What is the bond order of N_{2}+?

There are 14 electrons in N_{2} molecule, therefore we have the following arrangement of electrons

(σ1*s*)^{2} (σ*1*s*)^{2 }(σ2*s*)^{2} (σ*2*s*)^{2} (Π2p_{x} ^{2} = Π2p_{y} ^{2})(σ2*p* _{ z })^{2}

here bond order is = (N_{b}-N_{a}) / 2 = (10-4) / 2 = 3

Now because in N_{2}^{+}, there is one electron less in bonding orbital as compared to electrons in N_{2}, therefore, there will be 1 electron less in the σ2*p*_{ z }* *orbital. Therefore we have the following arrangement of electrons

(σ1*s*)^{2} (σ*1*s*)^{2 }(σ2*s*)^{2} (σ*2*s*)^{2} (Π2p_{x} ^{2} = Π2p_{y} ^{2})(σ2*p* _{ z })^{1}

Thus the bond order in N_{2}^{+} ion will be

= (9-4) / 2 = 2.5

**
**