The carbonyl carbon in an aldehyde or ketone is
A$sp^3$ hybridised, with bond angles of $\sim 109.5^\circ$
B$sp$ hybridised, with bond angles of $180^\circ$
C$sp^2$ hybridised, with bond angles of $\sim 120^\circ$
Dpurely $p$-orbital, with no hybridisation
Answer & Solution
Correct answer: C. $sp^2$ hybridised, with bond angles of $\sim 120^\circ$
1. Your syllabus: the carbonyl carbon has THREE atoms bonded to it (the doubly-bonded O and two other atoms — in formaldehyde, two H's; in acetaldehyde, one H and one C; in acetone, two C's).
2. Three sigma bonds around the central carbon → $sp^2$ HYBRIDISATION. Three sp$^2$ hybrids lie in a plane separated by $120^\circ$.
3. The fourth (unhybridised) $p$ orbital perpendicular to the plane forms the $\pi$ bond with the oxygen.
4. Hence the carbonyl group is PLANAR, with bond angles close to $120^\circ$. Real molecules show slight distortion (e.g. $\sim 121.7^\circ$ for $\angle\mathrm{HCO}$ in formaldehyde).
5. Option A is for tetrahedral $sp^3$ carbons (alkanes). Option B is the geometry of alkynes ($sp$). Option D is wrong — bonded atoms require hybridisation.
_Source: syllabus Chemistry Part 2, Ch 8, §8.1.2 (Structure of the Carbonyl Group), p. 4._
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