7.1 · MO & VB theory

Molecular orbital theory

Atomic orbitals combine into bonding and antibonding MOs.

By the end you should be able to:

Build molecular orbital diagrams for first- and second-row diatomics
Calculate bond order and relate it to bond length and strength
Predict magnetism from MO occupancy and explain s–p mixing

Interactive

MO diagram builder

Fill molecular orbitals for diatomic molecules and watch bond order and magnetism update.

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Key idea

Bonding and antibonding molecular orbitals

Atomic orbitals on the two atoms combine (a linear combination of atomic orbitals, LCAO) into the same number of molecular orbitals (MOs) that belong to the whole molecule:

Bonding MO (in-phase combination): electron density builds up between the nuclei; lower in energy than the atomic orbitals it came from.
Antibonding MO (out-of-phase combination, labelled $\sigma^*$ or $\pi^*$ ): a node between the nuclei; higher in energy.
σ MOs are symmetric around the bond axis (from s orbitals, or $2p_z$ orbitals pointing at each other). π MOs come from side-on overlap of $2p_x$ or $2p_y$ orbitals, have a nodal plane containing the bond axis, and come in degenerate pairs.

For second-row atoms, the two $2s$ orbitals give $\sigma_{2s}$ and $\sigma^*_{2s}$ , and the six $2p$ orbitals give $\sigma_{2p}$ , two $\pi_{2p}$ , two $\pi^*_{2p}$ and $\sigma^*_{2p}$ . Each MO holds two electrons.

Formula

Bond order

\text{bond order} = \tfrac{1}{2}\left(\text{bonding electrons} - \text{antibonding electrons}\right)

A higher bond order means a shorter, stronger bond: $\ce{N2}$ (bond order 3, ≈ 110 pm, 945 kJ/mol) > $\ce{O2}$ (2, ≈ 121 pm, 498 kJ/mol) > $\ce{F2}$ (1, ≈ 142 pm, 159 kJ/mol).
Bond order 0 means no stable molecule: $\ce{He2}$ , $\ce{Be2}$ and $\ce{Ne2}$ do not exist.
Half-integer bond orders are allowed: $\ce{H2+}$ and $\ce{He2+}$ each have bond order ½.
Core electrons fill both $\sigma_{1s}$ and $\sigma^*_{1s}$ and cancel, so counting only valence electrons gives the same bond order.

Key idea

MO energy order with and without s–p mixing

For B, C and N the $2s$ and $2p$ energies are close, so $\sigma_{2s}$ and $\sigma_{2p}$ interact (s–p mixing) and $\sigma_{2p}$ is pushed above $\pi_{2p}$ . For O, F and Ne the $2s$ – $2p$ gap is large, mixing is small and $\sigma_{2p}$ stays below $\pi_{2p}$ .

Level (low to high)	$\ce{B2}$ , $\ce{C2}$ , $\ce{N2}$ , $\ce{CO}$ , $\ce{CN-}$	$\ce{O2}$ , $\ce{F2}$ , $\ce{Ne2}$
1	$\sigma_{2s}$	$\sigma_{2s}$
2	$\sigma^*_{2s}$	$\sigma^*_{2s}$
3	$\pi_{2p}$ (two MOs)	$\sigma_{2p}$
4	$\sigma_{2p}$	$\pi_{2p}$ (two MOs)
5	$\pi^*_{2p}$ (two MOs)	$\pi^*_{2p}$ (two MOs)
6	$\sigma^*_{2p}$	$\sigma^*_{2p}$

Only levels 3 and 4 swap. The order is set by the atoms, so ions keep their parent's order ( $\ce{N2+}$ is mixed, $\ce{O2-}$ is not). The swap matters for magnetism: $\ce{B2}$ is observed to be paramagnetic, which only the mixed order predicts.

