Bond order $(\mathrm{BO})=\frac{1}{2}$ (number of electron in bonding MO's - number of electron in anti-bonding $\mathrm{MO}^{\prime} \mathrm{s}$ )
$\mathbf{C}_2^{2-}:$ Total number of electrons $=6+6+2=14$
Electronic configuration
$\sigma\left(1 s^2\right), \sigma^*\left(1 s^2\right), \sigma\left(2 s^2\right), \sigma^*\left(2 s^2\right), \pi 2 p_x^2=\pi 2 p_y^2, \sigma 2 p_z^2$
$\mathrm{BO}=\frac{1}{2} \times(10-4)=\frac{1}{2} \times 6=3$
$\mathbf{N}_{\mathbf{2}}:$ Total number of electrons $=7+7=14$
$\sigma\left(1 s^2\right), \sigma^*\left(1 s^2\right), \sigma\left(2 s^2\right), \sigma^*\left(2 s^2\right), \pi 2 p_x^2=\pi 2 p_y^2, \sigma 2 p_z^2$
$\mathrm{BO}=\frac{1}{2}(10-4)=\frac{6}{2}=3$
$\mathrm{O}_2^{2-}:$ Total number of electron $=8+8+2=18$
$\begin{array}{r}\sigma\left(1 s^2\right), \sigma^*\left(1 s^2\right), \sigma\left(2 s^2\right), \sigma^*\left(2 s^2\right), \sigma 2 p_z^2, \pi 2 p_x^2=\pi 2 p_y^2 \\ \pi^* 2 p_x^2=\pi^* 2 p_y^2\end{array}$
$\mathrm{BO}=\frac{1}{2}(10-8)$
$=\frac{1}{2} \times 2=1$
Option (a) is incorrect.
$\mathbf{O}_2^{+}$Total numbẹr of electrons $=8+8-1=15$
$\begin{aligned} \sigma\left(1 s^2\right), \sigma^*\left(1 s^2\right), \sigma\left(2 s^2\right), \sigma^*\left(2 s^2\right), \sigma\left(2 p_z^2\right), & \pi\left(2 p_x^2\right) \\ & =\pi\left(2 p_y^2\right), \pi^* 2 p_x^1\end{aligned}$
$\mathrm{BO}=\frac{1}{2}(10-5)=\frac{1}{2} \times 5=2.5$
$\mathrm{O}_2^{-}$Total number of electrons $=8+8+1=17$
$\mathrm{BO}=\frac{1}{2}(10-7)=\frac{1}{2} \times 3=1.5$
$\mathbf{N}_2^{+}$Total number of electron $=7+7-1=13$
$\sigma\left(1 s^2\right), \sigma^*\left(1 s^2\right), \sigma\left(2 s^2\right), \sigma^*\left(2 s^2\right), \pi\left(2 p_x^2\right)=\pi\left(2 p_y^2\right), \sigma\left(2 p_z^1\right)$
$\mathrm{BO}=\frac{1}{2}(9-4)=\frac{1}{2} \times 5=2.5$
Option (b) is correct.
Similarly; bond order of $\mathrm{O}_2=\frac{1}{2}(10-6)=\frac{1}{2} \times 4=2$
Bond order of $\mathrm{Li}_2=\frac{1}{2}(4-2)=1$
Bond order of $\mathrm{H}_2^{+}=\frac{1}{2}(1-0)=0.5$
Bond order of $C_2=\frac{1}{2}(8-4)=\frac{1}{2} \times 4=2$
Rest other options are incorrect, as they don't have bond order in fraction.
