Class 10 Science · Chapter 4

Carbon and its Compounds

201 Class 10 Science questions on Carbon and its Compounds from the NCERT textbook, NCERT Exemplar and CBSE board papers 2024–2026, each with its answer and a step-by-step solution.

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Questions · page 4 of 11

1. Q61 Exemplar · Short Answer Questions
   A compound X is formed by the reaction of a ​carboxylic acid ${\text{C}}_2{\text{H}}_4{\text{O}}_2$ and an alcohol in presence of a few drops of ${\text{H}}_2{\text{SO}}_4$. The alcohol on oxidation ​with ​alkaline ${\text{KMnO}}_4$ followed by acidification gives the same carboxylic acid as used in this ​reaction. Give the names and structures of

   1. (a)carboxylic acid, ​
   2. (b)alcohol and
   3. (c)the compound ​X. Also write the reaction.
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   (a) Ethanoic acid, (b) ethanol and (c) ethyl ethanoate, formed when the acid and alcohol react with conc. sulphuric acid.

   (a) The carboxylic acid is ethanoic acid, ${\text{CH}}_3\text{COOH}$, structure ${\text{H}}_3\text{C}-\text{COOH}$ with the carboxyl carbon carrying a doubly bonded oxygen and an $-\text{OH}$ group; (b) The alcohol is ethanol, ${\text{CH}}_3{\text{CH}}_2\text{OH}$, structure ${\text{H}}_3\text{C}-{\text{CH}}_2-\text{OH}$; (c) The compound X is ethyl ethanoate, ${\text{CH}}_3{\text{COOC}}_2{\text{H}}_5$, structure ${\text{H}}_3\text{C}-\text{CO}-\text{O}-{\text{CH}}_2-{\text{CH}}_3$. The reaction is ${\text{CH}}_3\text{COOH}+{\text{CH}}_3{\text{CH}}_2\text{OH}\stackrel{{\text{conc. H}}_2{\text{SO}}_4}{\to }{\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+{\text{H}}_2\text{O}$.

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2. Q62 Exemplar · Short Answer Questions
   Why ​detergents are ​better ​cleansing ​agents than soaps? ​Explain.
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   The calcium and magnesium salts of detergents are soluble, so they lather and clean in hard water, whereas soap forms insoluble scum.

   Because their calcium and magnesium salts are soluble. Soap is the sodium salt of a long-chain carboxylic acid, and in hard water the ${\text{Ca}}^{2+}$ and ${\text{Mg}}^{2+}$ ions convert it into insoluble calcium and magnesium salts, $2\text{RCOONa}+{\text{Ca}}^{2+}\to (\text{RCOO}{)}_2\text{Ca}\downarrow +2{\text{Na}}^{+}$, which separate as a curdy scum. The soap is used up before it can clean, no lather forms until all the ions have been precipitated, and the scum settles on the cloth. A detergent is the ammonium or sulphonate salt of a long-chain carboxylic acid, and its calcium and magnesium salts stay dissolved, so it lathers and cleans in hard water without any waste. A detergent also works in slightly acidic water, in which soap, being the salt of a weak acid, would be turned back into the free fatty acid.

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3. Q63 Exemplar · Short Answer Questions
   Name ​the functional groups present in the following ​compounds ​

   1. (a)${\text{CH}}_3\text{CO}{\text{CH}}_2{\text{CH}}_2{\text{CH}}_2{\text{CH}}_3$ ​
   2. (b)${\text{CH}}_3{\text{CH}}_2{\text{CH}}_2\text{COOH}$ ​
   3. (c)${\text{CH}}_3{\text{CH}}_2{\text{CH}}_2{\text{CH}}_2\text{CHO}$ ​
   4. (d)${\text{CH}}_3{\text{CH}}_2\text{OH}$
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   (a) Ketone, (b) carboxylic acid, (c) aldehyde, (d) alcohol.

   (a) Ketone, the carbonyl group $>\text{C}=\text{O}$ inside the chain (hexan-2-one); (b) Carboxylic acid, the carboxyl group $-\text{COOH}$ (butanoic acid); (c) Aldehyde, the group $-\text{CHO}$ (pentanal); (d) Alcohol, the hydroxyl group $-\text{OH}$ (ethanol).

