electrons to form positive ions. At this point an equilibrium is established. The equation for the equilibrium is:
Metal atoms Metal ions
M(s) Mn+(aq) + ne–
A potential difference is created between the metal rod and the positively charged ions in the solution.
The concentration of electrons on the metal rod is measured by a quantity called the electrode potential.
The half-cell can be represented as: metal | metal ion. The vertical line represents the phase boundary where a potential difference develops. For example, a zinc half-cell is represented as;
Zn(s) | Zn2+(aq)
The tendency of metals to form ions when in contact with their ions differs from one metal to another.
The Electrochemical Cell.
An electrochemical cell is obtained whenthe half-cells of two different metals are connected to form a complete cell so that the difference between the potential of the half cells can be measured.
The electrodes of the two half-cells are connected by metallic wires while the solutions are connected through a salt-bridge.
The salt-bridge is in the form of a filter paper soaked in a saturated solution of potassium nitrate or sodium nitrate.
The salts chosen for a salt-bridge must not react with either of the salt solutions in the half cells.
Electrons flow along the wire from the electrode with a higher concentration of electrons to the electrode with a lower concentration of electrons.
The difference between the electrode potentials of the two electrodes is called the electromotive force (e.m.f.) of the cell. The e.m.f. is measured in volts using a voltmeter.
Functions of the Salt Bridge
The functions of the salt-bridge are:
- Complete the circuit by making contact between the two solutions (electrolytes).
- Maintains balance of charges in electrolytes by providing ions to replace those ions that are used up or those that are formed.
Relative tendency of metals to Ionize.
The tendency of metals to form ions when in contact with their ions differs from one metal to another. This property can be used to obtain electrochemical cells, for example the zinc-copper electrochemical cellshown below.
When a copper-copper ions half-cell, Cu(s) | Cu2+ (aq) is connected to a zinc-zinc ions half-cell, Zn(s) | Zn2+(aq), the following observations are made:
- The zinc rod in the zinc-zinc ions half-cell wears out.
- The intensity of the blue colour of copper(II) sulphate solution decreases and red-brown deposits appear on the copper rod in the copper-copper ions half-cell.
- A voltage of 10V is registered by the voltmeter.
In the Zn(s) | Zn 2+| | Cu2+| Cu(s)cell, the following equations represent what happens in the two half cells:
Zinc electrode (anode)
Zn (s) Zn2+(aq) + 2e–
Copper electrode (cathode)
Cu2+(aq) + 2e– Cu(s)
In the Zn (s) | Zn2+(aq) half-cell the oxidation number of zinc increases from 0 to +2.
In the Cu(s) | Cu2+ (aq)half-cell the oxidation number of copper decreases from +2 to 0.
The anode is defined as the electrode at which oxidation takes place while the cathode is the electrode at which reduction takes place.
Oxidation occurs at the Zn(s) | Zn2+(aq) half-cell where electrons are released.
Reduction takes place at the Cu(s) | Cu2+(aq) where electrons are gained.
These reactions show that the zinc electrode has a higher tendency to form ions than the copper electrode when the metals are placed in solutions of their ions.
The zinc electrode has a higher accumulation of electrons and is more negative compared to the copper electrode which has a lower accumulation of electrons.
Therefore, the zinc terminal is relatively more NEGATIVE with respect to the copper terminal.
When the two half cells are connected, electrons flow FROM the zinc terminal through the connecting wire TO the copper terminal.
Electrons lost by the zinc electrode are gained by the copper(II) ions.
When the two ionic half equations are combined, the ionic equation for the electrochemical cell is obtained.
Zn2+(s) + Cu(aq) Zn2+(aq) + Cu(s)
The ionic equation and the e.m.f. of the electrochemical cell can be summed up in what is called a cell notation.
Zn(s) | Zn22+(aq) | | Cu2+(aq) | Cu(s) E = + 1.10 V
The single vertical line represents phase boundaries in the half-cells while the two vertical parallel lines represent the salt-bridge.
The half-cell in which electrons are released (oxidation takes place) is always on the left– hand side of the cell diagram, i.e., Zn(s) | Zn2+(aq).
Electrons flow FROM the lefthand half-cell TO the right hand half-cell.
Standard Electrode Potentials
The standard electrode potential of any element|element ions half-cell is taken as the difference between its potential and that of hydrogen|hydrogen ions half-cell.
The chosen standard electrode is hydrogen and is usually referred to as the standard hydrogen half cell. Being the reference electrode, it is assigned an electrode potential 0.00 V.
The standard conditions for measuring electrode potential are:
(i) Temperature of 25°C.
(ii) All solutions have a concentration of 1 mole per litre (1 M).
(iii) Pressure of 1 atmosphere.
Platinised platinum electrode is used as the electrode when the half-cell does not include a metal, e.g., H2(g) | H+(aq).
The platinised platinum electrode has three functions:
(i) It acts as an inert metal connection to the H | H+(aq) system.
(ii) It provides a surface area on which dissociation of hydrogen molecules can take place, i.e.,
H2(g) 2H+(aq) + 2e–
(iii) It serves as an electrical conductor to the external circuit.
The hydrogen electrode consists of an inert platinum electrode which is immersed in a 1.0 mole per litre solution of hydrogen ions, H+ Hydrogen gas is bubbled on the platinum electrode which is dipped in a solution containing 1 M hydrogen ions.
½ H2(g) H+(aq) + e–
The standard hydrogen half-cell is presented as:
Pt(s), H2(g) | H+(aq).
While the standard electrode potential, E for this reference half-cell which is zero can be represented as:
½ H2(g) H+(aq) + e– E = 0.00 V
The standard electrode potential difference isthe potential difference for a cell comprising a particular element in contact with one molar solution of its ions and the standard hydrogen electrode. It is denoted by the symbol, Eθ.
If an element has a greater tendency to lose electrons than hydrogen, the electrode potential of its half-cell is negative with respect to the hydrogen half-cell. (e.g., zinc)
The ionic half equations for the reactions occurring at the electrode are:
Zn (s) Zn2+(aq) + 2e– Eθ= 0.76 V
2H+(aq) + 2e– H2(g) Eθ = 0.00 V
The overall ionic equation is:
Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) Eθ = – 0.76 V
Similarly, if a F | F–(aq) half-cell is connected with the Pt(s)H(g) | H+(aq) half-cell, the e.m.f. registered for the cell is + 2.87 V. The half-cell reaction is as follows:
F2(g) + 2e– 2F–(aq)
H2(g) 2H+(aq) + 2e–
The overall ionic equation being:
H2(g) + F2(g) 2F–(aq) + 2H+(aq) Eθ = + 2.87 V
On the other hand, if the tendency of an electrode to lose electrons is lower than the hydrogen electrode, the electrode is positive with respect to hydrogen electrode and its potential is positive, e.g., copper.
Standard electrode potentials are sometimes referred to as standard reduction potential because they relate to the reduction reactions.
The tableof standard electrode potentials for some elements is arranged so that the strongest oxidising agent, fluorine, which has the most positive value for E° is at the top of the list. The weakest oxidising agent, lithium ions, Li+, with the most negative value of E° is at the bottom.
Similarly, fluoride ions is the weakest reducing agent while lithium is the strongest reducing agent.
Uses of Standard Electrode Potentials
Standard electrode potentials are used in:
- Comparing the oxidising and reducing powers of substances.
- Determining the e.m.f. of a cell.
- Predicting whether or not a reaction will take place.
Comparing Oxidising and Reducing Power
Lithium with Eθ = – 3.04 V has the highest tendency to lose electrons and therefore it is the strongest reducing agent.The more negative the Eθ value the greater the reducing power. Lithium has the least tendency to accept electrons hence it is the weakest oxidising agent.
Fluorine with Eθ of + 2.87 V has the highest tendency to accept electrons and therefore the strongest oxidising agent. The more positive the Eθ value, the greater the oxidising power. Conversely, fluorine is the weakest reducing agent since it has the least tendency to lose electrons.
Using Standard Electrode Potentials to calculate the e.m.f of a Cell
The e.m.f of a cell is obtained by changing the sign of the electrode potential of the half-cell thatundergoes oxidation and then adding to the electrode potential of the half-cell that undergoes reduction.
Eθcell = Ereduction – Eoxidation
Ereduction is the electrode potential of the half-cell that undergoes reduction.
Eoxidation is the electrode potential of the half-cell that undergoes oxidation.
Example 1
The reduction potentials of Mg(s) | Mg2+(aq) and Zn(s) | Zn2+(aq) half-cells are:
Mg2+(aq) + 2e– Mg(s) Eθ = – 2.37V
Zn2+(aq) + 2e– Zn(s) Eθ= – 0.76 V
Using the electrode potentials, answer the following questions.
(i) Write an ionic equation for a cell made by combining the two half cells.
(ii) Calculate the e.m.f. of the cell formed in part (i).
(iii) Write the cell notation for the cell.
Solution
- Since the electrode potential of magnesium is more negative, magnesium is the stronger reducing agent. Zinc undergoes reduction while magnesium is oxidised. The sign of the Eθvalue of Mg changes from negative to positive.
Mg(s) Mg2+ (aq) + 2e–+ 2.37 V
Zn2+ (aq) + 2e Zn(s)–0.76 V
Mg(s) + Zn2+(aq) Mg2+(aq) + Zn(s) + 1.61 V
- The e.m.f. of the cell is obtained by changing the sign of the electrode potential of magnesium and adding them, i.e. + 2.37 V + (– 0.76) = + 1.61 V.
- Mg(s) | Mg2+(aq) | | Zn2+ (aq) | Zn (s) Eθcell = + 1.61V
Example 2
Calculate the e.m.f. for the electrochemical cell represented below:
Al(s) | Al3+(aq) | | Zn 2+(aq) | Zn(s)
Given that:
Al3+(aq) + 3e– Al(s) E° = – 1.66 V
Zn2+(aq) + 2e– Zn (s) E° = –0.76 V
Solution
Al(s) Al3+(aq) + 3e– + 0.66 V
Zn2+ (aq) +2e– Zn(s) –0.76 V
To combine the two half equations, the number of electrons should be equal. To do this, we multiply the first equation by 2 and the second equation by 3. The electrode potentials remain the same. This is because the voltage does not depend on the number of electrons flowing.
2A(s) 2Al3+(aq) + 6e– + 1.66 V
3Zn2+ (aq) + 6e– 3Zn(s) –0.76 V
2Al(s) + 3Zn2+(aq) 2Al3+(aq) + 3Zn (s) + 0.90 V
Or
Eθcell = EθR.H.S – EθL.H.S
= – 0.76 V – (– 1.66 V)
= – 0.76 V + 1.66 V
= + 0.90 V
Example 3
Use the cell representation below to answer the question that follows
Cu(s) | Cu2+(aq) | | Ag+(aq) | | Ag(s) Eθcell = + 0.46 V
Given that the E value for Ag+(aq) | Ag(s) is + 0.80 V, calculate the Eθ value for Cu(s) | Cu2+(aq).
Solution
EθCell = EθR.H.S – EθL.H.S
Substituting
0.46 V = 0.180 V – EL.H.S
0.46 V – 0.80 V = –EL.H.S
– 0.34 = – EL.H.S
EθL.H.S = + 0.34 V
Example 4
Use the standard electrode potentials for elements A, B, C, D and E given below to answer the questions that follow. The letters do not represent actual symbols of elements.
| Eθ (volts) | |
| A2+(aq) + 2e– A(s) | –2.37 |
| B2+(aq) + 2e– B(s) | –0.76 |
| C+(aq) +e– ½ C2(g) | 0.00 |
| D2+(aq) +2e D2(s) | +0.34 |
| ½ E2(g) + e– E–(aq) | + 1.36 |
(i) What is the Eθ value of the strongest oxidising agent? Explain.
Answer:+ 1.36 most positive.
(ii) Which two of the above elements would produce the largest e.m.f or potential difference in an electrochemical cell Explain.
Answer:A and E, the elements with the most positive Eθ and the most negative Eθ.
(iii) What would be the initial potential difference of the cell chosen in(ii) above?
Answer:
E.m.f of cell = E reduction – R oxidation
= 1.36 – (–2.37)
= 1.36 V + 2.37 V
= + 3.73 V
(vi) Write the cell representation for the electrochemical cell formed.
Answer:A(s) | A2+(aq) | | E2(g) | E–(aq), Pt Eθ = 3.73 V
Using standard Electrode potential to predict it a Reduction will take place
Previously it was established that Zn reduces Cu2+
| Zn(s) Zn2+(aq) + 2e | + 0.76 V |
| Cu2+ + 2e– Cu(s) | + 0.34 V |
| Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) | + 1.10 V |
The cell potential or e.m.f., + 1.10 V is positive showing the reaction takes place. The e.m.f. for the reverse reaction:
Cu(s) + Zn2+(aq) Cu2+(aq) + Zn(s) is Eθ = – 1.10 V
The negative value implies that the reaction is unlikely to occur. In general reactions with an overall positive e.m.f can take place; while those with negative: e.m.f. cannot.
Example
Predict whether a reaction will occur between iodine and chloride ions.
I2(aq) + 2e– 2I– (aq) Eθ = + 1.36 V
Cl2 (aq) 2e–2Cl–(aq) Eθ= + 1.51 V
Solution
Adding the ionic half equations;
| I2(aq) + 2e– 2I–(aq) | Eθ = + 1.36 V |
| 2Cl(aq) Cl2(aq) + 2e | Eθ = –1.51 V |
| I2(aq) + 2Cl–(aq) 2I–(aq) + Cl2(g) | Eθ = –0.15 V |
The overall Eθ of the cell is negative. Therefore, iodine cannot displace chlorine from a chloride solution.
Uses of Electrochemical Cells
Electrochemical cells are used as a source of energy.
Dry Cells
Dry cells are used in a wide range of electrical appliances such as radios, watches, clocks, flashlights and electric bells. The dry cells are cheap and convenient to use because they contain the electrolyte in form of a paste rather than a liquid therefore cannot spill or leak.
An example of a dry cell is the Le’ Clanche cell shown below.
It consists of a zinc can which forms the negative terminal and a graphite rod which is the positive terminal.
The graphite rod is surrounded by a paste of ammonium chloride and zinc chloride, and powdered manganese(IV) oxide mixed with carbon. The powder increases the surface area of the positive terminal.
The function of manganese(IV) oxide is tooxidise the hydrogen produced at the electrode to water thus preventing any bubbles from coating the carbon terminal which would reduce its efficiency.
At the negative terminal:
Zn(s) Zn 2+(aq) + 2e–
At the positive terminal, ammonium ions are converted to ammonia and hydrogen gases
2NH4+(aq) + 2e– 2NH3(g) + H2(g)
The gases produced do not escape but are immediately used up in other reactions. The hydrogen is oxidised to water by the manganese(IV) oxide.Ammonia forms a complex with the zinc chloride in the paste.
A simple dry cell can produce a potential of 1.5 V.
Dry ammonium chloride does not conduct an electric current hence a paste, which is an electrolyte is used.
