Class 10 Science · Chapter 4

Carbon and its Compounds

442 Class 10 Science questions on Carbon and its Compounds from the NCERT textbook, NCERT Exemplar and CBSE board papers 2017–2026, each with its answer and a step-by-step solution.

CBSE Class 10 Science previous year papers, with answers →

Questions · page 19 of 23

1. Q361 CBSE 2017
   Write the molecular ​formula of first ​two members ​of ​homologous ​series having functional ​group ​–Cl.
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   The first two members are ${\text{CH}}_3\text{Cl}$ (chloromethane) and ${\text{C}}_2{\text{H}}_5\text{Cl}$ (chloroethane).

   The series is that of the chloroalkanes, general formula ${\text{C}}_n{\text{H}}_{2n+1}\text{Cl}$. Its first two members are chloromethane, ${\text{CH}}_3\text{Cl}$, and chloroethane, ${\text{C}}_2{\text{H}}_5\text{Cl}$ (that is, ${\text{CH}}_3{\text{CH}}_2\text{Cl}$).

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2. Q362 CBSE 2017
   Write the ​molecular ​formula ​of ​first two members of homologous ​series having functional group ​–Br.
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   Bromomethane, ${\text{CH}}_3\text{Br}$, and bromoethane, ${\text{C}}_2{\text{H}}_5\text{Br}$.

   The series is that of the bromoalkanes, general formula ${\text{C}}_n{\text{H}}_{2n+1}\text{Br}$. First member: bromomethane, ${\text{CH}}_3\text{Br}$. Second member: bromoethane, ${\text{C}}_2{\text{H}}_5\text{Br}$ (that is, ${\text{CH}}_3{\text{CH}}_2\text{Br}$).

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3. Q363 CBSE 2017
   Write the ​molecular ​formula of first two members ​of homologous series ​having ​functional ​group ​–OH.
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   The first two alcohols have molecular formulae ${\text{CH}}_4\text{O}$ (methanol) and ${\text{C}}_2{\text{H}}_6\text{O}$ (ethanol).

   The functional group –OH belongs to the alcohols; the series built on the alkanes has the general formula ${\text{C}}_n{\text{H}}_{2n+1}\text{OH}$. The first two members are: (1) Methanol, ${\text{CH}}_3\text{OH}$ (molecular formula ${\text{CH}}_4\text{O}$); (2) Ethanol, ${\text{C}}_2{\text{H}}_5\text{OH}$ (molecular formula ${\text{C}}_2{\text{H}}_6\text{O}$).

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4. Q364 CBSE 2017
   Write the molecular formula of the first ​two ​members of ​the homologous series ​having ​functional ​group $-\text{COOH}$.
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   The first two carboxylic acids are ${\text{CH}}_2{\text{O}}_2$ (methanoic acid) and ${\text{C}}_2{\text{H}}_4{\text{O}}_2$ (ethanoic acid).

   The $-\text{COOH}$ group marks the carboxylic acids. First member: methanoic acid, $\text{HCOOH}$, molecular formula ${\text{CH}}_2{\text{O}}_2$. Second member: ethanoic acid, ${\text{CH}}_3\text{COOH}$, molecular formula ${\text{C}}_2{\text{H}}_4{\text{O}}_2$.

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5. Q365 CBSE 2017
   Write the molecular formula ​of ​the first two members of the homologous ​series ​whose ​functional ​group ​is $-\text{CHO}$.
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   Methanal, ${\text{CH}}_2\text{O}$, and ethanal, ${\text{C}}_2{\text{H}}_4\text{O}$.

   First member: methanal, $\text{HCHO}$ — molecular formula ${\text{CH}}_2\text{O}$. Second member: ethanal, ${\text{CH}}_3\text{CHO}$ — molecular formula ${\text{C}}_2{\text{H}}_4\text{O}$.

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6. Q366 CBSE 2017
   Write the ​molecular formula of ​the first two ​members of the ​homologous ​series ​having functional group $>\text{C}=\text{O}$.
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   The first two ketones are propanone, ${\text{C}}_3{\text{H}}_6\text{O}$, and butanone, ${\text{C}}_4{\text{H}}_8\text{O}$.

   The series whose functional group is $>\text{C}=\text{O}$, the ketone group, is the ketones. Its first two members are propanone (acetone), ${\text{CH}}_3{\text{COCH}}_3$, molecular formula ${\text{C}}_3{\text{H}}_6\text{O}$, and butanone, ${\text{CH}}_3{\text{COCH}}_2{\text{CH}}_3$, molecular formula ${\text{C}}_4{\text{H}}_8\text{O}$.

