Class 10 Chemistry Worksheet on Chapter 4 Carbon and Its compounds ... - Free Printable
Educational worksheet: Class 10 Chemistry Worksheet on Chapter 4 Carbon and Its compounds .... Download and print for classroom or home learning activities.
JPG
631×817
58.4 KB
Free · Personal Use
Quality Assured by Worksheets Library Team
Reviewed for educational accuracy and age-appropriateness
ID: #1623612
⭐
Show Answer Key & Explanations
Step-by-step solution for: Class 10 Chemistry Worksheet on Chapter 4 Carbon and Its compounds ...
▼
Show Answer Key & Explanations
Step-by-step solution for: Class 10 Chemistry Worksheet on Chapter 4 Carbon and Its compounds ...
Here is the complete solution to all the questions in the image:
---
Q-5:
a) What are heteroatoms? What is their importance?
✔ Heteroatoms are atoms other than carbon and hydrogen that are present in organic molecules. Common examples include oxygen (O), nitrogen (N), sulfur (S), halogens (F, Cl, Br, I), phosphorus (P), etc.
✔ Importance of heteroatoms:
- They introduce polarity into molecules, affecting solubility, boiling point, and reactivity.
- They are responsible for the functional groups that define the chemical behavior of organic compounds (e.g., -OH in alcohols, -COOH in carboxylic acids).
- They play crucial roles in biological systems — for example, nitrogen in amino acids and DNA, oxygen in carbohydrates, sulfur in cysteine.
- They enable diverse chemical reactions and are key in pharmaceuticals, agrochemicals, and materials science.
---
b) Which of the molecules below are homoatomic and which are heteroatomic?
Given molecules:
Br₂, SO₂, CH₄, O₃, NH₃, N₂, S₈, HCl
✔ Homoatomic molecules: Composed of only one type of atom.
- Br₂ → Bromine (only Br atoms)
- O₃ → Ozone (only O atoms)
- N₂ → Nitrogen gas (only N atoms)
- S₈ → Sulfur (only S atoms)
✔ Heteroatomic molecules: Composed of two or more different types of atoms.
- SO₂ → Sulfur and Oxygen
- CH₄ → Carbon and Hydrogen
- NH₃ → Nitrogen and Hydrogen
- HCl → Hydrogen and Chlorine
> ✔ Note: Although CH₄ contains only C and H, it is still considered *heteroatomic* in many contexts because “homoatomic” typically refers to elements composed of a single element (like O₂, N₂, S₈). However, strictly speaking, since H and C are different elements, CH₄ is heteroatomic. In organic chemistry, "homoatomic" usually means diatomic or polyatomic molecules of a single element, so CH₄ is definitely heteroatomic.
---
Q-6: Fill in the blanks
a) Butane and isobutane are ___
✔ isomers (specifically, structural isomers — same molecular formula C₄H₁₀, different connectivity).
---
b) Sodium palmitate is a ___ while dodecylbenzenesulfonate is a ___
✔ Sodium palmitate is a soap (carboxylate salt from fatty acid — sodium salt of palmitic acid).
✔ Dodecylbenzenesulfonate is a detergent (synthetic surfactant with sulfonate head group).
> So:
Sodium palmitate is a _soap_ while dodecylbenzenesulfonate is a _detergent_.
---
c) Ethene is an ___ compound and decolourises the brown colour of Br₂ in CCl₄ to form ___.
✔ Ethene is an unsaturated compound (contains a C=C double bond).
✔ It reacts with bromine (Br₂) via electrophilic addition to form 1,2-dibromoethane (CH₂Br–CH₂Br).
> So:
Ethene is an _unsaturated_ compound and decolourises the brown colour of Br₂ in CCl₄ to form _1,2-dibromoethane_.
---
d) The C-C bond length in graphite is ___ and in diamond is ___
✔ In graphite, carbon atoms are sp² hybridized, forming planar layers with delocalized π electrons. The C–C bond length is approximately 1.42 Å.
