Chapter 4 · Carbon and its Compounds

NCERT Class 10 Science Chapter 4 Exercises

All 15 questions of NCERT Class 10 Science Chapter 4 Exercises (Carbon and its Compounds), in the book's order, each with its answer and a step-by-step solution.

1. Q1
   Ethane, ​with the ​molecular ​formula ${\text{C}}_2{\text{H}}_6$ has

   1. (a)6 covalent bonds.
   2. (b)7 covalent bonds.
   3. (c)8 covalent bonds.
   4. (d)9 covalent bonds.
   Show answer

   (b) 7 covalent bonds.

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2. Q2
   Butanone ​is ​a four-carbon compound with ​the ​functional ​group

   1. (a)carboxylic acid.
   2. (b)aldehyde.
   3. (c)ketone.
   4. (d)alcohol.
   Show answer

   (c) ketone.

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3. Q3
   While cooking, if the ​bottom of ​the ​vessel ​is getting blackened on ​the outside, it ​means that

   1. (a)the food is not cooked completely.
   2. (b)the fuel is not burning completely.
   3. (c)the fuel is wet.
   4. (d)the fuel is burning completely.
   Show answer

   (b) the fuel is not burning completely.

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4. Q4
   Explain the ​nature ​of ​the covalent bond using ​the ​bond formation ​in ${\text{CH}}_3\text{Cl}$.
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   A covalent bond is a shared electron pair; in ${\text{CH}}_3\text{Cl}$ carbon shares four pairs and every atom reaches a noble-gas arrangement.

   A covalent bond is formed by the **mutual sharing** of a pair of electrons, one electron contributed by each of the two atoms, so that both reach a noble-gas arrangement. Carbon ($2,4$) has four valence electrons and needs four more for an octet, but it can neither lose four electrons nor gain four; so in ${\text{CH}}_3\text{Cl}$ it shares one pair with each of the three hydrogen atoms and one pair with the chlorine atom. Each hydrogen then holds $2$ electrons (a duplet), chlorine holds $8$ (its three lone pairs together with the shared pair) and carbon holds $8$. The molecule's $4+3+7=14$ valence electrons appear in the structure as $4$ shared pairs and $3$ lone pairs. Because the electrons are shared and not transferred, no ions are formed: the bonds inside the molecule are strong, but the forces between the molecules are weak, so covalent compounds such as ${\text{CH}}_3\text{Cl}$ have low melting and boiling points and are poor conductors of electricity.

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5. Q5
   Draw the electron dot ​structures ​for

   1. (a)​ethanoic acid.
   2. (b)${\text{H}}_2\text{S}$.
   3. (c)propanone.
   4. (d)${\text{F}}_2$.
   Show answer

   Shared + lone pairs: ethanoic acid 8 + 4, ${\text{H}}_2\text{S}$ 2 + 2, propanone 10 + 2, ${\text{F}}_2$ 1 + 6; all valence electrons are shown.

   (a) Ethanoic acid, ${\text{CH}}_3\text{COOH}$ — $24$ valence electrons, that is $12$ pairs: three $\text{C}-\text{H}$ shared pairs, one $\text{C}-\text{C}$ shared pair, **two** shared pairs in the $\text{C}=\text{O}$, one $\text{C}-\text{O}$ shared pair and one $\text{O}-\text{H}$ shared pair, with two lone pairs on each of the two oxygen atoms ($8$ shared $+4$ lone). (b) Hydrogen sulphide, ${\text{H}}_2\text{S}$ — $8$ valence electrons, that is $4$ pairs: two $\text{S}-\text{H}$ shared pairs and two lone pairs on the sulphur. (c) Propanone, ${\text{CH}}_3{\text{COCH}}_3$ — $24$ valence electrons, that is $12$ pairs: six $\text{C}-\text{H}$ shared pairs, two $\text{C}-\text{C}$ shared pairs, **two** shared pairs in the $\text{C}=\text{O}$ and two lone pairs on the oxygen ($10$ shared $+2$ lone). (d) Fluorine, ${\text{F}}_2$ — $14$ valence electrons, that is $7$ pairs: one shared pair between the two atoms and three lone pairs on each of them. The four structures are drawn in the figure.

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6. Q6
   What ​is ​an homologous ​series? Explain ​with ​an ​example.
   Show answer

   A group of compounds with the same functional group and general formula, each member differing from the next by $-{\text{CH}}_2-$, such as the alkanes.