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7. Q367 CBSE 2017
   Write ​the ​molecular formula of ​the $2^{\text{nd}}$ and the $3^{\text{rd}}$ ​member of the homologous ​series whose ​first member is methane.
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   The second and third members are ethane, ${\text{C}}_2{\text{H}}_6$, and propane, ${\text{C}}_3{\text{H}}_8$.

   The series that begins with methane is the alkanes, ${\text{C}}_n{\text{H}}_{2n+2}$. Second member: ethane, ${\text{C}}_2{\text{H}}_6$. Third member: propane, ${\text{C}}_3{\text{H}}_8$.

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8. Q368 CBSE 2017
   Write the molecular ​formula ​of ​the $2^{\text{nd}}$ ​and $3^{\text{rd}}$ ​member of the homologous series whose first ​member is ​ethene.
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   The second and third members are propene, ${\text{C}}_3{\text{H}}_6$, and butene, ${\text{C}}_4{\text{H}}_8$.

   Second member: ${\text{C}}_3{\text{H}}_6$ (propene). Third member: ${\text{C}}_4{\text{H}}_8$ (butene).

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9. Q369 CBSE 2017
   Write the ​molecular formula of the 2nd ​and 3rd ​member of the homologous ​series where the first ​member ​is ethyne.
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   Propyne, ${\text{C}}_3{\text{H}}_4$, is the 2nd member and butyne, ${\text{C}}_4{\text{H}}_6$, the 3rd member of the alkyne series.

   2nd member: ${\text{C}}_3{\text{H}}_4$ (propyne); 3rd member: ${\text{C}}_4{\text{H}}_6$ (butyne).

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10. Q370 CBSE 2017
    Write the name ​of ​the first two member of homologous ​series having ​functional ​group ​–OH.
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    The first two alcohols are methanol, ${\text{CH}}_3\text{OH}$, and ethanol, ${\text{C}}_2{\text{H}}_5\text{OH}$.

    The –OH group is the functional group of the alcohols. The first two members of this homologous series are methanol, ${\text{CH}}_3\text{OH}$, and ethanol, ${\text{C}}_2{\text{H}}_5\text{OH}$.

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11. Q371 CBSE 2017
    Write the ​molecular formula ​of ​

    1. (i)Methane and ​
    2. (ii)​Ethanol.
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    Methane is ${\text{CH}}_4$ and ethanol is ${\text{C}}_2{\text{H}}_5\text{OH}$ (${\text{C}}_2{\text{H}}_6\text{O}$).

    (i) Methane: ${\text{CH}}_4$ (ii) Ethanol: ${\text{C}}_2{\text{H}}_5\text{OH}$ (the same atoms as ${\text{C}}_2{\text{H}}_6\text{O}$: two carbon, six hydrogen and one oxygen atom)

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12. Q372 CBSE 2017
    What is a saturated ​hydrocarbon ? ​Write ​the formula of ​any one saturated hydrocarbon.
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    A saturated hydrocarbon has only single covalent bonds; for example, ethane, ${\text{C}}_2{\text{H}}_6$.

    A saturated hydrocarbon is a compound of carbon and hydrogen only in which all the bonds are single covalent bonds: the carbon atoms are linked to one another only by single bonds, and each carbon uses its four valencies in single bonds. Example: ethane, ${\text{C}}_2{\text{H}}_6$ (or methane, ${\text{CH}}_4$).

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13. Q373 CBSE 2017
    Name ​the ​chemical reaction ​used ​to convert ​ethanol to ethanoic acid.
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    Oxidation: ethanol is oxidised to ethanoic acid by an oxidising agent such as alkaline potassium permanganate.

    Oxidation. Ethanol is oxidised to ethanoic acid by heating it with an oxidising agent such as alkaline potassium permanganate or acidified potassium dichromate: ${\text{CH}}_3{\text{CH}}_2\text{OH}\underset{{\text{or acidified K}}_2{\text{Cr}}_2{\text{O}}_7+\text{heat}}{\overset{{\text{alkaline KMnO}}_4+\text{heat}}{\to }}{\text{CH}}_3\text{COOH}$

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14. Q374 CBSE 2017
    Write ​the structural formula of propane ​and determine the number ​of ​covalent bonds in ​its ​molecule.
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    Propane is ${\text{CH}}_3-{\text{CH}}_2-{\text{CH}}_3$; it has 2 C-C and 8 C-H bonds, 10 covalent bonds in all.

