CBSE Class 10 · Chemistry
CBSE Class 10 Chemistry Carbon and its Compounds Previous Year Questions
Help your child master CBSE Class 10 Chemistry Carbon and its Compounds Previous Year Questions with this curated collection sourced from real board papers spanning 2019–2025. Every question comes with a detailed step-by-step solution, helping your child confidently tackle covalent bonding, homologous series, organic reactions, and naming of compounds — topics that consistently carry marks in the board exam.
CBSE Class 10 Chemistry Carbon and its Compounds — Questions with Solutions
Assertion (A): Propanal and propanone are structural isomers.
Reason (R): Propanal and propanone both have the same molecular formula.
Reason (R): Propanal and propanone both have the same molecular formula.
Solution
Answer: Option (A) is correct.
Assertion (A) is true: Propanal (an aldehyde) and propanone (a ketone) are indeed structural isomers.
Reason (R) is true: They both have the same molecular formula, $C_{3}H_{6}O$.
Reason (R) is the correct explanation: The definition of structural isomers is that they have the same molecular formula but different structures. The different structures of propanal ($CH_{3}CH_{2}CHO$) and propanone ($CH_{3}COCH_{3}$) make them structural isomers, and the reason explains this fundamental principle.
Assertion (A) is true: Propanal (an aldehyde) and propanone (a ketone) are indeed structural isomers.
Reason (R) is true: They both have the same molecular formula, $C_{3}H_{6}O$.
Reason (R) is the correct explanation: The definition of structural isomers is that they have the same molecular formula but different structures. The different structures of propanal ($CH_{3}CH_{2}CHO$) and propanone ($CH_{3}COCH_{3}$) make them structural isomers, and the reason explains this fundamental principle.
Given below are the structures of some hydrocarbons. Select the two structures which are related to each other from the given options:
Solution
Answer: Option (C) is correct.
(i) CH₃–CH₂–CH=CH₂ → But-1-ene (Alkene: contains a double bond)
(ii) CH₃–CH₂–CH₂–CH₃ → Butane (Alkane: only single bonds)
(iii) Isobutane (Branched alkane) → C₄H₁₀, isomer of butane
(iv) CH₃–CH₂–CH=CH–Cl → Chloro butene (Alkene with halogen substitution)
Correct Pair: (ii) and (iii)
Both are alkanes with the same molecular formula (C₄H₁₀). They differ in structure but not in formula — they are structural isomers.
(i) CH₃–CH₂–CH=CH₂ → But-1-ene (Alkene: contains a double bond)
(ii) CH₃–CH₂–CH₂–CH₃ → Butane (Alkane: only single bonds)
(iii) Isobutane (Branched alkane) → C₄H₁₀, isomer of butane
(iv) CH₃–CH₂–CH=CH–Cl → Chloro butene (Alkene with halogen substitution)
Correct Pair: (ii) and (iii)
Both are alkanes with the same molecular formula (C₄H₁₀). They differ in structure but not in formula — they are structural isomers.
Choose the incorrect statement about the common reaction used in hydrogenation of vegetable oils.
Solution
Answer: Option (D) is correct (the incorrect statement).
Hydrogenation of vegetable oils is an addition reaction where hydrogen (H₂) is added to unsaturated fatty acids (oils) to convert them into saturated fats.
This reaction is carried out in the presence of a metal catalyst like nickel (Ni) or palladium (Pd) — not an acid catalyst. Option (D) incorrectly states it uses an acid catalyst.
$$HC \equiv CH + 2H_2 \xrightarrow{Pd} CH_3\text{–}CH_3$$
Hydrogenation of vegetable oils is an addition reaction where hydrogen (H₂) is added to unsaturated fatty acids (oils) to convert them into saturated fats.
This reaction is carried out in the presence of a metal catalyst like nickel (Ni) or palladium (Pd) — not an acid catalyst. Option (D) incorrectly states it uses an acid catalyst.
$$HC \equiv CH + 2H_2 \xrightarrow{Pd} CH_3\text{–}CH_3$$
Select from the following a hydrocarbon having C−C bond and C=C bond.
Solution
Answer: Option (A) is correct.
Benzene (C₆H₆) is a cyclic hydrocarbon that has alternating single (C–C) and double (C=C) bonds in its ring structure.
• (B) Cyclohexane: Contains only single bonds (C–C), no double bonds.
• (C) Butyne: Contains a triple bond (C≡C) and single bonds, but no double bond (C=C).
• (D) Propyne: Contains a triple bond (C≡C) and no double bond (C=C).
Benzene (C₆H₆) is a cyclic hydrocarbon that has alternating single (C–C) and double (C=C) bonds in its ring structure.
• (B) Cyclohexane: Contains only single bonds (C–C), no double bonds.
• (C) Butyne: Contains a triple bond (C≡C) and single bonds, but no double bond (C=C).
• (D) Propyne: Contains a triple bond (C≡C) and no double bond (C=C).
If we make a carbon skeleton with four carbon atoms, the two different possible skeletons will be:
- ✗(A)
- ✓(B)
- ✗(C)
- ✗(D)
Solution
Answer: Option (B) is correct.
When forming carbon skeletons using four carbon atoms, there are two basic structural possibilities:
1. Straight-chain structure — all carbon atoms connected in a continuous line (n-butane).
2. Branched-chain structure — one carbon bonded as a side branch (isobutane).
Option (B) correctly shows these two structural skeletons.