Such a cell is called a primary cellbecauseonce the cell is used to supply energy the chemicals are used up and therefore the cell has to be discarded (it cannot be recharged).
Some cells can be reused after being recharged. These are called secondary cells. The lead acid accumulator is the most common secondary cells.
Accumulators
The main features of a lead-acid accumulator are the lead plate which is the negative terminal and the lead (VI) oxide plate which is the positive terminal. Both of these electrodes dip into an aqueous solution of sulphuric(VI) acid.
At the negative terminal, lead atoms lose electrons to form lead(II) ions.
Pb(s) Pb2+(aq) + 2e–
At the positive terminal,lead(IV) oxide reacts with hydrogen ions in sulphuric (VI) acid forming lead(II) Ions:
PbO2(s) + 4H+(aq) + 2e– Pb2+(aq) + 2H2O(l)
The lead (II) ions formed react instantly with the sulphate ions to form lead(II) sulphate which is insoluble and adheres to the electrodes.
Pb2+(aq) + SO42–(aq) PbSO4(s)
The net reaction that takes place is:
As the battery discharges, lead and lead(IV) oxide are depleted and the concentration of sulphuric(VI) acid decreases.
Since the density of the aqueous solution depends on the concentration of sulphuric(VI) acid, measurement of its density can be used as means of telling how far the battery is discharged.
During recharging of the battery, the electrode reactions shown are reversed so as to restore its original reactants:
Fuel Cells
Fuel cells are electrochemical cells which convert the chemical energy of a fuel directly to electrical energy such as the hydrogen oxygen cell shown below.
At the negative terminal, hydrogen reacts with hydroxide ions to form water and electrons are released.
2H2(g) + 4OH–(aq) 4H2O(l) + 4e–
At the positive terminal, oxygen and water acquire electrons to form hydroxide ions.
O2(g) + H2O(l) + 4 e–(aq) 4OH(aq)
The overall reaction in the hydrogen | oxygen cell is
2H2(g) + O2(g) 2H2O(l)
The cell goes on producing electricity as long as hydrogen and oxygen are fed into it, so that it does not become exhausted like a primary cell.
The fuel cell unlike a secondary cell does not store energy.
The electrode, other than completing the circuit also catalyses the reactions which increase the output of the cell.
Electrolysis
Electrolysis is the process in which electrical energy is used to cause non-spontaneous chemical reactions to occur.
In these reactions, the substance undergoes chemical decomposition.
Preferential Discharge of Ions During Electrolysis
In the aqueous solution, there are more than two ions since water also ionises. During electrolysis, only one of the anions and one of the cations can be discharged. Preferential discharge therefore takes place, according to the factors discussed below.
Factors Affecting Preferential Discharge During Electrolysis
- The concentration of the electrolyte
A cation or anion whose concentration is high is preferentially discharged if the ions are close in the electrochemical series.
- The product obtained at the electrode depends on the nature of electrode used.
- Position in the Electrochemical Series
The ease of reduction of cations and oxidation anions depends on their position in the electrochemical series. The cations high in the series require more energy to be reduced. Anions high in the series require more energy to be oxidised.
1. Electrolysisof dilute sodium chloride
Sodium chloride solution contains sodium ions (Na+) and chloride ions (Cl–) from sodium chloride, hydrogen ions (H+) and hydroxide ions (OH–) from water.
When an electric current is passed through the solution, chloride ions (Cl–) and hydroxide ions (OH–) migrate to the anode.
Hydroxide ions (Eθ= + 0.04 V) have a greater tendency to lose electrons compared to the chloride ions. (Eθ= +1.36 V).
Reaction at Anode
4OH–(aq) 4OH(aq) + 4e–
4OH(aq) 2H2O(l) + O2(g)
Overall;
4OH–(aq) 2H2O(l) + O2(g) + 4e–
Sodium ions (Na+) and hydrogen ions (H+) migrate to the cathode but hydrogen ions (Eθ= 0.00 volts) are preferentially discharged because they have a greater tendency to gain electrons than sodium ions (Eθ = –2.71 volts).
Reactions at Cathode
4H+(aq) +4e 4H(aq)
4H(aq) 2H2(g)
Overall;
4H+(aq) + 4e– 2H2(g)
Electrolysis of dilute sodium chloride evolves oxygen gas at the anode and hydrogen gas at the cathode. This is essentially the electrolysis of water since oxygen and hydrogen are the constituents of water.
2. Electrolysis of concentrated sodium chloride solution (Brine)
Brine is concentrated sodium chloride solution. The solution therefore contains the same ions as dilute sodium chloride.
When an electric current is passed through brine, chloride ions (Cl–) and hydroxide ions
(OH–) migrate to the anode.The chloride ions (Eθ = + 1.36 volts) are preferentially discharged because of their relatively high concentrations.
Reaction at Anode
2Cl–(aq) 2Cl(g) + 2e–
2Cl(g) Cl2(g)
Overall;
2Cl–(aq) Cl2 (g) + 2e–
Sodium ions (Na+) and hydrogen ions (H+) migrate to the cathode.
At the cathodes, sodium ions are not preferentially discharged in spite of their high concentration. This is because the tendency of hydrogen ions (Eθ = 0.000 volts) to gain electrons is much higher than that of sodium ions(Eθ = –2.71 volts).
Therefore,hydrogen ions are preferentially discharged.
Reactions at Cathode
2H+(aq) + 2e– 2H(g)
2H(g) H2(g)
Overall;
2H+(aq) +2e– H2(g)
The discharge of hydrogen ions leads to an increase in hydroxide (OH–) ions concentration and the solution becomes alkaline.
As the electrolysis process continues, the concentration of chloride ions decreases and eventually hydroxide ions are oxidised to water and oxygen gas.
3. Electrolysis of dilute sulphuric (VI) acid.
Dilute sulphuric(VI) acid contains sulphate (SO42–), hydroxide (OH–) and hydrogen (H+) ions.
When an electric current is passed through the dilute acid, sulphate and hydroxide ions migrate to the anode while the hydrogen ions migrate to the cathode.
Reaction at anode
The hydroxide ions(Eθ = + 0.40 volts) are preferentially discharged because they have a greater tendency to lose electrons than sulphate ions (Eθ = + 2.01 volts).
4OH–(aq) 4OH(aq) + e–
4OH(aq) 2H2O(l) + O2(g)
Overall;
4OH–(aq) 2H2O(l) + O2(g) + 4e–
Reaction at Cathode
4H+(aq) + 4e– 4H(g)
4H(g) 2H2(g)
Overall;
4H+(aq) + 4e– 2H2(g)
The four electrons lost by hydroxide ions to form one mole of oxygen molecules are gained by the four hydrogen ions to form two moles of hydrogen molecules. For every mole of oxygen gas produced at the anode two moles of hydrogen are formed at the cathode.
The volume of hydrogen is therefore twice that of oxygen.
The amount of water in the electrolyte decreases as the electrolysis process continues. This causes an increase in the concentration of the acid.
4. Electrolysisof aqueous magnesium sulphate
The ions present in magnesium sulphate solution are magnesium (Mg2+), sulphate (SO42–), hydrogen (H+) and hydroxide (OH–).
When an electric current is passed through the solution, hydroxide ions (Eθ = + 0.40 volts) and sulphate ions (Eθ = + 2.01 volts) migrate to the anode.
Hydroxide ions are preferentially discharged because of their greater tendency to lose electrons.
Reaction at Anode
4OH–(aq) 4OH(aq) +4e–
4OH(aq) 2H2O(l) + O2(g)
Overall;
4OH–(aq) 2H2O(l) + O2(g) + 4e
Both magnesium (Eθ = –2.38 volts) and hydrogen ions (Eθ = 0.00 volts) migrate to the cathode.
Hydrogen ions are preferentially discharged because of their greater tendency to gain electrons.
Reactions at Cathode
4H+(aq) +4e– 4H(g)
4H(g) 2H2(g)
Overall;
4H+ (aq) + 4e– 2H2(g)
It is observed that the volume of oxygen gas produced at the anode and hydrogen gas at the cathode are in the ratios of 1:2 respectively.
5. Electrolysis of copper(II) sulphate solution using different electrodes.
- Using inert electrodes (carbon or platinum)
Copper(II) sulphate solution contains copper(II) (Cu2+), Sulphate (SO42–), hydrogen ions (H+), and hydroxide (OH–) ions.
When the solution is electrolysed using carbon or platinum electrodes, sulphate ions (Eθ = + 2.01 volts) and hydroxide ions (Eθ= 0.20 volts) migrate to the anode while the copper(II) (Eθ= +0.34 volts) and hydrogen ions (Eθ = 0.00 volts) migrate to the cathode.
Reactions at Anode
The hydroxide ions have a greater tendency to lose electrons and therefore are preferentially discharged.
4OH–(aq) 4OH(aq) + 4e–
4OH(aq) 2H2O(l) + O2(g)
Overall;
4OH (aq) 2H2O(l) + O2(g) + 4e–
Reactions at Cathode
The copper(II) ions have greater tendency to gain electrons that the hydrogen ions and is therefore preferentially discharged. The atoms are deposited on the cathode as red-brown coating.
As a result, the mass of the cathode increases while that of the anode remains the same.
Cu2+(aq) + 2e Cu(s)
The concentration of copper(II) ions in solution decreases and the blue colour of the copper(II) sulphate solution becomes pale and finally colourless.
Hydrogen ions accumulate in the solutions and therefore the solution becomes acidic.
(b) Using copper electrodes
When copper electrodes are used in the electrolysis of copper(II) sulphate solution, the mass of the anode decreases, while that of the cathode increases.
Sulphate and hydroxide ions migrate to the anode, but none of them is discharged; instead the copper anode is gradually oxidised and goes into solution.
Cu(s) Cu2+(aq) + 2e
This explains the loss in mass of the anode. Less energy is needed for the copper anode to lose electrons than hydroxide ions.
Hydrogen and copper(II) ions migrate to the cathode where copper(II) ions are preferentially discharged because they have greater tendency to accept electrons.
Cu2+(aq) + 2e Cu(s)
The cathode is thus coated with a red-brown deposit of copper metal.
The amount of copper oxidised at the anode is equal to the amount of copper deposited on the cathode and therefore the concentration of copper(II) ions in the solution remains the same. The colour of the blue solution does not fade.
Carbon or platinum electrodes allow passage of an electric current into and out of the electrolyte without wearing out such electrodes.
Summary of Electrolysis
- Electrolyte: dilute sodium chloride
| Ions Present | |
| Cations | Anions |
| Sodium, Na+(aq) and hydrogen H+(aq) ions.
Hydrogen, H+(aq) ions discharged at the cathode, because sodium is above hydrogen in electro-chemical series. Hydrogen gas evolved. |
Chloride Cl–(aq) and hydroxide OH–(aq) ions.
Hydroxide (OH–)(aq) discharged at the anode, because hydrogen ions required less energy to discharge than chloride Cl–(aq) ions. Oxygen gas evolved. |
- Electrolyte:Brine (Concentrated Sodium chloride)
| Ions Present | |
| Cations | Anions |
| Sodium, Na+(aq) and hydrogen H+(aq) ions.
Hydrogen, H+(aq) ion discharged at the cathode, because sodium is above hydrogen in electro-chemical series. Hydrogen gas evolved. |
Chloride Cl–(aq) and hydroxide OH–(aq) ions
Hydroxide (OH–)(aq) discharged at the anode, because hydrogen ions required less energy. To discharge than chloride Cl–(aq) ions. Chlorine gas evolved. |
- Electrolyte: dilute sulphuric acid
| Ions Present | |
| Cations | Anions |
| Hydrogen, H+(aq)
Hydrogen, H+(aq) ions discharged at the cathode, because no other cations are present. Hydrogen gas evolved. |
Sulphate, SO42–(aq) and hydroxide, OH–(aq)
Hydroxide (OH–)(aq) discharged at the anode, because hydroxyl ions require less energy to discharge than sulphate ions. Oxygen gas evolved. |
- Electrolyte: Magnesium sulphate
| Ions Present | |
| Cations | Anions |
| Magnesium, Mg2+(aq) and hydrogen, H+ (aq) ions.
Hydrogen ions discharges at the cathode, because magnesium is above hydrogen in the electro-chemical series. Hydrogen gas evolved. |
Sulphate, SO42–(aq) and Hydroxide OH–(aq) ions.
Hydroxide ions discharged at the anode because hydroxide ions require less energy to discharge than sulphate ions. Oxygen gas evolved. |
- Electrolyte:Copper(II) sulphate Electrode:Carbon rods
| Ions Present |
|
| Cations | Anions |
| Copper, Cu2+(aq) and Hydrogen H+(aq) ions.
Copper ions discharged at the cathode, because copper is below hydrogen in the electro-chemical series. Colour of solution eventually fades. |
Sulphate, SO42–(aq) and Hydroxide OH–(aq) ions.
Hydroxide ions discharged at the anode because hydroxide ions require less energy to discharged than sulphate ions. Oxygen gas evolved. |
- Electrolyte:copper(II) sulphate Electrode:Carbon
| Ions Present | |
| Cations | Anions |
| Copper, Cu2+ (aq) and Hydrogen H+ (aq) ions.
Copper ions discharged at the cathode, because copper is below hydrogen in the electro-chemical series. Colour of solution does not fades. |
Sulphate, SO42– (aq) and Hydroxide OH– (aq) ions.
No ion discharged at the anode instead the electrode dissolves, and goes into solution as copper (II) ions. |
Quantitative Treatment of Electrolysis
An electric current is measured in amperes. The quantity of electric charge(Q) is measured in coulombs(C).
- A coulomb is the quantity of electricity passed when a current (I) of one ampere flows for a time(t) of one second.
The relationship between the mass of substance produced and the quantity of electricity passed is the basis of Faraday’s Law of electrolysis which states that the mass of a substance produced during electrolysis is directly proportional to the quantity of electricity passed.
The quantity of electricity carried by one mole of electrons is a constant called a Faraday (F) and is equivalent to 96,487 coulombs.
- 1 Faraday is equivalent to one mole of electrons. The number of electrons required to deposit a given ion is equivalent to the charge on the ion.
Worked Examples
- What mass of copper would be deposited on the cathode when a steady current of one ampere flows for 30 minutes through copper(II) sulphate solution?
(Cu = 63.5) Faraday constant = 96,487 C mol–1)
Solution
Reaction at the cathode
Cu2+(aq) + 2e Cu(s)
One mole of Cu ions required 2 moles of electrons.
Quantity of electricity (Q) = 1 × 30 × 60 coulombs.
1 mole of electrons carries a charge of 96,487 coulombs.
2 moles of electrons will carry 2 × 96,487 coulombs.
2 × 96,487 coulombs deposit 63.5 g of copper at the cathode.
Therefore 1 × 30 × 60 coulombs deposits:
- What volume of oxygen will be liberated at the anode when a current of 3 amperes is passed through magnesium sulphate solution for 45 minutes and 30 seconds?
