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NCERT Exemplar · Class 10 Science Carbon and its Compounds

57 questions · 57 still being checked

Short Answer Questions 40–46 (part 5 of 6)

  1. Exercise 40

    Carbon, Group (14\displaystyle 14) element in the Periodic Table, is known to form compounds with many elements.
    Write an example of a compound formed with
    (a)
    chlorine (Group 17\displaystyle 17 of Periodic Table)
    (b)
    oxgygen (Group 16\displaystyle 16 of Periodic Table)

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    NCERT’s answer
    (a)
    Carbon tetrachloride \(\displaystyle \left(\mathrm{CCl}_4\right)\)
    (b)
    Carbon dioxide \(\displaystyle \left(\mathrm{CO}_2\right)\)
    \[\text{valency: C}=4,\quad \text{Cl}=1,\quad \text{O}=2 \] \[4 = 4\times 1\ (\text{four Cl}),\qquad 4 = 2\times 2\ (\text{two O}) \] Answer: (a) \(\displaystyle \mathrm{CCl_4}\), (b) \(\displaystyle \mathrm{CO_2}\).

  2. Exercise 41

    In electron dot structure, the valence shell electrons are represented by crosses or dots.
    (a)
    The atomic number of chlorine is 17. Write its electronic configuration
    (b)
    Draw the electron dot structure of chlorine molecule.

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    NCERT’s answer
    (a)
    K, L, M
    $\displaystyle 2$, $\displaystyle 8$, $\displaystyle 7$
    (b)
    NCERT_Solution_Class10_Science_Exemplar_Ch4_Q41_ncert
    (a) \[\text{Cl}\ (Z=17):\quad K=2,\ L=8,\ M=7 \] (b) Seven valence electrons leave each Cl one short of an octet, so the two atoms share one pair. \[7 + 1 = 8\ \text{electrons around each Cl} \]Answer: (a) \(\displaystyle 2, 8, 7\). (b) \(\displaystyle \mathrm{Cl{:}Cl}\), one shared pair and three lone pairs on each Cl.

  3. Exercise 42

    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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    NCERT’s answer
    Carbon exhibits catenation much more than silicon or any other element due to its smaller size which makes the C-C bonds strong while the \(\displaystyle \mathrm{Si}-\mathrm{Si}\) bonds are comparatively weaker due to its large size.
    Both C and Si have four valence electrons, so both can catenate, but the strength differs sharply. \[E_{\mathrm{C-C}} \approx 348\ \mathrm{kJ\,mol^{-1}}, \qquad E_{\mathrm{Si-Si}} \approx 226\ \mathrm{kJ\,mol^{-1}} \] \[r_{\mathrm{Si}} > r_{\mathrm{C}} \;\Rightarrow\; \text{orbital overlap}\downarrow \;\Rightarrow\; E_{\mathrm{Si-Si}}\downarrow \] Carbon's small size gives strong orbital overlap and stable long chains, branches and rings. Silicon's larger atoms overlap poorly, so it catenates only up to short chains such as \(\displaystyle \mathrm{Si_6H_{14}}\), which are also less stable. Answer: Carbon's catenation is far stronger than silicon's — its C–C bonds are shorter, more stable, and less reactive.

  4. Exercise 43

    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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    NCERT’s answer
    Hint— The two can be distinguished by subjecting them to the flame. Saturated hydrocarbons generally give a clear flame while unsaturated hydrocarbons give a yellow flame with lots of black smoke.
    Flame test: burn each gas in air. \[\mathrm{C_2H_6}\ (80\%\ \mathrm{C}) \to \text{clean blue flame}, \qquad \mathrm{C_2H_4}\ (85.7\%\ \mathrm{C}) \to \text{yellow sooty flame} \] Bromine water, kept out of sunlight: pass each gas through it.\[\mathrm{CH_2{=}CH_2} + \mathrm{Br_2} \longrightarrow \mathrm{CH_2Br{-}CH_2Br} \] Ethene decolourises it at once; ethane cannot add bromine, so the colour stays. In sunlight ethane would substitute instead and spoil the test. Answer: Ethane burns with a clean flame, ethene with a yellow sooty flame; ethene also decolourises bromine water, ethane does not.

