The bond order of dioxygen is ' $m$ '. The bond order values of $\mathrm{N}_2^{+}$and $\mathrm{C}_2^{2-}$…
The bond order of dioxygen is ' $m$ '. The bond order values of $\mathrm{N}_2^{+}$and $\mathrm{C}_2^{2-}$ are respectively
$\frac{5 m}{4}, \frac{3 m}{2}$
$\frac{3 m}{2}, \frac{5 m}{4}$
$\frac{m}{2}, \frac{m}{3}$
$\frac{2 m}{3}, \frac{m}{2}$
Solution
Given that, the bond order of dioxygen is $m$. Electronic configuration of $\mathrm{O}_2(Z=16)=$
$\begin{array}{r}{[\sigma(1 s)]^2\left[\sigma^*(1 s)\right]^2[\sigma(2 s)]^2\left[\sigma\left({ }^* 2 s\right)\right]^2\left[\sigma\left(2 p_z\right)\right]^2\left[\pi\left(2 p_x\right)\right]^2} \\ {\left[\pi\left(2 p_y\right)\right]^2\left[\pi^*\left(2 p_x\right)\right]\left[\pi^*\left(2 p_y\right)\right]}\end{array}$
Bond order of oxygen $=\frac{1}{2}(10-6)=2$
Thus, value of $m=2$
Now, electronic configuration of $\mathrm{N}_2^{+}$is
$\begin{array}{r}{[\sigma(1 s)]^2\left[\sigma^*(1 s)\right]^2[\sigma(2 s)]^2\left[\sigma\left({ }^* 2 s\right)\right]^2\left[\pi\left(2 p_x\right)\right]^2\left[\pi\left(2 p_y\right)\right]^2} \\ {\left[\sigma\left(2 p_z\right]\right.}\end{array}$
Bond order of $\mathrm{N}_2^{+}=\frac{9-4}{2}=\frac{5}{2}=2.5$
and electronic configuration of $\mathrm{C}_2^{2-}$ is
$\begin{aligned} {\left.[\sigma(1 s)]^2\left[\sigma^*(1 s)\right]^2[\sigma(2 s)]^2\left[\sigma^* 2 s\right)\right]^2 } & {\left[\pi\left(2 p_x\right)\right]^2 } \\ & {\left[\pi\left(2 p_y\right)\right]^2\left[\sigma\left(2 p_z\right)\right]^2 }\end{aligned}$
Bond order of $\mathrm{C}_2^{2-}=\frac{10-4}{2}=\frac{6}{2}=3$
The bond order of $\mathrm{N}_2^{+}$and $\mathrm{C}_2^{2-}$ in term of $m$ is $\frac{5}{4} m$, $\frac{3}{2} m$ :