Frank Brothers Solutions for ICSE Class 10 Chemistry Chapter 11c Unsaturated Hydrocarbons

ICSE Solutions Frank Brothers Class 10 Chemistry Chapter 11c Unsaturated Hydrocarbons have been provided below and is also available in Pdf for free download. The Frank Brothers ICSE solutions for Class 10 Chemistry have been prepared as per the latest syllabus and ICSE books and examination pattern suggested in Class 10. Questions given in ICSE Frank Brothers book for Class 10 Chemistry are an important part of exams for Class 10 Chemistry and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for ICSE Class 10 Chemistry and also download more latest study material for all subjects. Chapter 11c Unsaturated Hydrocarbons is an important topic in Class 10, please refer to answers provided below to help you score better in exams

Frank Brothers Chapter 11c Unsaturated Hydrocarbons Class 10 Chemistry ICSE Solutions

Class 10 Chemistry students should refer to the following ICSE questions with answers for Chapter 11c Unsaturated Hydrocarbons in Class 10. These ICSE Solutions with answers for Class 10 Chemistry will come in exams and help you to score good marks

Chapter 11c Unsaturated Hydrocarbons Frank Brothers ICSE Solutions Class 10 Chemistry

Unsaturated Hydrocarbons

 

Solution 1.
(i) The n signifies Number of carbon atoms where as 2n signifies Number of hydrogen atoms.
(ii) Butene
(iii) \( C_3H_6 \)
(iv) Molecular formula of first member of alkene is: \( C_2H_4 \)
Structural formula:
H    H
 \  /
  C=C
 /  \
H    H
(v) Lower Homologues of alkene which contains four carbon atoms is : \( C_3H_6 \)
Higher Homologues of alkene which contains four carbon atoms is : \( C_5H_{10} \)

📝 Teacher's Note: Use simple examples like counting fingers to explain the n and 2n relationship. Draw the structural formula step by step on the board to help students visualize the double bond clearly.

🎯 Exam Tip: Always write both molecular and structural formulas when asked about alkenes. Remember the general formula \( C_nH_{2n} \) for quick verification.

 

Solution 2.
Ethylene is prepared in the laboratory by dehydration of ethyl alcohol.
Reaction:
\( CH_3CH_2OH \xrightarrow[\text{conc. H}_2\text{SO}_4]{170°C} CH_2 = CH_2 + H_2O \)

Procedure:
Take one part of ethyl alcohol in a flask. Add two parts of concentrated sulphuric acid and heat to 170°C. Ethylene is evolved and small amount of aluminium sulphate is also added to the flask to avoid frothing.
Gas is prepared by reacting ethyl alcohol in presence of sulphuric acid and is collected by downward displacement of water.
The gas is dried by passing the gas through sodium hydroxide solution.

📝 Teacher's Note: Emphasize the exact temperature (170°C) and explain that aluminium sulphate prevents violent bubbling. Demonstrate the setup with a diagram for better understanding.

🎯 Exam Tip: Always mention the temperature, catalyst, and method of gas collection. Write the balanced chemical equation clearly with proper conditions above the arrow.

 

Solution 3.
(a) Ethene into 1,2-dibromoethene:
When bromine gas is passed into inert solvent like \( CCl_4 \) containing dissolved ethene at room temperature, one molecule of \( Br_2 \) adds across the double bond to give ethylene dibromide.
\( CH_2 = CH_2 + Br_2 \rightarrow CH_2Br - CH_2Br \)

(b) Ethene to bromoethane:
Ethene reacts with halogen acids to form alkyl halides.
\( CH_2 = CH_2 + HBr \rightarrow CH_3 - CH_2Br \)

📝 Teacher's Note: Explain that these are addition reactions - the double bond opens up to accommodate new atoms. Use the analogy of opening a zipper to add new pieces.

🎯 Exam Tip: Always show the breaking of the double bond clearly in your equations. Mention the type of reaction (addition) for full marks.

 

Solution 4.
Physical properties:
• State: colourless gas, neutral to litmus, faint sweet odour
• Solubility: slightly soluble in water, highly soluble in organic solvents.
• Density: slightly lighter than air
• Melting point: -169°C
• Boiling point: -104°C

📝 Teacher's Note: Connect the physical properties to everyday observations - like why ethylene gas rises up and why it dissolves better in oil than water.

🎯 Exam Tip: List at least 4-5 physical properties including state, odour, solubility, and temperature points. Always mention "colourless gas" as the first property.

 

Solution 5.
(a) Ethanol to ethene:
\( CH_3CH_2OH \xrightarrow[\text{conc. H}_2\text{SO}_4]{170°C} CH_2 = CH_2 + H_2O \)

(b) Ethene to ethanol:
\( CH_2 = CH_2 + H_2O \xrightarrow{H^+} CH_3 - CH_2OH \)

📝 Teacher's Note: Point out that these are reverse reactions of each other. Explain dehydration (removal of water) vs hydration (addition of water) using simple examples.

🎯 Exam Tip: Always write the conditions clearly - temperature for dehydration and acid catalyst for hydration. Show the water molecule being removed/added explicitly.

