NCERT Exemplar Solutions Class 10 Science Carbon and its Compounds — Multiple Choice Questions Q1–15

NCERT Exemplar Problems, Class 10 Science — Carbon and its Compounds, Multiple Choice Questions 1 to 15. All 15 questions solved.

Carbon’s exemplar band runs to twenty-nine multiple-choice questions, and this page takes the first fifteen; Questions 16 to 29 continue into nomenclature, homologous series and the reactions of ethanol and ethanoic acid.

Organic chemistry at this level rests on one fact and three consequences:

• Carbon is tetravalent. Every carbon atom forms exactly four bonds, no more and no fewer. That single rule decides Questions 5, 11 and 15 outright, and it is the first thing to check when a structure is offered for inspection: count the bonds at every carbon.
• Carbon bonds to carbon — catenation — so chains, branches and rings are all possible, which is why one molecular formula can have several structures.
• Carbon compounds are covalent, so they are poor conductors and have low melting and boiling points; and they are built around functional groups, which is where their chemistry lives.

Key insight. Most of the wrong options in this half are structures that break the tetravalency rule — a carbon with three bonds or five, or a hydrogen written with a subscript as though it could form three. Counting bonds at each atom disposes of them faster than recognising the right answer does.

Choose the correct answer from the given four options in the following questions:

Question 1

Carbon exists in the atmosphere in the form of

Solution. The atmosphere contains carbon dioxide at about 0.04 per cent, which is small but is the form in which most atmospheric carbon exists — and it is the gas plants take up in photosynthesis.

It also contains carbon monoxide in traces, produced by the incomplete combustion of fuels in vehicles and stoves and by some natural processes. It is present in far smaller quantity than carbon dioxide, which is exactly what “in traces” records.

So both are present, in the proportions option (b) describes. Options (a) and (c) each name one and exclude the other.

Option (d) fails on a different ground: coal is a solid mineral found in the earth’s crust, not a constituent of the atmosphere at all. Carbon’s other crustal forms — diamond, graphite, carbonate rocks — are in the same position.

Question 2

Which of the following statements are usually correct for carbon compounds? These

(i) are good conductors of electricity (ii) are poor conductors of electricity (iii) have strong forces of attraction between their molecules (iv) do not have strong forces of attraction between their molecules

Solution. Both halves follow from one fact: carbon compounds are covalent.

Covalent bonding means electrons are shared, not transferred, so the molecules carry no charge and there are no free ions or free electrons anywhere. With no mobile charge carriers there is nothing to conduct a current, so carbon compounds are poor conductors — statement (ii). (Graphite is the famous exception, but graphite is an element, not a carbon compound.)

The bonds within a molecule are strong, but the forces between separate molecules are weak — statement (iv) — and it is those intermolecular forces that have to be overcome to melt or boil a substance. That is why carbon compounds generally have low melting and boiling points, in sharp contrast to the ionic compounds of the previous unit.

Keeping the two apart is the point of the question. Strong bonds inside the molecule, weak attraction between molecules.

Question 3

A molecule of ammonia ($\mathrm{NH_3}$) has

Solution. Count what each atom needs.

Nitrogen has five outer electrons and needs three more to complete its octet. Each hydrogen has one electron and needs one more. So nitrogen shares one electron pair with each of three hydrogen atoms:

$$\mathrm{H} – \mathrm{N} – \mathrm{H}, \quad \text{with a third } \mathrm{H} \text{ and a lone pair on the nitrogen}$$

Three single bonds, and one lone pair left over on the nitrogen — a total of eight electrons around it.

A double or triple bond to hydrogen is impossible, and that is the quickest way to see the answer. Hydrogen has only one electron and needs only one, so it can never share more than a single pair. Any option claiming a double or triple bond in a compound of hydrogen is wrong before the other atom is considered at all, which disposes of (b), (c) and (d) together.

Question 4

Buckminsterfullerene is an allotropic form of

Solution. Buckminsterfullerene is $\mathrm{C_{60}}$ — sixty carbon atoms arranged in a hollow cage of twenty hexagons and twelve pentagons, exactly the pattern of a football. It is an allotrope of carbon, the third the syllabus names alongside diamond and graphite.

The name is worth knowing because it explains itself: the cage resembles the geodesic domes designed by the architect Buckminster Fuller.

Phosphorus and sulphur do have allotropes of their own — white and red phosphorus, rhombic and monoclinic sulphur — and tin has grey and white forms, so none of the other options is absurd. But the specific name belongs to carbon.

The three carbon allotropes make a useful trio, because their properties differ so widely despite being the same element: diamond is the hardest natural substance and an insulator, graphite is soft and conducts electricity, and fullerene is a discrete molecule rather than a giant structure.

Question 5

Which of the following are correct structural isomers of butane?