When forming carbon skeletons using four carbon atoms, there are two basic structural possibilities:
1. Straight-chain structure — all carbon atoms connected in a continuous line (n-butane).
2. Branched-chain structure — one carbon bonded as a side branch (isobutane).
Option (B) correctly shows these two structural skeletons.
A hydrocarbon which does not belong to the same homologous series as the other three is:
Solution
Answer: Option (C) is correct.
For alkanes, the general formula is CₙH₂ₙ₊₂.
• (A) C₄H₁₀ → fits CₙH₂ₙ₊₂ → Alkane ✓
• (B) C₆H₁₄ → fits CₙH₂ₙ₊₂ → Alkane ✓
• (C) C₇H₁₄ → fits CₙH₂ₙ → This is an alkene, not an alkane ✗
• (D) C₁₀H₂₂ → fits CₙH₂ₙ₊₂ → Alkane ✓
Therefore, C₇H₁₄ does not belong to the same homologous series (alkanes) as the other three.
For alkanes, the general formula is CₙH₂ₙ₊₂.
• (A) C₄H₁₀ → fits CₙH₂ₙ₊₂ → Alkane ✓
• (B) C₆H₁₄ → fits CₙH₂ₙ₊₂ → Alkane ✓
• (C) C₇H₁₄ → fits CₙH₂ₙ → This is an alkene, not an alkane ✗
• (D) C₁₀H₂₂ → fits CₙH₂ₙ₊₂ → Alkane ✓
Therefore, C₇H₁₄ does not belong to the same homologous series (alkanes) as the other three.
Assertion (A): Carbon and its compounds are our major sources of fuels.
Reason (R): Most of the carbon compounds on burning release a large amount of heat and light.
Reason (R): Most of the carbon compounds on burning release a large amount of heat and light.
Solution
Answer: Option (A) is correct.
Carbon compounds such as petrol, diesel, kerosene, natural gas, and coal are widely used as fuels. They are preferred because on combustion, they release a large amount of heat and light, which is why they serve as efficient energy sources.
Hence, both the assertion and reason are true, and the reason correctly explains the assertion.
Carbon compounds such as petrol, diesel, kerosene, natural gas, and coal are widely used as fuels. They are preferred because on combustion, they release a large amount of heat and light, which is why they serve as efficient energy sources.
Hence, both the assertion and reason are true, and the reason correctly explains the assertion.
Carbon compounds:
(i) are good conductors of electricity.
(ii) are bad conductors of electricity.
(iii) have strong forces of attraction between their molecules.
(iv) have weak forces of attraction between their molecules.
(i) are good conductors of electricity.
(ii) are bad conductors of electricity.
(iii) have strong forces of attraction between their molecules.
(iv) have weak forces of attraction between their molecules.
Solution
Answer: Option (C) is correct.
Carbon compounds are bad conductors of electricity: Most organic carbon compounds do not conduct electricity because they do not have free electrons or ions. (Exception: Graphite conducts electricity.)
Carbon compounds have weak forces of attraction between their molecules: This is why many organic compounds have low melting and boiling points and exist as gases or liquids at room temperature.
Carbon compounds are bad conductors of electricity: Most organic carbon compounds do not conduct electricity because they do not have free electrons or ions. (Exception: Graphite conducts electricity.)
Carbon compounds have weak forces of attraction between their molecules: This is why many organic compounds have low melting and boiling points and exist as gases or liquids at room temperature.
The name and formula of the third member of the homologous series of alkynes is:
Solution
Answer: Option (D) is correct.
The general formula of alkynes is CₙH₂ₙ₋₂.
Members of the homologous series of alkynes:
1. Ethyne – C₂H₂
2. Propyne – C₃H₄
3. Butyne – C₄H₆ ← Third member
So, Butyne (C₄H₆) is the third member of the alkyne series.
The general formula of alkynes is CₙH₂ₙ₋₂.
Members of the homologous series of alkynes:
1. Ethyne – C₂H₂
2. Propyne – C₃H₄
3. Butyne – C₄H₆ ← Third member
So, Butyne (C₄H₆) is the third member of the alkyne series.
Consider the following statements about homologous series of carbon compounds:
(a) All succeeding members differ by –CH₂ unit.
(b) Melting point and boiling point increases with increasing molecular mass.
(c) The difference in molecular masses between two successive members is 16 u.
(d) C₂H₂ and C₃H₄ are NOT the successive members of alkyne series.
The correct statements are:
(a) All succeeding members differ by –CH₂ unit.
(b) Melting point and boiling point increases with increasing molecular mass.
(c) The difference in molecular masses between two successive members is 16 u.
(d) C₂H₂ and C₃H₄ are NOT the successive members of alkyne series.
The correct statements are:
Solution
Answer: Option (A) is correct.
• (a) True: In a homologous series, each successive member differs by a –CH₂ group (14 u in mass).
• (b) True: As molecular mass increases, melting and boiling points increase due to stronger van der Waals forces.
• (c) False: The difference in molecular mass between two members is 14 u, not 16 u.
• (d) False: C₂H₂ (ethyne) and C₃H₄ (propyne) are successive members of the alkyne series.
• (a) True: In a homologous series, each successive member differs by a –CH₂ group (14 u in mass).
• (b) True: As molecular mass increases, melting and boiling points increase due to stronger van der Waals forces.
• (c) False: The difference in molecular mass between two members is 14 u, not 16 u.
• (d) False: C₂H₂ (ethyne) and C₃H₄ (propyne) are successive members of the alkyne series.
(a) Select from the following the members of the same homologous series:
(b) What happens to (i) the melting point, and (ii) the solubility of compounds as the molecular mass of the compounds in a homologous series increases?