(Molar gas volume at r.t.p. = 24.0 litres, Faraday constant = 96,500 coulombs).
Solution
Reaction equation at the anode
4OH–(aq) 2H2O(l) + O2(g) + 4e
1 mole of electrons carry 96,500 coulombs.
4 moles of electrons carry 4 × 96,500 coulombs.
4 × 96,500 coulombs liberate 24 litres of oxygen.
3 × (45 × 60) +30) coulombs will liberate
= 0.50922 litres.
- In an experiment to electrolysecopper(II) sulphate solution using copper electrodes, 0.2 amperes were passed through the solution for 1,930 seconds. The mass of copper cathode increased from 6.35 to 6.478 g. Find the charge on a copper ion. (1 Faraday = 96,500 coulombs, Cu = 64).
Solution
Mass of copper deposited = (6.478 – 6.350) g = 0.128 g
Quantity of electricity passed = 0.2 × 1,930 = 386 C
0.128 g copper was deposited by 386 C
64 g of copper would be deposited by C = 193,000 C
1 mole of copper atoms (Mass 64 g) require 193,000 C
Number of Faradays required to deposit 64 g of copper at cathode is therefore
Applications of Electrolysis
Extraction of Reactive Elements
Electrolysis is used in the extraction of reactive elements such as sodium, magnesium, aluminum and chlorine.
Electroplating
This is the process of using electricity to coat one metal with another. This is done to protect some metals from corrosion. Electroplating is also done to make an article look attractive. Gold plated watches, silver utensils are common items.
Sacrificial metal (cathodic protection)
Iron or steel structures are protected from corrosion through sacrificial protection either by galvanising or cathodic protection.
Cathodic protection
Corrosion involves loss of electrons by an element to form ions. If it is a less reactive metal it is connected to a more reactive metal by a conductor when the conditions for causing corrosion are present, the more reactive metal ionises at the expenses of the less reactive. The more reactive metal is sacrificed and the method is sacrificial protection.
Galvanising
When the galvanised surface is scratched and iron is exposed, zinc passes into solution as zinc ions rather than Fe2+ ions. This is possible because zinc is easily oxidised than iron.
Purification of Metals
Copper and other less reactive metals are purified by electrolysis. The impure metal is made the anode and pure metal the cathode. The electrolyte contains the cation of the metals being purified.
Manufacture of Sodium Hydroxide and Chlorine from Electrolysis of Concentrated Sodium Chloride (Brine)
Sodium hydroxide and chlorine are manufactured by the electrolysis of brine by use of mercury cell, shown alongside.
The electrolyte in the mercury cell is a concentrated solution of sodium chloride (Brine). The anode in the cell is made of carbon or titanium because they do not react with chlorine gas. While the cathode is a moving film of mercury.
When an electric current is passed through concentrated sodium chloride solution, chloride (Cl–) ions and hydroxide (OH–) ions migrate to the anode. Chloride ions are preferentially discharged because of their relatively high concentration.
2Cl–(aq) 2Cl(g) + 2e–
2Cl(g) Cl2(g)
Both sodium (Na+) ions and hydrogen (H+) ions migrate to the cathode (moving film of mercury).Hydrogen ions are not discharged because of the over-potential(excess power) required to discharge it. Sodium ions are preferentially discharged instead.
2Na+(aq) + 2e– 2Na(l)
The sodium atoms formed dissolve in the hot mercury to form sodium amalgam (Na Hg)
Na(l) + Hg(l) NaHg(l)
The sodium amalgam is passed through a trough in the cell that contains distilled water. The sodium in the amalgam then reacts with water to form a solution of sodium hydroxide and hydrogen gas. Hydrogen is pumped out to the required place, while mercury is regenerated and recycled.
2NaHg(l) + 2H2O(l) 2NaOH(aq) + 2Hg(l) + H2(g)
The sodium hydroxide obtained this way is about fifty per cent pure. Pure sodium hydroxide is obtained by evaporating the water in the aqueous sodium hydroxide solution to get pellets or flakes.
The process is expensive due to the high cost of mercury and the safety measures applied since mercury is poisonous.
Review Exercises
- 2006 Q 15 P1
Study the standard reduction potential given and answer the questions that follow. (The letters are not the actual symbols of the elements).
Eθ (volts)
M2+ (aq) + 2e– M(s) -0.76
N2+ (aq) + 2e– N(s) -2.37
P+ (aq) + e– P(s) +0.80
Q2+ (aq) + 2e– Q(s): -0.14
- The standard reduction potential for Fe 2+ (aq) is -0.44 volts. Select the element which would best protect iron from rusting. (1 mark)
- Calculate the EѲ value for the cell represented as M(s)/ M2+ (aq) // P+(aq)/P(s). (2 marks)
- 2006 Q 1 P2
- What is an electrolyte? (1 mark)
- State how the following substances conduct electricity.
- Molten calcium chloride. (1 mark)
- (1 mark)
- The diagram below shows a set up that was used to electrolyse aqueous magnesium sulphate.
- On the diagram above, using an arrow, show the direction of flow of electrons. (1 mark)
- Identify the syringe in which hydrogen gas would be collected. Explain (1 mark)
- Explain why the concentration of magnesium sulphate was found to have increased at the end of the experiment. (2 marks)
- During the electrolysis, a current of 0.72 A was passed through the electrolyte for 15 minutes. Calculate the volume of gas produced at the anode. (1 Faraday = 96 500 coulombs; molar gas volume is 24000 cm3 at room temperature). (4 marks)
- 2006 Q 2b P2
Use the reduction potentials given below to explain why a solution containing copper ions should not be stored in a container made of zinc.
Zn2+ (aq) + 2e–Zn(s); Eθ = -0.76 V
Cu2+ (aq) + 2e‑Cu(s); Eθ = +0.34 V
(2 marks)
- 2007 Q 7 P1
- Use the information given below to draw a labelled diagram of an electrochemical cell that can be constructed to measure the electromotive force between G and J. (2 marks)
G2+ (aq) + 2e– G(s); Eθ = – 0.74 V
J2+ (aq) + 2e– J(s); Eθ = -0.14 V
- Calculate the Eθ value for the cell constructed in (a) above. (1 mark)
- 2007 Q 21 P1
- When brine is electrolyzed using inert electrodes, chlorine gas is liberated at the anode instead of oxygen. Explain this observation.
(2 marks)
- Name the product formed at the cathode. (1 mark)
- 2007 Q 28 P1
During the electrolysis of aqueous silver nitrate, a current of 5.0 A was passed through the electrolysis for 3 hours.
- Write the equation for reaction which took place at the anode. (1 mark)
- Calculate the mass of silver deposited (Ag = 108; 1 F=96500 C) (2 marks)
- 2008 Q 19 P1
Select a letter which represents a mono atomic gas.
Eθ (Volts)
Zn2+ (aq) + 2e–Zn (s) -0.76
Pb2+ (aq) +2e– Pb (s) -0.13
Ag+ (aq) + 2e‑ Ag (s) +0.80
Cu2+ (aq) + 2e–Cu (s) +0.30
- Write the cell representation for the electrochemical cell that would give the highest E (1 mark)
- State and explain the observations made when a copper rod is placed in a beaker containing silver nitrate solution. (2 marks)
- 2008 Q 21
The diagram below represents an experiment that was set up to investigate movement of ions during electrolysis.
When the circuit was completed, it was noticed that a blue colour spread towards the right.
- Explain this observation (2 marks)
- Write the equation for the reaction that occurred at the anode. (1 mark)
- 2008 Q 6 P2
The diagram below represents a set up that can be used to electrolyze aqueous copper (II) sulphate.
- (i) Describe how oxygen gas is produced during the electrolysis.
(2 marks)
(ii) Explain why copper electrodes are not suitable for this electrolysis. (2 marks)
- Impure copper is purified by an electrolytic process
- Name one ore from which copper is obtained (1 mark)
- Write the equation for the reaction that occur at the cathode during the purification of copper. (1 mark)
- In an experiment to electroplate a copper spoon with silver, a current of 0.5 A was passed for 18 minutes. Calculate the amount of silver deposited on the spoon (n = 96500 coulombs, Ag = 108) (3 marks)
- Give two reasons why some metals are electroplated (2 marks)
- 2009 Q 7 P1
When aluminium oxide was electrolysed, 1800kg of aluminium metal was obtained.
(a) Write an equation for the formation of aluminium metal. (1 mark)
(b) Calculate the quantity of electricity in faradays used. (Al = 27) (2 marks)
- 2009 Q 10 P1
Hydrogen and oxygen can be obtained by electrolysis of acidified water. Using equations for the reactions at the electrodes, explain why the volume of hydrogen obtained is twice that of oxygen. (2 marks)
- 2009 Q 12 P1
The standard reduction potentials of two half cells are:
Ag+ (aq) + e‑ Ag(s) Eθ = 0.80V
2H2O (l) + 2e–H2 (g) + 2OH– (aq); Eθ = 0.83V
Draw a labelled diagram of an electrochemical cell that can be constructed using the two half cells. (3 marks)
- 2009 Q 3 P2
The set-up below (figure 2) was used to electrolyse a bromide of metal D, DBr2
- Write the equation for the reaction at the:
(i) Cathode (1 mark)
(ii) Anode (1 mark)
- The electrodes used in the experiment were made of carbon and metal D which of the two electrodes was used as the anode? Give a reason. (2 marks)
- Give a reason why this experiment is carried in a fume cupboard. (1 mark)
- When a current of 0.4A is passed for 90 min, 2.3g of metal D were deposited.
- Describe how the amount D deposited was determined. (3marks)
- Calculate the relative atomic mass of metal D. (1 Faraday=96500 coulombs) (3marks)
- 2010 Q 7 P1
Complete the table below by writing the product formed at the electrodes during the electrolysis of the electrolytes given in the table. (3 marks)
| Electrolyte | Product at anode | Product at cathode |
| Aqueous sodium sulphate using inert electrodes | (½mark) | (½mark) |
| Aqueous copper (II) sulphate using copper electrodes. | (1 mark) | (1 mark) |
- 2010 Q 19 P1
The half equations involved in a cell are:
2H2O (l) + 2e– H2 (g) + 2OH-(aq): Eθ = – 0.83V
O2 (g) + 2H2O (l) + 4e–4OH–(aq): Eθ = + 0.40V
- Write the overall equation for the electrochemical cell. (1 mark)
- Calculate the e.m.f. generated by a battery consisting of ten cells.
(1 mark)
- State one environment advantage of using these cells in spacecrafts. (1 mark)
- 2010 Q 1 P2
- Which one of the following compounds; urea, ammonia, sugar and copper (II) chloride will conduct an electric current when dissolved in water? Give reasons. (2 marks)
- The diagram below shows an electrochemical cell. Study it and answer the questions that follows.
Given the following;
Fe2+ (aq) + 2e Fe (s); Eθ = – 0.44V
Zn2+ (aq) + 2e Zn (s); Eθ = – 0.76 V
- Show on the diagram using an arrow, the direction of flow of electrons (1 mark)
- Name two substances that are used to fill the part labelled L
(2 marks)
- In an experiment to electroplate iron with silver, a current of 0.5 amperes was passed through a solution of silver nitrate for one hour.
- Give two reasons why it is necessary to electroplate iron with silver. (2 marks)
- Calculate the mass of silver that was deposited on iron (Ag = 108, 1 Faraday= 96,500 coulombs) (3 marks)
- 2011 Q 12 P1, 2016 Q15 P1
Sodium hydroxide can be prepared by the following methods; I and II.
- Sodium metal Sodium hydroxide + Hydrogen
- Concentrated Sodium hydroxide + Chlorine + Hydrogen
sodium chloride
(a) Name one precaution that needs to be taken in method I. (1 mark)
(b) Give the name of process A. (1 mark)
(c) Give one use of sodium hydroxide. (1 mark)
- 2011 Q 2 P2
The set-up below was used by a student to investigate the products formed when aqueous copper (II) chloride was electrolysed using carbon electrodes.
- (i) Write the equation for the reaction that takes place at the cathode. (1 mark)
- Name and describe a chemical test for the product initially formed at the anode when a highly concentrated solution of copper (II) chloride is electrolysed. (3 marks)
- How would the mass of the anode change if the carbon anode was replaced with copper metal? Explain. (2 marks)
- 6g of metal B were deposited when a current of 0.45A was passed through an electrolyte for 72 minutes. Determine the charge on the ion of metal B. (Relative atomic mass of B = 59, 1 Faraday = 96 500 coulombs) (3 marks)
- The electrode potentials for cadmium and zinc are given below:
Cd2+ (aq) + 2e– Cd(s); Eθ = – 0 .4V
Zn2+ (aq) + 2e‑ ⇌Zn(s); Eθ = -0 .76V
Explain why it is not advisable to store a solution of cadmium nitrate in a container made of zinc. (2 marks)
- 2012 Q15 P1
Below is a representation of an electrochemical cell.
Pb(s) / Pb2+ (aq) // Ag+(aq) / Ag(s)
- What does // represent? (1 mark)
- Given the following:
Eθ V
Pb2+ (aq) + 2e– Pb(s); – 0.13
Ag+ (aq) + e– Ag(s); – 0.80
Calculate the E.M.F of the electrochemical cell. (2 marks)
- 2012 Q28 P1
The apparatus shown in the diagram below were used to investigate the products formed when concentrated sodium chloride was electrolysed using inert electrodes.
- Write the equation for the reaction that takes place at electrode A. (1 mark)
- If the concentrated sodium chloride was replaced with dilute sodium chloride, what product would be formed at electrode A? Explain. (2 marks)
- 2012 Q5 P2
- The set up below was used to investigate the products formed at the electrodes during electrolysis of aqueous magnesium sulphate using inert electrodes. Use it to answer the questions that follow.
- During the electrolysis, hydrogen gas was formed at electrode Y. Identify the anode. Give a reason for your answer. (2 marks)
- Write the equation for the reaction which takes place at electrode X (1 mark)
- Why is the concentration of magnesium sulphate expected to increase during electrolysis? (2 marks)
- What will be observed if red and blue litmus papers were dipped into the solution after electrolysis? (2 marks)
- During electrolysis of magnesium sulphate, a current of 0.3a was passed for 30 minutes. Calculate the volume of gas produced at the anode
(Molar gas volume = 24dm3; 1 faraday = 96,500C) (3 marks)
- State two applications of electrolysis (1 mark)
- 2013 Q6 P1
- A student electroplated a spoon with copper metal. Write an equation for the process that took place at the cathode. (1 mark)
- Calculate the time in minutes required to deposit 1.184g of copper if a current of 2 amperes was used. (1 Faraday = 96500 coulombs, Cu=63.5). (2 marks)
- 2013 Q4 P2, 2016 Q4 P2.
- The set below can be used to produce sodium hydroxide by electrolyzing brine
- Identify gas Y. (1 mark)
- Describe how aqueous sodium hydroxide is formed in the above set-up.
