Import Question JSON

\text{HINT: Bond order} = \frac{1}{2} (\text{N}_b - \text{N}_a) \text{, where } \text{N}_a \text{ is number of antibonding electrons and } \text{N}_b \text{ is number of bonding electrons.} \text{Explanation:} \text{The MOT diagram of given ions is as follows:} \text{1.) Electronic configuration of } \text{O}_2 \text{ (16 electrons)} = \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} \text{Bond order} = \frac{1}{2} (\text{N}_b - \text{N}_a) = \frac{1}{2} (10 - 6) = 2 \text{2.) Electronic configuration of } \text{O}_2^{+} \text{ (15 electrons)} = \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}}^{*0} \text{Bond order} = \frac{1}{2} (\text{N}_b - \text{N}_a) = \frac{1}{2} (10 - 5) = 2.5 \text{3.) Electronic configuration of } \text{O}_2^{-} \text{ (17 electrons)} = \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}}^{*2} = \pi_{2p_{y}}^{*1} \text{Bond order} = \frac{1}{2} (\text{N}_b - \text{N}_a) = \frac{1}{2} (10 - 7) = 1.5 \text{4.) Electronic configuration of } \text{N}_2 \text{ (14 electrons)} = \sigma_{1s}^{2}, \sigma_{1s}^{*2}, \sigma_{2s}^{2}, \sigma_{2s}^{*2}, \pi_{2p_{x}}^{2} = \pi_{2p_{y}}^{2}, \sigma_{2p_{z}}^{2} \text{Bond order} = \frac{1}{2} (\text{N}_b - \text{N}_a) = \frac{1}{2} (10 - 4) = 3 \begin{array}{|c|c|c|c|c|} \hline \text{Molecule} & \text{O}_2^{-} & \text{O}_2 & \text{O}_2^{+} & \text{N}_2 \\ \hline \text{Bond order} & 1.5 & 2.0 & 2.5 & 3.0 \\ \hline \end{array} \text{Arranging in increasing order of bond order: } \text{O}_2^{-} < \text{O}_2 < \text{O}_2^{+} < \text{N}_2 \text{Therefore, the correct option is 2.}