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4. Q64 Exemplar · Short Answer Questions
   How is ethene ​prepared ​from ​ethanol? ​Give the reaction ​involved in ​it.
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   Ethanol is dehydrated by heating at 443 K with excess concentrated sulphuric acid, giving ethene and water.

   Ethene is prepared by the dehydration of ethanol: ethanol is heated at 443 K with excess concentrated sulphuric acid, which removes a molecule of water from it. ${\text{CH}}_3{\text{CH}}_2\text{OH}\underset{443\text{K}}{\overset{{\text{conc. H}}_2{\text{SO}}_4}{\to }}{\text{CH}}_2={\text{CH}}_2+{\text{H}}_2\text{O}$

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5. Q65 Exemplar · Short Answer Questions
   Intake ​of small ​quantity ​of ​methanol can be ​lethal. ​Comment.
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   The body oxidises methanol to methanal, which coagulates cell protein and attacks the optic nerve, causing blindness; methanoic acid then may cause death.

   Methanol itself is not the poison; what the body makes from it is. In the liver methanol is oxidised to methanal, ${\text{CH}}_3\text{OH}\to \text{HCHO}$. Methanal reacts very rapidly with the components of the cell protoplasm and coagulates their protein, killing the cells; the optic nerve is attacked in particular, so blindness follows even after a small quantity. Methanal is oxidised further to methanoic acid, $\text{HCOOH}$, which makes the blood strongly acidic, and death can follow. For this reason the ethanol supplied for industrial use is deliberately made unfit to drink by adding methanol and a dye to it, giving denatured spirit.

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6. Q66 Exemplar · Short Answer Questions
   A gas is evolved ​when ethanol reacts ​with sodium. Name ​the gas ​evolved ​and also write ​the balanced chemical equation ​of the reaction ​involved.
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   Hydrogen is evolved: $2{\text{CH}}_3{\text{CH}}_2\text{OH}+2\text{Na}\to 2{\text{CH}}_3{\text{CH}}_2\text{ONa}+{\text{H}}_2$

   The gas evolved is hydrogen, ${\text{H}}_2$; it burns with a pop. Sodium displaces the hydrogen of the hydroxyl group and takes its place, giving sodium ethoxide. $2{\text{CH}}_3{\text{CH}}_2\text{OH}+2\text{Na}\to 2{\text{CH}}_3{\text{CH}}_2\text{ONa}+{\text{H}}_2$

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7. Q67 Exemplar · Short Answer Questions
   Ethene is formed when ​ethanol at ​443 K is heated ​with excess of concentrated ​sulphuric acid. What is ​the role of ​sulphuric ​acid in this reaction? Write the balanced chemical ​equation of this reaction.
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   Concentrated sulphuric acid is a dehydrating agent that removes water from ethanol to give ethene.

   Concentrated sulphuric acid acts as a dehydrating agent: it removes a molecule of water from the ethanol and holds it, converting the alcohol into the alkene. It is not consumed, so it is written above the arrow. ${\text{CH}}_3{\text{CH}}_2\text{OH}\underset{443\text{K}}{\overset{{\text{conc. H}}_2{\text{SO}}_4}{\to }}{\text{CH}}_2={\text{CH}}_2+{\text{H}}_2\text{O}$

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8. Q68 Exemplar · Short Answer Questions
   Carbon, Group (14) element ​in the Periodic Table, is ​known to form compounds with many elements. Write an ​example of ​a ​compound ​formed with

   1. (a)chlorine (Group 17 of Periodic Table)
   2. (b)oxgygen (Group ​16 of Periodic ​Table)
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   (a) Carbon tetrachloride, ${\text{CCl}}_4$; (b) carbon dioxide, ${\text{CO}}_2$.

   (a) With chlorine: carbon tetrachloride, ${\text{CCl}}_4$. Chlorine is in group 17 and has a valency of one, so four chlorine atoms are needed to use up carbon's four bonds, and the carbon shares one electron pair with each. (b) With oxygen: carbon dioxide, ${\text{CO}}_2$, written $\text{O}=\text{C}=\text{O}$. Oxygen is in group 16 and has a valency of two, so two oxygen atoms are enough, each joined to the carbon by a double bond.