(b) What happens to (i) the melting point, and (ii) the solubility of compounds as the molecular mass of the compounds in a homologous series increases?
Answer
(a) The given compounds are:
1. Propanal → Aldehyde 2. Propanone → Ketone 3. Propanoic acid → Carboxylic acid 4. Butanal → Aldehyde
Propanal and Butanal are members of the same homologous series (Aldehyde series). Both have the functional group –CHO, and their molecular formulae differ by one –CH₂– unit.
(b)(i) Melting Point: As molecular size and mass increase, van der Waals forces strengthen. Therefore, melting point increases gradually down the homologous series.
(b)(ii) Solubility: As molecular mass increases, the non-polar hydrocarbon chain becomes longer, reducing polarity. Hence, solubility in water decreases. Solubility in organic solvents generally increases with molecular mass.
1. Propanal → Aldehyde 2. Propanone → Ketone 3. Propanoic acid → Carboxylic acid 4. Butanal → Aldehyde
Propanal and Butanal are members of the same homologous series (Aldehyde series). Both have the functional group –CHO, and their molecular formulae differ by one –CH₂– unit.
(b)(i) Melting Point: As molecular size and mass increase, van der Waals forces strengthen. Therefore, melting point increases gradually down the homologous series.
(b)(ii) Solubility: As molecular mass increases, the non-polar hydrocarbon chain becomes longer, reducing polarity. Hence, solubility in water decreases. Solubility in organic solvents generally increases with molecular mass.
What are micelles? Draw the structure.
Answer
A micelle is an aggregate of soap molecules formed in water.
Structure of a Micelle:
• Hydrophobic tails cluster in the centre of the micelle.
• Hydrophilic heads remain on the outside, interacting with water.
• The hydrophobic parts trap dirt/grease in the core.
• The hydrophilic parts keep the micelle suspended in water.
• Dirt is easily rinsed away when water is poured.
Structure of a Micelle:
• Hydrophobic tails cluster in the centre of the micelle.
• Hydrophilic heads remain on the outside, interacting with water.
• The hydrophobic parts trap dirt/grease in the core.
• The hydrophilic parts keep the micelle suspended in water.
• Dirt is easily rinsed away when water is poured.
Write the chemical name and the formula of a carbon compound having two carbon atoms in its molecule and belonging to the family with functional group –OH. What happens when this compound reacts with the following? (Give chemical equation and the name of the main product formed in each case)
(i) Excess conc. sulphuric acid ($H_2SO_4$) at 443 K (ii) Ethanoic acid in the presence of an acid (iii) Acidified potassium dichromate (iv) Sodium metal
(i) Excess conc. sulphuric acid ($H_2SO_4$) at 443 K (ii) Ethanoic acid in the presence of an acid (iii) Acidified potassium dichromate (iv) Sodium metal
Answer
The compound is Ethanol — Formula: C₂H₅OH
(i) Reaction with excess conc. H₂SO₄ at 443 K (Dehydration): $$C_2H_5OH \xrightarrow{\text{H}_2\text{SO}_4,\, 443\,K} C_2H_4 + H_2O$$ Main product: Ethene (C₂H₄)
(ii) Reaction with Ethanoic acid in presence of acid (Esterification): $$C_2H_5OH + CH_3COOH \xrightarrow{\text{acid}} CH_3COOC_2H_5 + H_2O$$ Main product: Ethyl acetate (Ethyl ethanoate)
(iii) Reaction with Acidified potassium dichromate (Oxidation): $$C_2H_5OH + [O] \xrightarrow{\text{acidified } K_2Cr_2O_7} CH_3COOH$$ Main product: Ethanoic acid (CH₃COOH)
(iv) Reaction with Sodium metal: $$2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2$$ Main product: Sodium ethoxide (C₂H₅ONa)
(i) Reaction with excess conc. H₂SO₄ at 443 K (Dehydration): $$C_2H_5OH \xrightarrow{\text{H}_2\text{SO}_4,\, 443\,K} C_2H_4 + H_2O$$ Main product: Ethene (C₂H₄)
(ii) Reaction with Ethanoic acid in presence of acid (Esterification): $$C_2H_5OH + CH_3COOH \xrightarrow{\text{acid}} CH_3COOC_2H_5 + H_2O$$ Main product: Ethyl acetate (Ethyl ethanoate)
(iii) Reaction with Acidified potassium dichromate (Oxidation): $$C_2H_5OH + [O] \xrightarrow{\text{acidified } K_2Cr_2O_7} CH_3COOH$$ Main product: Ethanoic acid (CH₃COOH)
(iv) Reaction with Sodium metal: $$2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2$$ Main product: Sodium ethoxide (C₂H₅ONa)
A hydrocarbon which is different from the other three is:
Solution
Answer: Option (B) is correct.
$C_5H_{10}$ is an alkene (general formula $C_nH_{2n}$).
• (A) $C_7H_{16}$: Alkane (heptane) — $C_nH_{2n+2}$
• (C) $C_2H_6$: Alkane (ethane) — $C_nH_{2n+2}$
• (D) $CH_4$: Alkane (methane) — $C_nH_{2n+2}$
All others are alkanes; $C_5H_{10}$ belongs to the alkene series — the odd one out.
$C_5H_{10}$ is an alkene (general formula $C_nH_{2n}$).
