Key Idea Bond order $=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}$
where, $\mathrm{N}_{\mathrm{b}}=$ number of electrons in bonding $M O$
$\mathrm{N}_{\mathrm{a}}=$ number of electrons in anti bonding $M O$
$$
\begin{aligned}
& \mathrm{N}_2(7+7=14)=\sigma 1 \mathrm{~s}^2, \dot{*} 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2, \stackrel{\dot{*}}{\sigma} 2 \mathrm{~s}^2, \pi 2 \mathrm{p}_{\mathrm{x}}^2 \\
& \approx \pi 2 \mathrm{p}_{\mathrm{y}}^2, \sigma 2 \mathrm{p}_{\mathrm{z}}^2
\end{aligned}
$$

$$
\begin{aligned}
& \mathrm{BO}= \frac{10-4}{2}=3 \\
& \mathrm{~N}_2(7+7+1=15) \\
&= \sigma 1 \mathrm{~s}^2, \dot{*} 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2, \\
& \stackrel{*}{\sigma} 2 \mathrm{~s}^2, \sigma 2 \mathrm{p}_z^2, \pi 2 \mathrm{p}_{\mathrm{x}}^2 \approx \pi 2 \mathrm{p}_{\mathrm{y}}^2, \pi 2 \mathrm{p}_{\mathrm{x}}^1
\end{aligned}
$$

$$
\begin{aligned}
& \mathrm{BO}=\frac{10-5}{2}=2.5 \\
& \mathrm{~N}_2^{2-}(7+7+2=16)
\end{aligned}
$$
$\begin{aligned} & =\sigma 1 \mathrm{~s}^2, \star 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2, \star 2 \mathrm{~s}^2, \sigma 2 \mathrm{p}_{\mathrm{z}}^2, \pi 2 \mathrm{p}_{\mathrm{z}}^2 \\ & \approx \pi 2 \mathrm{p}_{\mathrm{y}}^2, \pi 2 \mathrm{p}_{\mathrm{x}}^1 \approx \pi 2 \mathrm{p}_{\mathrm{y}}^1 \\ & \end{aligned}$

$$
\mathrm{BO}=\frac{10-6}{2}=2
$$
Hence, the increasing order of B.O is,
$$
\mathrm{N}_2^{2-} < \mathrm{N}_2^{-} < \mathrm{N}_2
$$