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9. Q69 Exemplar · Short Answer Questions
   In electron dot structure, the valence shell electrons ​are represented by crosses ​or dots. ​

   1. (a)The atomic number of chlorine is ​17. ​Write its electronic configuration ​
   2. (b)Draw the ​electron dot structure of chlorine molecule.
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   Chlorine is 2, 8, 7; in ${\text{Cl}}_2$ the two atoms share one pair, so each has 1 shared and 3 lone pairs (8 electrons).

   (a) Chlorine has 17 electrons, filled as $2$ in the K shell, $8$ in the L shell and $7$ in the M shell, so its electronic configuration is $2,8,7$. (b) Each chlorine atom has 7 valence electrons and is one short of an octet, so two atoms share one electron pair between them. In the electron dot structure the two chlorine symbols are joined by that one shared pair, and three lone pairs are drawn around each atom; every chlorine then has $1$ shared pair and $3$ lone pairs, that is $8$ electrons. The structural formula is $\text{Cl}-\text{Cl}$.

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10. Q70 Exemplar · Short Answer Questions
    Catenation is ​the ability of an atom ​to ​form bonds with other ​atoms of the same element. It ​is exhibited by both ​carbon and silicon. Compare the ability ​of ​catenation of the two elements. Give reasons.
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    Carbon catenates far more than silicon because its smaller atom forms short, strong C-C bonds, whereas silicon's longer Si-Si bonds are weak.

    Both elements catenate, because both are in group 14 and have four valence electrons, but they do so to very different extents. Carbon shows catenation to an enormous extent: it forms straight chains, branched chains and closed rings of practically any length, and this is why there are millions of carbon compounds. Silicon catenates only slightly, forming chains of about seven or eight atoms at most, and those chains are readily attacked by oxygen of the air and by water. The reason is the size of the atom. A carbon atom is much smaller than a silicon atom, so the two nuclei in a $\text{C}-\text{C}$ bond are closer together and hold the shared electron pair more firmly; the bond is short and strong. Silicon atoms are larger, so the $\text{Si}-\text{Si}$ bond is longer, the shared pair is held more loosely and the bond is much weaker. Strong bonds keep carbon chains together and stable; weak ones let silicon chains break.

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11. Q71 Exemplar · Short Answer Questions
    Unsaturated hydrocarbons contain ​multiple bonds ​between the ​two C-atoms ​and show addition reactions. ​Give the test to ​distinguish ethane ​from ​ethene.
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    Pass each gas through bromine water: ethene decolourises it, but ethane leaves it orange-red.

    Use the bromine water test. Take orange-red bromine water in two test tubes and pass ethene through one and ethane through the other. Ethene decolourises the bromine water, because it is unsaturated and adds the bromine across its carbon-carbon double bond: ${\text{CH}}_2={\text{CH}}_2+{\text{Br}}_2⟶{\text{CH}}_2\text{Br}-{\text{CH}}_2\text{Br}$ (1,2-dibromoethane, colourless). Ethane is saturated, has no double bond to add across, and leaves the bromine water orange-red. The gas that discharges the colour is therefore ethene, and the one that does not is ethane.

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12. Q72 Exemplar · Short Answer Questions
    Match ​the reactions given ​in Column ​(A) with the names ​given ​in column ​(B).
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    (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii). That is, the reaction of methanol with ethanoic acid is an esterification reaction, the reaction of ethene with hydrogen over nickel is an addition reaction, the reaction of methane with chlorine in sunlight is a substitution reaction, and the reaction of ethanoic acid with sodium hydroxide is a neutralisation reaction.

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13. Q73 Exemplar · Short Answer Questions
    Write ​the ​structural ​formulae of all ​the isomers ​of ​hexane.
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    Hexane has 5 isomers: n-hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane and 2,3-dimethylbutane.