   A homologous series is a group of carbon compounds that have the **same** general formula and the **same** functional group, in which each member differs from the one before it by a $-{\text{CH}}_2-$ unit, that is by $14\text{u}$ in molecular mass. All the members can be prepared by similar methods and show similar chemical properties, because the functional group is the same throughout, while their physical properties such as melting point, boiling point and solubility change gradually as the carbon chain lengthens. Example — the alkanes, general formula ${\text{C}}_n{\text{H}}_{2n+2}$: methane ${\text{CH}}_4$ ($16\text{u}$), ethane ${\text{C}}_2{\text{H}}_6$ ($30\text{u}$), propane ${\text{C}}_3{\text{H}}_8$ ($44\text{u}$), butane ${\text{C}}_4{\text{H}}_{10}$ ($58\text{u}$) and pentane ${\text{C}}_5{\text{H}}_{12}$ ($72\text{u}$). Each member is one ${\text{CH}}_2$ unit — $14\text{u}$ — heavier than the one before it, and the first four are gases while the later members are liquids, which is the gradual change in physical properties the definition describes.

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7. Q7
   How can ​ethanol and ​ethanoic acid ​be ​differentiated on the basis ​of their ​physical and chemical ​properties?
   Show answer

   Ethanoic acid smells of vinegar, tastes sour, turns blue litmus red and fizzes with sodium carbonate; ethanol does none of these.

   **Physical properties**. Ethanol has a pleasant smell and a burning taste, melts at $156\text{K}$ and boils at $351\text{K}$, so it stays liquid in the cold. Ethanoic acid has the sharp smell of vinegar and a sour taste, freezes at $290\text{K}$ — so pure ethanoic acid solidifies to an ice-like mass in a cold room, which is why it is called glacial acetic acid — and boils at $391\text{K}$. **Chemical properties**. (i) Litmus: ethanoic acid turns blue litmus red, while ethanol leaves both papers unchanged. (ii) Sodium carbonate or sodium hydrogencarbonate: ethanoic acid gives brisk effervescence of carbon dioxide, which turns lime water milky, whereas ethanol gives no reaction at all — $2{\text{CH}}_3\text{COOH}+{\text{Na}}_2{\text{CO}}_3\to 2{\text{CH}}_3\text{COONa}+{\text{H}}_2\text{O}+{\text{CO}}_2$. (iii) Sodium hydroxide: ethanoic acid is neutralised to sodium ethanoate and water, while ethanol does not react. The carbonate test is the decisive one, because only the acid releases the gas.

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8. Q8
   Why does micelle ​formation take ​place when soap is added ​to ​water? Will a micelle be formed ​in other ​solvents such ​as ​ethanol also?
   Show answer

   In water, the hydrophobic tails cluster inside a micelle with ionic heads outside; in ethanol no micelle forms because ethanol dissolves both ends.

   Micelle formation takes place in water because a soap molecule has two ends of opposite character: a long hydrocarbon tail, which is hydrophobic and will not mix with water, and an ionic head $-{\text{COO}}^{-}{\text{Na}}^{+}$, which is hydrophilic and dissolves readily. Water takes the ionic heads into solution but will not take the tails, and the tails cannot leave the water, so the molecules arrange themselves into a spherical cluster — a micelle — with all the hydrocarbon tails pointing inwards, away from the water, and all the ionic heads on the outside in contact with it. Any oily dirt present dissolves in the hydrocarbon interior of the micelle and is carried away with it. **No**, a micelle will not be formed in ethanol. Ethanol is an organic solvent and dissolves the hydrocarbon chain just as readily as it dissolves the ionic end, so no part of the soap molecule has to be shielded from the solvent. With nothing left to hide there is no driving force for the molecules to cluster, and the soap simply dissolves in ethanol as separate molecules.

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9. Q9
   Why ​are ​carbon and ​its compounds ​used ​as ​fuels for most ​applications?
   Show answer

   They give much heat and light with a clean flame, have high calorific value, and are easy to store, transport and ignite.

