Given below are two statements : Statement I : One mole of propyne reacts with excess of sodium to liberate half a mole of $$H_2 $$ gas. Statement II : Four g of propyne reacts with $$ NaNH_2 \text{ to liberate } NH_3$$ gas which occupies 224 mL at STP. In the light of the above statements, choose the most appropriate answer from the options given below:

We need to evaluate two statements about propyne ($$CH_3-C \equiv CH$$, molecular weight = 40 g/mol).

Statement I: One mole of propyne reacts with excess of sodium to liberate half a mole of H₂ gas.

Propyne has one terminal acidic hydrogen (on the $$\equiv CH$$). Only terminal alkynes react with sodium metal:

$$2 CH_3C \equiv CH + 2Na \rightarrow 2 CH_3C \equiv CNa + H_2$$

So 2 moles of propyne give 1 mole of H₂. Therefore, 1 mole of propyne gives $$\frac{1}{2}$$ mole of H₂.

Statement I is CORRECT.

Statement II: Four g of propyne reacts with NaNH₂ to liberate NH₃ gas which occupies 224 mL at STP.

4 g of propyne = $$\frac{4}{40} = 0.1$$ mol

$$CH_3C \equiv CH + NaNH_2 \rightarrow CH_3C \equiv CNa + NH_3$$

0.1 mol propyne produces 0.1 mol NH₃.

At STP, 0.1 mol of gas occupies $$0.1 \times 22400 = 2240$$ mL.

The statement says 224 mL, which would correspond to 0.01 mol. This is incorrect.

Statement II is INCORRECT.

The correct answer is Option 3: Statement I is correct but Statement II is incorrect.