    Hexane, ${\text{C}}_6{\text{H}}_{14}$, has five isomers. (i) ${\text{CH}}_3-{\text{CH}}_2-{\text{CH}}_2-{\text{CH}}_2-{\text{CH}}_2-{\text{CH}}_3$, n-hexane; (ii) ${\text{CH}}_3-\text{CH}({\text{CH}}_3)-{\text{CH}}_2-{\text{CH}}_2-{\text{CH}}_3$, 2-methylpentane, a five-carbon chain with a methyl group on carbon 2; (iii) ${\text{CH}}_3-{\text{CH}}_2-\text{CH}({\text{CH}}_3)-{\text{CH}}_2-{\text{CH}}_3$, 3-methylpentane, a five-carbon chain with a methyl group on carbon 3; (iv) ${\text{CH}}_3-\text{C}({\text{CH}}_3{)}_2-{\text{CH}}_2-{\text{CH}}_3$, 2,2-dimethylbutane, a four-carbon chain with two methyl groups on carbon 2; (v) ${\text{CH}}_3-\text{CH}({\text{CH}}_3)-\text{CH}({\text{CH}}_3)-{\text{CH}}_3$, 2,3-dimethylbutane, a four-carbon chain with a methyl group on carbon 2 and another on carbon 3.

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14. Q74 Exemplar · Short Answer Questions
    What ​is ​the role ​of metal or reagents written ​on arrows in the ​given ​chemical reactions?

    1. (a)$({\text{CH}}_3{)}_2\text{C}=\text{C}({\text{CH}}_3{)}_2+{\text{H}}_2\stackrel{\text{Ni}}{\to }({\text{CH}}_3{)}_2\text{CH}-\text{CH}({\text{CH}}_3{)}_2$ ​
    2. (b)${\text{CH}}_3\text{COOH}+{\text{CH}}_3{\text{CH}}_2\text{OH}\stackrel{{\text{Conc. H}}_2{\text{SO}}_4}{\to }{\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+{\text{H}}_2\text{O}$
    3. (c)${\text{CH}}_3{\text{CH}}_2\text{OH}\underset{\text{Heat}}{\overset{{\text{Alk. KMnO}}_4}{\to }}{\text{CH}}_3\text{COOH}$
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    Ni is a catalyst; conc. sulphuric acid is a catalyst and dehydrating agent; alkaline KMnO4 is an oxidising agent.

    (a) Nickel is a catalyst. It speeds up the addition of hydrogen across the carbon-carbon double bond and is recovered unchanged, which is why it is written above the arrow and appears on neither side of the equation. (b) Concentrated sulphuric acid is a catalyst and, above all, a dehydrating agent. It catalyses the esterification and absorbs the water formed, so the ester is not hydrolysed back and the yield is high. (c) Alkaline potassium permanganate is an oxidising agent. It supplies the oxygen that converts the $-{\text{CH}}_2\text{OH}$ group of ethanol into the $-\text{COOH}$ group of ethanoic acid, and is itself reduced, its purple colour fading.

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15. Q75 Exemplar · Long Answer Questions
    A salt X is formed and a gas is evolved when ethanoic acid ​reacts with sodium ​hydrogencarbonate. Name the salt X and the gas evolved. Describe an activity and draw the ​diagram of the apparatus ​to prove that the ​evolved gas is ​the one which you have ​named. Also, write chemical equation of ​the reaction involved.
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    X is sodium ethanoate and the gas is carbon dioxide, shown by lime water turning milky.

    The salt X is sodium ethanoate (sodium acetate), ${\text{CH}}_3\text{COONa}$, and the gas evolved is carbon dioxide, ${\text{CO}}_2$. Activity: take about 1 g of sodium hydrogencarbonate in a test tube A clamped to a stand and close it with a cork carrying a thistle funnel and a delivery tube. Pour about 2 mL of ethanoic acid down the thistle funnel, whose lower end dips below the liquid so that no gas escapes back through it; a brisk effervescence begins. Lead the delivery tube into a second test tube B containing freshly prepared lime water. The lime water turns milky, which proves that the gas is carbon dioxide; if the gas is passed in excess the milkiness disappears again as soluble calcium hydrogencarbonate is formed. The reactions are ${\text{CH}}_3\text{COOH}+{\text{NaHCO}}_3⟶{\text{CH}}_3\text{COONa}+{\text{H}}_2\text{O}+{\text{CO}}_2$ in tube A, and ${\text{Ca(OH)}}_2+{\text{CO}}_2\to {\text{CaCO}}_3\downarrow +{\text{H}}_2\text{O}$ in tube B.

