Which one of the following compound forms salt on reaction with NaNH₂?

The correct option is C₂H₂.

Sodium amide (${\mathrm{NaNH}}_2$) is a very strong base. It is strong enough to deprotonate compounds that have relatively acidic hydrogens, such as terminal alkynes (alkynes with a hydrogen atom bonded to a triply bonded carbon).

Let's analyze the acidity of the given hydrocarbons:
1. Ethyne ($C_2{H}_2$ or $\mathrm{HC}\equiv \mathrm{CH}$): The carbon atoms in ethyne are $\mathrm{sp}$ hybridized. An $\mathrm{sp}$ hybridized carbon has 50% s-character, making it highly electronegative. Consequently, the $C-H$ bond is quite polar, and the hydrogen atom is weakly acidic.
2. Ethene ($C_2{H}_4$): The carbons are ${\mathrm{sp}}^2$ hybridized (33.3% s-character), which makes them less electronegative and significantly less acidic than ethyne.
3. Ethane ($C_2{H}_6$): The carbons are ${\mathrm{sp}}^3$ hybridized (25% s-character), making them the least acidic.
4. Benzene ($C_6{H}_6$): The carbons are ${\mathrm{sp}}^2$ hybridized, so it is also not acidic enough to react with sodium amide under standard conditions.

Due to the high acidity of the terminal hydrogen in ethyne ($C_2{H}_2$), it reacts readily with ${\mathrm{NaNH}}_2$ to form a sodium salt (sodium acetylide) along with the evolution of ammonia gas. The chemical reaction is represented as follows:

$\mathrm{HC}\equiv \mathrm{CH}+{\mathrm{NaNH}}_2\to \mathrm{HC}\equiv C^{-}{\mathrm{Na}}^{+}+{\mathrm{NH}}_3$