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 20 of 23

1. Q381 CBSE 2017
   Complete the ​following ​chemical ​equations ​:

   1. (i)${\text{CH}}_3\text{COOH}+{\text{Na}}_2{\text{CO}}_3⟶$
   2. (ii)${\text{CH}}_4+{\text{O}}_2⟶$
   3. (iii)${\text{C}}_2{\text{H}}_5\text{OH}+\text{Na}⟶$
   Show answer

   Ethanoic acid gives carbon dioxide with sodium carbonate, methane burns to carbon dioxide and water, and ethanol gives hydrogen with sodium.

   (i) $2{\text{CH}}_3\text{COOH}+{\text{Na}}_2{\text{CO}}_3⟶2{\text{CH}}_3\text{COONa}+{\text{H}}_2\text{O}+{\text{CO}}_2$ (sodium ethanoate, water and carbon dioxide) (ii) ${\text{CH}}_4+2{\text{O}}_2⟶{\text{CO}}_2+2{\text{H}}_2\text{O}+\text{heat and light}$ (iii) $2{\text{C}}_2{\text{H}}_5\text{OH}+2\text{Na}⟶2{\text{C}}_2{\text{H}}_5\text{ONa}+{\text{H}}_2$ (sodium ethoxide and hydrogen)

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2. Q382 CBSE 2017
   Complete ​the ​following chemical ​equations ​:

   1. (i)${\text{C}}_2{\text{H}}_5\text{OH}+{\text{O}}_2⟶$
   2. (ii)${\text{C}}_2{\text{H}}_5\text{OH}\underset{\text{443K}}{\overset{{\text{Conc.H}}_2{\text{SO}}_4}{\to }}$
   3. (iii)${\text{CH}}_3\text{COOH}+{\text{NaHCO}}_3⟶$
   Show answer

   Ethanol burns to give ${\text{CO}}_2$ and water, dehydrates to ethene at 443 K, and ethanoic acid reacts with sodium hydrogencarbonate to give sodium ethanoate.

   (i) Combustion: ${\text{C}}_2{\text{H}}_5\text{OH}+3{\text{O}}_2\to 2{\text{CO}}_2+3{\text{H}}_2\text{O}$ + heat + light (ii) Dehydration: ${\text{C}}_2{\text{H}}_5\text{OH}\underset{443\text{K}}{\overset{{\text{conc. H}}_2{\text{SO}}_4}{\to }}{\text{CH}}_2={\text{CH}}_2+{\text{H}}_2\text{O}$ (ethene and water) (iii) ${\text{CH}}_3\text{COOH}+{\text{NaHCO}}_3\to {\text{CH}}_3\text{COONa}+{\text{H}}_2\text{O}+{\text{CO}}_2$ (sodium ethanoate, water and carbon dioxide)

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3. Q383 CBSE 2017
   Name two ​oxidising agents that are used for the conversion ​of alcohols to acids. Distinguish between ​ethanol and ​ethanoic acid ​on the basis ​of ​

   1. (i)litmus test, and
   2. (ii)reaction with sodium hydrogen carbonate.
   Show answer

   Alkaline ${\text{KMnO}}_4$ and acidified ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$ are oxidising agents; only ethanoic acid turns blue litmus red and gives effervescence with sodium hydrogencarbonate.

   Oxidising agents: alkaline potassium permanganate (${\text{KMnO}}_4$) and acidified potassium dichromate (${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$), used with heating. (i) Litmus test: ethanoic acid turns blue litmus red, as it is acidic; ethanol does not change the colour of litmus, as it is neutral. (ii) Reaction with sodium hydrogencarbonate: ethanoic acid reacts with brisk effervescence, giving off carbon dioxide, which turns lime water milky (${\text{CH}}_3\text{COOH}+{\text{NaHCO}}_3\to {\text{CH}}_3\text{COONa}+{\text{H}}_2\text{O}+{\text{CO}}_2$); ethanol does not react, so there is no effervescence.

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4. Q384 CBSE 2017
   Write the structural formula of ethanol. ​What happens when it ​is heated with excess of conc. ${\text{H}}_2{\text{SO}}_4$ ​at ​443 K ? Write ​the chemical equation for the reaction stating the ​role of conc. ${\text{H}}_2{\text{SO}}_4$ in this ​reaction.
   Show answer

   Ethanol heated with excess conc. ${\text{H}}_2{\text{SO}}_4$ at 443 K is dehydrated to ethene; the acid acts as a dehydrating agent.