• (A) $C_7H_{16}$: Alkane (heptane) — $C_nH_{2n+2}$
• (C) $C_2H_6$: Alkane (ethane) — $C_nH_{2n+2}$
• (D) $CH_4$: Alkane (methane) — $C_nH_{2n+2}$
All others are alkanes; $C_5H_{10}$ belongs to the alkene series — the odd one out.
All compounds of carbon which contain only carbon and hydrogen are called hydrocarbons. Among these, saturated hydrocarbons are called ‘alkanes’, unsaturated compounds with one or more double bonds are called ‘alkenes’ and compounds containing one or more triple bonds are called ‘alkynes’.
(a) Identify the following compound and write its name: $CH_3 – C \equiv CH$
(b) How do saturated and unsaturated carbon compounds differ in terms of the flame produced by them on burning?
(c)(i) Which type of hydrocarbons undergo addition reactions? Show with an example.
OR
(c)(ii) What are structural isomers? Draw two structural isomers of butane ($C_4H_{10}$).
(a) Identify the following compound and write its name: $CH_3 – C \equiv CH$
(b) How do saturated and unsaturated carbon compounds differ in terms of the flame produced by them on burning?
(c)(i) Which type of hydrocarbons undergo addition reactions? Show with an example.
OR
(c)(ii) What are structural isomers? Draw two structural isomers of butane ($C_4H_{10}$).
Answer
(a) The compound $CH_3 – C \equiv CH$ is Propyne. It has 3 carbon atoms with a triple bond between the first and second carbon atoms, making it an alkyne.
(b)
• Saturated hydrocarbons (alkanes) produce a blue, clean flame when burned — complete combustion producing only CO₂ and water.
• Unsaturated hydrocarbons (alkenes and alkynes) burn with a yellow, smoky flame due to incomplete combustion. The double/triple bonds make combustion less efficient, producing carbon (soot).
(c)(i) Unsaturated hydrocarbons undergo addition reactions, where multiple bonds break to allow new atoms to be added.
Example — Ethene with hydrogen: $$C_2H_4 + H_2 \rightarrow C_2H_6$$ Ethene (alkene) adds hydrogen to become ethane (alkane).
OR
(c)(ii) Structural isomers are compounds with the same molecular formula but different structural arrangements.
Two structural isomers of butane ($C_4H_{10}$):
1. n-Butane: $CH_3\text{–}CH_2\text{–}CH_2\text{–}CH_3$ (straight chain)
2. Isobutane: Branched chain
(b)
• Saturated hydrocarbons (alkanes) produce a blue, clean flame when burned — complete combustion producing only CO₂ and water.
• Unsaturated hydrocarbons (alkenes and alkynes) burn with a yellow, smoky flame due to incomplete combustion. The double/triple bonds make combustion less efficient, producing carbon (soot).
(c)(i) Unsaturated hydrocarbons undergo addition reactions, where multiple bonds break to allow new atoms to be added.
Example — Ethene with hydrogen: $$C_2H_4 + H_2 \rightarrow C_2H_6$$ Ethene (alkene) adds hydrogen to become ethane (alkane).
OR
(c)(ii) Structural isomers are compounds with the same molecular formula but different structural arrangements.
Two structural isomers of butane ($C_4H_{10}$):
1. n-Butane: $CH_3\text{–}CH_2\text{–}CH_2\text{–}CH_3$ (straight chain)
2. Isobutane: Branched chain
The number of covalent bond(s) present in a nitrogen molecule is/are: (Atomic number of nitrogen is 7)
Solution
Answer: Option (B) is correct.
Atomic number of nitrogen = 7, so its electronic configuration is $1s^2\ 2s^2\ 2p^3$. Each nitrogen atom has 5 valence electrons and needs 3 more to complete its octet.
In N₂, each nitrogen atom shares 3 electrons with the other — forming three covalent bonds (a triple bond): $$\text{N} \equiv \text{N}$$
Atomic number of nitrogen = 7, so its electronic configuration is $1s^2\ 2s^2\ 2p^3$. Each nitrogen atom has 5 valence electrons and needs 3 more to complete its octet.
In N₂, each nitrogen atom shares 3 electrons with the other — forming three covalent bonds (a triple bond): $$\text{N} \equiv \text{N}$$
The number of single and double bonds present in a molecule of benzene ($\text{C}_6\text{H}_6$) respectively, are:
Solution
Answer: Option (A) is correct.
Benzene ($\text{C}_6\text{H}_6$) has a hexagonal ring with alternating single and double bonds. There are 9 single bonds (6 C–H bonds + 3 C–C single bonds) and 3 double bonds (C=C) in the molecule.
Benzene ($\text{C}_6\text{H}_6$) has a hexagonal ring with alternating single and double bonds. There are 9 single bonds (6 C–H bonds + 3 C–C single bonds) and 3 double bonds (C=C) in the molecule.
(i) Conversion of ethanol to ethanoic acid is an oxidation reaction. Why? Name the oxidizing agent in this conversion and write the chemical equation for this oxidation reaction. Explain how this reaction is different from the reaction in which ethanol burns in the presence of oxygen.
(ii) What is a homologous series of carbon compounds? Write the name of the first two members of a homologous series of compounds having functional group –COOH.
(ii) What is a homologous series of carbon compounds? Write the name of the first two members of a homologous series of compounds having functional group –COOH.
Answer
(i)
The conversion of ethanol to ethanoic acid is an oxidation reaction because ethanol loses hydrogen atoms (gains oxygen), which is the characteristic of oxidation.
Oxidizing Agent: Acidified potassium dichromate (K₂Cr₂O₇) or alkaline KMnO₄.
