Chemical Bonding Result Question 6
6. Two pi and half sigma bonds are present in
(2019 Main, 10 Jan I)
(a) $\mathrm{O}_2^{+}$
(b) $\mathrm{N}_2$
(c) $\mathrm{N}_2^{+}$
(d) $\mathrm{O}_2$
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Answer:
Correct Answer: 6. ( c )
Solution:
- The energy order of $MOs$ of the given species are as follows:
$\begin{aligned} \mathrm{O}_2\left(16 e^{-\prime} s\right) & =\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \\ \pi 2 p_x^2 & =\pi 2 p_y^2, \pi^* 2 p_x^1=\pi^* 2 p_y^1, \\ \mathrm{O}_2^{+}\left(15 e^{-\prime} s\right) & =\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \\ \pi 2 p_x^2 & =\pi 2 p_y^2, \pi^* 2 p_x^1 \approx \pi^* 2 p_y^0 \\ \mathrm{~N}_2\left(14 e^{-} \mathrm{s}\right) & =\sigma 1 s^2 \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2 \\ \pi 2 p_x^2 & =\pi 2 p_y^2, \sigma 2 p_z^2 \\ \mathrm{~N}_2^{+}\left(13 e^{-} \mathrm{s}\right) & =\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \\ \pi 2 p_x^2 & =\pi 2 p_z^2 \sigma 2 p_z^1\end{aligned}$
Thus, in case of $\mathrm{N}_2^{+}$, two $\pi$ -bonds and half $\sigma$ -bond are present in the bonding $MOs$.
BETA
