Question

which of these is the correct lewis structure for hydrogen cyanide, hcn?
h — c — n
h = c = n
h — c = n
h — c ≡ n

Explanation:

Step1: Calculate Valence Electrons

H has 1, C has 4, N has 5. Total: \(1 + 4 + 5 = 10\) valence electrons.

Step2: Analyze Each Structure

• First structure: Single bonds (H - C, C - N) + lone pairs. Count electrons: \(2 + 2 + (4 + 6) = 14\) (too many).
• Second structure: Double bonds (H = C, C = N) + lone pairs. Count: \(4 + 4 + 6 = 14\) (too many).
• Third structure: H - C (2e), C = N (4e), lone pairs. Count: \(2 + 4 + (2 + 4) = 12\) (too many).
• Fourth structure: H - C (2e), C ≡ N (6e), N lone pair (2e). Total: \(2 + 6 + 2 = 10\) (matches total valence electrons). Also, C and N have octets (C: 4 + 32 = 10? Wait, C is bonded to H (1) and N (3), so 4 electrons (1 single, 1 triple: 1 + 32 = 7? Wait, no: Lewis structure for HCN: H - C ≡ N, with N having 1 lone pair (2 electrons). Valence electrons: H (1), C (4: 1 from H, 3 from N triple bond), N (5: 3 from triple bond, 2 from lone pair). Total: 1 + 4 + 5 = 10. The fourth structure has H - C (single, 2e), C ≡ N (triple, 6e), N lone pair (2e): 2 + 6 + 2 = 10. Correct.

Answer:

The fourth option: H—C≡N: (the last option with the triple bond between C and N and a lone pair on N)