Chapter 4 · Carbon and its Compounds

NCERT Exemplar Class 10 Science Chapter 4 Short Answer Questions

All 17 questions of NCERT Exemplar Class 10 Science Chapter 4 Short Answer Questions (Carbon and its Compounds), in the book's order, each with its answer and a step-by-step solution.

1. Q30
   Draw ​the ​electron ​dot structure ​of ethyne and also draw its ​structural ​formula.
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   Each H shares one pair with a C and the two C atoms share three pairs, giving $\text{H}-\text{C}\equiv \text{C}-\text{H}$.

   In the electron dot structure each hydrogen shares one electron pair with a carbon atom and the two carbon atoms share three pairs with each other, using all $10$ valence electrons as $5$ shared pairs, with no lone pair anywhere. Replacing each shared pair by a line gives the structural formula $\text{H}-\text{C}\equiv \text{C}-\text{H}$.

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2. Q31
   Write the ​names of the following compounds ​(a), ​(b), ​(c) and (d), ​whose structural formulae are given ​in the ​figure.

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   The four compounds are (a) pentanoic acid, (b) but-1-yne, (c) heptanal and (d) pentan-1-ol.

   (a) Pentanoic acid; (b) But-1-yne (butyne); (c) Heptanal; (d) Pentan-1-ol (pentanol).

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3. Q32
   Identify and name ​the functional groups ​present ​in the ​following compounds (a), (b), (c) ​and (d), whose structural formulae ​are given in ​the ​figure.

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   (a) Alcohol, (b) carboxylic acid, (c) ketone, (d) alkene.

   (a) Alcohol, the hydroxyl group $-\text{OH}$ (propan-1-ol); (b) Carboxylic acid, the carboxyl group $-\text{COOH}$ (propanoic acid); (c) Ketone, the carbonyl group $>\text{C}=\text{O}$ inside the chain (pentan-3-one); (d) Alkene, a carbon-carbon double bond $\text{C}=\text{C}$ (but-1-ene).

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4. Q33
   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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5. Q34
   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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6. Q35
   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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7. Q36
   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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8. Q37
   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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9. Q38
   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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10. Q39
    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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11. Q40
    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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12. Q41
    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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13. Q42
    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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14. Q43
    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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15. Q44
    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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16. Q45
    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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17. Q46
    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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