✔ In diamond, carbon atoms are sp³ hybridized, forming a tetrahedral 3D network. The C–C bond length is approximately 1.54 Å.
> So:
The C-C bond length in graphite is _1.42 Å_ and in diamond is _1.54 Å_.
---
e) Methane molecule is tetrahedral but ethene and benzene are ___
✔ Methane (CH₄) has tetrahedral geometry due to sp³ hybridization.
✔ Ethene (C₂H₄) has trigonal planar geometry around each carbon (sp² hybridization).
✔ Benzene (C₆H₆) is planar with all carbons sp² hybridized — also trigonal planar geometry.
> So:
Methane molecule is tetrahedral but ethene and benzene are _planar_ (or specifically, _trigonal planar_ around each carbon).
---
Q-7: Match Column I with Column II
| Column I | Column II |
|------------------------|---------------------------------------------------------------------------|
| A) Neutralisation Reaction | ii) CH₃COOH + NaOH → CH₃COONa + H₂O |
| B) Acid Hydrolysis | i) CH₃COOC₂H₅ + H₂O →[dil. H₂SO₄, Δ] CH₃COOH + C₂H₅OH |
| C) Dehydration | iv) CH₃CH₂OH →[conc. H₂SO₄, 443K] CH₂=CH₂ + H₂O |
| D) Hydrogenation | iii) Vegetable oil + H₂ →[Ni, Δ] Fat |
✔ Explanation:
- A) Neutralisation Reaction: Acid + Base → Salt + Water → matches ii
- B) Acid Hydrolysis: Ester + Water → Acid + Alcohol (in presence of acid catalyst) → matches i
- C) Dehydration: Removal of water from alcohol to form alkene → matches iv
- D) Hydrogenation: Addition of H₂ to unsaturated compounds (like vegetable oils → saturated fats) → matches iii
---
✔ Final Answers Summary:
Q-5a: Heteroatoms = non-C/H atoms; important for reactivity, polarity, function in biomolecules.
Q-5b:
- Homoatomic: Br₂, O₃, N₂, S₈
- Heteroatomic: SO₂, CH₄, NH₃, HCl
Q-6:
a) isomers
b) soap, detergent
c) unsaturated, 1,2-dibromoethane
d) 1.42 Å, 1.54 Å
e) planar
Q-7:
A → ii
B → i
C → iv
D → iii
---
Let me know if you need diagrams or further explanations!
---
Q-5:
a) What are heteroatoms? What is their importance?
✔ Heteroatoms are atoms other than carbon and hydrogen that are present in organic molecules. Common examples include oxygen (O), nitrogen (N), sulfur (S), halogens (F, Cl, Br, I), phosphorus (P), etc.
✔ Importance of heteroatoms:
- They introduce polarity into molecules, affecting solubility, boiling point, and reactivity.
- They are responsible for the functional groups that define the chemical behavior of organic compounds (e.g., -OH in alcohols, -COOH in carboxylic acids).
- They play crucial roles in biological systems — for example, nitrogen in amino acids and DNA, oxygen in carbohydrates, sulfur in cysteine.
- They enable diverse chemical reactions and are key in pharmaceuticals, agrochemicals, and materials science.
---
b) Which of the molecules below are homoatomic and which are heteroatomic?
Given molecules:
Br₂, SO₂, CH₄, O₃, NH₃, N₂, S₈, HCl
✔ Homoatomic molecules: Composed of only one type of atom.
- Br₂ → Bromine (only Br atoms)
- O₃ → Ozone (only O atoms)
- N₂ → Nitrogen gas (only N atoms)
- S₈ → Sulfur (only S atoms)
✔ Heteroatomic molecules: Composed of two or more different types of atoms.
- SO₂ → Sulfur and Oxygen
- CH₄ → Carbon and Hydrogen
- NH₃ → Nitrogen and Hydrogen
- HCl → Hydrogen and Chlorine
> ✔ Note: Although CH₄ contains only C and H, it is still considered *heteroatomic* in many contexts because “homoatomic” typically refers to elements composed of a single element (like O₂, N₂, S₈). However, strictly speaking, since H and C are different elements, CH₄ is heteroatomic. In organic chemistry, "homoatomic" usually means diatomic or polyatomic molecules of a single element, so CH₄ is definitely heteroatomic.