  5. Exercise 44

    Match the reactions given in Column (A) with the names given in column (B).
    Column (A)Column (B)
    (a) CH3OH+CH3COOH→H+CH3COOCH3+H2O\displaystyle \mathrm{CH}_3 \mathrm{OH}+\mathrm{CH}_3 \mathrm{COOH} \xrightarrow{\mathrm{H}^{+}} \mathrm{CH}_3 \mathrm{COOCH}_3+\mathrm{H}_2 \mathrm{O}(i) Addition reaction
    (b) CH2=CH2+H2→NiCH3−CH3\displaystyle \mathrm{CH}_2=\mathrm{CH}_2+\mathrm{H}_2 \xrightarrow{\mathrm{Ni}} \mathrm{CH}_3-\mathrm{CH}_3(ii) Substitution reaction
    (c) CH4+Cl2→ Sunlight CH3Cl+HCl\displaystyle \mathrm{CH}_4+\mathrm{Cl}_2 \xrightarrow{\text { Sunlight }} \mathrm{CH}_3 \mathrm{Cl}+\mathrm{HCl}(iii) Neutralisation reaction
    (d) CH3COOH+NaOH⟶CH3COONa+H2O\displaystyle \mathrm{CH}_3 \mathrm{COOH}+\mathrm{NaOH} \longrightarrow \mathrm{CH}_3 \mathrm{COONa}+\mathrm{H}_2 \mathrm{O}(iv) Esterification reaction

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    NCERT’s answer
    (a)
    —(iv)
    (b)
    — (i) (c) — (ii)
    (d)
    — (iii)
    Each pair is fixed by what happens to the bonds.
    (a)
    Alcohol and acid form an ester and water: esterification.
    (b)
    The \(\displaystyle \mathrm{C{=}C}\) bond gains \(\displaystyle \mathrm{H_2}\) over Ni: addition.
    (c)
    An H of \(\displaystyle \mathrm{CH_4}\) is replaced by Cl in sunlight: substitution.
    (d)
    Acid and base give a salt and water: neutralisation.
    Answer: (a)–(iv), (b)–(i), (c)–(ii), (d)–(iii).

  6. Exercise 45

    Write the structural formulae of all the isomers of hexane.

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    NCERT’s answer
    (a)
    (b)
    (c)
    (d)
    (e)
    NCERT_Solution_Class10_Science_Exemplar_Ch4_Q45_ncert
    NCERT_Solution_Class10_Science_Exemplar_Ch4_Q45_ncert_2
    NCERT_Solution_Class10_Science_Exemplar_Ch4_Q45_ncert_3
    NCERT_Solution_Class10_Science_Exemplar_Ch4_Q45_ncert_4
    NCERT_Solution_Class10_Science_Exemplar_Ch4_Q45_ncert_5
    All isomers share \(\displaystyle \mathrm{C_6H_{14}}\): one straight chain and four branched skeletons. \[\text{n-hexane:}\quad \mathrm{CH_3{-}CH_2{-}CH_2{-}CH_2{-}CH_2{-}CH_3} \] \[\text{2-methylpentane:}\quad \mathrm{CH_3{-}CH(CH_3){-}CH_2{-}CH_2{-}CH_3} \] \[\text{3-methylpentane:}\quad \mathrm{CH_3{-}CH_2{-}CH(CH_3){-}CH_2{-}CH_3} \] \[\text{2,2-dimethylbutane:}\quad \mathrm{CH_3{-}C(CH_3)_2{-}CH_2{-}CH_3} \] \[\text{2,3-dimethylbutane:}\quad \mathrm{CH_3{-}CH(CH_3){-}CH(CH_3){-}CH_3} \] Answer: Hexane has $\displaystyle 5$ structural isomers, listed above.

  7. Exercise 46

    What is the role of metal or reagents written on arrows in the given chemical reactions?
    (a)
    (b)
    CH3COOH+CH3CH2OH→ Conc. H2SO4CH3COOC2H5+H2O\displaystyle \mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \xrightarrow{\text { Conc. } \mathrm{H}_2 \mathrm{SO}_4} \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5+\mathrm{H}_2 \mathrm{O}
    (c)
    CH3CH2OH→ Heat  Alk. KMnO4CH3COOH\displaystyle \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \xrightarrow[\text { Heat }]{\text { Alk. } \mathrm{KMnO}_4} \mathrm{CH}_3 \mathrm{COOH}
    NCERT_Question_Class10_Science_Exemplar_Ch4_Q46

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    NCERT’s answer
    Hint— (a) Ni acts as a catalyst
    (b)
    Concentrated \(\displaystyle \mathrm{H}_2 \mathrm{SO}_4\) acts as a catalyst
    (c)
    Alkaline \(\displaystyle \mathrm{KMnO}_4\) acts as an oxidising agent
    (a)
    \(\displaystyle \mathrm{H_2}\) adds across \(\displaystyle \mathrm{C{=}C}\) on the Ni surface, which speeds up the addition.
    (b)
    It also takes up the \(\displaystyle \mathrm{H_2O}\) formed, so the ester does not hydrolyse back.
    (c)
    It supplies oxygen to the alcohol; heat drives the oxidation.
    \[\mathrm{CH_3CH_2OH} + 2[\mathrm{O}] \longrightarrow \mathrm{CH_3COOH} + \mathrm{H_2O} \]
    Answer: (a) Ni is a catalyst; (b) conc. \(\displaystyle \mathrm{H_2SO_4}\) is a catalyst and dehydrating agent; (c) alk. \(\displaystyle \mathrm{KMnO_4}\) is an oxidising agent.