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Molecular orbitals

Molecular orbital diagram of CO

Bond order 3, diamagnetic. 14 electrons: Nb = 10, Na = 4.

OC
  • C
  • O

COCarbon monoxide

Linear

C: 1 sigma bond, 1 lone pair

i

Two atoms always lie on a line. The dipole moment of CO is very small.

  • Hybridisationsp
  • Polaritypolar
  • Bond order3

Molecular orbitals

14 electrons
3bond order
diamagnetic

(Nb − Na) / 2 = (10 − 4) / 2

1s1s2s2s2p2pσ1sσ*1sσ2sσ*2sπ2pσ2pzπ*2pσ*2pzC atomMoleculeO atomEnergy

σ1s² σ*1s² σ2s² σ*2s² (π2px² = π2py²) σ2pz²

Why this shape

  1. C has 4 valence electrons. In the Lewis structure it makes one triple bond to O and keeps one lone pair, which leaves it a formal charge of −1.
  2. Steric number = one sigma bond + one lone pair = 2, so C is sp hybridised. A double or triple bond counts once: only its sigma bond uses a hybrid orbital.
  3. Two electron pairs point in opposite directions, along a line (linear electron geometry). With only two atoms, the molecule itself is a straight line.
  4. Two atoms always lie on a straight line, so there is no bond angle.
  5. The two atoms are different, so the bond is polar and so is CO.

Lewis structure

10 valence electrons
i

A triple bond with a lone pair on each atom: the formal charges are -1 on C and +1 on O.

CO−+

Formal charges: C −1, O +1

Working

Filling the diagram

  1. CO has 14 electrons: 6 from C and 8 from O.
  2. With 14 electrons or fewer, the π2p pair lies below σ2pz and fills first.
  3. Configuration: σ1s² σ*1s² σ2s² σ*2s² (π2px² = π2py²) σ2pz²
  4. Bonding electrons Nb = 10, antibonding electrons Na = 4.
  5. Bond order = (Nb − Na) / 2 = (10 − 4) / 2 = 3.
  6. Every electron is paired, so CO is diamagnetic.

Class 11, Chapter 4 Chemical Bonding and Molecular Structure.

Questions

MO diagram of CO: quick answers

What is the molecular orbital configuration of CO?

σ1s² σ*1s² σ2s² σ*2s² (π2px² = π2py²) σ2pz². With 14 electrons or fewer, the π2p pair lies below σ2pz and fills first.

What is the bond order of CO from its MO diagram?

3: Nb = 10 and Na = 4, so (10 - 4) / 2 = 3. It is diamagnetic.

Why is σ2pz above π2p in CO?

Up to N2 the 2s and 2p orbitals are close in energy and mix, which pushes σ2pz above the π2p pair.

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