   Structural formula of ethanol, ${\text{C}}_2{\text{H}}_5\text{OH}$: $$\begin{array}{ccccccccc}& & \text{H}& & \text{H}& & & & \\ & & |& & |& & & & \\ \text{H}& -& \text{C}& -& \text{C}& -& \text{O}& -& \text{H}\\ & & |& & |& & & & \\ & & \text{H}& & \text{H}& & & & \end{array}$$ When ethanol is heated with excess concentrated sulphuric acid at $443$ K it is dehydrated: each molecule loses one molecule of water, and ethene, an unsaturated hydrocarbon, is formed. $${\text{CH}}_3{\text{CH}}_2\text{OH}\underset{443\text{K}}{\overset{{\text{hot conc. H}}_2{\text{SO}}_4}{\to }}{\text{CH}}_2={\text{CH}}_2+{\text{H}}_2\text{O}$$ Role of conc. ${\text{H}}_2{\text{SO}}_4$: it acts as a dehydrating agent — it removes water from ethanol.

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5. Q385 CBSE 2017
   Two carbon compounds A and B ​have ​the ​molecular formula ${\text{C}}_4{\text{H}}_{10}$ and ${\text{C}}_3{\text{H}}_6$. Which of the ​two ​is most likely to show addition reaction and why ? Write a ​chemical equation ​to show an addition reaction. Where is this reaction used ?
   Show answer

   B, ${\text{C}}_3{\text{H}}_6$, is unsaturated and shows addition: propene + hydrogen gives propane; this hydrogenation makes vegetable ghee from oils.

   B (${\text{C}}_3{\text{H}}_6$) is more likely to show an addition reaction. It fits the general formula ${\text{C}}_n{\text{H}}_{2n}$ of the alkenes; taken as the open-chain compound the question intends, it is propene, an unsaturated hydrocarbon with a carbon–carbon double bond, across which hydrogen can add. A (${\text{C}}_4{\text{H}}_{10}$) fits ${\text{C}}_n{\text{H}}_{2n+2}$: it is a saturated alkane with only single bonds, and it shows substitution, not addition. Equation: ${\text{CH}}_3-\text{CH}={\text{CH}}_2+{\text{H}}_2\stackrel{\text{Ni}}{\to }{\text{CH}}_3-{\text{CH}}_2-{\text{CH}}_3$ (propene to propane). Use: this addition of hydrogen (hydrogenation) is used to convert vegetable oils into vegetable ghee, with a nickel catalyst.

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6. Q386 CBSE 2017
   Complete ​the ​following chemical ​equations

   1. (i)${\text{CH}}_3\text{COOH}+{\text{Na}}_2{\text{CO}}_3⟶$
   2. (ii)${\text{CH}}_3\text{COOH}+\text{NaOH}⟶$
   3. (iii)${\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+\text{NaOH}⟶$
   Show answer

   Ethanoic acid gives sodium ethanoate with sodium carbonate and with sodium hydroxide; ethyl ethanoate with sodium hydroxide gives sodium ethanoate and ethanol.

   (i) $2{\text{CH}}_3\text{COOH}+{\text{Na}}_2{\text{CO}}_3\to 2{\text{CH}}_3\text{COONa}+{\text{H}}_2\text{O}+{\text{CO}}_2$ (sodium ethanoate, water and carbon dioxide) (ii) ${\text{CH}}_3\text{COOH}+\text{NaOH}\to {\text{CH}}_3\text{COONa}+{\text{H}}_2\text{O}$ (sodium ethanoate and water) (iii) ${\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+\text{NaOH}\to {\text{CH}}_3\text{COONa}+{\text{C}}_2{\text{H}}_5\text{OH}$ (sodium ethanoate and ethanol; this is saponification)

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7. Q387 CBSE 2017
   Complete ​the ​following ​chemical ​equations ​:

   1. (i)${\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+\text{NaOH}⟶$
   2. (ii)${\text{CH}}_3\text{COOH}+\text{NaOH}⟶$
   3. (iii)${\text{C}}_2{\text{H}}_5\text{OH}+{\text{CH}}_3\text{COOH}\stackrel{{\text{Conc. H}}_2{\text{SO}}_4}{\to }$
   Show answer

   Saponification gives sodium ethanoate and ethanol, neutralisation gives sodium ethanoate and water, and esterification gives ethyl ethanoate and water.