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16. Q76 Exemplar · Long Answer Questions
    1. (a)What are hydrocarbons? Give examples.
    2. (b)Give the structural ​differences between ​saturated and ​unsaturated hydrocarbons with two ​examples each.
    3. (c)What ​is a functional group? ​Give examples of ​four different functional groups.
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    Hydrocarbons contain only carbon and hydrogen; saturated ones have single bonds, unsaturated ones a double or triple bond; functional groups give characteristic properties.

    (a) Hydrocarbons are compounds made up of carbon and hydrogen only, with no other element present. Examples: methane ${\text{CH}}_4$, ethane ${\text{C}}_2{\text{H}}_6$, ethene ${\text{C}}_2{\text{H}}_4$, ethyne ${\text{C}}_2{\text{H}}_2$ and benzene ${\text{C}}_6{\text{H}}_6$. (b) In a saturated hydrocarbon every carbon-carbon bond is a single bond, so each carbon carries as many hydrogen atoms as it can; general formula ${\text{C}}_n{\text{H}}_{2n+2}$; examples methane ${\text{CH}}_4$ and ethane ${\text{CH}}_3-{\text{CH}}_3$. In an unsaturated hydrocarbon at least one pair of carbon atoms is joined by a double or a triple bond, so those carbons carry fewer hydrogen atoms; examples ethene ${\text{CH}}_2={\text{CH}}_2$, an alkene of formula ${\text{C}}_n{\text{H}}_{2n}$, and ethyne $\text{HC}\equiv \text{CH}$, an alkyne of formula ${\text{C}}_n{\text{H}}_{2n-2}$. Saturated hydrocarbons give substitution reactions and burn with a clean blue flame; unsaturated ones give addition reactions, decolourise bromine water and burn with a yellow sooty flame. (c) A functional group is an atom or group of atoms present in a molecule that gives the compound its characteristic chemical properties, whatever the length of its carbon chain. Four examples: the alcohol group $-\text{OH}$, as in ethanol; the aldehyde group $-\text{CHO}$, as in ethanal; the ketone group $>\text{C}=\text{O}$, as in propanone; and the carboxylic acid group $-\text{COOH}$, as in ethanoic acid.

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17. Q77 Exemplar · Long Answer Questions
    Name the ​reaction which is commonly used in the ​conversion ​of ​vegetable oils to fats. ​Explain ​the reaction involved ​in detail.
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    Hydrogenation: hydrogen adds across the double bonds of vegetable oil with a nickel catalyst at about 473 K, giving a solid saturated fat.

    The reaction is hydrogenation, an addition reaction; industrially it is called the hardening of oils. Vegetable oils are esters of glycerol with long-chain carboxylic acids whose chains contain carbon-carbon double bonds, so they are unsaturated and are liquid at room temperature. When hydrogen gas is bubbled through the oil in the presence of a catalyst of finely divided nickel at about 473 K, hydrogen adds across each double bond, one atom going to each of the two carbon atoms, and the double bond becomes a single bond: $\text{R}-\text{CH}=\text{CH}-{\text{R}}^{\prime }+{\text{H}}_2\underset{473\text{K}}{\overset{\text{Ni}}{\to }}\text{R}-{\text{CH}}_2-{\text{CH}}_2-{\text{R}}^{\prime }$. Nothing leaves the molecule, which is why this is an addition and not a substitution. Once the chains are saturated the substance is solid at room temperature; the product is vanaspati ghee. The nickel is a catalyst and is recovered unchanged.

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18. Q78 Exemplar · Long Answer Questions
    1. (a)Write ​the formula and draw electron dot ​structure ​of carbon tetrachloride.
    2. (b)​What is saponification? ​Write ​the ​reaction involved in this process.
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    ${\text{CCl}}_4$ has carbon sharing a pair with each of four chlorines; saponification is hydrolysis of an ester by an alkali, giving soap from fats.