 

Solution 6.
When bromine solution is added to ethane and ethene, In case of ethene the orange colour of bromine disappears where as in case of ethane colour does not disappear.

📝 Teacher's Note: This is a simple test to distinguish between saturated and unsaturated hydrocarbons. Demonstrate this with actual bromine water if possible for visual impact.

🎯 Exam Tip: Mention the color change clearly - orange/brown bromine water becomes colorless with alkenes but remains orange with alkanes.

 

Solution 7.
The addition of hydrogen across the double bond is called hydrogenation.
Ethene reacts with hydrogen gas when heated in presence of catalyst like nickel, to give ethane.
\( CH_2 = CH_2 + H_2 \xrightarrow[\text{Ni}]{300°C} CH_3 - CH_3 \)

Use of hydrogenation:
Hydrogenation of Oils: The oils can be converted into ghee which are semi solid at room temperature.

📝 Teacher's Note: Relate this to the making of vanaspati ghee from liquid oils. This helps students connect chemistry to daily life and understand practical applications.

🎯 Exam Tip: Define hydrogenation first, then give the reaction with proper conditions. Always mention the practical application in the oil industry for complete marks.

 

Solution 8.
Two equations:
(i) \( CH_2 = CH_2 + H_2 \xrightarrow[\text{Ni}]{300°C} CH_3 - CH_3 \)
(ii) \( CH_2 = CH_2 + H_2O \xrightarrow{H^+} CH_3 - CH_2OH \)

📝 Teacher's Note: Explain that both are addition reactions but with different reagents. Help students see the pattern - the double bond always opens up to accommodate new atoms.

🎯 Exam Tip: Write both equations clearly with proper conditions. The first is hydrogenation, the second is hydration - mentioning the reaction names helps secure full marks.

 

Solution 9.
Addition Reaction: The reactions in which molecules of the attacking reagent add across the double or triple bond of an unsaturated compound to yield saturated compound.
Ethene is a reactive compound due to presence of double bond. Hence undergo addition reactions.
Two examples:
(i) Addition of hydrogen: \( CH_2 = CH_2 + H_2 \xrightarrow[\text{Ni}]{300°C} CH_3 - CH_3 \)
(ii) Addition of halogens: \( CH_2 = CH_2 + Br_2 \rightarrow CH_2Br - CH_2Br \)

📝 Teacher's Note: Start with the definition, then explain why alkenes are reactive. Use the analogy of a weak spot (double bond) that easily breaks to accommodate new atoms.

🎯 Exam Tip: Always start with the definition of addition reaction. Give two different types of examples - one with hydrogen and one with halogens for complete coverage.

 

Solution 10.
Three uses of ethene are:
1. It is used for the manufacture of polythene
2. For the artificial ripening of fruits.
3. As a general anaesthetic.

📝 Teacher's Note: Connect these uses to students' daily experiences - plastic bags (polythene), banana ripening, and hospital procedures. This makes the chemistry more relatable.

🎯 Exam Tip: List exactly three uses as asked. Include the manufacture of polythene as it's the most important industrial use of ethylene.

 

Solution 11.
(i) When ethene is passed in bromine solution in \( CCl_4 \), the orange colour of bromine disappears due to formation of colourless ethylene bromide.
\( CH_2 = CH_2 + Br_2 \rightarrow CH_2Br - CH_2Br \)
(ii) When Baeyer's reagent reacts with ethene, colour of potassium permanganate gets discharged in this reaction.
\( CH_2 = CH_2 + H_2O + O \xrightarrow[\text{KMnO}_4]{25°C-30°C} CH_2OH - CH_2OH \)

📝 Teacher's Note: Explain that both are tests for unsaturation. The color changes are visual proof that a reaction has occurred. Baeyer's test uses cold, dilute KMnO4.

🎯 Exam Tip: Mention the color changes clearly - bromine water decolorizes, and pink KMnO4 becomes colorless. Both are important tests for identifying alkenes.

ICSE Frank Brothers Solutions Class 10 Chemistry Chapter 11c Unsaturated Hydrocarbons

Students can now access the detailed Frank Brothers Solutions for Chapter 11c Unsaturated Hydrocarbons on our portal. These solutions have been carefully prepared as per latest ICSE Class 10 syllabus. Each solution given above has been updated based on the current year pattern to ensure Class 10 students have the most updated Chemistry content.

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Our subject experts have provided detailed explanations for all the questions found in the Frank Brothers textbook for Class 10 Chemistry. We have focussed on making the concepts easy for you in Chapter 11c Unsaturated Hydrocarbons so that students can understand the concepts behind every answer. For all numerical problems and theoretical concepts these solutions will help in strengthening your analytical skill required for the ICSE examinations.

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Yes, our solutions for Chapter 11c Unsaturated Hydrocarbons are designed as per new 2026 ICSE standards. 40% competency-based questions required for Class 10, are included to help students understand application-based logic behind every Chemistry answer.

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Yes, every exercise in Chapter 11c Unsaturated Hydrocarbons from the Frank Brothers textbook has been solved step-by-step. Class 10 students will learn Chemistry conceots before their ICSE exams.

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