2023 H2 Chem P1 Q16 — Introduction to Organic Solution

The answer is C (23).
An \(sp^3\) carbon hybridises its 2s orbital with three 2p orbitals, an \(sp^2\) carbon with two 2p orbitals, and an \(sp\) carbon with one 2p orbital.
• Propane, \(\text{CH}_3\text{CH}_2\text{CH}_3\): 3 \(sp^3\) carbons, \(3 \times 3 = 9\) p orbitals.
• Benzene, \(\text{C}_6\text{H}_6\): 6 \(sp^2\) carbons, \(6 \times 2 = 12\) p orbitals.
• Ethyne, H-C\(\equiv\)C-H: 2 \(sp\) carbons, \(2 \times 1 = 2\) p orbitals.
Total \(= 9 + 12 + 2 = 23\).
• A is wrong: 20 is 3 too few; it matches treating the propane carbons as \(sp^2\) (\(6 + 12 + 2 = 20\)), but propane carbons form four single bonds and are \(sp^3\).
• B is wrong: 21 is \(9 + 12\), which leaves out the 2 p orbitals hybridised in ethyne.
• D is wrong: 33 is \(11 \times 3\), counting all three 2p orbitals of every carbon; in benzene and ethyne some 2p orbitals stay unhybridised to form the \(\pi\) bonds.