Chemical Equation: $$\text{C}_2\text{H}_5\text{OH} + [O] \xrightarrow{\text{K}_2\text{Cr}_2\text{O}_7} \text{CH}_3\text{COOH} + \text{H}_2\text{O}$$ Difference from burning of ethanol:
When ethanol burns, it undergoes complete combustion: $$\text{C}_2\text{H}_5\text{OH} + 3O_2 \rightarrow 2CO_2 + 3H_2O$$ In oxidation, ethanol is converted to ethanoic acid (a useful product). In combustion, it is fully broken down to CO₂ and water, releasing energy as heat and light.
(ii)
A homologous series is a group of organic compounds with the same functional group, similar chemical properties, and each successive member differing by a –CH₂– group.
First two members of the homologous series with –COOH (carboxylic acids):
1. Methanoic acid (Formic acid) – HCOOH
2. Ethanoic acid (Acetic acid) – CH₃COOH
The conversion of ethanol to ethanoic acid is an oxidation reaction because ethanol loses hydrogen atoms (gains oxygen), which is the characteristic of oxidation.
Oxidizing Agent: Acidified potassium dichromate (K₂Cr₂O₇) or alkaline KMnO₄.
Chemical Equation: $$\text{C}_2\text{H}_5\text{OH} + [O] \xrightarrow{\text{K}_2\text{Cr}_2\text{O}_7} \text{CH}_3\text{COOH} + \text{H}_2\text{O}$$ Difference from burning of ethanol:
When ethanol burns, it undergoes complete combustion: $$\text{C}_2\text{H}_5\text{OH} + 3O_2 \rightarrow 2CO_2 + 3H_2O$$ In oxidation, ethanol is converted to ethanoic acid (a useful product). In combustion, it is fully broken down to CO₂ and water, releasing energy as heat and light.
(ii)
A homologous series is a group of organic compounds with the same functional group, similar chemical properties, and each successive member differing by a –CH₂– group.
First two members of the homologous series with –COOH (carboxylic acids):
1. Methanoic acid (Formic acid) – HCOOH
2. Ethanoic acid (Acetic acid) – CH₃COOH
Select ionic compounds from the following:
(a) $\text{CCl}_4$ (b) $KCl$ (c) $HCl$ (d) $NaCl$
(a) $\text{CCl}_4$ (b) $KCl$ (c) $HCl$ (d) $NaCl$
Solution
Answer: Option (A) is correct.
(b) KCl: Ionic compound — K (metal) loses an electron to form K⁺, Cl gains it to form Cl⁻.
(d) NaCl: Ionic compound — Na loses an electron to form Na⁺, Cl gains it to form Cl⁻.
(a) CCl₄: Covalent compound — carbon and chlorine (both non-metals) share electrons.
(c) HCl: Covalent compound — hydrogen and chlorine share electrons in a covalent bond.
(b) KCl: Ionic compound — K (metal) loses an electron to form K⁺, Cl gains it to form Cl⁻.
(d) NaCl: Ionic compound — Na loses an electron to form Na⁺, Cl gains it to form Cl⁻.
(a) CCl₄: Covalent compound — carbon and chlorine (both non-metals) share electrons.
(c) HCl: Covalent compound — hydrogen and chlorine share electrons in a covalent bond.
The products obtained when sodium reacts with ethanol are:
Solution
Answer: Option (A) is correct.
When sodium (Na) reacts with ethanol (C₂H₅OH), it displaces the hydrogen from the –OH group, forming sodium ethoxide and hydrogen gas.
$$2\text{Na}(s) + 2\text{C}_2\text{H}_5\text{OH}(l) \rightarrow 2\text{C}_2\text{H}_5\text{ONa}(aq) + \text{H}_2(g)$$
When sodium (Na) reacts with ethanol (C₂H₅OH), it displaces the hydrogen from the –OH group, forming sodium ethoxide and hydrogen gas.
$$2\text{Na}(s) + 2\text{C}_2\text{H}_5\text{OH}(l) \rightarrow 2\text{C}_2\text{H}_5\text{ONa}(aq) + \text{H}_2(g)$$
The functional group present in butanone is:
Solution
Answer: Option (D) is correct.
Butanone (C₄H₈O) is a ketone. It contains a carbonyl group (–C=O) where the carbonyl carbon is bonded to two other carbon atoms — the characteristic functional group of all ketones.
Butanone (C₄H₈O) is a ketone. It contains a carbonyl group (–C=O) where the carbonyl carbon is bonded to two other carbon atoms — the characteristic functional group of all ketones.
(i) Give reason why carbon can neither form $\text{C}^{4+}$ cations nor $\text{C}^{4-}$ anions but forms covalent compounds.
(ii) What is a homologous series of carbon compounds? Write the molecular formula of any two consecutive members of a homologous series of aldehydes.
(iii) Write the number of (1) single covalent bonds and (2) double covalent bonds in a molecule of benzene.
(ii) What is a homologous series of carbon compounds? Write the molecular formula of any two consecutive members of a homologous series of aldehydes.
(iii) Write the number of (1) single covalent bonds and (2) double covalent bonds in a molecule of benzene.
Answer
(i)
Carbon has 4 valence electrons.
• To form $\text{C}^{4+}$: losing 4 electrons requires very high energy — energetically unfavourable.
• To form $\text{C}^{4-}$: gaining 4 electrons is difficult as the small nucleus cannot hold 10 electrons.
• Therefore, carbon shares electrons with other atoms, forming covalent compounds.
(ii)
A homologous series is a group of organic compounds with the same functional group, similar chemical properties, each successive member differing by a –CH₂– group.
