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Bond Order from Molecular Orbital Diagrams

Bonding & Molecular StructureAdvanced7 min read
On this page
  1. The method
  2. Example 1: H₂⁺ vs H₂
  3. Example 2: N₂, N₂⁺ and N₂⁻
  4. Example 3: O₂ and its ions
  5. Example 4: nitric oxide, NO
  6. Example 5: cyanide, CN⁻
  7. Example 6: C₂ and the importance of s–p mixing
  8. Summary of trends
  9. Common mistakes
  10. Practice
  11. Key takeaways

Bond order calculations from molecular orbital (MO) diagrams are a standard advanced exam question, and they reward a careful, systematic approach. Once you’ve filled the diagram correctly, you can predict not only the bond order but also relative bond lengths and strengths and whether the species is paramagnetic. This article gives the method and then works through six examples, including ions and molecules made of two different atoms. For the theory, see molecular orbital theory for beginners.

The method

  1. Count the valence electrons in the species. Add one for each negative charge; subtract one for each positive charge. (Core 1s electrons fill σ1s and σ*1s equally and cancel out, so they can be ignored for period 2.)
  2. Choose the correct energy order:
    • For species with up to 14 total electrons based on B, C, N (s–p mixing): σ2s, σ*2s, π2p (×2), σ2p, π*2p (×2), σ*2p.
    • For species based on O, F, Ne: σ2s, σ*2s, σ2p, π2p (×2), π*2p (×2), σ*2p.
    • For heteronuclear species like NO and CO, use the N₂-type order (it gives the right answer for bond order in these examples).
  3. Fill the orbitals from lowest energy, two electrons per orbital, following Hund’s rule for the degenerate (equal-energy) π pairs.
  4. Count bonding (σ2s, π2p, σ2p) and antibonding (σ*2s, π*2p, σ*2p) electrons.
  5. Bond order = ½(bonding − antibonding).
  6. Unpaired electrons? Any → paramagnetic; none → diamagnetic.

A time-saver: the σ2s and σ*2s orbitals are always both full for these species, so they cancel. You can just count from σ2p/π2p upward.

Example 1: H₂⁺ vs H₂

H₂⁺ has 1 electron: (σ1s)¹.

  • Bond order = ½(1 − 0) = ½. One unpaired electron → paramagnetic.

H₂ has 2 electrons: (σ1s)².

  • Bond order = 1, diamagnetic.

Prediction: H₂ has a stronger, shorter bond. Data: H₂ bond energy 436 kJ mol⁻¹, length 74 pm; H₂⁺ about 256 kJ mol⁻¹, length 106 pm. ✓

Example 2: N₂, N₂⁺ and N₂⁻

N₂: 10 valence electrons.

  • (σ2s)²(σ*2s)²(π2p)⁴(σ2p)²
  • Bonding: 2 + 4 + 2 = 8. Antibonding: 2.
  • Bond order = ½(8 − 2) = 3. All paired → diamagnetic.

N₂⁺: 9 electrons — remove one from the highest occupied orbital, σ2p.

  • (σ2s)²(σ*2s)²(π2p)⁴(σ2p)¹
  • Bond order = ½(7 − 2) = 2½. One unpaired electron → paramagnetic.

N₂⁻: 11 electrons — the extra electron goes into π*2p.

  • Bond order = ½(8 − 3) = 2½. Paramagnetic.

Prediction: both ions have weaker, longer bonds than N₂. Removing an electron from a bonding orbital weakens the bond — as does adding one to an antibonding orbital. N₂’s bond order of 3 is the highest of any period 2 diatomic, which fits its enormous bond energy (945 kJ mol⁻¹) and chemical inertness.

Example 3: O₂ and its ions

This is the most famous series. For oxygen species, use the O₂-type order.

Species Valence e⁻ Configuration (from σ2p) Bonding Antibonding Bond order Unpaired Magnetism
O₂⁺ 11 σ2p² π2p⁴ π*2p¹ 8 3 2½ 1 Paramagnetic
O₂ 12 σ2p² π2p⁴ π*2p² 8 4 2 2 Paramagnetic
O₂⁻ (superoxide) 13 σ2p² π2p⁴ π*2p³ 8 5 1½ 1 Paramagnetic
O₂²⁻ (peroxide) 14 σ2p² π2p⁴ π*2p⁴ 8 6 1 0 Diamagnetic

(Counts include σ2s² and σ*2s², which cancel: bonding = 2 + 2 + 4 = 8, antibonding = 2 + π* electrons.)

Prediction: bond length increases and bond strength decreases in the order O₂⁺ < O₂ < O₂⁻ < O₂²⁻.

