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NCERT Exemplar · Q58

Q.Match the species in Column I with the bond order in Column II.
Column I

(i) NO
(ii) CO
(iii) O2^-
(iv) O2
Column II
(a) 1.5
(b) 2.0
(c) 2.5
(d) 3.0
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Bond order measures the number of electron pairs holding two atoms together. Using molecular orbital theory, we count bonding and antibonding electrons to find: NO → 2.5, CO → 3.0, O₂⁻ → 1.5, O₂ → 2.0.

Bond order is the net number of chemical bonds between a pair of atoms. In molecular orbital theory, electrons occupy orbitals that span both nuclei—some strengthen the bond (bonding orbitals), others weaken it (antibonding orbitals). The formula captures this tug-of-war:

Bond Order=12(electrons in bonding MOs−electrons in antibonding MOs)\text{Bond Order} = \frac{1}{2}\left(\text{electrons in bonding MOs} - \text{electrons in antibonding MOs}\right)

A higher bond order means a stronger, shorter bond. Let's work through each species by filling molecular orbitals according to energy and counting carefully.

1. NO (Nitric oxide): 15 electrons total

Nitrogen contributes 7 electrons, oxygen contributes 8. For diatomic molecules with Z≤14Z \leq 14 (where ZZ is the total nuclear charge), the σ2p\sigma_{2p} orbital lies above the π2p\pi_{2p} pair. The filling order is:

σ1s2 σ1s∗2 σ2s2 σ2s∗2 π2p4 σ2p2 π2p∗1\sigma_{1s}^2 \, \sigma_{1s}^*{}^2 \, \sigma_{2s}^2 \, \sigma_{2s}^*{}^2 \, \pi_{2p}^4 \, \sigma_{2p}^2 \, \pi_{2p}^*{}^1

  • Bonding electrons: 2+2+4+2=102 + 2 + 4 + 2 = 10
  • Antibonding electrons: 2+2+1=52 + 2 + 1 = 5

Bond Order=12(10−5)=2.5\text{Bond Order} = \frac{1}{2}(10 - 5) = 2.5

2. CO (Carbon monoxide): 14 electrons total

Carbon contributes 6, oxygen contributes 8. Same orbital ordering as NO:

σ1s2 σ1s∗2 σ2s2 σ2s∗2 π2p4 σ2p2\sigma_{1s}^2 \, \sigma_{1s}^*{}^2 \, \sigma_{2s}^2 \, \sigma_{2s}^*{}^2 \, \pi_{2p}^4 \, \sigma_{2p}^2

  • Bonding electrons: 2+2+4+2=102 + 2 + 4 + 2 = 10
  • Antibonding electrons: 2+2=42 + 2 = 4

Bond Order=12(10−4)=3.0\text{Bond Order} = \frac{1}{2}(10 - 4) = 3.0

This triple bond makes CO extraordinarily stable.

3. O₂⁻ (Superoxide ion): 17 electrons total

Neutral O₂ has 16 electrons; the anion adds one more. For oxygen (Z=16>14Z = 16 > 14), the σ2p\sigma_{2p} orbital drops below the π2p\pi_{2p} pair:

σ1s2 σ1s∗2 σ2s2 σ2s∗2 σ2p2 π2p4 π2p∗3\sigma_{1s}^2 \, \sigma_{1s}^*{}^2 \, \sigma_{2s}^2 \, \sigma_{2s}^*{}^2 \, \sigma_{2p}^2 \, \pi_{2p}^4 \, \pi_{2p}^*{}^3

  • Bonding electrons: 2+2+2+4=102 + 2 + 2 + 4 = 10
  • Antibonding electrons: 2+2+3=72 + 2 + 3 = 7

Bond Order=12(10−7)=1.5\text{Bond Order} = \frac{1}{2}(10 - 7) = 1.5

4. O₂ (Dioxygen): 16 electrons total …

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