    Structural formula of propane (${\text{C}}_3{\text{H}}_8$): ${\text{CH}}_3-{\text{CH}}_2-{\text{CH}}_3$, that is, three carbon atoms joined by single bonds, with three hydrogen atoms on each end carbon and two on the middle carbon (drawn in full in the figure). Number of covalent bonds: 2 C–C bonds + 8 C–H bonds = 10 covalent bonds.

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15. Q375 CBSE 2017
    Write ​the molecular ​formula of ​

    1. (i)​ethane and ​
    2. (ii)​ethanoic ​acid
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    Ethane is ${\text{C}}_2{\text{H}}_6$; ethanoic acid is ${\text{C}}_2{\text{H}}_4{\text{O}}_2$.

    (i) Ethane: ${\text{C}}_2{\text{H}}_6$. (ii) Ethanoic acid: ${\text{C}}_2{\text{H}}_4{\text{O}}_2$, usually written ${\text{CH}}_3\text{COOH}$ to show its $-\text{COOH}$ group.

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16. Q376 CBSE 2017
    Why is ​the conversion of ethanol to ethanoic acid an oxidation reaction ? ​How ​is it different from combustion ​of ethanol in oxygen ​? Write ​the chemical ​equations for both reactions.
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    Ethanol gains oxygen and loses hydrogen, so its conversion to ethanoic acid is oxidation; unlike combustion, it is partial and controlled.

    Ethanol (${\text{C}}_2{\text{H}}_6\text{O}$) changes into ethanoic acid (${\text{C}}_2{\text{H}}_4{\text{O}}_2$) by gaining one oxygen atom and losing two hydrogen atoms, the oxygen being added by an oxidising agent (alkaline potassium permanganate or acidified potassium dichromate). Gain of oxygen and loss of hydrogen are oxidation, so the conversion is an oxidation reaction: ${\text{CH}}_3{\text{CH}}_2\text{OH}+2[\text{O}]\to {\text{CH}}_3\text{COOH}+{\text{H}}_2\text{O}$ (with alkaline ${\text{KMnO}}_4$ or acidified ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$, on heating) Difference: this is a partial, controlled oxidation. Only the $-{\text{CH}}_2\text{OH}$ end changes to $-\text{COOH}$, the two carbon atoms stay joined, and the product is another carbon compound. In combustion, ethanol burns in oxygen and the whole molecule is oxidised completely to carbon dioxide and water, with the release of a large amount of heat and light: ${\text{C}}_2{\text{H}}_5\text{OH}+3{\text{O}}_2\to 2{\text{CO}}_2+3{\text{H}}_2\text{O}+\text{heat and light}$

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17. Q377 CBSE 2017
    What is a functional group ​in carbon ​compounds ? Write the ​structural ​formula and ​the functional ​group present in

    1. (i)propanol and
    2. (ii)​chloropropane.
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    A functional group is an atom or group of atoms that gives a compound its characteristic properties; propanol has $-\text{OH}$ and chloropropane has $-\text{Cl}$.

    A functional group is an atom or a group of atoms (a heteroatom such as oxygen, nitrogen, sulphur or a halogen, or a group containing it) that replaces a hydrogen atom in a hydrocarbon chain and gives the compound its characteristic chemical properties, whatever the length and nature of the carbon chain. (i) Propanol: ${\text{CH}}_3-{\text{CH}}_2-{\text{CH}}_2-\text{OH}$; functional group: the alcohol (hydroxyl) group, $-\text{OH}$. (ii) Chloropropane: ${\text{CH}}_3-{\text{CH}}_2-{\text{CH}}_2-\text{Cl}$; functional group: the halo (chloro) group, $-\text{Cl}$. (The full structural formulas, with every bond drawn, are in the figure.)

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18. Q378 CBSE 2017
    With the help ​of an example explain ​the process of hydrogenation. Mention the essential ​conditions for ​this reaction and also state ​the change ​in ​the physical property with the formation of the product.
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    Hydrogenation is the addition of hydrogen to an unsaturated compound with a catalyst; it turns liquid vegetable oil into solid vanaspati ghee.

