Chemical Bonding And Molecular Structure Ques 15
15. According to MO theory which of the following lists ranks the nitrogen species in terms of increasing bond order?
[2009]
(a) $N_2^{2-}<N_2^{-}<N_2$
(b) $N_2<N_2^{2-}<N_2^{-}$
(c) $N_2^{-}<N_2^{2-}<N_2$
(d) $N_2^{-}<N_2<N_2^{2-}$
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Answer:
Correct Answer: 15. (a)
Solution:
(a) Molecular orbital configuration of
$N_2^{2-}=\sigma 1 s^{2} \sigma^{} 1 s^{2} \sigma 2 s^{2} \sigma^{} 2 s^{2}-$
$ \begin{cases} \pi 2 p_y^2 \\ \pi 2 p_z^2 \end{cases} \sigma 2 p_x^2 \begin{cases} \pi^* 2 p_y^1 \\ \pi^* 2 p_z^1 \end{cases} $
Bond order $=\frac{10-6}{2}=2$
$N_2^-=\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2$
$ \begin{cases} \pi 2 p_y^2 \\ \pi 2 p_z^2 \end{cases} \sigma 2 p_x^2 \begin{cases} \pi^* 2 p_y^1 \\ \pi^* 2 p_z^0 \end{cases} $
Bond order $=\frac{10-5}{2}=2.5$
$N_2=\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \begin{cases} \pi 2 p_y^{2} \\ \pi 2 p_z^2 \end{cases} , \sigma 2 p_x^2$
Bond order $=\frac{10-4}{2}=3$
$\therefore \quad$ The correct order is $=N_2^{2-}<N_2^{-}<N_2$
BETA