---
Q-6: Fill in the blanks
a) Butane and isobutane are ___
✔ isomers (specifically, structural isomers — same molecular formula C₄H₁₀, different connectivity).
---
b) Sodium palmitate is a ___ while dodecylbenzenesulfonate is a ___
✔ Sodium palmitate is a soap (carboxylate salt from fatty acid — sodium salt of palmitic acid).
✔ Dodecylbenzenesulfonate is a detergent (synthetic surfactant with sulfonate head group).
> So:
Sodium palmitate is a _soap_ while dodecylbenzenesulfonate is a _detergent_.
---
c) Ethene is an ___ compound and decolourises the brown colour of Br₂ in CCl₄ to form ___.
✔ Ethene is an unsaturated compound (contains a C=C double bond).
✔ It reacts with bromine (Br₂) via electrophilic addition to form 1,2-dibromoethane (CH₂Br–CH₂Br).
> So:
Ethene is an _unsaturated_ compound and decolourises the brown colour of Br₂ in CCl₄ to form _1,2-dibromoethane_.
---
d) The C-C bond length in graphite is ___ and in diamond is ___
✔ In graphite, carbon atoms are sp² hybridized, forming planar layers with delocalized π electrons. The C–C bond length is approximately 1.42 Å.
✔ In diamond, carbon atoms are sp³ hybridized, forming a tetrahedral 3D network. The C–C bond length is approximately 1.54 Å.
> So:
The C-C bond length in graphite is _1.42 Å_ and in diamond is _1.54 Å_.
---
e) Methane molecule is tetrahedral but ethene and benzene are ___
✔ Methane (CH₄) has tetrahedral geometry due to sp³ hybridization.
✔ Ethene (C₂H₄) has trigonal planar geometry around each carbon (sp² hybridization).
✔ Benzene (C₆H₆) is planar with all carbons sp² hybridized — also trigonal planar geometry.
> So:
Methane molecule is tetrahedral but ethene and benzene are _planar_ (or specifically, _trigonal planar_ around each carbon).
---
Q-7: Match Column I with Column II
| Column I | Column II |
|------------------------|---------------------------------------------------------------------------|
| A) Neutralisation Reaction | ii) CH₃COOH + NaOH → CH₃COONa + H₂O |
| B) Acid Hydrolysis | i) CH₃COOC₂H₅ + H₂O →[dil. H₂SO₄, Δ] CH₃COOH + C₂H₅OH |
| C) Dehydration | iv) CH₃CH₂OH →[conc. H₂SO₄, 443K] CH₂=CH₂ + H₂O |
| D) Hydrogenation | iii) Vegetable oil + H₂ →[Ni, Δ] Fat |
✔ Explanation:
- A) Neutralisation Reaction: Acid + Base → Salt + Water → matches ii
- B) Acid Hydrolysis: Ester + Water → Acid + Alcohol (in presence of acid catalyst) → matches i
- C) Dehydration: Removal of water from alcohol to form alkene → matches iv
- D) Hydrogenation: Addition of H₂ to unsaturated compounds (like vegetable oils → saturated fats) → matches iii
---
✔ Final Answers Summary:
Q-5a: Heteroatoms = non-C/H atoms; important for reactivity, polarity, function in biomolecules.
Q-5b:
- Homoatomic: Br₂, O₃, N₂, S₈
- Heteroatomic: SO₂, CH₄, NH₃, HCl
Q-6:
a) isomers
b) soap, detergent
c) unsaturated, 1,2-dibromoethane
d) 1.42 Å, 1.54 Å
e) planar
Q-7:
A → ii
B → i
C → iv
D → iii
---
Let me know if you need diagrams or further explanations!
Parent Tip: Review the logic above to help your child master the concept of carbon compounds worksheet.