   Carbon and its compounds are used as fuels because they burn in air to give carbon dioxide, water and a great deal of heat and light: ${\text{C}}_x{\text{H}}_y+{\text{O}}_2⟶{\text{CO}}_2+{\text{H}}_2\text{O}+\text{heat and light}$. Four properties make them the practical choice. (i) They have a high calorific value — about $55$ kJ per gram for LPG and $45$ for petrol, against $17$–$22$ for wood and only $6$–$8$ for a cow-dung cake — so a small mass gives a large amount of heat. (ii) With a sufficient supply of air they burn with a clean, smokeless blue flame and leave practically no ash or residue, because their products are a gas and a vapour. (iii) Most of them are liquids or easily liquefied gases, so they are easy to store in a cylinder or tank, to transport and to handle safely. (iv) They do not react with air at ordinary temperature and so must be ignited, which makes the burning easy to start, to regulate and to stop at will. They are also abundant and cheap, being available as coal, petroleum, natural gas and biomass.

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10. Q10
    Explain ​the formation ​of ​scum ​when ​hard water ​is treated with soap.
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    Calcium and magnesium ions in hard water react with soap to form an insoluble white curdy precipitate, the scum.

    Hard water contains dissolved calcium and magnesium salts, so it supplies ${\text{Ca}}^{2+}$ and ${\text{Mg}}^{2+}$ ions. Soap is the sodium salt of a long-chain carboxylic acid, and while the sodium and potassium salts of such acids are soluble in water, their calcium and magnesium salts are not. When soap is added to hard water, each doubly charged calcium or magnesium ion takes two soap anions and precipitates with them: $2{\text{C}}_{17}{\text{H}}_{35}\text{COONa}(aq)+{\text{Ca}}^{2+}(aq)⟶({\text{C}}_{17}{\text{H}}_{35}\text{COO}{)}_2\text{Ca}(s)+2{\text{Na}}^{+}(aq)$, and similarly with ${\text{Mg}}^{2+}$. This insoluble calcium or magnesium salt of the fatty acid separates out as a white curdy solid — the scum — which sticks to the cloth and to the vessel. Because the soap is used up in this precipitation, it cannot form micelles, so a great deal of soap is wasted and no lather appears until every calcium and magnesium ion has been precipitated.

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11. Q11
    What ​change ​will you ​observe ​if ​you test ​soap with litmus paper (red ​and blue)?
    Show answer

    Red litmus turns blue and blue litmus stays blue, because soap solution is basic.

    Red litmus paper turns blue, and blue litmus paper shows no change — it stays blue. The reason is that soap is the sodium salt of a long-chain carboxylic acid, that is, the salt of a strong base (sodium hydroxide) and a weak acid. In water its carboxylate ion takes a proton from a water molecule, ${\text{C}}_{17}{\text{H}}_{35}{\text{COO}}^{-}+{\text{H}}_2\text{O}\rightleftharpoons {\text{C}}_{17}{\text{H}}_{35}\text{COOH}+{\text{OH}}^{-}$, leaving hydroxide ions in excess, so the solution is basic with a pH of about $9$–$10$. A basic solution turns red litmus blue and leaves blue litmus unchanged. The soap must be in solution, or the paper moistened, for the test to work at all.

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12. Q12
    What ​is ​hydrogenation? What is ​its ​industrial ​application?
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    Hydrogenation is the addition of hydrogen to an unsaturated compound over a catalyst; it is used to harden vegetable oils into vanaspati ghee.

    Hydrogenation is the addition of hydrogen to an unsaturated compound in the presence of a catalyst such as nickel, palladium or platinum, by which the compound is converted into the corresponding saturated compound. It is an addition reaction: the multiple bond opens and one hydrogen atom attaches to each carbon, with nothing leaving the molecule. In general ${\text{R}}_1-\text{CH}=\text{CH}-{\text{R}}_2+{\text{H}}_2\stackrel{\text{Ni}}{\to }{\text{R}}_1-{\text{CH}}_2-{\text{CH}}_2-{\text{R}}_2$; for example ${\text{CH}}_2={\text{CH}}_2+{\text{H}}_2\stackrel{\text{Ni}}{\to }{\text{CH}}_3-{\text{CH}}_3$. Its chief industrial application is the hardening of oils. Vegetable oils are liquid because their long fatty-acid chains are unsaturated; when hydrogen is passed through the warm oil over a finely divided nickel catalyst, those chains become saturated and the oil sets to a semi-solid fat, which is sold as vanaspati ghee (and, elsewhere, as margarine). Note that this makes the product nutritionally more like an animal fat, since it is exactly the unsaturated chains that are removed.

