Calculate the precise bond order of the molecular ion using Molecular Orbital Theory. — Chemical Bonding Chemistry Question

💡 Solution & Explanation

$N_2$ has 14 electrons and a bond order of 3 ($\frac{10 - 4}{2}$). Ionizing it to $N_2^+$ removes 1 electron from the Highest Occupied Molecular Orbital (HOMO), which is the bonding $\sigma 2p_z$ orbital. The new configuration has 9 bonding and 4 antibonding electrons. Bond Order = $\frac{9 - 4}{2} = 2.5$.