Bond order and magnetism

Bond order

Each bonding electron helps hold the atoms together, and each antibonding electron works against it. The bond order counts what is left over, in bonds:

• Bond order $= \dfrac{1}{2}\,(N_b - N_a)$

Here $N_b$ is the number of electrons in bonding molecular orbitals, and $N_a$ the number in antibonding ones.

A bond order of zero means the antibonding electrons cancel the bonding ones. $\text{He}_2$, $\text{Be}_2$ and $\text{Ne}_2$ are not stable molecules.

Bond order, length and enthalpy

The higher the bond order, the more electron density holds the nuclei together. So the bond is stronger and shorter.

• $\text{N}_2$: bond order $3$, bond length $110\ \text{pm}$, bond enthalpy $946\ \text{kJ mol}^{-1}$.
• $\text{O}_2$: bond order $2$, bond length $121\ \text{pm}$, bond enthalpy $498\ \text{kJ mol}^{-1}$.
• $\text{F}_2$: bond order $1$, bond length $144\ \text{pm}$, bond enthalpy $155\ \text{kJ mol}^{-1}$.

A half bond order is possible too. $\text{H}_2^+$ has one bonding electron and none antibonding, so its bond order is $\tfrac{1}{2}$. It exists, though its bond is weak.

Magnetism

• A species with all electrons paired is diamagnetic: a magnetic field pushes it away slightly.
• A species with unpaired electrons is paramagnetic: a magnetic field pulls it in.

In oxygen, the last two electrons go into the two $\pi^*$ orbitals of the same energy. By Hund's rule they stay unpaired, one in each. That is why liquid oxygen clings to a magnet. Boron, $\text{B}_2$, is paramagnetic for the same reason: its two $\pi 2p$ electrons are unpaired. Nitrogen and carbon, $\text{N}_2$ and $\text{C}_2$, are diamagnetic.

Ions

Removing or adding electrons changes the bond order. Look at which orbital the electron leaves or enters.

• $\text{O}_2^+$ loses an antibonding $\pi^*$ electron: bond order $2.5$, stronger and shorter than in $\text{O}_2$.
• $\text{O}_2^-$, the superoxide ion, gains one: bond order $1.5$.
• $\text{O}_2^{2-}$, the peroxide ion, gains two: bond order $1$, with all electrons paired.
• $\text{N}_2^+$ loses a bonding electron: its bond order falls from $3$ to $2.5$.

Practise with new numbers

Try it in Python

The program works out the bond order and the unpaired electrons of oxygen and four of its ions.

Common mistakes

• Forgetting to halve. Two electrons make one bond, so the difference is divided by $2$.
• Thinking every ion is weaker than its molecule. $\text{O}_2^+$ has a stronger bond than $\text{O}_2$, because it lost an antibonding electron.
• *Pairing the two $\pi^$ electrons of oxygen.** They stay unpaired, which makes oxygen paramagnetic.
• Treating bond order zero as a weak bond. It means no stable molecule at all.

Remember these

• Bond order $= \dfrac{1}{2}(N_b - N_a)$. Zero means no molecule.
• Higher bond order: stronger bond, shorter bond.
• Unpaired electrons: paramagnetic. All paired: diamagnetic.
• $\text{O}_2$ and $\text{B}_2$ are paramagnetic. $\text{N}_2$ and $\text{C}_2$ are diamagnetic.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. $\text{He}_2^+$ has two bonding electrons and one antibonding electron. What is its bond order, and can it exist?

2. Which has the longer bond, $\text{O}_2$ or $\text{O}_2^{2-}$?

3. Why is $\text{C}_2$ diamagnetic but $\text{B}_2$ paramagnetic?

4. What happens to the bond of $\text{N}_2$ when it loses an electron? And the bond of $\text{O}_2$?