Q.What is meant by the term bond order? Calculate the bond order of: $N_2$, $O_2$, $O_2^+$ and $O_2^-$.

Concept: Bond Order Strength – Bond order measures the number of chemical bonds between a pair of atoms. It is calculated as half the difference between the number of bonding electrons and antibonding electrons in the molecular orbital configuration.

Steps for each molecule/ion:

1. $N_2$ (14 electrons): MO configuration: \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2

   Bonding electrons = 10, Antibonding = 4 → Bond order = $\frac{10-4}{2} = 3$

2. $O_2$ (16 electrons): MO: \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi^*_{2p_x}^1 \pi^*_{2p_y}^1

   Bonding = 10, Antibonding = 6 → Bond order = $\frac{10-6}{2} = 2$ …

What is bond order? The core idea

Bond order is a number that tells you how strong and stable a chemical bond is. Think of it as the net number of bonding interactions between two atoms after cancelling out the destabilising antibonding ones. A higher bond order means a shorter, stronger bond and more energy needed to break it.

In molecular orbital (MO) theory, when two atomic orbitals combine, they form a bonding orbital (lower energy, stabilising) and an antibonding orbital (higher energy, destabilising). Electrons fill these orbitals from lowest energy upward. Bond order is simply:

Why divide by 2? Because each bonding pair contributes one net bond. A single bond (like in $H_2$) has bond order 1, a double bond has 2, a triple bond has 3.

Step-by-step calculations

We need the MO electron configurations for each molecule. For homonuclear diatomic molecules of second-period elements, the MO energy order depends on whether the $2p$ orbitals overlap strongly. For $N_2$ and $O_2$ (and their ions), the order is:

• For $N_2$ (and lighter): $\sigma_{1s} < \sigma_{1s}^* < \sigma_{2s} < \sigma_{2s}^* < \pi_{2p_x} = \pi_{2p_y} < \sigma_{2p_z} < \pi_{2p_x}^* = \pi_{2p_y}^* < \sigma_{2p_z}^*$
• For $O_2$ (and heavier): $\sigma_{1s} < \sigma_{1s}^* < \sigma_{2s} < \sigma_{2s}^* < \sigma_{2p_z} < \pi_{2p_x} = \pi_{2p_y} < \pi_{2p_x}^* = \pi_{2p_y}^* < \sigma_{2p_z}^*$

The core $1s$ orbitals (and their antibonding counterparts) are always filled and cancel out — they contribute nothing to bond order. So we only count valence electrons.

1. $N_2$ (7 electrons per atom = 14 total valence electrons)

Electron configuration (valence only): $\sigma_{2s}^2 \, \sigma_{2s}^{*2} \, \pi_{2p_x}^2 \, \pi_{2p_y}^2 \, \sigma_{2p_z}^2$

• Bonding electrons: $\sigma_{2s}$ (2) + $\pi_{2p_x}$ (2) + $\pi_{2p_y}$ (2) + $\sigma_{2p_z}$ (2) = 8
• Antibonding electrons: $\sigma_{2s}^*$ (2) = 2

This matches the triple bond in $N \equiv N$ — one sigma and two pi bonds.

2. $O_2$ (8 electrons per atom = 16 total valence electrons)

For $O_2$, the $\sigma_{2p_z}$ orbital is lower in energy than the $\pi_{2p}$ orbitals. Configuration: $\sigma_{2s}^2 \, \sigma_{2s}^{*2} \, \sigma_{2p_z}^2 \, \pi_{2p_x}^2 \, \pi_{2p_y}^2 \, \pi_{2p_x}^{*1} \, \pi_{2p_y}^{*1}$

• Bonding electrons: $\sigma_{2s}$ (2) + $\sigma_{2p_z}$ (2) + $\pi_{2p_x}$ (2) + $\pi_{2p_y}$ (2) = 8
• Antibonding electrons: $\sigma_{2s}^*$ (2) + $\pi_{2p_x}^*$ (1) + $\pi_{2p_y}^*$ (1) = 4

This is a double bond — but with two unpaired electrons (paramagnetic), which is unusual and correctly predicted by MO theory.

3. $O_2^+$ (15 valence electrons — remove one electron from $O_2$) …