Two consecutive members of the aldehyde series (functional group –CHO):
1. Methanal (Formaldehyde): CH₂O
2. Ethanal (Acetaldehyde): C₂H₄O
(iii)
Benzene (C₆H₆) has alternating single and double bonds and each carbon bonds to one hydrogen.
(1) Single covalent bonds: 3 (C–C) + 6 (C–H) = 9 single covalent bonds
(2) Double covalent bonds: 3 double bonds (C=C) in the ring
Carbon has 4 valence electrons.
• To form $\text{C}^{4+}$: losing 4 electrons requires very high energy — energetically unfavourable.
• To form $\text{C}^{4-}$: gaining 4 electrons is difficult as the small nucleus cannot hold 10 electrons.
• Therefore, carbon shares electrons with other atoms, forming covalent compounds.
(ii)
A homologous series is a group of organic compounds with the same functional group, similar chemical properties, each successive member differing by a –CH₂– group.
Two consecutive members of the aldehyde series (functional group –CHO):
1. Methanal (Formaldehyde): CH₂O
2. Ethanal (Acetaldehyde): C₂H₄O
(iii)
Benzene (C₆H₆) has alternating single and double bonds and each carbon bonds to one hydrogen.
(1) Single covalent bonds: 3 (C–C) + 6 (C–H) = 9 single covalent bonds
(2) Double covalent bonds: 3 double bonds (C=C) in the ring
(i) Write the name and the molecular formula of a commercially important carbon compound having functional group –OH.
(ii) Write chemical equations to show the reactions of the above mentioned compound with: 1. Excess conc. sulphuric acid 2. Ethanoic acid in presence of acid catalyst 3. Sodium metal 4. Alkaline potassium permanganate (KMnO₄)
Also write the name of the product formed in each reaction.
(ii) Write chemical equations to show the reactions of the above mentioned compound with: 1. Excess conc. sulphuric acid 2. Ethanoic acid in presence of acid catalyst 3. Sodium metal 4. Alkaline potassium permanganate (KMnO₄)
Also write the name of the product formed in each reaction.
Answer
(i) Ethanol (ethyl alcohol) — Molecular Formula: C₂H₅OH
(ii)
1. Reaction with excess conc. H₂SO₄ (Dehydration): $$\text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{C}_2\text{H}_4 + \text{H}_2\text{O}$$ Product: Ethene (Ethylene)
2. Reaction with Ethanoic acid in presence of acid catalyst (Esterification): $$\text{C}_2\text{H}_5\text{OH} + \text{CH}_3\text{COOH} \xrightarrow{\text{acid}} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$ Product: Ethyl acetate (ester)
3. Reaction with Sodium metal: $$2\text{C}_2\text{H}_5\text{OH} + 2\text{Na} \rightarrow 2\text{C}_2\text{H}_5\text{ONa} + \text{H}_2$$ Product: Sodium ethoxide (C₂H₅ONa)
4. Reaction with Alkaline KMnO₄ (Oxidation): $$\text{C}_2\text{H}_5\text{OH} + \text{KMnO}_4 \xrightarrow{\text{alkaline}} \text{CH}_3\text{COOH} + \text{MnO}_2 + \text{KOH}$$ Product: Acetic acid / Ethanoic acid (CH₃COOH)
(ii)
1. Reaction with excess conc. H₂SO₄ (Dehydration): $$\text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{C}_2\text{H}_4 + \text{H}_2\text{O}$$ Product: Ethene (Ethylene)
2. Reaction with Ethanoic acid in presence of acid catalyst (Esterification): $$\text{C}_2\text{H}_5\text{OH} + \text{CH}_3\text{COOH} \xrightarrow{\text{acid}} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$ Product: Ethyl acetate (ester)
3. Reaction with Sodium metal: $$2\text{C}_2\text{H}_5\text{OH} + 2\text{Na} \rightarrow 2\text{C}_2\text{H}_5\text{ONa} + \text{H}_2$$ Product: Sodium ethoxide (C₂H₅ONa)
4. Reaction with Alkaline KMnO₄ (Oxidation): $$\text{C}_2\text{H}_5\text{OH} + \text{KMnO}_4 \xrightarrow{\text{alkaline}} \text{CH}_3\text{COOH} + \text{MnO}_2 + \text{KOH}$$ Product: Acetic acid / Ethanoic acid (CH₃COOH)
(i) Define a homologous series of carbon compounds.
(ii) Why is the melting and boiling point of C₄H₈ higher than that of C₃H₆ or C₂H₄?
(iii) Why do we NOT see any gradation in chemical properties of homologous series compounds?
(iv) Write the name and structures of (a) aldehyde and (b) ketone with molecular formula C₃H₆O.
(ii) Why is the melting and boiling point of C₄H₈ higher than that of C₃H₆ or C₂H₄?
(iii) Why do we NOT see any gradation in chemical properties of homologous series compounds?
(iv) Write the name and structures of (a) aldehyde and (b) ketone with molecular formula C₃H₆O.
Answer
(i) A homologous series is a series of organic compounds that have the same functional group, similar chemical properties, and differ by a –CH₂– group in their molecular formula.
(ii) As the size of the molecule increases, the van der Waals forces between molecules become stronger, requiring more energy to overcome them. Therefore, the melting and boiling points of C₄H₈ are higher than those of C₃H₆ or C₂H₄.
(iii) We do not see any gradation in chemical properties because all members contain the same functional group, which determines their chemical reactivity. Their chemical properties are therefore similar despite differences in molecular size.