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13. Q13
    Which of ​the following ​hydrocarbons ​undergo ​addition ​reactions: ${\text{C}}_2{\text{H}}_6$, ${\text{C}}_3{\text{H}}_8$, ${\text{C}}_3{\text{H}}_6$, ${\text{C}}_2{\text{H}}_2$ and ${\text{CH}}_4$.
    Show answer

    ${\text{C}}_3{\text{H}}_6$ and ${\text{C}}_2{\text{H}}_2$ undergo addition reactions because they are unsaturated; the other three are saturated.

    Only the unsaturated hydrocarbons undergo addition reactions, so the answer is ${\text{C}}_3{\text{H}}_6$ (propene, which has one $\text{C}=\text{C}$ bond) and ${\text{C}}_2{\text{H}}_2$ (ethyne, which has one $\text{C}\equiv \text{C}$ bond). The test is the saturated quota ${\text{C}}_n{\text{H}}_{2n+2}$: methane ${\text{CH}}_4$ ($n=1$, quota $4$), ethane ${\text{C}}_2{\text{H}}_6$ ($n=2$, quota $6$) and propane ${\text{C}}_3{\text{H}}_8$ ($n=3$, quota $8$) all sit exactly at their quota and are saturated, so every carbon already holds four single bonds and nothing can be added; they undergo substitution instead. Propene has $6$ hydrogens against a quota of $8$ — two short, one double bond — and ethyne has $2$ against a quota of $6$ — four short, one triple bond. Their additions are ${\text{CH}}_2=\text{CH}-{\text{CH}}_3+{\text{H}}_2\stackrel{\text{Ni}}{\to }{\text{CH}}_3-{\text{CH}}_2-{\text{CH}}_3$ and $\text{CH}\equiv \text{CH}+2{\text{H}}_2\stackrel{\text{Ni}}{\to }{\text{CH}}_3-{\text{CH}}_3$, the triple bond taking two molecules of hydrogen because it opens twice.

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14. Q14
    Give ​a ​test that can be used to differentiate ​between saturated and unsaturated ​hydrocarbons.
    Show answer

    Add bromine water: unsaturated hydrocarbons decolourise it, while saturated ones leave its reddish-brown colour unchanged.

    Use the bromine water test. Take a little of each hydrocarbon in a separate test tube and add bromine water — a reddish-brown solution of bromine in water — drop by drop with shaking, keeping both tubes out of sunlight. The unsaturated hydrocarbon decolourises the bromine water: the reddish-brown colour disappears, because bromine is added across the multiple bond to give a colourless product, ${\text{CH}}_2={\text{CH}}_2+{\text{Br}}_2⟶{\text{CH}}_2\text{Br}-{\text{CH}}_2\text{Br}$. The saturated hydrocarbon gives no change and the reddish-brown colour persists, because it can react with bromine only by substitution and that reaction requires sunlight. A secondary test is to burn them: a saturated hydrocarbon gives a clean blue flame while an unsaturated one gives a yellow, sooty and smoky flame owing to its higher percentage of carbon. This flame test is less reliable, since a saturated hydrocarbon burnt in too little air is also sooty, so bromine water is the decisive test.

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15. Q15
    Explain ​the ​mechanism ​of ​the cleaning ​action ​of ​soaps.
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    Soap molecules form micelles that trap oily dirt at their centre; their charged surface keeps the dirt suspended until it is rinsed away.

    Most dirt is oily or greasy, and grease does not dissolve in water, so water alone cannot remove it. A soap molecule has two different ends: a long hydrocarbon chain, which is hydrophobic (water-repelling) and dissolves in oil, and an ionic group $-{\text{COO}}^{-}{\text{Na}}^{+}$, which is hydrophilic (water-loving) and dissolves in water. When soap is put into water the hydrocarbon tails cannot be accommodated by the water, so the molecules cluster with their tails turned inwards and their ionic heads facing the water, forming spherical aggregates called micelles. When greasy dirt is present the tails bury themselves in the droplet of grease while the ionic heads project outwards into the water, so each droplet is trapped in the centre of a micelle. The outer surface of every micelle is therefore made of negatively charged heads; since all of them carry the same charge, the micelles repel one another and cannot merge, so the droplets stay dispersed in the water as an emulsion instead of settling back on the cloth. Agitating the clothes helps the grease break away from the fibres, and rinsing carries the micelles, with the dirt inside them, away with the water — leaving the cloth clean. Note that the grease is never dissolved: it is wrapped, given an ionic surface and carried off in suspension.

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