The changes in bond length with respect to $\mathrm{N}-\mathrm{N}$ and $\mathrm{O}-\mathrm{O}$, when…
The changes in bond length with respect to $\mathrm{N}-\mathrm{N}$ and $\mathrm{O}-\mathrm{O}$, when $\mathrm{N}_2$ becomes $\mathrm{N}_2^{+}$and $\mathrm{O}_2$ becomes $\mathrm{O}_2^{+}$are respectively
increases, decreases
decreases, increases
increases, increases
decreases, decreases
Solution
$
\begin{aligned}
\mathrm{N}_2(14) \rightarrow \sigma 1 s^2 < \stackrel{*}{\sigma} 1 s^2 < \sigma 2 s^2 < \stackrel{*}{\sigma} 2 s^2 & < \pi 2 p_x^2 \\
& \approx \pi 2 p_y^2 < \sigma 2 p_z^2
\end{aligned}
$
Bond order $=\frac{10-4}{2}=3$
$
\begin{aligned}
\mathrm{N}_2^{+}(13) \rightarrow \sigma l s^2 < \stackrel{*}{\sigma} \mathrm{l} s^2 < \sigma 2 s^2 < *{ }_\sigma 2 s^2 < & \pi 2 p_x^2 \\
& \approx \pi 2 p_y^2 < \sigma 2 p_z^1
\end{aligned}
$
Bond order $=\frac{9-4}{2}=2.5$
$
\begin{aligned}
& \mathrm{O}_2(16) \rightarrow \sigma 1 s^2 < \stackrel{*}{\sigma} 1 s^2 < \sigma 2 s^2 < \stackrel{*}{\sigma} 2 s^2 < \sigma 2 p_z^2 < \pi 2 p_x^2 \\
& \approx \pi 2 p_y^2 < \pi 2 p_x^1 \approx \pi 2 p_y^1 \\
& \text { Bond order }=\frac{10-6}{2}=2 \\
& \mathrm{O}_2^{+}(15) \rightarrow \sigma l s^2 < \stackrel{*}{\sigma} \mathrm{l} s^2 < \sigma 2 s^2 < \stackrel{*}{\sigma} 2 s^2 < \sigma 2 p_z^2 < \pi 2 p_x^2 \\
& =\pi 2 p_y^2 < \stackrel{*}{\pi} 2 p_x^1=\stackrel{*}{\pi} 2 p_y^0 \\
& \text { Bond order }=\frac{10-5}{2}=2.5 \\
&
\end{aligned}
$
$\because$ Bond order $\propto \frac{1}{\text { Bond length }}$
$\therefore$ In case of $\mathrm{N}_2^{+}$bond length increases as bond order decreases. While in $\mathrm{O}_2^{+}$bond order increases and hence bond length decreases