(2 marks)
- One of the uses of sodium hydroxide is in the manufacturing of soaps. State one other use of sodium hydroxide. (1 mark)
- Study the information given in the table below and answer the question that follows
| Half reaction | Electrode potential Eθ (V) |
| D2+ (aq)+2e– D (s)
E+ (aq)+e– E (s) F (aq) + e– F‑ aq G2+(aq) +2e G(s) H2+(aq) +2e–H (s) J+(aq) + e– J(s) |
-0.13
+0.80 +0.68 -2.87 +0.34 -2.71 |
- Construct an electrochemical cell that will produce the largest e.m.f (3 marks)
- Calculate the emf of the cell constructed in(i) above (2 marks)
- Why is it not advisable to store a solution containing E+ ions in a container made of H? (2 marks)
- 2014 Q24 P1
- A student electrolyzed dilute sodium chloride solution using inert carbon electrodes. Name the products at:
- Anode:
- Cathode: (2 marks)
- If the experiment was repeated using concentrated sodium chloride instead of dilute sodium chloride solution, write the half equation at the anode. (1 mark)
- 2015 Q11 P1
Dilute sulphuric (VI) acid was electrolysed using platinum electrodes.
Name the product formed at the anode and give a reason for your answer. (2 marks)
- 2015 Q18 P1
Study the standard electrode potentials in the table below and answer the questions that follow.
Eθ (volts)
Cu2+ (aq) + 2e– Cu (s) + 0.34
Mg2+ (aq) + 2e–Mg (s); – 2.38
Ag+ (aq) + e–Ag (s); + 0-80
Ca2+ (aq) + 2e– Ca (s); -2-87
- Which of the metals is the strongest reducing agent? (1 mark)
- What observations will be made if a silver coin was dropped into an aqueous solution of copper (II) sulphate? Explain. (2 marks)
- 2015 Q4 P2
- The diagram below represents a dry cell. Use it to answer the questions that follows.
- Which of the letters represent;
- Carbon electrode? (1 mark)
- The electrolyte? (1 mark)
- One of the substances used in a dry cell is manganese (IV) oxide.
State two roles of manganese (IV) oxide in the dry cells. (2 marks)
- Below is simplified electrolytic cell used for purification of copper. Study it and answer the questions that follows.
- Identify the cathode. (1 mark)
- Write the equation for the reaction at the anode. (1 mark)
- What name is given to L? (1 mark)
- A current of 0.6 A was passed Through the electrolyte for 2 hours. Determine the amount of copper deposited.
(Cu=63.5; 1 Faraday = 96,500 coulombs) (3 marks)
- State two uses of copper metal (1 mark)
- 2017 P1 Q3.
The diagram in Figure 1 shows a section of a dry cell. Study it and answer the questions that follow.
- Name the part labelled (1 mark)
- The part labelled A is a paste. Give a reason why it is not used in dry form. (1 mark)
- What is the purpose of the zinc container? (1 mark)
- 2017 P1 Q22.
(a) What is an inert electrode? (1 mark)
(b) State the products formed when brine is electrolysed using inert electrodes.
Anode: ……………. (1 mark)
Cathode: …………… (1 mark)
- 2017 P2 Q2(b)
- Copper (II) sulphate solution was electrolysed using the set up in Figure 1.
- State the observations made during electrolysis. (1½ marks)
- Write the equation for the reaction that occurs at the anode. (1 mark)
- State the expected change in pH of the electrolyte after electrolysis. (½mark)
- The experiment was repeated using copper electrodes instead of carbon electrodes. Describe the observations made at each electrode. (1 mark)
- Electroplating is an important industrial process.
- What is meant by electroplating. (1 mark)
- State the purpose of electroplating. (1 mark)
- During electroplating of an iron spoon, a current of 0.6 amperes was passed through aqueous silver nitrate solution for 11/2 hours. Calculate the mass of silver that was deposited on the spoon. (3 marks)
(Ag = 108.0; 1 F = 96,500 C mol-1)
- 2018 P1 Q 16.
Metals X and Y have standard electrode potentials of -0.13 V and -0.76V respectively. The metals were connected to form a cell as shown in Figure 4.
- Name the part labelled (1 mark)
- State one function of the part labelled Z. (1 mark)
- Calculate the e.m.f. of the cell. (1 mark)
- 2018 P2 Q3(c)
Use the standard electrode potentials in Table 2 to answer the questions that follow.
- Write the half-cell representation for the element whose electrode potential is for hydrogen. (1 mark)
- Arrange the elements in order of reducing power, starting with the weakest reducing agent. (1 mark)
- Select two half cells which combine to give a cell with the least e.m.f. (1 mark)
- Calculate the e.m.f of the half cells identified in (iii) I. (1 mark)
- 2019 P1 Q 19.
Given that the Eθ of Cu(s)/Cu2+(aq) is + 0.34V and that 0f Zn(s)/Zn2+(aq) is – 0.76V, draw a labelled diagram of zinc and copper electrochemical cell.
(3 marks)
- 2019 P2 Q6.
- What is meant by standard electrode potential of an element? (1 mark)
- Use the standard electrode potentials given below to answer the questions that follow.
- State whether acidified MnO-4can oxidise M2+. Give a reason. (2 marks)
- Select two half-cells which when combined will give the highest e.m.f. (1 mark)
- Write the cell representation for the cell formed in b (ii). (1 mark)
- Calculate the Eθ value for the cell formed in b (iii). (1 mark)
- A mass of 1.24g of a divalent metal was deposited when a current of 6A was passed through a solution of a metal sulphate for 12 minutes. Determine the relative atomic mass of the metal.
(1 Faraday = 96,500 C mol -1) (3 marks)
- State two applications of electrolysis. (1 mark)
STANDARD 8 ENGLISH SCHEMES OF WORK TERM 1-3
English schemes of work
Standard Eight Term I
References
- New progressive primary English teacher’s guide book 8
- New progressive primary English pupil’s book 8
- Primary English pupil’s book 8
- Primary English teacher’s guide book 8
| WEEK | LESSON | TOPIC | OBJECTIVES | TEACHER;S ACTIVITIES | LEARNER’S ACTIVITIES | RESOURCES/ | REFERENCES | ASSESSMENT | REMARKS |
| 1 | REPORTING AND OREPERATION | ||||||||
| 2 | 1 | Oral work | By the end of the lesson the pupil should be able to construct correct sentences using the pattern
-clauses introduced by who -clauses introduced by whose -verb followed by to + infinitive -verbs + object + infinitive |
– Explaining
– Pronouncing words – Asking questions – Describing events |
– Explaining
– Pronouncing words – Asking questions – Describing events |
– poster
– pictures from the book – Handkerchief |
NPPE PB8 Pg3-7
NPPE 8 Pg2-8 PE PB8 Pg2-8 PE TG8 P2-6 |
ORAL EXERCISE | |
| 2 | Oral work | By the end of the lesson the pupil should be able to,pronounce and construct correct sentence using the new words | – Explaining
– Pronouncing words – Asking questions – Describing events |
– Explaining
– Pronouncing words – Asking questions – Describing events |
OPDEA | NPPE PB8 Pg3-7
NPPE 8 Pg2-8 PE PB8 Pg2-8 PE TG8 P2-6 |
Oral questions | ||
| 3-4 | Reading | By the end of the lesson the pupil should be able to, read passage ‘Social values’,fluently and accurately and answer oral questions.
|
– Explaining
– Pronouncing words – Asking questions – Describing events – Reading |
– Explaining
– Pronouncing words – Asking questions – Describing events – Reading |
– poster
– pictures from the book
|
NPPE PB8 Pg3-7
NPPE 8 Pg2-8 PE PB8 Pg2-8 PE TG8 P2-6 |
Oral questions | ||
| 5 | Writing | By the end of the lesson the pupil should be able to write correct sentences using the pattern
-the-ing form a subject -question tags -using present progressive question |
– Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg3-7
NPPE 8 Pg2-8 PE PB8 Pg2-8 PE TG8 P2-6 |
Witten exercise | ||
| 6 | Writing | By the end of the lesson the pupil should be able to write composition to a dry a hot place | – Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg3-7
NPPE 8 Pg2-8 PE PB8 Pg2-8 PE TG8 P2-6 |
Writing composition | ||
| 7 | Library | By the end of the lesson the pupil should be able to read
-the story in ‘fun spot’ for comprehension and enjoyment a story book for comprehension and for enjoyment |
– Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg7-12
NPPE 8 Pg8-13 PE PB8 Pg8-14 PE TG8 P6-11 |
Oral questions | ||
| 3 | 1 | Oral work | By the end of the lesson the pupil should be able to construct correct sentences using the pattern
–present simple tense describing general truth -present simple tense for habitual action -present continuous tense for future arrangements |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg7-12
NPPE 8 Pg8-13 PE PB8 Pg8-14 PE TG8 P6-11 |
ORAL EXERCISE | |
| 2 | Oral work | By the end of the lesson the pupil should be able to, construct correct sentences using the new words given | – Pronouncing words
– Asking questions – Describing events Explanation |
– Pronouncing words
– Asking questions – Describing events Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg7-12
NPPE 8 Pg8-13 PE PB8 Pg8-14 PE TG8 P6-11 |
Oral questions | ||
| 3-4 | Reading | By the end of the lesson the pupil should be able to read the passage ‘ A little boys dream’ correctly and fluently | – Pronouncing words
– Asking questions – Describing events – Reading |
– Pronouncing words
– Asking questions – Describing events – Reading |
– poster
– pictures from the book
|
NPPE PB8 Pg7-12
NPPE 8 Pg8-13 PE PB8 Pg8-14 PE TG8 P6-11 |
Oral questions | ||
| 5-6 | Writing | By the end of the lesson the pupil should be able to, write correct sentence using
-present continuous tense -present simple tense -determiners, this, that, these, those, some, any, no, enough |
– Pronouncing words
– Asking questions – Describing events – Writing |
– Pronouncing words
– Asking questions – Describing events – Writing |
– poster
– pictures from the book
|
NPPE PB8 Pg7-12
NPPE 8 Pg8-13 PE PB8 Pg8-14 PE TG8 P6-11 |
Witten exercise | ||
| 7 | Library | By the end of the lesson the pupil should be able to read
-the story in ‘fun spot’ for comprehension and enjoyment a story book for comprehension and for enjoyment |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg7-12
NPPE 8 Pg8-13 PE PB8 Pg8-14 PE TG8 P6-11 |
Writing composition | ||
| 4 | 1 | Oral work | By the end of the lesson the pupil should be able to construct correct sentences using
-phrasal verb with get -present simple tense -past perfect tense present continuous tense |
– Explaining
– Pronouncing words – Asking questions – Describing events |
– Explaining
– Pronouncing words – Asking questions – Describing events |
– poster
– pictures from the book – Handkerchief |
NPPE PB8 Pg12-18
NPPE 8 Pg13-19 PE PB8 Pg14-20 PE TG8 P11-17 |
Oral questions | |
| 2 | Oral work | By the end of the lesson the pupil should be able to,use pronounce and construct sentence using new words | – Explaining
– Pronouncing words – Asking questions – Describing events |
– Explaining
– Pronouncing words – Asking questions – Describing events |
OPDEA | NPPE PB8 Pg12-18
NPPE 8 Pg13-19 PE PB8 Pg14-20 PE TG8 P11-17 |
ORAL EXERCISE | ||
| 3-4 | Reading | By the end of the lesson the pupil should be able to, read passage ‘mutulu,fluently and accurately and answer oral questions.