Method

Filling an MO diagram

Count valence electrons: add both atoms' valence electrons, then subtract the charge ( $\ce{O2^2-}$ : 6 + 6 + 2 = 14).
Choose the level order: s–p mixed for diatomics of B, C and N (and CO, CN⁻); unmixed for O, F and Ne.
Fill from the lowest MO up, two electrons per MO with opposite spins (Pauli). In a degenerate pair ( $\pi_{2p}$ or $\pi^*_{2p}$ ), put one electron in each before pairing (Hund).
Bond order = ½(bonding − antibonding).
Count unpaired electrons: any unpaired means paramagnetic, none means diamagnetic.

Key idea

Heteronuclear diatomics: CO, CN⁻ and NO

When the two atoms differ, the more electronegative atom's atomic orbitals lie lower in energy. Bonding MOs are then weighted toward that atom and antibonding MOs toward the other.

$\ce{CO}$ has 4 + 6 = 10 valence electrons, the same as $\ce{N2}$ : $(\sigma_{2s})^2(\sigma^*_{2s})^2(\pi_{2p})^4(\sigma_{2p})^2$ , bond order 3, diamagnetic.
The bonding MOs lie closer to O, but the HOMO ( $\sigma_{2p}$ ) is concentrated on C, which is why CO binds to metals (such as the iron in hemoglobin) through carbon.
$\ce{CN-}$ and $\ce{NO+}$ are also isoelectronic with $\ce{N2}$ (10 electrons, bond order 3). $\ce{NO}$ has 11 valence electrons: bond order 2.5 with one unpaired $\pi^*_{2p}$ electron, so it is paramagnetic.

Common mistake

MO diagram traps

Wrong: using the $\ce{O2}$ level order for $\ce{B2}$ . Right: B₂, C₂ and N₂ have $\pi_{2p}$ below $\sigma_{2p}$ . The unmixed order would put B₂'s last two electrons in $\sigma_{2p}$ (diamagnetic); the mixed order puts one in each $\pi_{2p}$ , matching the observed paramagnetism.
Wrong: $\ce{O2+}$ has fewer electrons, so its bond is weaker. Right: the electron removed was antibonding ( $\pi^*_{2p}$ ), so the bond order rises from 2 to 2.5 and the bond gets shorter and stronger.
Wrong: forgetting the charge. Right: an anion adds electrons and a cation removes them before you fill ( $\ce{O2-}$ has 13 valence electrons).
Wrong: the Lewis structure shows O=O with all electrons paired, so O₂ is diamagnetic. Right: O₂ is paramagnetic; this is the classic success of MO theory over Lewis structures.

Worked example

Worked example: O₂ and its ions

Liquid oxygen sticks to the poles of a magnet. $\ce{O2}$ has 12 valence electrons in the unmixed order: $(\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p})^2(\pi_{2p})^4(\pi^*_{2p})^2$ . The last two electrons occupy the two degenerate $\pi^*_{2p}$ orbitals singly, so O₂ has 2 unpaired electrons and bond order ½(8 − 4) = 2.

Adding or removing electrons changes only the $\pi^*_{2p}$ count (8 bonding electrons in every case):

Species	Valence electrons	$\pi^*_{2p}$ electrons	Bond order	Unpaired	Magnetism
$\ce{O2+}$	11	1	2.5	1	paramagnetic
$\ce{O2}$	12	2	2	2	paramagnetic
$\ce{O2-}$ (superoxide)	13	3	1.5	1	paramagnetic
$\ce{O2^2-}$ (peroxide)	14	4	1	0	diamagnetic

Bond length: $\ce{O2+} < \ce{O2} < \ce{O2-} < \ce{O2^2-}$ . Bond energy runs the other way.

Compare $\ce{N2+}$ : 9 valence electrons in the mixed order, $(\sigma_{2s})^2(\sigma^*_{2s})^2(\pi_{2p})^4(\sigma_{2p})^1$ . A bonding electron was removed, so the bond order drops from 3 to ½(7 − 2) = 2.5; one unpaired electron, paramagnetic.

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