(iv)
(a) Aldehyde — Propanal: $$\text{CH}_3\text{CH}_2\text{CHO}$$ (b) Ketone — Propanone (Acetone): $$\text{CH}_3\text{CO}\text{CH}_3$$
(ii) As the size of the molecule increases, the van der Waals forces between molecules become stronger, requiring more energy to overcome them. Therefore, the melting and boiling points of C₄H₈ are higher than those of C₃H₆ or C₂H₄.
(iii) We do not see any gradation in chemical properties because all members contain the same functional group, which determines their chemical reactivity. Their chemical properties are therefore similar despite differences in molecular size.
(iv)
(a) Aldehyde — Propanal: $$\text{CH}_3\text{CH}_2\text{CHO}$$ (b) Ketone — Propanone (Acetone): $$\text{CH}_3\text{CO}\text{CH}_3$$
3 mL of ethanol is taken in a test tube and warmed gently in a water bath. A 5% solution of alkaline potassium permanganate is added first drop by drop to this solution, then in excess.
(a) How is 5% solution of KMnO₄ prepared?
(b) State the role of alkaline potassium permanganate in this reaction. What happens on adding it in excess?
(c) Explain by writing the reaction.
(a) How is 5% solution of KMnO₄ prepared?
(b) State the role of alkaline potassium permanganate in this reaction. What happens on adding it in excess?
(c) Explain by writing the reaction.
Answer
(a) Preparation of 5% KMnO₄ Solution:
Dissolve 5 g of KMnO₄ in 100 mL of distilled water. Stir until completely dissolved.
(b) Role of Alkaline KMnO₄:
It acts as a strong oxidizing agent, oxidizing ethanol (C₂H₅OH) to ethanoic acid (CH₃COOH).
When added in excess: The oxidation goes to completion, converting all ethanol into ethanoic acid.
(c) Chemical Reaction: $$\text{C}_2\text{H}_5\text{OH} + 2[O] \xrightarrow[\text{alk.}]{\text{KMnO}_4} \text{CH}_3\text{COOH} + \text{H}_2\text{O}$$ Ethanol is oxidized by KMnO₄ (in alkaline medium) to form ethanoic acid and water.
Dissolve 5 g of KMnO₄ in 100 mL of distilled water. Stir until completely dissolved.
(b) Role of Alkaline KMnO₄:
It acts as a strong oxidizing agent, oxidizing ethanol (C₂H₅OH) to ethanoic acid (CH₃COOH).
When added in excess: The oxidation goes to completion, converting all ethanol into ethanoic acid.
(c) Chemical Reaction: $$\text{C}_2\text{H}_5\text{OH} + 2[O] \xrightarrow[\text{alk.}]{\text{KMnO}_4} \text{CH}_3\text{COOH} + \text{H}_2\text{O}$$ Ethanol is oxidized by KMnO₄ (in alkaline medium) to form ethanoic acid and water.
List two chemical properties on the basis of which ethanol and ethanoic acid may be differentiated and explain how.
Answer
1. Reaction with Sodium Metal (Na):
• Ethanol reacts slowly: $$2\text{C}_2\text{H}_5\text{OH} + 2\text{Na} \rightarrow 2\text{C}_2\text{H}_5\text{ONa} + \text{H}_2$$ • Ethanoic acid reacts vigorously: $$2\text{CH}_3\text{COOH} + 2\text{Na} \rightarrow 2\text{CH}_3\text{COONa} + \text{H}_2$$ Observation: Both liberate H₂, but ethanoic acid reacts much faster and more vigorously.
2. Reaction with Sodium Carbonate (Na₂CO₃):
• Ethanol does not react with Na₂CO₃.
• Ethanoic acid reacts: $$2\text{CH}_3\text{COOH} + \text{Na}_2\text{CO}_3 \rightarrow 2\text{CH}_3\text{COONa} + \text{H}_2\text{O} + \text{CO}_2$$ Observation: Effervescence (CO₂ gas) is seen only with ethanoic acid, not with ethanol.
• Ethanol reacts slowly: $$2\text{C}_2\text{H}_5\text{OH} + 2\text{Na} \rightarrow 2\text{C}_2\text{H}_5\text{ONa} + \text{H}_2$$ • Ethanoic acid reacts vigorously: $$2\text{CH}_3\text{COOH} + 2\text{Na} \rightarrow 2\text{CH}_3\text{COONa} + \text{H}_2$$ Observation: Both liberate H₂, but ethanoic acid reacts much faster and more vigorously.
2. Reaction with Sodium Carbonate (Na₂CO₃):
• Ethanol does not react with Na₂CO₃.
• Ethanoic acid reacts: $$2\text{CH}_3\text{COOH} + \text{Na}_2\text{CO}_3 \rightarrow 2\text{CH}_3\text{COONa} + \text{H}_2\text{O} + \text{CO}_2$$ Observation: Effervescence (CO₂ gas) is seen only with ethanoic acid, not with ethanol.
How are covalent bonds formed?
Answer
Covalent bonds are formed when two atoms share one or more pairs of electrons to achieve a stable noble gas electronic configuration. This bonding usually occurs between non-metal atoms. Each shared pair of electrons forms one covalent bond.
