Identify the compound

Propyne is a terminal alkyne: $\mathrm{CH_3-C\equiv CH}$

Terminal alkynes contain one acidic hydrogen attached to the carbon of the triple bond.

Check Statement I

Reaction of a terminal alkyne with sodium metal: $2\,\mathrm{RC\equiv CH} + 2\,\mathrm{Na} \rightarrow 2\,\mathrm{RC\equiv CNa} + \mathrm{H_2}$

Thus, 2 moles of terminal alkyne give 1 mole of $\mathrm{H_2}$.

Therefore, 1 mole of propyne gives: $\frac{1}{2}\text{ mole of } \mathrm{H_2}$

So, Statement I is correct.

Check Statement II

Reaction of propyne with sodium amide: $\mathrm{CH_3-C\equiv CH + NaNH_2 \rightarrow CH_3-C\equiv CNa + NH_3}$

Stoichiometry shows: $1\text{ mole propyne } \rightarrow 1\text{ mole } \mathrm{NH_3}$

Moles of propyne in 4 g

Molar mass of propyne $\mathrm{C_3H_4}$: $3(12) + 4(1) = 36 + 4 = 40\,\mathrm{g/mol}$

Moles of propyne: $\frac{4}{40} = 0.1\,\mathrm{mol}$

Hence moles of $\mathrm{NH_3}$ liberated: $0.1\,\mathrm{mol}$

Volume of $\mathrm{NH_3}$ at STP

At STP, 1 mole gas occupies: $22.4\,\mathrm{L} = 22400\,\mathrm{mL}$

So volume of $0.1$ mol $\mathrm{NH_3}$: $0.1 \times 22.4 = 2.24\,\mathrm{L} = 2240\,\mathrm{mL}$

But the statement says 224 mL, which is incorrect.

Therefore, Statement II is incorrect.

Final conclusion

• Statement I: Correct
• Statement II: Incorrect

Therefore, the correct option is: $\boxed{\text{D}}$