Solution. Structural isomers of butane must satisfy two conditions at once: the molecular formula must be $\mathrm{C_4H_{10}}$, and every carbon must have exactly four bonds. Check each structure against both.

(i) Four carbons in a row. The two end carbons carry three hydrogens each and the two middle carbons two each: $3 + 2 + 2 + 3 = 10$ hydrogens, so $\mathrm{C_4H_{10}}$, and every carbon has four bonds. This is n-butane, and it is valid.

(ii) Three carbons in a row with a fourth hanging below the middle one. The central carbon is bonded to three carbons and one hydrogen — four bonds — and each of the three outer carbons carries three hydrogens: $3 + 1 + 3 + 3 = 10$. Again $\mathrm{C_4H_{10}}$. This is 2-methylpropane (isobutane), the only other structural isomer of butane, and it is valid.

(iii) Count the bonds at the branch carbon, drawn at the lower right. It has one bond up to the chain, one hydrogen to its right and one below — three bonds. A carbon with three bonds is impossible, so this structure is not a molecule at all. And even if the missing hydrogen were supplied, the skeleton would be a three-carbon chain with a methyl branch, which is 2-methylpropane again — the same compound as (ii), not a third isomer.

(iv) Four carbons joined in a ring, each carrying two hydrogens. That gives $\mathrm{C_4H_8}$, not $\mathrm{C_4H_{10}}$. Cyclobutane is a real compound, but a different molecular formula means it is not an isomer of butane at all.

So (i) and (ii), option (c) — and it is worth noting that butane has exactly two structural isomers and no more, which is why no option offering three could have been right.

Question 6

$$\mathrm{CH_3-CH_2-OH} \xrightarrow{\text{Alkaline } \mathrm{KMnO_4} + \text{Heat}} \mathrm{CH_3-COOH}$$

In the above given reaction, alkaline $\mathrm{KMnO_4}$ acts as

Solution. Compare the two organic formulae atom by atom. Ethanol is $\mathrm{C_2H_6O}$ and ethanoic acid is $\mathrm{C_2H_4O_2}$: the molecule has gained an oxygen and lost two hydrogens. Both of those are oxidation, so the ethanol has been oxidised, and whatever did it is the oxidising agent.

Option (a) is the reverse and can be dismissed on the same evidence. Option (d) is wrong because a dehydrating agent removes water — that is what concentrated sulphuric acid does to ethanol, giving ethene, a different reaction entirely.

Option (c) is the subtle one. A catalyst speeds a reaction up without being consumed; the permanganate here is consumed, being reduced as it oxidises the alcohol, and the purple colour disappears as it goes. You can see it being used up, which is the practical test.

Alkaline potassium permanganate and acidified potassium dichromate are the two oxidising agents this chapter uses, and both convert an alcohol to the corresponding carboxylic acid. Because they add oxygen to the molecule, they are often called oxidising agents in this context.

Question 7

Oils on treating with hydrogen in the presence of palladium or nickel catalyst form fats. This is an example of

Solution. Vegetable oils are unsaturated — their carbon chains contain double bonds — and fats are saturated, with only single bonds. Hydrogen is added across those double bonds, and nothing leaves the molecule:

$$\mathrm{-CH=CH-} + \mathrm{H_2} \xrightarrow{\text{Ni}} \mathrm{-CH_2-CH_2-}$$

A reaction in which atoms are added to a molecule without anything being removed is an addition reaction, and it is the characteristic reaction of unsaturated compounds.

Substitution, option (b), is the opposite arrangement — an atom in the molecule is replaced by another and something leaves. That is the characteristic reaction of saturated compounds, and Question 13’s chlorination of an alkane is the example.

The clue that identifies an addition is the catalyst: nickel or palladium is required, and those are the hydrogenation catalysts. Question 7 and Question 13 are a deliberate pair — unsaturated adds, saturated substitutes.

This particular reaction is hydrogenation, used industrially to turn vegetable oil into vanaspati.

Question 8

In which of the following compounds, — OH is the functional group?

Solution. All four names begin with butan-, so all four have a four-carbon chain and the stem tells you nothing. The suffix carries the functional group, and this question is really a test of those four endings:

Suffix	Functional group	Name of the class
-ol	$\mathrm{-OH}$	alcohol
-al	$\mathrm{-CHO}$	aldehyde
-one	$\mathrm{>C=O}$	ketone
-oic acid	$\mathrm{-COOH}$	carboxylic acid

So butanol is the alcohol, and its functional group is the hydroxyl group $\mathrm{-OH}$.

Butanoic acid is the near miss and is worth a sentence, because its $\mathrm{-COOH}$ group does contain an $\mathrm{O}$ and an $\mathrm{H}$. But the functional group there is the whole carboxyl group, not a hydroxyl; the behaviour of the compound — acidity, ester formation — comes from the carboxyl as a unit, and it is nothing like an alcohol’s.

