The answer is A (\(\text{HC} \equiv \text{CH}\)).
Hybridisation of a C atom follows from the number of \(\sigma\) bonds it forms (no lone pairs on these C atoms): 2 \(\sigma\) bonds = sp, 3 = sp\(^2\), 4 = sp\(^3\). A C atom in a triple bond forms 2 \(\sigma\) bonds (sp); a C atom in a C=C double bond forms 3 \(\sigma\) bonds (sp\(^2\)). H atoms are not hybridised.
• A: \(\text{H}-\text{C} \equiv \text{C}-\text{H}\): each C forms 2 \(\sigma\) bonds, so there are exactly 2 sp atoms.
• B is wrong: in ethene each C forms 3 \(\sigma\) bonds, so there are 2 sp\(^2\) atoms and no sp atoms.
• C is wrong: in \(\text{HC} \equiv \text{C}-\text{C} \equiv \text{CH}\) all four C atoms are in triple bonds and form 2 \(\sigma\) bonds each, so there are 4 sp atoms, not 2.
• D is wrong: in buta-1,3-diene all four C atoms form 3 \(\sigma\) bonds, so there are 4 sp\(^2\) atoms and no sp atoms.
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