   (i) ${\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+\text{NaOH}⟶{\text{CH}}_3\text{COONa}+{\text{C}}_2{\text{H}}_5\text{OH}$ (saponification: sodium ethanoate and ethanol) (ii) ${\text{CH}}_3\text{COOH}+\text{NaOH}⟶{\text{CH}}_3\text{COONa}+{\text{H}}_2\text{O}$ (neutralisation: sodium ethanoate and water) (iii) ${\text{C}}_2{\text{H}}_5\text{OH}+{\text{CH}}_3\text{COOH}\stackrel{{\text{Conc. H}}_2{\text{SO}}_4}{\to }{\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+{\text{H}}_2\text{O}$ (esterification: ethyl ethanoate and water)

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8. Q388 CBSE 2017
   Two carbon compounds X and Y have the molecular formula ${\text{C}}_3{\text{H}}_6$ and ${\text{C}}_4{\text{H}}_{10}$ ​respectively. ​Which one of the two ​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.
   Show answer

   X, ${\text{C}}_3{\text{H}}_6$, is an unsaturated alkene and shows addition; propene adds hydrogen to form propane.

   X (${\text{C}}_3{\text{H}}_6$) is the one most likely to show an addition reaction. Justification: ${\text{C}}_3{\text{H}}_6$ fits ${\text{C}}_n{\text{H}}_{2n}$, the general formula of the alkenes, so X is most likely propene, an unsaturated compound with a carbon–carbon double bond, and addition takes place across a double bond. Y, ${\text{C}}_4{\text{H}}_{10}$, fits ${\text{C}}_n{\text{H}}_{2n+2}$: it is an alkane, saturated, with only single bonds, so it cannot add anything. Equation: ${\text{CH}}_3\text{CH}={\text{CH}}_2+{\text{H}}_2\stackrel{\text{Ni catalyst}}{\to }{\text{CH}}_3{\text{CH}}_2{\text{CH}}_3$ (propene and hydrogen give propane).

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9. Q389 CBSE 2017
   The molecular formula of two carbon ​compounds are ${\text{C}}_4{\text{H}}_8$ ​and ${\text{C}}_3{\text{H}}_8$. Which ​one of ​the two 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.
   Show answer

   ${\text{C}}_4{\text{H}}_8$ is an unsaturated alkene, so it shows addition with hydrogen to form butane; ${\text{C}}_3{\text{H}}_8$ is saturated propane.

   ${\text{C}}_4{\text{H}}_8$ is the one most likely to show an addition reaction. Justification: ${\text{C}}_4{\text{H}}_8$ fits the general formula of the alkenes, ${\text{C}}_n{\text{H}}_{2n}$ ($n=4$), so it is most likely an unsaturated hydrocarbon (a butene) with a carbon–carbon double bond across which atoms can add. ${\text{C}}_3{\text{H}}_8$ fits ${\text{C}}_n{\text{H}}_{2n+2}$ ($n=3$): it is propane, a saturated alkane with only single bonds, which reacts by substitution rather than addition. Equation (addition of hydrogen, nickel catalyst): ${\text{CH}}_2=\text{CH}-{\text{CH}}_2-{\text{CH}}_3+{\text{H}}_2\stackrel{\text{Ni catalyst}}{\to }{\text{CH}}_3-{\text{CH}}_2-{\text{CH}}_2-{\text{CH}}_3$, that is, ${\text{C}}_4{\text{H}}_8+{\text{H}}_2\to {\text{C}}_4{\text{H}}_{10}$ (butane).

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10. Q390 CBSE 2017
    Explain, giving reason, why carbon neither forms ${\text{C}}^{4+}$ cations nor ${\text{C}}^{4-}$ anions, but forms ​covalent compounds which ​are ​bad ​conductors of electricity and have low ​melting ​point and low boiling point.
    Show answer

    Carbon shares its four electrons instead of gaining or losing them, so its covalent compounds conduct poorly and have low melting and boiling points.

    Carbon (atomic number 6) has the configuration 2, 4, with four valence electrons, so it would have to gain or lose four electrons to attain a noble gas configuration. It does not form ${\text{C}}^{4-}$: gaining four electrons would be difficult, because a nucleus with six protons cannot easily hold on to ten electrons. It does not form ${\text{C}}^{4+}$: losing four electrons would require a large amount of energy, and would leave six protons holding on to just two electrons. Instead, carbon shares its four valence electrons with other atoms and forms covalent bonds. Since the electrons are shared and no ions are formed, such covalent compounds are poor conductors of electricity. The bonds within their molecules are strong, but the forces of attraction between the molecules are weak, so they have low melting and boiling points.

