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Q.
The bond order of $N _{2}^{+}$on the basis of molecular orbital theory is [Atomic number of $N =7$ ]
J & K CETJ & K CET 2013Chemical Bonding and Molecular Structure
Solution:
We know that N3 molecule is having $14$ electrons so.
$N _{2}=14$
$N_{2}^{+}=14-1=13,$
according to MOT (Molecular orbital theory) the configruation will be
$N_{2}^{+}=K K \sigma 2 S^{2} \sigma^{*} 2 S^{2} \pi 2 p_{x}^{2} \pi 2 p_{y}^{2} \sigma 2 p_{z}^{1}$
Bond order $=\frac{1}{2}$ [Bonding - Antibonding]
$=\frac{1}{2}[7-2]=2 \frac{1}{2}=2.5$