    (a) The formula of carbon tetrachloride is ${\text{CCl}}_4$. Carbon has 4 valence electrons and each chlorine has 7, so the molecule holds $4+4\times 7=32$ valence electrons, that is 16 pairs. In the electron dot structure the carbon atom is at the centre and shares one electron pair with each of the four chlorine atoms, which uses 4 pairs; the remaining 12 pairs are drawn as 3 lone pairs on each chlorine atom. The carbon then has 4 shared pairs, that is 8 electrons, and each chlorine has 1 shared pair and 3 lone pairs, also 8 electrons, so every atom has an octet. (b) Saponification is the hydrolysis of an ester by an alkali: an ester heated with a solution of sodium hydroxide gives the sodium salt of the carboxylic acid together with the alcohol. ${\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+\text{NaOH}\stackrel{\text{heat}}{\to }{\text{CH}}_3\text{COONa}+{\text{C}}_2{\text{H}}_5\text{OH}$. When the ester is a fat or a vegetable oil, an ester of glycerol with long-chain carboxylic acids, the salt formed is soap and the alcohol is glycerol, which is how the process gets its name: fat or oil $+3\text{NaOH}\to$ soap $+$ glycerol.

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19. Q79 Exemplar · Long Answer Questions
    Esters ​are ​sweet-smelling substances and are ​used in making perfumes. Suggest ​some activity and the ​reaction involved for ​the preparation of an ester with well ​labeled diagram.
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    Warm ethanol and glacial ethanoic acid with conc. sulphuric acid in a water bath; a fruity smell shows ethyl ethanoate formed.

    Activity: take about 1 mL of ethanol and 1 mL of glacial ethanoic acid in a clean dry test tube and add a few drops of concentrated sulphuric acid. Clamp the tube in a beaker of water so that the water surrounds the mixture, and warm this water bath over a burner for about five minutes; the tube is never heated directly in the flame, because ethanol and the ester are volatile and highly inflammable. Then pour the contents into a beaker holding 20 to 50 mL of water and smell it: a sweet fruity smell shows that the ester ethyl ethanoate has been formed. The reaction is ${\text{CH}}_3\text{COOH}+{\text{CH}}_3{\text{CH}}_2\text{OH}\stackrel{{\text{conc. H}}_2{\text{SO}}_4}{\to }{\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+{\text{H}}_2\text{O}$, with the concentrated sulphuric acid acting as catalyst and dehydrating agent.

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20. Q80 Exemplar · Long Answer Questions
    A compound ​C (molecular formula, ${\text{C}}_2{\text{H}}_4{\text{O}}_2$) reacts with Na - metal to form ​a compound ​R and evolves a gas which burns with a pop ​sound. Compound C on treatment with an alcohol A in presence of an acid forms a sweet smelling compound S (molecular formula, ${\text{C}}_3{\text{H}}_6{\text{O}}_2$). On addition of NaOH to C, it ​also gives R and water. S on treatment with NaOH solution gives ​back R and A. ​Identify C, R, A, S and ​write down the reactions involved.
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    C is ethanoic acid, R sodium ethanoate, A methanol and S methyl ethanoate, with four reactions written from the clues.

    C is ethanoic acid, ${\text{CH}}_3\text{COOH}$; R is sodium ethanoate, ${\text{CH}}_3\text{COONa}$; A is methanol, ${\text{CH}}_3\text{OH}$; and S is methyl ethanoate, ${\text{CH}}_3{\text{COOCH}}_3$. The reactions are: $2{\text{CH}}_3\text{COOH}+2\text{Na}\to 2{\text{CH}}_3\text{COONa}+{\text{H}}_2$; ${\text{CH}}_3\text{COOH}+{\text{CH}}_3\text{OH}\stackrel{{\text{conc. H}}_2{\text{SO}}_4}{\to }{\text{CH}}_3{\text{COOCH}}_3+{\text{H}}_2\text{O}$; ${\text{CH}}_3\text{COOH}+\text{NaOH}\to {\text{CH}}_3\text{COONa}+{\text{H}}_2\text{O}$; and ${\text{CH}}_3{\text{COOCH}}_3+\text{NaOH}\to {\text{CH}}_3\text{COONa}+{\text{CH}}_3\text{OH}$.

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