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11. Q391 CBSE 2017
    What are hydrocarbons ? ​Write the general formula of ​

    1. (i)​saturated hydrocarbons, and
    2. (ii)unsaturated hydrocarbons ​and draw the ​structure of ​one hydrocarbon of each ​type.
    Show answer

    Hydrocarbons contain carbon and hydrogen only; saturated alkanes are ${\text{C}}_n{\text{H}}_{2n+2}$; unsaturated alkenes are ${\text{C}}_n{\text{H}}_{2n}$ and alkynes ${\text{C}}_n{\text{H}}_{2n-2}$.

    Hydrocarbons are compounds of carbon and hydrogen only. (i) Saturated (open-chain) hydrocarbons, the alkanes: ${\text{C}}_n{\text{H}}_{2n+2}$. Example: ethane, ${\text{C}}_2{\text{H}}_6$, ${\text{CH}}_3-{\text{CH}}_3$, with single bonds only (structure in the figure). (ii) Unsaturated hydrocarbons: alkenes, with one carbon–carbon double bond, ${\text{C}}_n{\text{H}}_{2n}$; alkynes, with one carbon–carbon triple bond, ${\text{C}}_n{\text{H}}_{2n-2}$. Example: ethene, ${\text{C}}_2{\text{H}}_4$, ${\text{CH}}_2={\text{CH}}_2$ (structure in the figure).

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12. Q392 CBSE 2017
    Distinguish between ​esterification and ​saponification reactions with the ​help ​of the chemical equations for ​each. ​State ​one use of each

    1. (i)esters, and
    2. (ii)saponification ​process.
    Show answer

    Esterification forms an ester from acid and alcohol; saponification splits an ester with alkali; esters are used in perfumes, saponification in making soap.

    Esterification: a carboxylic acid reacts with an alcohol, in the presence of an acid catalyst (a few drops of concentrated sulphuric acid) on warming, to form an ester and water. $${\text{CH}}_3\text{COOH}+{\text{C}}_2{\text{H}}_5\text{OH}\stackrel{{\text{conc. H}}_2{\text{SO}}_4,\text{warm}}{\to }{\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+{\text{H}}_2\text{O}$$ Saponification: an ester reacts with an alkali (sodium hydroxide) and is converted back to the alcohol and the sodium salt of the carboxylic acid. $${\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+\text{NaOH}⟶{\text{CH}}_3\text{COONa}+{\text{C}}_2{\text{H}}_5\text{OH}$$ Differences: reactants — acid + alcohol in esterification, ester + alkali in saponification; condition — an acid catalyst in esterification, an alkali in saponification; products — ester + water in esterification, alcohol + sodium salt of the acid in saponification; esterification forms an ester, saponification breaks one down. Uses: (i) esters are sweet-smelling and are used in making perfumes and as flavouring agents; (ii) saponification is used in the preparation of soap.

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13. Q393 CBSE 2017
    An organic compound ‘A’ (${\text{C}}_2{\text{H}}_6\text{O}$) on reacting with ​acidified ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$ ​forms another compound ‘B’. On adding ​a piece ​of sodium to ‘B’ a gas ‘X’ evolves which burns with a pop ​sound. Identify ‘A’, ​‘B’ and ‘X’ and ​also write the chemical equations for the ​reactions involved.
    Show answer

    A is ethanol, B is ethanoic acid and X is hydrogen gas.

    'A' is ethanol, ${\text{C}}_2{\text{H}}_5\text{OH}$; 'B' is ethanoic acid (acetic acid), ${\text{CH}}_3\text{COOH}$; 'X' is hydrogen gas, ${\text{H}}_2$. Oxidation of A to B: ${\text{CH}}_3{\text{CH}}_2\text{OH}+2[\text{O}]\stackrel{{\text{acidified K}}_2{\text{Cr}}_2{\text{O}}_7,\text{heat}}{\to }{\text{CH}}_3\text{COOH}+{\text{H}}_2\text{O}$ B with sodium: $2{\text{CH}}_3\text{COOH}+2\text{Na}\to 2{\text{CH}}_3\text{COONa}+{\text{H}}_2$ The hydrogen burns with a pop: $2{\text{H}}_2+{\text{O}}_2\to 2{\text{H}}_2\text{O}$

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14. Q394 CBSE 2017
    Write the molecular formula of ​benzene ​and ​draw its structure. List in tabular form the ​two properties ​in ​which ​covalent compounds differ from ​ionic compounds.
    Show answer

    Benzene is ${\text{C}}_6{\text{H}}_6$, a six-carbon ring; covalent compounds have low melting points and conduct poorly, unlike ionic compounds.

