Question

Statement 1: Acetylene on reacting with sodamide gives sodium acetylide and ammonia.
Statement 2: $sp$-hybridised carbon atoms of acetylene are considerably electronegative.

• A. If both Statement $1$ and Statement $2$ are true and the statement $2$ is the correct explanation of the statement $1$
• B. If both Statement $1$ and Statement $2$ are true but statement $2$ is not the correct explanation of the statement $1$
• C. If Statement $1$ is true but Statement $2$ is false
• D. If Statement $1$ is false but Statement $2$ is true

Answer 1

Answer: (A)

Acetylene on reaction with sodamide gives sodium acetylide and ammonia.
$CH\equiv CH\stackrel{NaN{H}_2}{\to }\underset{\stackrel{\text{monosodium}}{\mathrm{acetylide}}}{CH\equiv \stackrel{-}{C}\stackrel{+}{\mathrm{Na}}}+N{H}_3$
Here formation of sodium acetylide can be explained by $sp$ hydridisation of carbon atom. As we know that an electron in $s$-orbital is more tightly held than in a $p$-orbital. In $sp$ hydridisation, $5$-character is $50\%$ as compared to $s{p}^2(33\%)$ or $s{p}^3(25\%)$. So, due to large s-character the carbon atom is quite electronegative and hence $Na$ of $NaN{H}_2$ can replace $H^{+}$ of $C-H$ bond.
Hence statement $1$ and statement $2$ both are true and statement $2$ is the correct explanation of statement $1$.