Butanal has $\mathrm{-CHO}$ at the end of the chain; butanone has the carbonyl in the middle.

Question 9

The soap molecule has a

Solution. A soap is the sodium salt of a long-chain carboxylic acid, $\mathrm{CH_3(CH_2)_{16}COO^-Na^+}$, and it has two ends with opposite affinities — which is precisely what makes it work.

The head is the ionic $\mathrm{-COO^-Na^+}$ group. Being charged, it is attracted to the polar water molecules, so it is hydrophilic — water-loving. This is the short end and is conventionally drawn as the head.

The tail is the long hydrocarbon chain. Being non-polar, it is repelled by water but dissolves readily in oil and grease, so it is hydrophobic — water-hating.

Options (c) and (d) have both ends the same and can be rejected without any chemistry: a molecule with the same affinity at both ends could not bridge oil and water, and bridging them is the entire function of a detergent. Option (b) has the two the right way round but attached to the wrong ends.

This is what makes a micelle possible — Question 14 — with the tails buried in the grease and the ionic heads facing out into the water.

Question 10

Which of the following is the correct representation of electron dot structure of nitrogen?

Solution. Work out what the structure must look like before comparing.

Nitrogen has five valence electrons and needs eight, so it must gain a share in three more. Two nitrogen atoms achieve that by sharing three pairs between them — a triple bond — which leaves each atom with one lone pair of its own:

$$\mathrm{:N} \equiv \mathrm{N:}$$

That gives two checks to apply. The bond must be triple, and the total number of valence electrons must be ten, five from each atom.

Option (d) has three pairs between the atoms and one lone pair on each: $6 + 2 + 2 = 10$ electrons, with each nitrogen surrounded by eight. Both checks pass.

Option (a) has only one shared pair, a single bond, so neither nitrogen reaches an octet even though the electron count happens to come to ten. Option (b) has two shared pairs, a double bond, and again leaves each nitrogen one electron short. Option (c) has a single bond and nine electrons in all, so it fails the count as well.

The electron count is the faster of the two tests, but the octet check is the one that separates (a) and (b) from the answer, so both are worth doing.

Question 11

Structural formula of ethyne is

Solution. The name tells you the structure. The stem eth- means two carbon atoms, and the suffix -yne means a carbon–carbon triple bond. So ethyne is two carbons joined by a triple bond, with one hydrogen on each to complete the four bonds:

which is option (a), $\mathrm{C_2H_2}$.

Option (b) is impossible as written. $\mathrm{H_3}$ is not a chemical fragment: hydrogen has one electron and forms exactly one bond, so three hydrogen atoms cannot join into a group that then bonds to carbon. A subscript on a hydrogen attached to a single bond is always an error.

Option (c) is ethene, $\mathrm{C_2H_4}$ — two carbons joined by a double bond with two hydrogens each. Right number of carbons, wrong bond.

Option (d) is ethane, $\mathrm{C_2H_6}$ — a single bond with three hydrogens on each carbon.

So (b), (c) and (d) between them cover the three ways of getting it wrong: an impossible formula, and the two other members of the two-carbon family. The suffix is the discriminator: -ane single, -ene double, -yne triple.

Question 12

Identify the unsaturated compounds from the following

(i) Propane    (ii) Propene    (iii) Propyne    (iv) Chloropropane

Solution. A compound is unsaturated if it contains at least one carbon–carbon double or triple bond. All four compounds here have three carbons, so once again the suffix decides everything.

• Propane, $\mathrm{C_3H_8}$ — the -ane ending means all single bonds. Saturated.
• Propene, $\mathrm{C_3H_6}$ — -ene means a double bond. Unsaturated.
• Propyne, $\mathrm{C_3H_4}$ — -yne means a triple bond. Unsaturated.
• Chloropropane, $\mathrm{C_3H_7Cl}$ — the stem is still propane, so the carbon skeleton has only single bonds. Saturated.

So (ii) and (iii), option (d).

Chloropropane is the interesting distractor. It is a substituted hydrocarbon — a chlorine has replaced a hydrogen — and that might loosely be described as not being a plain hydrocarbon, but it changes nothing about the bonds between the carbons. Saturation is a statement about carbon–carbon bonds only, and a substituent does not affect it.

Question 13

Chlorine reacts with saturated hydrocarbons at room temperature in the

Solution. Saturated hydrocarbons are notably unreactive — every bond is a strong single bond and there is no double bond for a reagent to attack. Their characteristic reaction is substitution, in which a chlorine atom replaces a hydrogen:

$$\mathrm{CH_4} + \mathrm{Cl_2} \xrightarrow{\text{Sunlight}} \mathrm{CH_3Cl} + \mathrm{HCl}$$

and it needs energy to get started, because the chlorine molecule has to be split into atoms before anything can happen. Sunlight supplies exactly that: ultraviolet light breaks the $\mathrm{Cl-Cl}$ bond, and the chlorine atoms produced then attack the alkane.