    Molecular formula of benzene: ${\text{C}}_6{\text{H}}_6$. Structure: a hexagonal ring of six carbon atoms joined by alternate single and double bonds, with one hydrogen atom on each carbon (as drawn in the figure). The two properties in which covalent compounds differ from ionic compounds: Melting and boiling points — covalent compounds made of separate molecules: generally low; ionic compounds: high. Electrical conductivity — covalent compounds: generally poor conductors, as their bonding gives no ions; ionic compounds: conduct electricity when molten or dissolved in water (not as solids).

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15. Q395 CBSE 2017
    What happens when (write ​chemical equation ​in each case)

    1. (a)ethanol is burnt in air ?
    2. (b)ethanol is ​heated with excess conc. ${\text{H}}_2{\text{SO}}_4$ at $443$ K ​?
    3. (c)​a piece ​of sodium ​is dropped into ethanol ?
    Show answer

    Ethanol burns to carbon dioxide and water, dehydrates to ethene with hot conc. ${\text{H}}_2{\text{SO}}_4$, and gives hydrogen and sodium ethoxide with sodium.

    (a) Ethanol burns in air with a clean blue flame, forming carbon dioxide and water and releasing heat and light: ${\text{C}}_2{\text{H}}_5\text{OH}+3{\text{O}}_2⟶2{\text{CO}}_2+3{\text{H}}_2\text{O}+\text{heat and light}$ (b) Ethanol is dehydrated to ethene; the hot concentrated sulphuric acid acts as a dehydrating agent: ${\text{CH}}_3{\text{CH}}_2\text{OH}\underset{443\text{K}}{\overset{\text{hot conc. }{\text{H}}_2{\text{SO}}_4}{\to }}{\text{CH}}_2={\text{CH}}_2+{\text{H}}_2\text{O}$ (c) Sodium reacts with ethanol: hydrogen gas is given off (bubbles) and sodium ethoxide is formed: $2{\text{CH}}_3{\text{CH}}_2\text{OH}+2\text{Na}⟶2{\text{CH}}_3{\text{CH}}_2\text{ONa}+{\text{H}}_2$

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16. Q396 CBSE 2017
    Give reason why ​carbon can neither form ${\text{C}}^{4+}$ cations ​nor ${\text{C}}^{4-}$ anions, but ​forms covalent ​compounds ​which are ​bad conductors of electricity and have ​low melting and boiling points ?
    Show answer

    Carbon cannot gain or lose four electrons easily, so it shares them; covalent compounds have no ions and weak forces between molecules.

    Carbon has the configuration $2,4$, so it would have to gain or lose four electrons to attain a noble-gas configuration. (i) It does not form ${\text{C}}^{4-}$: gaining four electrons would leave $6$ protons holding $10$ electrons, and it would be difficult for the nucleus to hold on to the four extra electrons. (ii) It does not form ${\text{C}}^{4+}$: removing four electrons would need a large amount of energy, leaving $6$ protons holding only $2$ electrons. So carbon shares its four valence electrons with other atoms and forms covalent bonds. (iii) In such covalent compounds (for example methane) no ions are formed, so there are no charged particles to carry a current: they are bad conductors of electricity. The bonds within each molecule are strong but the forces between molecules are weak, so little energy is needed to separate the molecules: their melting and boiling points are low.

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17. Q397 CBSE 2017
    What are ​structural isomers ​? Why are isomers of first three members of alkane series ​not possible ​? Write the all ​possible ​structures of isomer ​of fourth member of this series.
    Show answer

    Structural isomers share a molecular formula but differ in structure; methane, ethane and propane have only one structure each, and butane has two isomers.