|
– Explaining
– Pronouncing words – Asking questions – Describing events – Reading |
– Explaining
– Pronouncing words – Asking questions – Describing events – Reading |
– poster
– pictures from the book
|
NPPE PB7 Pg7-12
NPPE TG7 Pg6-11 PE PB7 Pg8-14 PE TG7 Pg5-9 |
Oral questions | ||
| 5 | Writing | By the end of the lesson the pupil should be able to write correct sentences
-using past perfect tense -past continuous tense -modal verbs |
– Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg12-18
NPPE 8 Pg13-19 PE PB8 Pg14-20 PE TG8 P11-17 |
Oral questions | ||
| 6 | Writing | By the end of the lesson the pupil should be able to write a formal letter | – Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg12-18
NPPE 8 Pg13-19 PE PB8 Pg14-20 PE TG8 P11-17 |
Witten exercise | ||
| 7 | Library | By the end of the lesson the pupil should be able to read
-the story in ‘fun spot’ for comprehension and enjoyment a story book for comprehension and for enjoyment |
– Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg12-18
NPPE 8 Pg13-19 PE PB8 Pg14-20 PE TG8 P11-17 |
Writing composition | ||
| 5 | 1 | Oral work | By the end of the lesson the pupil should be able to construct correct sentences using
-determiners with singular nouns -determiner with plural noun present simple tense |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg12-18
NPPE 8 Pg13-19 PE PB8 Pg14-20 PE TG8 P11-17 |
Oral questions | |
| 2 | Oral work | By the end of the lesson the pupil should be able to, pronounce and construct correct sentences using the new words | – Pronouncing words
– Asking questions – Describing events – Explanation |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg12-18
NPPE 8 Pg13-19 PE PB8 Pg14-20 PE TG8 P11-17 |
ORAL EXERCISE | ||
| 3-4 | Reading | By the end of the lesson the pupil should be able to read the passage ‘ life skills’ correctly and fluently | – Pronouncing words
– Asking questions – Describing events – Reading |
– Pronouncing words
– Asking questions – Describing events – Reading |
– poster
– pictures from the book
|
NPPE PB8 Pg12-18
NPPE 8 Pg13-19 PE PB8 Pg14-20 PE TG8 P11-17 |
Oral questions | ||
| 5-6 | Writing | By the end of the lesson the pupil should be able to, write correct sentence using
-adverbs of degree -the infinitive with/without to -order of adjectives |
– Pronouncing words
– Asking questions – Describing events – Writing |
– Pronouncing words
– Asking questions – Describing events – Writing |
– poster
– pictures from the book
|
NPPE PB8 Pg12-18
NPPE 8 Pg13-19 PE PB8 Pg14-20 PE TG8 P11-17 |
Oral questions | ||
| 7 | Library | By the end of the lesson the pupil should be able to read
-the story in ‘fun spot’ for comprehension and enjoyment a story book for comprehension and for enjoyment -a class debate |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg12-18
NPPE 8 Pg13-19 PE PB8 Pg14-20 PE TG8 P11-17 |
Witten exercise | ||
| 6 | 1 | Oral work | By the end of the lesson the pupil should be able to construct correct sentences using the pattern
-determiners in affirmative, interrogative -past simple tense to narrate past events -past continuous tense for actions in progressive -past continuous tense for extended action in the past |
– Explaining
– Pronouncing words – Asking questions – Describing events |
– Explaining
– Pronouncing words – Asking questions – Describing events |
– poster
– pictures from the book – Handkerchief |
NPPE PB8 Pg12-18
NPPE 8 Pg13-19 PE PB8 Pg14-20 PE TG8 P11-17 |
Writing composition | |
| 2 | Oral work | By the end of the lesson the pupil should be able to,use of
-hope by sentences with possible condition -pronounce and construct sentence using new words |
– Explaining
– Pronouncing words – Asking questions – Describing events |
– Explaining
– Pronouncing words – Asking questions – Describing events |
OPDEA | NPPE PB8 Pg12-18
NPPE 8 Pg13-19 PE PB8 Pg14-20 PE TG8 P11-17 |
Oral questions | ||
| 3-4 | Reading | By the end of the lesson the pupil should be able to, read passage ‘The child labourer’,fluently and accurately and answer oral questions. | – Explaining
– Pronouncing words – Asking questions – Describing events Reading |
– Explaining
– Pronouncing words – Asking questions – Describing events Reading |
– poster
– pictures from the book
|
NPPE PB7 Pg12-18
NPPE TG7 Pg-11-17 PE PB7 Pg8-14-19 PE TG7 Pg11-16 |
ORAL EXERCISE | ||
| 5 | Writing | By the end of the lesson the pupil should be able to write correct sentences using
-the colon -the passives -countable and uncountable nouns |
– Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg12-18
NPPE 8 Pg13-19 PE PB8 Pg14-20 PE TG8 P11-17 |
Oral questions | ||
| 6 | Writing | By the end of the lesson the pupil should be able to write a guided composition | – Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg12-18
NPPE 8 Pg13-19 PE PB8 Pg14-20 PE TG8 P11-17 |
Oral questions | ||
| 7 | Library | By the end of the lesson the pupil should be able to read
-the story in ‘fun spot’ for comprehension and enjoyment a story book for comprehension and for enjoyment |
– Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg12-18
NPPE 8 Pg13-19 PE PB8 Pg14-20 PE TG8 P11-17 |
Witten exercise | ||
| 7 | MID TERM EXAMINATIONS | ||||||||
| 8 | 1 | Oral work | By the end of the lesson the pupil should be able to construct correct sentences using the pattern
-past continuous tense for action in progression at specific time -verbs followed by present participle -use brackets at punctuation marks
|
– Pronouncing words
– Asking questions – Describing events – Explanation |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg35-44
NPPE 8 Pg33-38 PE PB8 Pg31-36 PE TG8 P26-34 |
Oral questions | |
| 2 | Oral work | By the end of the lesson the pupil should be able to, pronounce construct correct sentences using the new words given | – Pronouncing words
– Asking questions – Describing events Explanation |
– Pronouncing words
– Asking questions – Describing events Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg35-44
NPPE 8 Pg33-38 PE PB8 Pg31-36 PE TG8 P26-34 |
ORAL EXERCISE | ||
| 3-4 | Reading | By the end of the lesson the pupil should be able to read the passage ‘ The East African parliament’ correctly and fluently | – Pronouncing words
– Asking questions – Describing events – Reading |
– Pronouncing words
– Asking questions – Describing events – Reading |
– poster
– pictures from the book
|
NPPE PB8 Pg35-44
NPPE 8 Pg33-38 PE PB8 Pg31-36 PE TG8 P26-34 |
Oral questions | ||
| 5 | Writing | By the end of the lesson the pupil should be able to, write correct sentence using
i. speech mark ii. the hyphen iii. the bracket |
– Pronouncing words
– Asking questions – Describing events – Writing |
– Pronouncing words
– Asking questions – Describing events – Writing |
– poster
– pictures from the book
|
NPPE PB8 Pg35-44
NPPE 8 Pg33-38 PE PB8 Pg31-36 PE TG8 P26-34 |
Written questions | ||
| 6 | Writing | By the end of the lesson the pupil should be able to write an interesting story given an ending | – Pronouncing words
– Asking questions – Describing events – Writing |
– Pronouncing words
– Asking questions – Describing events – Writing |
– poster
– pictures from the book
|
NPPE PB8 Pg35-44
NPPE 8 Pg33-38 PE PB8 Pg31-36 PE TG8 P26-34 |
Witten exercise | ||
| 7 | Library | By the end of the lesson the pupil should be able to read
-the story in ‘fun spot’ for comprehension and enjoyment a story book for comprehension and for enjoyment |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg35-44
NPPE 8 Pg33-38 PE PB8 Pg31-36 PE TG8 P26-34 |
Writing composition | ||
| 9 | 1 | Oral work | By the end of the lesson the pupil should be able to construct correct sentences using the pattern
-determiner with uncountable nouns -verb + noun/pronoun + infinitives |
– Explaining
– Pronouncing words – Asking questions – Describing events |
– Explaining
– Pronouncing words – Asking questions – Describing events |
– poster
– pictures from the book – Handkerchief |
NPPE PB8 Pg35-44
NPPE 8 Pg33-38 PE PB8 Pg31-36 PE TG8 P26-34 |
Oral questions | |
| 2 | Oral work | By the end of the lesson the pupil should be able to,use of
-use of whether or not -use auxiliary verbs pronounce and construct sentence using new words |
– Explaining
– Pronouncing words – Asking questions – Describing events |
– Explaining
– Pronouncing words – Asking questions – Describing events |
OPDEA | NPPE PB8 Pg35-44
NPPE 8 Pg33-38 PE PB8 Pg31-36 PE TG8 P26-34 |
Written questions | ||
| 3-4 | Reading | By the end of the lesson the pupil should be able to, read passage, ‘Aunt Tara’s shop’ fluently and accurately and answer oral questions. | – Explaining
– Pronouncing words – Asking questions – Describing events Reading |
– Explaining
– Pronouncing words – Asking questions – Describing events Reading |
– poster
– pictures from the book
|
NPPE PB8 Pg35-44
NPPE 8 Pg33-38 PE PB8 Pg31-36 PE TG8 P26-34 |
Witten exercise | ||
| 5-6 | Writing | By the end of the lesson the pupil should be able to write correct sentences using
● past perfect continuous tense ● present simple tense ● write an informal letter to a friend |
– Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg35-44
NPPE 8 Pg33-38 PE PB8 Pg31-36 PE TG8 P26-34 |
Writing composition | ||
| 7 | Library | By the end of the lesson the pupil should be able to read
-the story in ‘fun spot’ for comprehension and enjoyment a story book for comprehension and for enjoyment |
– Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg35-44
NPPE 8 Pg33-38 PE PB8 Pg31-36 PE TG8 P26-34 |
Oral questions | ||
| 10 | 1 | Oral work | By the end of the lesson the pupil should be able to construct correct sentences using the pattern
-present perfect continuous tense -clause introduced by who -present simple tense |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg43-50
NPPE 8 Pg42-48 PE PB8 Pg41-47 PE TG8 Pg38-44 |
Written questions | |
| 2 | Oral work | By the end of the lesson the pupil should be able to, construct correct sentences using the new words given | – Pronouncing words
– Asking questions – Describing events Explanation |
– Pronouncing words
– Asking questions – Describing events Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg43-50
NPPE 8 Pg42-48 PE PB8 Pg41-47 PE TG8 Pg38-44 |
Witten exercise | ||
| 3-4 | Reading | By the end of the lesson the pupil should be able to read the passage ‘ The trade exhibition’ correctly and fluently | – Pronouncing words
– Asking questions – Describing events – Reading |
– Pronouncing words
– Asking questions – Describing events – Reading |
– poster
– pictures from the book
|
NPPE PB8 Pg43-50
NPPE 8 Pg42-48 PE PB8 Pg41-47 PE TG8 Pg38-44 |
Writing composition | ||
| 5 | Writing | By the end of the lesson the pupil should be able to, write correct sentence using
–the dash -regular and irregular verbs -the hyphen |
– Pronouncing words
– Asking questions – Describing events – Writing |
– Pronouncing words
– Asking questions – Describing events – Writing |
– poster
– pictures from the book
|
NPPE PB8 Pg43-50
NPPE 8 Pg42-48 PE PB8 Pg41-47 PE TG8 Pg38-44 |
Oral questions | ||
| 6 | Writing | By the end of the lesson the pupil should be able to, write a composition about an exhibition | – Pronouncing words
– Asking questions – Describing events – Writing |
– Pronouncing words
– Asking questions – Describing events – Writing |
– poster
– pictures from the book
|
NPPE PB8 Pg43-50
NPPE 8 Pg42-48 PE PB8 Pg41-47 PE TG8 Pg38-44 |
Written questions | ||
| 7 | Library | By the end of the lesson the pupil should be able to read
-the story in ‘fun spot’ for comprehension and enjoyment a story book for comprehension and for enjoyment |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg43-50
NPPE 8 Pg42-48 PE PB8 Pg41-47 PE TG8 Pg38-44 |
Witten exercise | ||
| 11 | REVISION | ||||||||
| 12 | EXAMINATIONS | ||||||||
| 13 | MARKING | ||||||||
| 14 | CLOSING THESCHOOL | ||||||||
English schemes of work
Standard Eight Term II
Year 2016
References
- New progressive primary English teacher’s guide book 8
- New progressive primary English pupil’s book 8
- Primary English pupil’s book 8
- Primary English teacher’s guide book 8
| WEEK | LESSON | TOPIC | SUB-TOPIC | OBJECTIVES | ACTIVITIES | ACTIVITIES | RESOURCES/ | REFERENCES | ASSESSMENT | REMARKS | ||
| 1 | REPORTING AND PREPRATIONS | |||||||||||
| 2 | 1 | social values | Oral work | By the end of the lesson the pupil should be able to construct correct sentences using
-appropriate determiners with uncountable nouns -past simple passive -use of ‘as well as’ as a connector meaning ‘also’ -verb + ing as a nominal |
– Explaining
– Pronouncing words – Asking questions – Describing events |
– Explaining
– Pronouncing words – Asking questions – Describing events |
– poster
– pictures from the book – Handkerchief |
NPPE PB8 Pg58-63
NPPE 8 Pg56-62 PE PB8 Pg53-61 PE TG8 Pg53-58 |
Oral questions | |||
| 2 | Oral work | By the end of the lesson the pupil should be able to, pronounce and construct correct sentences using new words, | – Explaining
– Pronouncing words – Asking questions – Describing events |
– Explaining
– Pronouncing words – Asking questions – Describing events |
OPDEA | NPPE PB8 Pg58-63
NPPE 8 Pg56-62 PE PB8 Pg53-61 PE TG8 Pg53-58 |
Oral questions | |||||
| 3-4 | Reading | By the end of the lesson the pupil should be able to, read passage, ‘the Olympic games’,fluently and accurately and answer oral questions.
|
– Explaining
– Pronouncing words – Asking questions – Describing events – Reading |
– Explaining
– Pronouncing words – Asking questions – Describing events – Reading |
– poster
– pictures from the book
|
NPPE PB8 Pg58-63
NPPE 8 Pg56-62 PE PB8 Pg53-61 PE TG8 Pg53-58 |
Oral questions | |||||
| 5 | Writing | By the end of the lesson the pupil should be able to write correct sentences using
–determiners – write a conversation accurately |
– Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg58-63
NPPE 8 Pg56-62 PE PB8 Pg53-61 PE TG8 Pg53-58 |
Written exercise | |||||
| 7 | Library | By the end of the lesson the pupil should be able to read
-the story in ‘fun spot’ for comprehension and enjoyment a story book for comprehension and for enjoyment |
– Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg58-63
NPPE 8 Pg56-62 PE PB8 Pg53-61 PE TG8 Pg53-58 |
Written exercise | |||||
| 3 | 1-2 | Career | Oral work | By the end of the lesson the pupil should be able to construct correct sentences using
–when/after/as soon as -determiners a, an, the -comparatives and superlatives construct and pronounce sentences using new words |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg58-63
NPPE 8 Pg56-62 PE PB8 Pg53-61 PE TG8 Pg53-58 |
Oral questions | |||
| 3-4 | Reading | By the end of the lesson the pupil should be able to read the passage ‘ Haile Gerbrselassie’ correctly and fluently and answer oral question | – Pronouncing words
– Asking questions – Describing events – Reading |
– Pronouncing words
– Asking questions – Describing events – Reading |
– poster
– pictures from the book
|
NPPE PB8 Pg58-63
NPPE 8 Pg56-62 PE PB8 Pg53-61 PE TG8 Pg53-58 |
Oral questions | |||||
| 5 | Writing | By the end of the lesson the pupil should be able to, write correct sentence using
i. correct order of adjectives |
– Pronouncing words
– Asking questions – Describing events – Writing |
– Pronouncing words
– Asking questions – Describing events – Writing |
– poster
– pictures from the book
|
NPPE PB8 Pg58-63
NPPE 8 Pg56-62 PE PB8 Pg53-61 PE TG8 Pg53-58 |
Oral questions | |||||
| 7 | Library | By the end of the lesson the pupil should be able to read
-the story in ‘fun spot’ for comprehension and enjoyment a story book for comprehension and for enjoyment |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg58-63
NPPE 8 Pg56-62 PE PB8 Pg53-61 PE TG8 Pg53-58 |
Written exercise | |||||
| 4 | 1 | Oral work | By the end of the lesson the pupil should be able to construct correct sentences using
– present simple tense for habitual action -present continuous tense for extended action -verb + to+ infinitives -the dash |
– Explaining
– Pronouncing words – Asking questions – Describing events |
– Explaining
– Pronouncing words – Asking questions – Describing events |
– poster
– pictures from the book – Handkerchief |
NPPE PB8 Pg58-63
NPPE 8 Pg56-62 PE PB8 Pg53-61 PE TG8 Pg53-58 |
Written exercise | ||||
| 2 | Oral work | By the end of the lesson the pupil should be able to,use pronounce and construct sentence using new words | – Explaining
– Pronouncing words – Asking questions – Describing events |
– Explaining
– Pronouncing words – Asking questions – Describing events |
OPDEA | NPPE PB8 Pg58-63
NPPE 8 Pg56-62 PE PB8 Pg53-61 PE TG8 Pg53-58 |
Oral questions | |||||
| 3-4 | Reading | By the end of the lesson the pupil should be able to, read passage ‘The young Farmers Club’,fluently and accurately and answer oral questions.