That is why the reaction is described as photochemical, and why it does not proceed in the dark — which disposes of option (a) directly.

Water and hydrochloric acid, options (c) and (d), do nothing to help. Neither supplies the energy needed, and hydrogen chloride is in fact a product of the reaction rather than a condition for it.

Notice the pairing with Question 7 again: an unsaturated compound adds with a metal catalyst, a saturated one substitutes in sunlight.

Question 14

In the soap micelles

Solution. A micelle forms in water, and that fact fixes the arrangement completely. Each end of the soap molecule goes where it is comfortable.

The ionic end is hydrophilic, so it faces outwards, into the surrounding water. The hydrocarbon tail is hydrophobic, so it turns inwards, away from the water, and the tails cluster together in the centre.

That is option (a), and it is the arrangement that makes soap work: the oily dirt dissolves in the hydrocarbon interior of the micelle, while the charged exterior keeps the whole cluster suspended in the water so it can be rinsed away.

Option (b) reverses both ends and would put the water-hating chains in contact with the water. Options (c) and (d) put both ends in the same place, which is impossible for a molecule whose two ends have opposite affinities — the same objection that disposed of options (c) and (d) in Question 9.

A detail worth knowing: because every micelle carries the same negative charge on its surface, micelles repel each other and do not coalesce, which is why the dirt stays suspended rather than settling back onto the cloth.

Question 15

Pentane has the molecular formula $\mathrm{C_5H_{12}}$. It has

Solution. Count the two kinds of bond separately.

Carbon–hydrogen bonds. Each of the twelve hydrogen atoms forms exactly one bond, so there are 12 of these.

Carbon–carbon bonds. Five carbon atoms joined in a chain need 4 links — one fewer than the number of atoms, the same way five fence posts in a line need four rails.

$$12 + 4 = 16$$

Option (c).

There is a second way to get there that also checks the answer. Every carbon forms four bonds and every hydrogen one, so the total number of bond ends is $5 \times 4 + 12 \times 1 = 32$; each bond has two ends, so there are $32 \div 2 = 16$ bonds. Two independent routes to the same number is worth the extra few seconds.

Option (b) counts only the hydrogens and forgets the carbon skeleton. Option (d) is the count you get from assuming five carbons need five links, which over-counts the chain by one — and it is the answer a ring would give, not a chain.

Note that the answer is the same for any isomer of $\mathrm{C_5H_{12}}$, branched or straight, since branching rearranges the four carbon–carbon bonds without changing how many there are.

Common mistakes

• Not counting bonds at every carbon, in Question 5. Structure (iii) there has a carbon with only three bonds, which makes it not a molecule at all. Tetravalency is the first test to apply to any offered structure, before considering isomerism.
• Treating cyclobutane as an isomer of butane, in Question 5. It is $\mathrm{C_4H_8}$, not $\mathrm{C_4H_{10}}$. Isomers must have the same molecular formula; a ring costs two hydrogens.
• Accepting $\mathrm{H_3}$ as a fragment, in Question 11. Hydrogen forms exactly one bond, so a subscript on a hydrogen bonded by a single line is always an error.
• Calling the permanganate a catalyst, in Question 6. It is consumed — the purple colour fades as it is reduced — so it is a reagent, specifically the oxidising agent.
• Confusing $\mathrm{-OH}$ with $\mathrm{-COOH}$, in Question 8. Butanoic acid contains an O and an H but its functional group is the whole carboxyl group, and it behaves nothing like an alcohol.
• Counting only the C–H bonds in Question 15. A chain of $n$ carbons has $n – 1$ carbon–carbon bonds. Forgetting them gives 12; assuming there are five gives 17.
• Calling chloropropane unsaturated, in Question 12. Saturation is about carbon–carbon bonds only; a substituent replaces a hydrogen and leaves the skeleton saturated.
• Reversing the soap molecule, in Questions 9 and 14. The ionic end is the water-loving one and faces outwards; the hydrocarbon tail is water-hating and hides in the middle. Options putting both ends in the same place can be rejected without any chemistry at all.

Practise next

• Exemplar Multiple Choice Questions Q16–29 — the second half of this band, on benzene, nomenclature, homologous series and the reactions of ethanol and ethanoic acid.
• Exemplar Short and Long Answer Questions — where the structures of Questions 5, 10 and 11 have to be drawn rather than recognised.
• NCERT textbook Chapter 4 — Carbon and its Compounds for the end-of-chapter exercise on bonding, functional groups and soaps.
• Textbook activities for the experiments behind Questions 9 and 14 — the behaviour of soap in hard and soft water is observed there rather than described.