Q4. How many σ and π bonds are there in benzene, C6H6 (a ring of six carbons, each with one H, written with three alternating C=C bonds)?
Six C–H σ bonds and six C–C σ bonds round the ring make 12σ. The three C=C double bonds add 3π. Answer: 12σ, 3π.
Q5. Count the σ and π bonds in acrylonitrile, CH2=CH–C≡N.
C–H: 3σ. C=C: 1σ + 1π. C–C: 1σ. C≡N: 1σ + 2π. Answer: 6σ, 3π.
Q6. Why is a σ bond stronger than a π bond between the same two atoms?
Bond strength depends on the extent of overlap. Head-on overlap along the internuclear axis puts the orbitals’ densest parts together, so the overlap is large. Sideways overlap of parallel p orbitals is smaller, so the π bond is weaker.
Q7. Taking z as the internuclear axis, say whether each pair gives positive, negative or zero overlap: (a) 2pz and 2pz with + lobes facing; (b) 2pz on one atom and 2px on the other; (c) 2px and 2px, side by side with like signs on the same side; (d) 2pz and 2pz with a + lobe facing a − lobe.
(a) Positive: like lobes meet head-on, giving a σ bond. (b) Zero: the + and − lobes of px meet the pz lobe equally, so the parts cancel. (c) Positive: like lobes overlap sideways, giving a π bond. (d) Negative overlap for the drawn phases: their sum is destructive. Orbital signs themselves are conventional; bonding versus antibonding depends on the relative phases and combination coefficients.
Q8. Oxygen is [He] 2s2 2px2 2py1 2pz1. By simple overlap with H 1s orbitals: (a) how many O–H bonds form, and from which orbitals? (b) Are they σ or π? (c) Why does oxygen need no promotion while carbon does?
(a) Two, from the half-filled 2py and 2pz; the 2s and 2px pairs stay as lone pairs. (b) Both σ (s–p head-on overlap). (c) Oxygen already has the two half-filled orbitals it needs; ground-state carbon has two but needs four.