|
– Explaining
– Pronouncing words – Asking questions – Describing events – Reading |
– Explaining
– Pronouncing words – Asking questions – Describing events – Reading |
– poster
– pictures from the book
|
NPPE PB8 Pg63-70
NPPE 8 Pg62-68 PE PB8 Pg61-67 PE TG8 Pg58-64 |
Oral questions | |||||
| 5-6 | Writing | By the end of the lesson the pupil should be able to write correct sentences using
-countable and uncountable nouns -the dash -adverb with a verb |
– Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg63-70
NPPE 8 Pg62-68 PE PB8 Pg61-67 PE TG8 Pg58-64 |
Oral questions | |||||
| 7 | Library | By the end of the lesson the pupil should be able to read
-the story in ‘fun spot’ for comprehension and enjoyment a story book for comprehension and for enjoyment |
– Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg63-70
NPPE 8 Pg62-68 PE PB8 Pg61-67 PE TG8 Pg58-64 |
Written exercise | |||||
| 5 | 1 | Oral work | By the end of the lesson the pupil should be able to construct correct sentences using
-not only …but also -determiners/cross-examine/hear + nominal -phrasal verbs |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg63-70
NPPE 8 Pg62-68 PE PB8 Pg61-67 PE TG8 Pg58-64 |
Written exercise | ||||
| 2 | Oral work | By the end of the lesson the pupil should be able to, construct correct sentences using the new words given | – Pronouncing words
– Asking questions – Describing events – Explanation |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg63-70
NPPE 8 Pg62-68 PE PB8 Pg61-67 PE TG8 Pg58-64 |
Oral questions | |||||
| 3-4 | Reading | By the end of the lesson the pupil should be able to read the passage ‘ A court of law’ correctly and fluently | – Pronouncing words
– Asking questions – Describing events – Reading |
– Pronouncing words
– Asking questions – Describing events – Reading |
– poster
– pictures from the book
|
NPPE PB8 Pg63-70
NPPE 8 Pg62-68 PE PB8 Pg61-67 PE TG8 Pg58-64 |
Oral questions | |||||
| 5 | Writing | By the end of the lesson the pupil should be able to, write correct sentence
-pronoun which /that |
– Pronouncing words
– Asking questions – Describing events – Writing |
– Pronouncing words
– Asking questions – Describing events – Writing |
– poster
– pictures from the book
|
NPPE PB8 Pg63-70
NPPE 8 Pg62-68 PE PB8 Pg61-67 PE TG8 Pg58-64 |
Oral questions | |||||
| 6 | Writing | By the end of the lesson the pupil should be able to, write composition a bout an accident | – Pronouncing words
– Asking questions – Describing events – Writing |
– Pronouncing words
– Asking questions – Describing events – Writing |
– poster
– pictures from the book
|
NPPE PB8 Pg63-70
NPPE 8 Pg62-68 PE PB8 Pg61-67 PE TG8 Pg58-64 |
Written exercise | |||||
| 7 | Library | By the end of the lesson the pupil should be able to read the story of ‘ fun spot for comprehension and enjoyment | – Pronouncing words
– Asking questions – Describing events – Explanation |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg63-70
NPPE 8 Pg62-68 PE PB8 Pg61-67 PE TG8 Pg58-64 |
Written exercise | |||||
| 6 | 1 | Oral work | By the end of the lesson the pupil should be able to construct correct sentences
-this is…/it is called…/it is used -relative pronoun that/this |
– Explaining
– Pronouncing words – Asking questions – Describing events |
– Explaining
– Pronouncing words – Asking questions – Describing events |
– poster
– pictures from the book – Handkerchief |
NPPE PB8 Pg63-70
NPPE 8 Pg62-68 PE PB8 Pg61-67 PE TG8 Pg58-64 |
Oral questions | ||||
| 2 | Oral work | By the end of the lesson the pupil should be able to,pronounce and construct sentences using new words | – Explaining
– Pronouncing words – Asking questions – Describing events |
– Explaining
– Pronouncing words – Asking questions – Describing events |
OPDEA | NPPE PB8 Pg63-70
NPPE 8 Pg62-68 PE PB8 Pg61-67 PE TG8 Pg58-64 |
Oral questions | |||||
| 3-4 | Reading | By the end of the lesson the pupil should be able to, read passage ‘A court drama’,fluently and accurately and answer oral questions. | – Explaining
– Pronouncing words – Asking questions – Describing events Reading |
– Explaining
– Pronouncing words – Asking questions – Describing events Reading |
– poster
– pictures from the book
|
NPPE PB8 Pg92-98
NPPE 8 Pg90-99 PE PB8 Pg93-101 PE TG8 Pg93-98 |
Oral questions | |||||
| 5-6 | Writing | By the end of the lesson the pupil should be able to write correct sentences
-using relative clauses with who/whom/whose/where/when |
– Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg92-98
NPPE 8 Pg90-99 PE PB8 Pg93-101 PE TG8 Pg93-98 |
Written exercise | |||||
| 7 | Library | By the end of the lesson the pupil should be able to read
-the story in ‘fun spot’ for comprehension and enjoyment a story book for comprehension and for enjoyment |
– Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg92-98
NPPE 8 Pg90-99 PE PB8 Pg93-101 PE TG8 Pg93-98 |
Written exercise | |||||
| 7 | MID TERM EXAMINATIONS | |||||||||||
| 8 | 1 | Oral work | By the end of the lesson the pupil should be able to construct correct sentences using
-present simple tense for habitual action -present continuous + or else -adverbs of degree |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg92-98
NPPE 8 Pg90-99 PE PB8 Pg93-101 PE TG8 Pg93-98 |
Oral questions | ||||
| 2 | Oral work | By the end of the lesson the pupil should be able to, construct correct sentences using the new words given | – Pronouncing words
– Asking questions – Describing events Explanation |
– Pronouncing words
– Asking questions – Describing events Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg92-98
NPPE 8 Pg90-99 PE PB8 Pg93-101 PE TG8 Pg93-98 |
Oral questions | |||||
| 3-4 | Reading | By the end of the lesson the pupil should be able to read the passage ‘ Don’t throw us in the sea’ correctly and fluently | – Pronouncing words
– Asking questions – Describing events – Reading |
– Pronouncing words
– Asking questions – Describing events – Reading |
– poster
– pictures from the book
|
NPPE PB8 Pg92-98
NPPE 8 Pg90-99 PE PB8 Pg93-101 PE TG8 Pg93-98 |
Written exercise | |||||
| 5 | Writing | By the end of the lesson the pupil should be able to, write correct sentence using
i. the semi colon ii. conjuctions |
– Pronouncing words
– Asking questions – Describing events – Writing |
– Pronouncing words
– Asking questions – Describing events – Writing |
– poster
– pictures from the book
|
NPPE PB8 Pg92-98
NPPE 8 Pg90-99 PE PB8 Pg93-101 PE TG8 Pg93-98 |
Written exercise | |||||
| 6 | Writing | By the end of the lesson the pupil should be able to write a story a journey by sea | – Pronouncing words
– Asking questions – Describing events – Writing |
– Pronouncing words
– Asking questions – Describing events – Writing |
– poster
– pictures from the book
|
NPPE PB8 Pg92-98
NPPE 8 Pg90-99 PE PB8 Pg93-101 PE TG8 Pg93-98 |
Oral questions | |||||
| 7 | Library | By the end of the lesson the pupil should be able to read
-the story in ‘fun spot’ for comprehension and enjoyment a story book for comprehension and for enjoyment |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg92-98
NPPE 8 Pg90-99 PE PB8 Pg93-101 PE TG8 Pg93-98 |
Oral questions | |||||
| 9 | 1 | Oral work | By the end of the lesson the pupil should be able to construct correct sentences using the pattern
i. the + proper noun ii. description of general truths iii. imaginary events iv. relative clauses introduced by who/whose
|
– Explaining
– Pronouncing words – Asking questions – Describing events |
– Explaining
– Pronouncing words – Asking questions – Describing events |
– poster
– pictures from the book – Handkerchief |
NPPE PB8 Pg92-98
NPPE 8 Pg90-99 PE PB8 Pg93-101 PE TG8 Pg93-98 |
||
| 3-4 | Reading | By the end of the lesson the pupil should be able to, read passage, ‘you are not about to die’ fluently and accurately and answer oral questions. | – Explaining
– Pronouncing words – Asking questions – Describing events Reading |
– Explaining
– Pronouncing words – Asking questions – Describing events Reading |
– poster
– pictures from the book
|
NPPE PB8 Pg102-108
NPPE 8 Pg100-109 PE PB8 Pg103-111 PE TG8 Pg103-108 |
|||
| 5-6 | Writing | By the end of the lesson the pupil should be able to write correct sentences using
● new word with suffixes including,- ful, -less, -ly, ment, -ish |
– Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg102-108
NPPE 8 Pg100-109 PE PB8 Pg103-111 PE TG8 Pg103-108 |
|||
| 7 | Library | By the end of the lesson the pupil should be able to read
-the story in ‘fun spot’ for comprehension and enjoyment a story book for comprehension and for enjoyment |
– Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg102-108
NPPE 8 Pg100-109 PE PB8 Pg103-111 PE TG8 Pg103-108 |
|||
| 10 | 1 | Oral work | By the end of the lesson the pupil should be able to construct correct sentences using the pattern
-will + 2nd and 3rd person subject shall + 1st person subject present perfect continuous tense |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg102-108
NPPE 8 Pg100-109 PE PB8 Pg103-111 PE TG8 Pg103-108 |
||
| 2 | Oral work | By the end of the lesson the pupil should be able to, construct correct sentences using the new words given | – Pronouncing words
– Asking questions – Describing events Explanation |
– Pronouncing words
– Asking questions – Describing events Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg102-108
NPPE 8 Pg100-109 PE PB8 Pg103-111 PE TG8 Pg103-108 |
|||
| 3-4 | Reading | By the end of the lesson the pupil should be able to read the passage ‘ Hi-tech phone’ correctly and fluently | – Pronouncing words
– Asking questions – Describing events – Reading |
– Pronouncing words
– Asking questions – Describing events – Reading |
– poster
– pictures from the book
|
NPPE PB8 Pg102-108
NPPE 8 Pg100-109 PE PB8 Pg103-111 PE TG8 Pg103-108 |
|||
| 5 | Writing | By the end of the lesson the pupil should be able to, write correct sentence using
-past continuous tense correct order of adjectives |
– Pronouncing words
– Asking questions – Describing events – Writing |
– Pronouncing words
– Asking questions – Describing events – Writing |
– poster
– pictures from the book
|
NPPE PB8 Pg102-108
NPPE 8 Pg100-109 PE PB8 Pg103-111 PE TG8 Pg103-108 |
|||
| 6 | Writing | By the end of the lesson the pupil should be able to, write a descriptive composition | – Pronouncing words
– Asking questions – Describing events – Writing |
– Pronouncing words
– Asking questions – Describing events – Writing |
– poster
– pictures from the book
|
NPPE PB8 Pg102-108
NPPE 8 Pg100-109 PE PB8 Pg103-111 PE TG8 Pg103-108 |
|||
| 7 | Library | By the end of the lesson the pupil should be able to read
-the story in ‘fun spot’ for comprehension and enjoyment a story book for comprehension and for enjoyment |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg102-108
NPPE 8 Pg100-109 PE PB8 Pg103-111 PE TG8 Pg103-108 |
|||
| 11 | REVISION | ||||||||
| 12 | EXAMINATIONS | ||||||||
| 13 | MARKING AND CLOSING THE SCHOOL | ||||||||
English schemes of work
Standard Eight Term III
Year 2016
References
- New progressive primary English teacher’s guide book 8
- New progressive primary English pupil’s book 8
- Primary English pupil’s book 8
| WEEK | LESSON | TOPIC | OBJECTIVES | TEACHER’S ACTIVITIES | LEARNER’S ACTIVITIES | RESOURCES/ | REFERENCES | ASSSESSMENT | REMARKS |
| 1 | REPORTING AND PREPARATIONS | ||||||||
| 2 | 1 | Oral work | By the end of the lesson the pupil should be able to construct correct sentences using the pattern
-time clauses with present simple + will/shall -use of will and shall to make prediction -use of verb followed by present participle -positive and negative statement |
– Explaining
– Pronouncing words – Asking questions – Describing events |
– Explaining
– Pronouncing words – Asking questions – Describing events |
– poster
– pictures from the book – Handkerchief |
NPPE PB8 Pg113-120
NPPE TG8 Pg111-114 PE PB8 Pg113-119 PE TG8 Pg108-114 |
Oral questions | |
| 2 | Oral work | By the end of the lesson the pupil should be able to,use construct sentence using the new word | – Explaining
– Pronouncing words – Asking questions – Describing events |
– Explaining
– Pronouncing words – Asking questions – Describing events |
OPDEA | NPPE PB8 Pg113-120
NPPE TG8 Pg111-114 PE PB8 Pg113-119 PE TG8 Pg108-114 |
Oral questions | ||
| 3-4 | Reading | By the end of the lesson the pupil should be able to, read passage, “space exploration,fluently and accurately and answer oral questions. | – Explaining
– Pronouncing words – Asking questions – Describing events Reading |
– Explaining
– Pronouncing words – Asking questions – Describing events Reading |
– poster
– pictures from the book
|
NPPE PB8 Pg113-120
NPPE TG8 Pg111-114 PE PB8 Pg113-119 PE TG8 Pg108-114 |
Oral exercise | ||
| 5 | Writing | By the end of the lesson the pupil should be able to write correct sentences using
–present perfect continuous tense |
– Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg113-120
NPPE TG8 Pg111-114 PE PB8 Pg113-119 PE TG8 Pg108-114 |
Written exercise | ||
| 6 | Writing | By the end of the lesson the pupil should be able to write a composition | – Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg113-120
NPPE TG8 Pg111-114 PE PB8 Pg113-119 PE TG8 Pg108-114 |
Writing composition | ||
| 7 | Library | By the end of the lesson the pupil should be able to read
-the story in ‘fun spot’ for comprehension and enjoyment a story book for comprehension and for enjoyment |
– Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg113-120
NPPE TG8 Pg111-114 PE PB8 Pg113-119 PE TG8 Pg108-114 |
Written exercise | ||
| 3 | 1 | Oral work | By the end of the lesson the pupil should be able to construct correct sentences using —structure that can follow find
■ present perfect continuous tense for the event that began in the pas but still continuous ■ contrast of present perfect and past simple with for ■ contrast of for and since with present perfect continuous ■ use of past participle and past perfect |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg122-128
NPPE 8 Pg120-129 PE PB8 Pg123-131 PE TG8 Pg113-118 |
Oral questions | |
| 2 | Oral work | By the end of the lesson the pupil should be able to, construct correct sentences using the new words given | – Pronouncing words
– Asking questions – Describing events Explanation |
– Pronouncing words
– Asking questions – Describing events Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg122-128
NPPE 8 Pg120-129 PE PB8 Pg123-131 PE TG8 Pg113-118 |
Oral questions | ||
| 4 | Reading | By the end of the lesson the pupil should be able to read the passage ‘ if you elect me…’ correctly and fluently | – Pronouncing words
– Asking questions – Describing events – Reading |
– Pronouncing words
– Asking questions – Describing events – Reading |
– poster
– pictures from the book
|
NPPE PB8 Pg122-128
NPPE 8 Pg120-129 PE PB8 Pg123-131 PE TG8 Pg113-118 |
Oral exercise | ||
| 5 | Writing | By the end of the lesson the pupil should be able to, write correct sentence using
suffixes –ness, -able, -ment, -ly, -ish, -y |
– Pronouncing words
– Asking questions – Describing events – Writing |
– Pronouncing words
– Asking questions – Describing events – Writing |
– poster
– pictures from the book
|
NPPE PB8 Pg122-128
NPPE 8 Pg120-129 PE PB8 Pg123-131 PE TG8 Pg113-118 |
Written exercise | ||
| 7 | Library | By the end of the lesson the pupil should be able to read
-the story in ‘fun spot’ for comprehension and enjoyment a story book for comprehension and for enjoyment |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg122-128
NPPE 8 Pg120-129 PE PB8 Pg123-131 PE TG8 Pg113-118 |
Writing composition | ||
| 4 | 1 | Oral work | By the end of the lesson the pupil should be able to construct correct sentences
-conditional with would and could + negative -understands/see + question -use of should for obligation -use of should for probability |
– Explaining
– Pronouncing words – Asking questions – Describing events |
– Explaining
– Pronouncing words – Asking questions – Describing events |
– poster
– pictures from the book – Handkerchief |
NPPE PB8 Pg122-128
NPPE 8 Pg120-129 PE PB8 Pg123-131 PE TG8 Pg113-118 |
Written exercise | |
| 2 | Oral work | By the end of the lesson the pupil should be able to,use pronounce and construct sentence using new words | – Explaining
– Pronouncing words – Asking questions – Describing events |
– Explaining
– Pronouncing words – Asking questions – Describing events |
OPDEA | NPPE PB8 Pg122-128
NPPE 8 Pg120-129 PE PB8 Pg123-131 PE TG8 Pg113-118 |
Oral questions | ||
| 3-4 | Reading | By the end of the lesson the pupil should be able to, read passage ‘debate on technology’,fluently and accurately and answer oral questions.