    Hydrogenation is the addition of hydrogen to an unsaturated carbon compound (one with a double or triple bond between carbon atoms) in the presence of a catalyst; hydrogen adds across the multiple bond and the compound becomes saturated (or less unsaturated). It is an addition reaction. Example: ${\text{CH}}_2={\text{CH}}_2+{\text{H}}_2\stackrel{\text{Ni}}{\to }{\text{CH}}_3-{\text{CH}}_3$ (ethene to ethane). The same reaction converts vegetable oils, which generally have long unsaturated carbon chains, into vegetable ghee (vanaspati). Essential conditions: an unsaturated compound, hydrogen gas and a catalyst such as nickel or palladium; for vegetable oils, hydrogen is passed through the heated oil containing finely divided nickel. Change in physical property: the vegetable oil is a liquid at room temperature, whereas the product, vanaspati ghee, is a solid (semi-solid) at room temperature; the melting point rises.

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19. Q379 CBSE 2017
    The general formula of ​the organic compounds ​‘A’, ‘B’ and ‘C’ is ${\text{C}}_n{\text{H}}_{2n}$. Their boiling ​points ​are $-162{}^{\circ }\text{C}$, $-42.2{}^{\circ }\text{C}$ ​and $-0.5{}^{\circ }\text{C}$ respectively. Based on this information answer the following :

    1. (a)Which type of compounds ‘A’, ‘B’ and ‘C’ are and why ?
    2. (b)Which of these has maximum number of carbon atoms ​in the molecule and why ​?
    3. (c)Write the name and structural formula of the second member of this series.
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    A, B and C are alkenes (general formula ${\text{C}}_n{\text{H}}_{2n}$); C has the most carbon atoms as it boils highest; the second member is propene.

    (a) A, B and C are alkenes: unsaturated hydrocarbons with one carbon-carbon double bond, members of one homologous series. The reason is their general formula, ${\text{C}}_n{\text{H}}_{2n}$, which is that of the alkene series: it has two hydrogen atoms fewer than the alkane with the same number of carbon atoms, ${\text{C}}_n{\text{H}}_{2n+2}$, which for an open chain means one C=C double bond (a ring compound such as cyclohexane, ${\text{C}}_6{\text{H}}_{12}$, also fits ${\text{C}}_n{\text{H}}_{2n}$, but the homologous series with this general formula in the textbook is the alkene series). (b) C has the maximum number of carbon atoms. In a homologous series the boiling point rises as the molecular mass, and so the number of carbon atoms, increases; C has the highest boiling point, $-0.5{}^{\circ }\text{C}$. (c) The second member is propene, ${\text{C}}_3{\text{H}}_6$, ${\text{CH}}_3-\text{CH}={\text{CH}}_2$. Its structural formula is $$\begin{array}{ccccccc}& & \text{H}& & \text{H}& & \text{H}\\ & & |& & |& & |\\ \text{H}& -& \text{C}& -& \text{C}& =& \text{C}\\ & & |& & & & |\\ & & \text{H}& & & & \text{H}\end{array}$$

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20. Q380 CBSE 2017
    Two carbon compounds X and Y have the molecular ​formula ${\text{C}}_4{\text{H}}_8$ and ${\text{C}}_5{\text{H}}_{12}$ ​respectively. Which one of these ​is ​most likely to ​show addition reaction ? Justify ​your answer. Also give ​the chemical equation to explain the process of addition reaction in this ​case.
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    X, ${\text{C}}_4{\text{H}}_8$, is an unsaturated alkene and shows addition; it adds hydrogen with nickel to form butane.

    X (${\text{C}}_4{\text{H}}_8$) is most likely to show an addition reaction. Justification: ${\text{C}}_4{\text{H}}_8$ fits the general formula of the alkenes, ${\text{C}}_n{\text{H}}_{2n}$ with $n=4$, so X is most likely an unsaturated hydrocarbon (a butene) with a carbon–carbon double bond across which atoms can add. Y, ${\text{C}}_5{\text{H}}_{12}$, fits ${\text{C}}_n{\text{H}}_{2n+2}$ with $n=5$: it is a saturated alkane (such as pentane) with only single bonds, so it has no double bond to add across and reacts by substitution instead. Equation (addition of hydrogen in the presence of a nickel catalyst): ${\text{CH}}_3-\text{CH}=\text{CH}-{\text{CH}}_3+{\text{H}}_2\stackrel{\text{Ni}}{\to }{\text{CH}}_3-{\text{CH}}_2-{\text{CH}}_2-{\text{CH}}_3$, that is, ${\text{C}}_4{\text{H}}_8+{\text{H}}_2\stackrel{\text{Ni}}{\to }{\text{C}}_4{\text{H}}_{10}$ (butane).

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