CBSE Class 10 Science 2022 Outside Delhi Term II Set 2

Question : 1 of 7

Marks: +1, -0

Consider the following molecular formulae of carbon compounds:
(i)

{\CH}_3 {\COOH}

(ii)

{\CH}_3 {\OH}

(iii)

{\C}_2 {\H}_6

(iv)

{\C}_3 {\H}_4

(iv)

{\C}_4 {\H}_8

(a) Which one of these compounds belongs to homologous series of alcohols?
(b) Identify the compound having triple bond between carbon - carbon atoms.
(c) Write the molecular formula of the first member of the homologous series to which

{\CH}_3 {\COOH}

belongs.
(d) Write the general formula of the series to which the compound

{\C}_4 {\H}_8

belongs.

Solution:

(a)

{\CH}_3 {\OH}

, methanol belongs to homologous series of alcohols and it has the general formula of

{\R}- {\OH}

.
(b)

{\C}_3 {\H}_4

Lewis Structure of

{\C}_3 {\H}_4

\H-\C↖{|↖{\H}}↙{|↙{\H}}-\C≡\C-\H

{\C}_n {\H}_{2 n+1} {\COOH}

. The molecular formula is

{\HCOOH}

(Methanoic acid).
(d) The general formula for alkenes can be written as

{\C}_n {\H}_{2 n^{'}}

where

n=2,3,4

and

{\C}_4 {\H}_8

belongs to alkene series.

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