|
– Explaining
– Pronouncing words – Asking questions – Describing events – Reading |
– Explaining
– Pronouncing words – Asking questions – Describing events – Reading |
– poster
– pictures from the book
|
NPPE PB8 Pg122-128
NPPE 8 Pg120-129 PE PB8 Pg123-131 PE TG8 Pg113-118 |
Oral questions | ||
| 5 | Writing | By the end of the lesson the pupil should be able to write correct sentences using
– preposition – correct order of adjective – correct tense |
– Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg122-128
NPPE 8 Pg120-129 PE PB8 Pg123-131 PE TG8 Pg113-118 |
Oral exercise | ||
| 6 | Writing | By the end of the lesson the pupil should be able debate on technology has done more harm than good | – Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg122-128
NPPE 8 Pg120-129 PE PB8 Pg123-131 PE TG8 Pg113-118 |
Written exercise | ||
| 7 | Library | By the end of the lesson the pupil should be able to read
-the story in ‘fun spot’ for comprehension and enjoyment a story book for comprehension and for enjoyment |
– Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg122-128
NPPE 8 Pg120-129 PE PB8 Pg123-131 PE TG8 Pg113-118 |
Writing composition | ||
| 5 | 1 | Oral work | By the end of the lesson the pupil should be able to construct correct sentences using
conjuctions – |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg122-128
NPPE 8 Pg120-129 PE PB8 Pg123-131 PE TG8 Pg113-118 |
Written exercise | |
| 2 | Oral work | By the end of the lesson the pupil should be able to, construct correct sentences using the new words given | – Pronouncing words
– Asking questions – Describing events – Explanation |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg122-128
NPPE 8 Pg120-129 PE PB8 Pg123-131 PE TG8 Pg113-118 |
Oral questions | ||
| 3 | Reading | By the end of the lesson the pupil should be able to read the passage ‘ Leopard is dead’ correctly and fluently | – Pronouncing words
– Asking questions – Describing events – Reading |
– Pronouncing words
– Asking questions – Describing events – Reading |
– poster
– pictures from the book
|
NPPE PB8 Pg122-128
NPPE 8 Pg120-129 PE PB8 Pg123-131 PE TG8 Pg113-118 |
Oral questions | ||
| 5 | Writing | By the end of the lesson the pupil should be able to, write correct sentence using
– irregular verbs – |
– Pronouncing words
– Asking questions – Describing events – Writing |
– Pronouncing words
– Asking questions – Describing events – Writing |
– poster
– pictures from the book
|
NPPE PB8 Pg122-128
NPPE 8 Pg120-129 PE PB8 Pg123-131 PE TG8 Pg113-118 |
Oral exercise | ||
| 6 | Writing | By the end of the lesson the pupil should be able to, write a composition of caring for our wildlife | – Pronouncing words
– Asking questions – Describing events – Writing |
– Pronouncing words
– Asking questions – Describing events – Writing |
– poster
– pictures from the book
|
NPPE PB8 Pg122-128
NPPE 8 Pg120-129 PE PB8 Pg123-131 PE TG8 Pg113-118 |
Written exercise | ||
| 7 | Library | By the end of the lesson the pupil should be able to read
-the story in ‘fun spot’ for comprehension and enjoyment a story book for comprehension and for enjoyment |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– Pronouncing words
– Asking questions – Describing events – Explanation |
– poster
– pictures from the book
|
NPPE PB8 Pg134-140
NPPE TG8 Pg135-141 PE PB8 Pg132-139 PE TG8 Pg118-124 |
Writing composition | ||
| 6 | 1 | Oral work | By the end of the lesson the pupil should be able to construct correct sentences using the pattern
-using of linking words e.g. so…that, too…to. prefer…to, very…but, no sooner…than, not only…but also, in spite, in fact -indirect speech |
– Explaining
– Pronouncing words – Asking questions – Describing events |
– Explaining
– Pronouncing words – Asking questions – Describing events |
– poster
– pictures from the book – Handkerchief |
NPPE PB8 Pg134-140
NPPE TG8 Pg135-141 PE PB8 Pg132-139 PE TG8 Pg118-124 |
Written exercise | |
| 2 | Oral work | By the end of the lesson the pupil should be able to,use of construct correct sentences using the new word
|
– Explaining
– Pronouncing words – Asking questions – Describing events |
– Explaining
– Pronouncing words – Asking questions – Describing events |
OPDEA | NPPE PB8 Pg134-140
NPPE TG8 Pg135-141 PE PB8 Pg132-139 PE TG8 Pg118-124 |
Oral questions | ||
| 3-4 | Reading | By the end of the lesson the pupil should be able to, read passage ‘The carjackers’,fluently and accurately and answer oral questions. | – Explaining
– Pronouncing words – Asking questions – Describing events Reading |
– Explaining
– Pronouncing words – Asking questions – Describing events Reading |
– poster
– pictures from the book
|
NPPE PB8 Pg134-140
NPPE TG8 Pg135-141 PE PB8 Pg132-139 PE TG8 Pg118-124 |
Oral questions | ||
| 5 | Writing | By the end of the lesson the pupil should be able to write correct sentences using
-use of dictionary |
– Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg134-140
NPPE TG8 Pg135-141 PE PB8 Pg132-139 PE TG8 Pg118-124 |
Oral exercise | ||
| 6 | Writing | By the end of the lesson the pupil should be able to write composition on carjacking | – Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg134-140
NPPE TG8 Pg135-141 PE PB8 Pg132-139 PE TG8 Pg118-124 |
Written exercise | ||
| 7 | Library | By the end of the lesson the pupil should be able to read
-the story in ‘fun spot’ for comprehension and enjoyment a story book for comprehension and for enjoyment |
– Pronouncing words
– Asking questions – Describing events – Writing
|
– Pronouncing words
– Asking questions – Describing events – Writing
|
– poster
– pictures from the book
|
NPPE PB8 Pg134-140
NPPE TG8 Pg135-141 PE PB8 Pg132-139 PE TG8 Pg118-124 |
Writing composition | ||
| 7 | MID TERM EXAMINATIONS | ||||||||
| 8 | FIRST KCPE TRIAL EXAMINATION | ||||||||
| 9 | SECOND KCPE TRIAL EXAMINATION | ||||||||
| 10 | THIRD KCPE TRIAL EXAMINATION/ REHEARSALS | ||||||||
| 11 | KCPE EXAMINATIONS | ||||||||
| 6 | – | – | |||||||
FORM 4 CHEMISTRY NOTES HANDBOOK
Grade 7 Term 3 Schemes of Work
Free PP2 (Pre-Primary) CBC Notes, Exams & Schemes of Work
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ENGLISH PP3 FORM 4 JOINT EXAM WITH ANSWERS IN PDF
101/3
FORM 4 ENGLISH
Paper 3
(Creative compositions and essays based on set texts)
Time: 2 ½ Hours
MWAKICAN JOINT EXAMINATIONS
ENGLISH
Paper 3
2 ½ Hours
Instructions to candidates
- Answer three questions only on the answer sheets provided
- question one and two are compulsory
- In question three, choose only one of the optional texts which you have prepared on.
- Each of your essays must not exceed 450 words
- Your answers must be written in English
- Where the candidate presents work on more than one optional texts, only the first one to.
- IMAGINATIVE COMPOSITION (COMPULSORY) (20 MARKS)
- Write a composition ending with the words:
From that day, I learnt never to trust anybody.
OR
- Write a story to illustrate the saying
Pride comes before a fall.
- COMPULSORY SET TEXT (Blossoms of the Savannah) (20 MARKS).
Our greatest enemies are those close to us; support this statement with illustrations from the text.
- OPTIONAL SET TEXTS (20 MARKS)
EITHER
- The short story: Memories we lost and other stories
Using adequate illustrations from the story ‘Almost Home’ by Barvy Mc kinley, write an essay with the title ‘The Challenges of illegal Immigration,’
OR
- The Novel: The Pearl – John Steinbeck
‘Great expectations make frustrated men’. Using illustrations from the novel, write an essay supporting this statement.
MWAKICAN JOINT EXAMINATIONS MARKING SCHEME
PAPER 3 FORM 4 ENGLISH
Q1 a. Deduct 4 marks AD if the candidate does not end with the given words.
- Deny marks if the meaning of the proverb is given. Meaning should be from the story.
TABLE OF CATEGORIZATION
D CLASS (01-05)
The candidate either does not communicate at all or his language ability is so minimal that the examiner practically has to guess what the candidate wants to communicate. The candidate fails to fit English words he knows into meaningful sentences.
The subject is glanced at or distorted. Practically no valid punctuation. All kinds of errors are “Broken English.”.
D- (01-01) Chaotic. Little meaning whatsoever. Question paper or some words from it simply copied.
D (03) Flow of thought almost impossible to follow. The errors are continuous.
D+(04-05) Although English is often broken and the essay is full of errors of all types one can at least guess what the candidate wants to communicate.
C CLASS (06-10) – Generally, there is difficulty in communication.
The candidate communicates understandably but only more or less clearly. He/she is not confident with their language. The language is often undeveloped. There may be some digressions. Unnecessary repetitions are frequent. The arrangement is weak and the flow is jerky. There is no economy of language mother tongue influence is felt in spelling; there is direct translation.
C- (06-07)- The candidate obviously finds it difficult to communicate his ideas. He is seriously hampered by his very limited knowledge of the language structure and vocabulary. This results in many gross errors of agreement, spelling, misuse of prepositions, tenses, verb agreement and sentence construction.
C 08 The candidate communicates but not with consistent clarity. His linguistic abilities being very limited, he cannot avoid frequent errors in sentence structure. There is little variety or originality. Very bookish English. Links are weak, incorrect and at times repeated.
C+ (09-10) The candidates communicates clearly but in the flat and uncertain manner. Simple concepts and sentences are often strained. There may be an overuse of clinches or unsuitable idioms. Proverbs are misquoted or misinterpreted. The flow is still jerky. There are some errors of agreement, tenses and spelling.
B CLASS (11-15) Generally, there is fluency in communications
This class is characterized by greater fluency and ease of expression. The candidate demonstrates that he/she can use English as a normal way expressing himself. Sentences are varied and usually well constructed. Some candidates become ambitious and even over ambitious i.e the candidate may use too much vocabulary in an effort to impress. There may be items of merit of the one word or one’s expression types. Many essays in this category may be just clean and unassuming but they still show that the candidate is at ease with the language.
B(11-12) The candidates communicates fairly and with some fluency. There may be little variety in sentence structure gross errors are occasional.
B 13 The sentences are varied but rather simple and straight forward. The candidate does not strain himself in an effort to impress. There is a fair range of vocabulary and idiom. Some items or merit, economy of language. The candidate seems to express themselves naturally and effortlessly.
B+ (14-15) The candidate communicates his ideas pleasantly and without strain. There are few errors and slips. Tenses, spellings and punctuations are quite good. A number of merit of “whole sentence” or the “whole expression” type are evident.
A CLASS (16-20) Communication is efficient
The candidate communicates not only fluently, but attractively, with originality and efficiency. He has the ability to make the reader share his deep feelings, emotions, enthusiasm. He expresses himself freely and without any visible constraint. The script gives evidence of maturity, good planning and a tinge of humour. Many items of merit which indicate that the candidate has complete command of the language. There is no strain, just pleasantness, clever arrangement and felicity of expressions.
A-(16-17) The candidates shows competence and fluency in using language. He may lack imagination or originality which usually provides the “spark” in such essays. Vocabulary, idioms, sentence structure, links and variety are impressive. Gross errors are very rare.
A18 Positive Ability.
A few errors that are felt to be slips. The story or arguments has a definite impact. No grammar problem. Variety of structures. A definite spark.
A+(19-20) The candidates communicates not only information and meaning, but also and especially the candidate’s whole self, his feelings, taste, points of view, youth and culture, this ability to communicate his deep self may express itself in many ways: wide range effective vocabulary, original approach, vivid and sustained account in case of a narrative, well developed and ordered argument in case of a debate or discussion. A very definite spark.
Q 2. Blossoms of the Savannah.
√ Ole Kaelo. as a father he is close to his daughters and should therefore advocate for their best interest. He however betrays them. He does not support their bisg dream of joining Egerton university. He marries Resian off to Oloisudori. He denies Taito a chance to go to Mombasa for a Music extravaganza. He loved which is ironical.
√ Olarinkoi. pretends to be a guardian angel when he rescues Taiyo and Resian from the vagabonds. However he betrays Reslains trust when he attempts to rape her after kidnapping her. He also wants to marry her against her wish.
√ Mama Milanoi. A mother is supposed to nurture and protect her children. Mama Milanoi does the opposite when she assists the Enkamuratani get access to Taiyo. She sides with the Ole kaelo’s idea of having to Oloisudori.
√ Ole supeyo. He is a close friend and mentor to Ole kaelo. When Ole kaelo goes for his help to evade Oloisudori’s demands, Ole supeyo declines to buy the stocks and contracts that had imprisoned Ole kaelo to Oloisudori “……..But Ole supeyo had declined the offer effectively throwing him back to the hyena.”
√Oloisudori. He is a close business associate to Ole Kaelo. He visits Ole Kaero’s home after their contract. His eyeing Ole Keiro daughter Resian and his opportunism to wanting to forcefully marry Resian is enough betrayal to Ole Kaelo. He also blackmails Ole Kaelo by demanding for his two daughters; one to go with his friend.
Introduction – 2MKS
Body – 3×paragraphs= 12MKS
Conclusion – 2MKS
Language – 4 MKS
- a) SHORT STORY
Individual who unlawfully move to a foreign county are bound to suffer certain challenges. By moving to Ireland illegally, Ali Mah fouz experiences a multitude of challenges.
