Molecular orbital diagram of CO
Bond order 3, diamagnetic. 14 electrons: Nb = 10, Na = 4.
Molecular orbitals
14 electrons3bond order
diamagnetic
diamagnetic
(Nb − Na) / 2 = (10 − 4) / 2
σ1s² σ*1s² σ2s² σ*2s² (π2px² = π2py²) σ2pz²
Why this shape
- 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.
- 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.
- Two electron pairs point in opposite directions, along a line (linear electron geometry). With only two atoms, the molecule itself is a straight line.
- Two atoms always lie on a straight line, so there is no bond angle.
- The two atoms are different, so the bond is polar and so is CO.
Lewis structure
10 valence electronsi
A triple bond with a lone pair on each atom: the formal charges are -1 on C and +1 on O.
Formal charges: C −1, O +1
Filling the diagram
- CO has 14 electrons: 6 from C and 8 from O.
- With 14 electrons or fewer, the π2p pair lies below σ2pz and fills first.
- Configuration: σ1s² σ*1s² σ2s² σ*2s² (π2px² = π2py²) σ2pz²
- Bonding electrons Nb = 10, antibonding electrons Na = 4.
- Bond order = (Nb − Na) / 2 = (10 − 4) / 2 = 3.
- Every electron is paired, so CO is diamagnetic.
Class 11, Chapter 4 Chemical Bonding and Molecular Structure.
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.