Examples:
1. Hydrogen molecule (H₂) — Single bond: $$\text{H} \cdot + \cdot \text{H} \rightarrow \text{H–H}$$ 2. Oxygen molecule (O₂) — Double bond: $$\text{O=O}$$ 3. Nitrogen molecule (N₂) — Triple bond: $$\text{N} \equiv \text{N}$$
Examples:
1. Hydrogen molecule (H₂) — Single bond: $$\text{H} \cdot + \cdot \text{H} \rightarrow \text{H–H}$$ 2. Oxygen molecule (O₂) — Double bond: $$\text{O=O}$$ 3. Nitrogen molecule (N₂) — Triple bond: $$\text{N} \equiv \text{N}$$
“Carbon prefers to share its valence electrons with other atoms of carbon or with atoms of other elements rather than gaining or losing the valence electrons in order to attain noble gas configuration.” Justify this statement.
Answer
The atomic number of carbon is 6, with electronic configuration K, L = 2, 4. Carbon has 4 valence electrons and needs 4 more to complete its octet.
• Forming $\text{C}^{4-}$ anion: Gaining 4 electrons — the nucleus with only 6 protons cannot effectively hold 10 electrons.
• Forming $\text{C}^{4+}$ cation: Losing 4 electrons — requires an enormous amount of energy, making it energetically unfavourable.
Therefore, carbon overcomes this problem by sharing its valence electrons, forming stable covalent compounds.
• Forming $\text{C}^{4-}$ anion: Gaining 4 electrons — the nucleus with only 6 protons cannot effectively hold 10 electrons.
• Forming $\text{C}^{4+}$ cation: Losing 4 electrons — requires an enormous amount of energy, making it energetically unfavourable.
Therefore, carbon overcomes this problem by sharing its valence electrons, forming stable covalent compounds.
(i) Write the name and draw the structure of a saturated hydrocarbon with four carbon atoms.
(ii) Write the number of single covalent bonds present in this compound.
(ii) Write the number of single covalent bonds present in this compound.
Answer
(i) The saturated hydrocarbon with four carbon atoms is Butane.
Structure: $$\text{CH}_3\text{–CH}_2\text{–CH}_2\text{–CH}_3$$ (ii) In butane: 3 C–C bonds + 10 C–H bonds = 13 single covalent bonds in total.
Structure: $$\text{CH}_3\text{–CH}_2\text{–CH}_2\text{–CH}_3$$ (ii) In butane: 3 C–C bonds + 10 C–H bonds = 13 single covalent bonds in total.
Rehmat classified the reaction between Methane and Chlorine in the presence of sunlight as a substitution reaction. Justify Rehmat’s view and illustrate the reaction with the help of a balanced chemical equation.
Answer
Saturated hydrocarbons are fairly unreactive, but in the presence of sunlight, chlorine reacts with methane in a very fast reaction. Chlorine replaces the hydrogen atoms one by one.
It is called a substitution reaction because one type of atom (chlorine) takes the place of another (hydrogen). This justifies Rehmat’s classification.
Balanced Chemical Equation: $$\text{CH}_4 + \text{Cl}_2 \xrightarrow{\text{sunlight}} \text{CH}_3\text{Cl} + \text{HCl}$$ Methane reacts with chlorine in sunlight to form chloromethane and hydrochloric acid — a substitution reaction.
It is called a substitution reaction because one type of atom (chlorine) takes the place of another (hydrogen). This justifies Rehmat’s classification.
Balanced Chemical Equation: $$\text{CH}_4 + \text{Cl}_2 \xrightarrow{\text{sunlight}} \text{CH}_3\text{Cl} + \text{HCl}$$ Methane reacts with chlorine in sunlight to form chloromethane and hydrochloric acid — a substitution reaction.
Explore More CBSE Class 10 Chemistry Chapters
Frequently Asked Questions
What does the Carbon and its Compounds chapter cover in CBSE Class 10 Chemistry? ⌄
The chapter covers covalent bonding, the versatility of carbon (catenation and tetravalency), homologous series, functional groups, nomenclature of organic compounds, chemical properties of ethanol and ethanoic acid, soaps and detergents, and the action of micelles. These topics appear regularly as MCQs, short answers, and long questions in CBSE board exams.
How many marks does Carbon and its Compounds carry in the CBSE Class 10 board exam? ⌄
Carbon and its Compounds is one of the most heavily weighted chapters in CBSE Class 10 Science, typically contributing 7–10 marks across 1-mark MCQs, 2-mark short answers, 3-mark questions, and 5-mark long-answer or case-study questions. Your child should be thorough with organic reactions and naming conventions to score full marks.
What are the most important topics to focus on in Carbon and its Compounds? ⌄
The highest-priority topics are: properties and reactions of ethanol and ethanoic acid (esterification, oxidation, dehydration, reaction with sodium), the concept and examples of homologous series, naming of organic compounds with functional groups, structural isomers, and the working of soaps through micelle formation. These appear in virtually every board paper.
What common mistakes do students make in Carbon and its Compounds questions? ⌄
A very common error is confusing the general formulae of alkanes (CₙH₂ₙ₊₂), alkenes (CₙH₂ₙ), and alkynes (CₙH₂ₙ₋₂), leading to mistakes in identifying homologous series members. Students also frequently mix up ethanol and ethanoic acid reactions, or forget to write balanced chemical equations with correct products. Regular practice with previous year questions helps build accuracy under exam pressure.
How does Angle Belearn help students master Carbon and its Compounds? ⌄
Angle Belearn’s CBSE specialists curate chapter-wise question banks drawn directly from real board papers, each paired with clear, step-by-step solutions written in exam-ready language. By practising these verified questions, your child learns both the correct answers and the structured approach needed to earn full marks — from writing chemical equations to explaining concepts precisely the way examiners expect.
