U2.6 · Resonance and Formal ChargePractice · 共振与形式电荷

Total valence electrons = 5 + 2 + 1 = 8. The two N–H bonds use 4, and the other 4 are two lone pairs on N. Formal charge = valence electrons − lone-pair electrons − bonds = 5 − 4 − 2 = −1. The whole charge of the ion sits on N; each H has a formal charge of 0.

$\ce{O3}$ has 2 equivalent resonance structures, with the double bond on either side. The real molecule is the resonance hybrid: both O–O bonds are identical, with an average bond order of (2 + 1) ÷ 2 = 1.5, so the length lies between the single and double bonds (121–148 pm). The measured value is about 128 pm.

Formal charges in the order N, N, O: N=N=O (−1, +1, 0), N≡N–O (0, +1, −1), N–N≡O (−2, +1, +1). In N≡N=O the middle N would have 10 electrons, so it fails. The first two both have a total |formal charge| of 2, less than N–N≡O. Then look at where the negative charge sits: it belongs on the more electronegative atom, and O (3.44) beats N (3.04), so N≡N–O is most reasonable.

$\ce{NO2-}$ has 2 resonance structures, with the N=O double bond on either side and a lone pair on N, whose formal charge is 0. Resonance structures are different drawings of one ion; the ion doesn’t switch between them. The real ion is the hybrid: both N–O bonds have a bond order of 1.5 and the same length. Two double bonds would give N 10 electrons.

Total valence electrons = 7 + 4 × 6 + 1 = 32. The four single bonds use 8, three lone pairs on each O use 24, and Cl has no lone pairs. Formal charge on Cl = 7 − 0 − 4 = +3; each O is 6 − 6 − 1 = −1. The sum, +3 − 4 = −1, equals the charge on the ion.