According to MO theory which of the following lists ranks the nitrogen species in terms of increasing bond…
According to MO theory which of the following lists ranks the nitrogen species in terms of increasing bond order ?
- $\mathrm{N}_2^{-} < \mathrm{N}_2 < \mathrm{N}_2^{2-}$
- $\mathrm{N}_2^{2-} < \mathrm{N}_2^{-} < \mathrm{N}_2$
- $\mathrm{N}_2 < \mathrm{N}_2^{2-} < \mathrm{N}_2^{-}$
- $\mathrm{N}_2^{-} < \mathrm{N}_2^{2-} < \mathrm{N}_2$
Solution
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,
$\mathrm{N}_2^{2-} < \mathrm{N}_2^{-} < \mathrm{N}_2$
Asked in: NEET 2009 (Screening)
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