Any other relevant introduction
- Impersonation – Ali impersonate a medical student. He told people he was a medical student. His face book picture should an eager young man standing outside the college of surgery with a bundle of books, two of which are telephone directories. He buys an iPod even though he doesn’t know how to load the device with music. – pg 74
- Odd jobs – He worked for two Egyptian brothers, slicing kebab meat into a half moon pan. He powers washed cars, scrubbed pots, wiped down tables, sold Christmas trees door to door, worked in meat packing plant.
- Arrests-he makes his first attempt to escape an arrest by jumping and running to a French woman who doesn’t help him. He alarms other passengers by calling out he is a terrorist. He in another escape attempt, ducked down and dashed under the body of a trailor, barely dogging the moving wheels. When he pushes between tourist buses and is caught by a tow bar, the pain is instant and crippling.
- Violence
Ali is hit by Tarrant – ….. and ran straight into Tarrant’s fist, a wall made of bone and skin. Ali licked his laps and tasted blood.
- Disillusionment
Although the “felt …. cut the Mediterranean like scissors through close, Ah, or remembering what awaits him back home himself ‘and now three years in jail ahead of him …. He swallowed a mouthful and then he swallowed some more…..”
- Conclusion
Those who illegally move to foreign countries experience a number of challenges.
Mark 3.3.3.3. introduction – 2 conclusion. 2 grammar 4
3 b)THE PEARL
- Kino and Juana’s expectations to have Coyotito treated.
When Coyotito is stung by the scorpion, Kino and Juana rush him to the doctor. Although Kina is aware that the doctor was not of his people, and could kill the doctor more easily than talk to him. (pg 26) Later when the doctor learns Kino had found the pearl of the world, he has great expectations of benefiting from the pearl. He even goes to treat Coyotito at home. The doctor frustrates them by not treating the boy. Kino hits the gate with frustration.
- Priest
When the father heard that Kino had found great fortune, a great pearl, he goes to Kino’s home and tells him that he hoped “they will remember to give thanks, my son, to him who has given them this treasure….” Pg48. Expectations were that if the pearl was not bought, it could have been given to the church. (pg 42,63,68)
- Beggars
When grapevine reaches beggars about the pearl of the world, they have great expectations for “they knew that there is no alms give in the world like a poor man who is suddenly lucky” pg 42, Every man suddenly became related to kino’s pearl and kino’s pearl went into the dreams, the speculations, the schemes, the plans, the future, the wishes, there heeds, the lusts, the hungers….” Pg 43
- Thieves
Attempts are made by thieves to make away with Kino’s pearl. The night before he goes to sell his pearl, a thief comes to steal it but both Kino and the thief are hurt. (pg 59-61) The thieves also pursue Kino and his family as they go to sell the pearl.
Before he leaves the house, he is attacked and searched. When Juana tries tothrow the pearl away, Kino attack her. He is as well attacked by an assailant whom he kills. (pg 86-89)
He is further pursued to the mountains where he kills his assailants and his son Coyotito is killed. (pg 115-116)
- Kino’s family
Kino and his family have great expectations from the pearl. In the pearl he saw Juana and Coyotito and himself standing and kneeling at the high alter…. being married, …how they were dressed. Juana in a shawl stiff with newness…” Then to come with little things Kino wanted …… pg 44-45 . He knew the pearl would see Coyotito join school.
- Pearl buyers
The pearl buyers had great expectations on Kino’s pearl. “They got out papers so that …….. they put their pearls in the desks for it is not good to let an inferior pearl be seen beside a beauty” pg 69 When they negotiate, ‘the dealers hand had become a personality.” pg 71 when Kino grabs the pearl from them, the dealers “knew they had played too hard they knew they would be disciplined for their failure …” pg 75
- Conclusion
When our expectations are high, so are our frustrations when they are not met.
Mark 3.3.3.3
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Form 1 Business Studies Exams and Marking Schemes Free
Form 1 Business Studies Exams and Marking Schemes Free
NAME: ………………………………………………… ADM NO.: ……………………
SCHOOL: ………………………… DATE: ………………… SIGN: ….……….
BUSINESS STUDIES
TIME: 2 HOURS
JOINT EXAMINATION – FORM ONE, TERM THREE 2023
INSTRUCTION TO STUDENTS
- Answer all the questions in the spaces provided
- This paper consists of 10 printed pages. Students should check the question paper to ensure that all pages are printed as indicated and no questions are missing.
FOR EXAMINER’S USE ONLY
| QUESTION NO. | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 |
| MARKS |
| QUESTION NO. | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 |
| MARKS |
GRAND TOTAL MARKS
- List FOUR disciplines studied in business studies. (4mks)
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- …………………………………………………………………………………………………………………………………………………………………………………………………………………
- State the type of utility created by each of the following activities. (4mks)
| ACTIVITY | LEVEL OF PRODUCTION |
| a) Keeping money in the bank | |
| b) Baking a loaf of bread | |
| c) Importing vehicles from Japan | |
| d) Selling medicine to patient |
- List FOUR characteristics of free goods. (4mks)
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- State FOUR types of business activities. (4mks)
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- Outline FOUR documents used in home trade. (4mks)
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- Muiga wants to start a business. State FOUR reasons why he would start one. (4mks)
- ………………………………………………………………………………………………………
- ………………………………………………………………………………………………………
- ………………………………………………………………………………………………………
- ……………………………………………………………………………………………………
- State FOUR equipment that facilitate mailing in an office. (4mks)
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- Highlight FOUR advantages of automatic vending machines. (4mks)
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………
………………………………………………………………………………………………………
- ………………………………………………………………………………………………………
……………………………………………………………………………………………………..
- Mention FOUR reasons why a consumer wouldsatisfy basic wants before secondary wants. (4mks)
i)………………………………………………………………………………………………………………………………………………………………………………………………………………
ii)………………………………………………………………………………………………………………………………………………………………………………………………………………
iii)………………………………………………………………………………………………………………………………………………………………………………………………………………
iv………………………………………………………………………………………………………………………………………………………………………………………………………………
- Name the factor of production that each of the following resources relate to. (4mks)
| Resource | Factor of production |
| a) Manager | |
| b) Motor Vehicle. | |
| c) Water | |
| d) Owner |
- Outline FOURsources of business idea. (4mks)
- ………………………………………………………………………………………………………
- ………………………………………………………………………………………………………
- ………………………………………………………………………………………………………
- ………………………………………………………………………………………………………
- List down FOUR external factors of environment that affect the business negatively. (4mks)
- ………………………………………………………………………………………………………
- ………………………………………………………………………………………………………
- ………………………………………………………………………………………………………
- ………………………………………………………………………………………………………
- State FOUR characteristicsof direct production . (4mks)
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- List down circumstances under which a cheque may be dishourned. (4mks)
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- Highlight FOUR importance of entrepreneurship to Kenyan Economy. (4mks)
- ………………………………………………………………………………………………………
- ………………………………………………………………………………………………………
- ……………………………………………………………………………………………………..
- …………………………………………………………………………………………………….
- Outline FOUR advantages of division of labour and specialization. (4mks)
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- State down FOUR advantages of engaging e-commerce in business activities. (4mks)
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………
- ………………………………………………………………………………………………………
- List FOUR causes of business failure. (4mks)
- …………………………………………………………………………………………………
- …………………………………………………………………………………………………
- …………………………………………………………………………………………………
- …………………………………………………………………………………………………
19.Highlight FOUR small scale retailers without shops. (4mks)
- ………………………………………………………………………………………………………
- ………………………………………………………………………………………………………
- ………………………………………………………………………………………………………
- ………………………………………………………………………………………………………
- List FOUR classification of economic resources (4mks)
- ………………………………………………………………………………………………………
- ………………………………………………………………………………………………………
- ………………………………………………………………………………………………………
- ………………………………………………………………………………………………………
21.Outline FOUR different rewards for labour as a factor of production. (4mks)
- ……………………………………………………………………………………………………………………………………………………………………………………………………
- ……………………………………………………………………………………………………………………………………………………………………………………………………
- ……………………………………………………………………………………………………………………………………………………………………………………………………
- …………………………………………………………………………………………………………………………………………………………………………………………………..
- Highlight FOUR measures that can be taken by an office to safeguard the property of the organization. (4mks)
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………………………………………………………………………………………………………………
- ………………………………………………………………………………………………………
23.Classify each of the following activities into appropriate level of production. (4mks)
| Activity | Level of production |
| a) Wholesaling | |
| b) Mining | |
| c) Maize milling | |
| d) Insurance |
24.State FOUR differences between goods and services. (4mks)
| GOODS | SERVICES |
25.Daudi sold crates of soda to Otis at sh.1000 on credit. He gave him a quantity discount of 10%. He also gave him cash discounts of;
10% if payment was made before 1 week.
5% if payment was made after 2 weeks
2% if payment was made after 1 month
Otis made the payment by the end of 2nd week. Calculate;
- Quantity discount (1mk)
- Cash discount (2mks)
c.Amount he paid (1mk)
__________________________________________________________________________________________________
BUSINESS STUDIES FORM 1 MARKING SCHEME TERM 3
- List FOUR disciplines studied in business studies. (4mks)
- Economics
- Commerce
- Accounting
- Office practice
- Entrepreneurship
- State the type of utility created by each of the following activities (4mks)
| ACTIVITY | LEVEL OF PRODUCTION |
| a) Keeping money in the bank | Time utility |
| b) Baking a loaf of bread | Form utility |
| c) Importing vehicles from Japan | Place utility |
| d) Selling medicine to patient | Possession utility |
- List FOUR characteristics of free goods. (4mks)
- They are gifts of nature
- they are plenty in supply/abundant
- they are priceless/have no money value
- they have utility
- State FOUR types of business activities (4mks)
- Manufacturing
- processing of raw materials
- Extraction
- Construction
- provision of services
- trade
- Distribution of goods
- Outline FOUR documents used in home trade (4mks)
- Credit note
- Debit note
- statement of account
- receipts
- letter of inquiry
- catalogue
- quotation
- price list
- order
- delivery note
- invoice
- pro-forma invoice
- Muiga wants to start a business. State FOUR reasons he would start one (4mks)
- to earn profit
- To provide goods and services
- to create employment
- to be own boss
- as an outlet for new innovation
- to utilize spare time
- to utilize available resources
- State FOUR equipment that facilitate mailing in an office (4mks)
- folding machine
- franking machine
- addressing machine
- sorting machine
- letter opener
- composite
- Highlight FOUR advantages of automatic vending machines (4mks)
- Commodities can be brought anytime because no attendant is required
- They save the owner the cost of employing a shop attendant
- They can be put strategically to boost sales e.g institution
- They are not affected by weather changes
- Provide good in cash basis protecting the owner from burden of bad debts
- They are fast and accurate
- Mention FOUR reasons why a consumer would satisfy basic wants before secondary wants (4mks)
- They cannot be postponed
- They are felt needs
- one cannot do without them/necessary for life/essential for survival
- They may be satisfied at individual level
- Name the factor of production that each of the following resources relate to (4mks)
| Resource | Factor of production |
| a) Manager | Labour |
| b) Motor Vehicle. | Capital |
| c) Water | Land |
| d) Owner | Enterpreneurship |
- Outline FOUR sources of a business idea
- Listening to what people say
- Hobbies
- Surveys
- Newspapers
- Journals
- Waste materials
- Spotting the market gap
- Shows and exhibition
- List down FOUR external factors of environment that affect the business negatively. (4mks)
- unfair competition
- Inadequate technology
- Natural calamities
- Unfair government policies
- Political instability
- Unfavorable socio-cultural practice
- poor infrastructure
- State FOUR characteristics of direct production (4mks)
- Production of low quality goods
- Encourage individualism
- lead to low standards of living
- can be very tiring
- Does not encourage invention and innovation
- cheap/simple tools are used in production
- it is mostly used on small scale
- the rate of production is low
- A lot of time is wasted from one job to another
- List down circumstances under which a cheque may be dishourned (4mks)
- In case of insufficient funds in the account
- When the signature of the account holder differs from the specimen in the bank
- When the cheque is post dated
- if the cheque has been altered
- if the account holder is dead ,insane or bankruptcy
- Highlight FOUR importance of entrepreneurship to Kenyan Economy (4mks)
- Creates employment
- Contributes to formation of capital
- Reduces rural-urban migration
- Raises standards of living
- improves infrastructure
- Reduces foreign dominance in the economy
- make use of local resources
- promotes technology
- promotes entrepreneurial culture
- saves on imports
- Outline FOUR advantages of division of labour and specialization. (4mks)
- Outpt per worker is greatly increased
- encourage invention and innovation
- makes it possible for greater use use of machine which makes production fast& efficient
- Enables worker to acquire skills in a particular field
- High quality goods and services are produced
- Worker spend less mental and physical effort
- saves on time
- State down FOUR advantages of engaging in e-commerce in business activities. (4mks)
- facilitates access to large market
- Creates conducive environment for business to transact without discriminations
- reduces paperwork since dealings are done online
- reduces cost of sending receiving and storing information
- Enables one to access important information that will enable him improve the running of his business
- it is a faster way of doing business as it facilitates urgent transactions
- List FOUR causes of business failure. (4mks)
- poor management of people
- Poor location/Lack of customers
- Lack of raw materials
- Inadequate finance
- Stiff competition
- Poor infrastructure
- Lack of commitment to the business
- Poor financial management
- poor debt management
- poor public relations
- poor market research
- low level of creativity and innovation
- Highlight four small scale retailers without shops (4mks)
- Itinerant traders
- Roadside sellers
- open air market traders
- Automatic vending machines
- List FOUR classification of economic resources (4mks)
- Natural resources
- Man-made resources
- Human resources
- Renewable resources
- Non –renewable resources
- Outline FOUR different rewards for labour as a factor of production (4mks)
- wages
- salaries
- commissions
- tips
- Highlight FOUR measures that can be taken by an office to safeguard the property of the organization. (4mks)
- Enhancing security against theft i.e by employing watchmen, installing security lights and fencing business premises
- Ensuring physical care by people using the property
- keeping records of movement and usage of the property
- Physical stock taking of each property
- Taking insurance cover
- proper repairs and servicing
- Classify each of the following activities into appropriate level of production (4mks)
| Activity | Level of production |
| Wholesaling | Tertiary |
| Mining | Primary |
| Maize milling | Secondary |
| Insurance | Tertiary |
- State FOUR differences between goods and services. (4mks)
| Goods | Services |
| Tangible | Intangible |
| Can be stored | Hard to be stored |
| Can be standardized | Difficult to be standardized |
| Can change possession | inseparable from the provider |
| Not all goods are perishable | Are highly perishable |
| Most can be seen | Services are invisible |
25.Daudi sold crates of soda to Otis at sh.1000 on credit.He gave him a quantity discount of 10%.He also gave him cash discounts of;
10% if payment was made before 1 week.
5% if payment was made after 2 weeks
2% if payment was made after 1 month
Otis made the payment by the end of 2ndweek.Calculate;
- Trade discount (1mk)
10/100×1000=sh 100
- Cash discount (2mks)
1000-100=900
2.5/100x900=sh.45
c.Amount he paid (1mk)
100+45=145
1000-145=sh 855