    Structural isomers are compounds that have the same molecular formula but different structures, that is, different arrangements of their atoms (for example, of their carbon chain). The first three alkanes are methane (${\text{CH}}_4$), ethane (${\text{C}}_2{\text{H}}_6$) and propane (${\text{C}}_3{\text{H}}_8$). In an alkane, with one, two or three carbon atoms there is only one way of joining the carbon atoms, as a single open chain; a branch needs a carbon atom bonded to three other carbon atoms, which requires at least four carbon atoms. So each of them has only one possible structure, and isomers are not possible. The fourth member, butane (${\text{C}}_4{\text{H}}_{10}$), has two structural isomers: n-butane, ${\text{CH}}_3-{\text{CH}}_2-{\text{CH}}_2-{\text{CH}}_3$ (straight chain), and isobutane or 2-methylpropane, ${\text{CH}}_3-\text{CH}({\text{CH}}_3)-{\text{CH}}_3$ (branched chain). Their full structural formulas are drawn in the figure.

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18. Q398 CBSE 2017
    Explain esterification reaction with ​the help of ​a ​chemical ​equation. Describe ​an ​activity to show ​esterification.
    Show answer

    A carboxylic acid and an alcohol react with a few drops of concentrated sulphuric acid to give a sweet-smelling ester and water.

    Esterification: a carboxylic acid reacts with an alcohol in the presence of an acid catalyst (a few drops of concentrated sulphuric acid) to form a sweet-smelling compound called an ester, and water. For ethanoic acid and ethanol: ${\text{CH}}_3\text{COOH}+{\text{CH}}_3{\text{CH}}_2\text{OH}\stackrel{\text{conc. }{\text{H}}_2{\text{SO}}_4}{\to }{\text{CH}}_3{\text{COOCH}}_2{\text{CH}}_3+{\text{H}}_2\text{O}$ — ethanoic acid and ethanol give ethyl ethanoate and water. Activity: take $1$ mL of ethanol (absolute alcohol) and $1$ mL of glacial acetic acid in a test tube and add a few drops of concentrated sulphuric acid. Warm the test tube in a water bath for at least five minutes. Pour the contents into a beaker containing $20$–$50$ mL of water and smell the mixture. A sweet, fruity smell shows that an ester has been formed.

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19. Q399 CBSE 2017
    What are ​covalent compounds ? ​How are they different ​from ​ionic compounds ? ​List any ​two ​properties of covalent compounds.
    Show answer

    Covalent compounds form by sharing electron pairs, ionic ones by transferring electrons; covalent compounds have low melting and boiling points and conduct poorly.

    Covalent compounds are compounds in which the atoms are held together by covalent bonds, that is, by the sharing of pairs of electrons between atoms — for example methane, ${\text{CH}}_4$, in which carbon shares one electron pair with each of four hydrogen atoms. Difference from ionic compounds: an ionic compound is formed by the transfer of electrons from a metal to a non-metal, which produces oppositely charged ions held together by strong electrostatic forces (as in sodium chloride). In a covalent compound the electrons are shared, no ions are formed, and compounds such as methane consist of separate molecules with only weak forces between them. So ionic compounds have high melting and boiling points and conduct electricity when molten or dissolved in water, while covalent compounds of this kind have low melting and boiling points and are generally poor conductors. Two properties of covalent compounds: 1. They generally have low melting and boiling points, because the forces between their molecules are weak (methane melts at $90$ K and boils at $111$ K). 2. They are generally poor conductors of electricity, because they contain no ions.

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20. Q400 CBSE 2017
    What ​is a homologous series of carbon ​compounds ? Write the formula of the fourth and ​fifth ​member of a homologous series having general formula ${\text{C}}_n{\text{H}}_{2n+1}\text{OH}$. List two differences in physical properties found in ​the two consecutive members ​of ​a homologous series.
    Show answer

    A homologous series has the same functional group and members differing by $-{\text{CH}}_2-$; the fourth and fifth members are ${\text{C}}_4{\text{H}}_9\text{OH}$ and ${\text{C}}_5{\text{H}}_{11}\text{OH}$.

    A homologous series is a series of carbon compounds that have the same functional group substituting for hydrogen in a carbon chain, and therefore similar chemical properties; any two successive members differ by a $-{\text{CH}}_2-$ unit ($14$ u in molecular mass), and all the members can be represented by one general formula. Fourth member ($n=4$): ${\text{C}}_4{\text{H}}_9\text{OH}$ (butanol). Fifth member ($n=5$): ${\text{C}}_5{\text{H}}_{11}\text{OH}$ (pentanol). Two differences in physical properties between consecutive members: (1) the higher member has a higher boiling point (melting points also rise with molecular mass in general); (2) their solubility in a particular solvent differs - for example, the solubility of the alcohols in water decreases as the chain gets longer.

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