Data (approximate O–O bond lengths): O₂⁺ 112 pm, O₂ 121 pm, O₂⁻ 133 pm, O₂²⁻ 149 pm. ✓

A striking result: removing an electron from O₂ to form O₂⁺ strengthens the bond, because the electron comes from an antibonding π* orbital. Lewis structures can’t predict this. The superoxide ion, O₂⁻, is the reactive species formed when electrons leak from the respiratory chain (see the chemistry of ageing); the peroxide ion, O₂²⁻, has a single O–O bond, as in hydrogen peroxide. For O₂’s magnetism in detail, see why is oxygen paramagnetic?.

Example 4: nitric oxide, NO

N (5) + O (6) = 11 valence electrons.

  • Using the N₂-type order: (σ2s)²(σ*2s)²(π2p)⁴(σ2p)²(π*2p)¹
  • Bonding: 8. Antibonding: 3.
  • Bond order = ½(8 − 3) = 2½. One unpaired electron → paramagnetic.

NO is a radical — a molecule with an unpaired electron — which is why it’s so reactive. It reacts rapidly with oxygen in air to form brown NO₂. In the body, NO acts as a signalling molecule that relaxes blood vessels.

NO⁺ (the nitrosonium ion, 10 electrons) loses that antibonding electron: bond order 3, diamagnetic, shorter bond (about 106 pm vs 115 pm in NO). It’s isoelectronic with N₂ and CO.

Example 5: cyanide, CN⁻

C (4) + N (5) + 1 (charge) = 10 electrons.

  • (σ2s)²(σ*2s)²(π2p)⁴(σ2p)²
  • Bond order = ½(8 − 2) = 3. Diamagnetic.

CN⁻ is isoelectronic with N₂, CO and NO⁺ — all have 10 valence electrons and bond order 3. The same MO diagram (with small energy shifts) describes all four, which is a nice example of how MO theory reveals patterns across different substances.

Neutral CN (9 electrons) has bond order 2½ and one unpaired electron: it’s the cyano radical, detected in interstellar space and comet tails.

Example 6: C₂ and the importance of s–p mixing

C₂ has 8 valence electrons.

  • With s–p mixing (correct for carbon): (σ2s)²(σ*2s)²(π2p)⁴
    • Bond order = ½(6 − 2) = 2. The four π electrons fill both π orbitals in pairs → diamagnetic. ✓ (C₂ is observed to be diamagnetic.)
  • Without s–p mixing (the O₂-type order, which would be wrong here): (σ2s)²(σ*2s)²(σ2p)²(π2p)²
    • Bond order still 2, but the two π electrons would be unpaired → paramagnetic ✗.

So the order of orbitals doesn’t change the bond order here, but it does change the magnetic prediction. Experimental magnetism is how chemists know which order is correct. (Modern calculations suggest C₂’s bonding is more subtle still, but bond order 2 and diamagnetism are the standard answers at this level.) B₂ is the other classic case: with s–p mixing, its two π electrons are unpaired, making it paramagnetic, as observed.

  • Higher bond order → shorter, stronger bond.
  • Removing an electron from a bonding MO lowers bond order; from an antibonding MO raises it.
  • Adding an electron to an antibonding MO lowers bond order.
  • Isoelectronic species (same number of electrons) have the same bond order.

Common mistakes

  • Forgetting the charge when counting electrons.
  • Using the wrong energy order for B₂, C₂ and N₂ (s–p mixing) versus O₂ and F₂.
  • Pairing electrons in the π* orbitals too early — Hund’s rule says fill singly first.
  • Counting σ*2s as bonding. Watch the stars.
  • Assuming “more electrons = stronger bond.” It depends on which orbitals they enter.

Practice

Calculate the bond order and state whether the species is paramagnetic:

  1. F₂ and F₂⁺
  2. CO
  3. B₂

Answers:

  1. F₂: 14 electrons, bond order ½(8 − 6) = 1, diamagnetic. F₂⁺: 13 electrons, one fewer antibonding electron, bond order 1½, paramagnetic — a stronger bond than F₂.
  2. CO: 10 electrons, bond order 3, diamagnetic.
  3. B₂: 6 electrons, (σ2s)²(σ*2s)²(π2p)², bond order 1, two unpaired electrons → paramagnetic.

Key takeaways

  • Bond order = ½(bonding − antibonding electrons).
  • Use the N₂-type order (π below σ2p) for B₂, C₂, N₂ and the O₂-type order for O₂, F₂.
  • Fill degenerate π orbitals singly first; unpaired electrons mean paramagnetism.
  • Bond order predicts bond length and strength trends, including the surprising O₂⁺ > O₂.
  • Isoelectronic species (N₂, CO, CN⁻, NO⁺) share bond order 3.

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