Molecular Orbital Theory for Diatomic Molecules

The s-p Mixing Phenomenon

A common point of confusion in chemistry is why the order of energy levels changes as you move across the second row of the periodic table. This is due to a phenomenon known as s-p mixing.

In "early" second-row elements, the energy gap between the $2s$ and $2p$ atomic orbitals is relatively small. This allows the $2s$-derived molecular orbitals and the $2p$-derived molecular orbitals to interact or "mix," which shifts their energy levels.

1. Early Period 2 Elements ($Li_2$ through $N_2$)

For these elements, s-p mixing is significant. This interaction pushes the $\sigma_{2p}$ orbital higher in energy, placing it above the $\pi_{2p}$ orbitals. The energy sequence is: $\sigma_{2s}^* < \sigma_{2s} < \pi_{2p} < \sigma_{2p} < \pi_{2p}^* < \sigma_{2p}^*$

2. Late Period 2 Elements ($O_2$ through $Ne_2$)

As the nuclear charge increases (moving toward the right of the periodic table), the $2s$ and $2p$ orbitals become more distinct in energy, and s-p mixing decreases. Consequently, the $\sigma_{2p}$ orbital drops back down to its "expected" position below the $\pi_{2p}$ orbitals. The energy sequence is: $\sigma_{2s}^* < \sigma_{2s} < \sigma_{2p} < \pi_{2p} < \pi_{2p}^* < \sigma_{2p}^*$

Exam Tip: Always check which "side" of the periodic table your molecule falls on before drawing your diagram. If you are dealing with $N_2$, the $\pi$ orbitals come first; if you are dealing with $O_2$, the $\sigma$ orbital comes first.

Predicting Stability: Bond Order

Once the molecular electron configuration is written, we can determine the stability of the molecule using the Bond Order formula. Bond order tells us the net number of bonds between the atoms; a higher bond order generally means a stronger, more stable bond.

$\text{Bond Order} = \frac{(\text{number of bonding } e^-) - (\text{number of antibonding } e^-)}{2}$

• Stable Molecules: Have a bond order $> 0$.
• Unstable Molecules: A bond order of $0$ indicates that the destabilizing effect of the antibonding electrons perfectly cancels the stabilizing effect of the bonding electrons. For example, $Be_2$ has a bond order of $0$ and is therefore not expected to form.

Predicting Magnetism

MO theory is uniquely powerful because it can predict a molecule's magnetic properties based on its electron configuration.

• Paramagnetic: A molecule is paramagnetic if it contains one or more unpaired electrons. These molecules are attracted to magnetic fields. A classic example is $O_2$, which has two unpaired electrons in its $\pi^*$ orbitals.
• Diamagnetic: A molecule is diamagnetic if all electrons are paired. These molecules are weakly repelled by magnetic fields. For example, $F_2$ is diamagnetic.

Worked Examples

Example 1: $C_2$

To find the properties of $C_2$, we look at its valence electrons. Since $C$ is an early element, we use the s-p mixing sequence.

1. Configuration: $\sigma_{2s}^2 \sigma_{2s}^{*2} \pi_{2p}^4$ (Note: $\sigma_{2p}$ is empty here).
2. Bond Order: $\frac{6 \text{ (bonding)} - 2 \text{ (antibonding)}}{2} = 2$.
3. Magnetism: All electrons are paired, so it is diamagnetic.

Example 2: $F\_2^+$ (Cation)

If we remove an electron from $F_2$ to form the $F\_2^+$ ion:

1. Bond Order: Removing an electron from an antibonding orbital ($\pi_{2p}^*$) actually increases the bond order. For $F\_2^+$, the bond order is $1.5$.
2. Magnetism: Because there is an odd number of electrons, there must be an unpaired electron, making it paramagnetic.