Chemical Bonding and Molecular Structure - Result Question 110
####113. 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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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{array} { l } { \pi 2 p _ { y } ^ { 2 } } \\ { \pi 2 p _ { z } ^ { 2 } } \end{matrix} \sigma \sigma 2 p _ { x } ^ { 2 } {\begin{matrix} \pi * 2 p_y^{1} \\ \pi * 2 p_z^{1} \end{matrix} .. $
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{array} { l } { \pi 2 p _ { y } ^ { 2 } } \\ { \pi 2 p _ { z } ^ { 2 } } \end{matrix} \quad \sigma 2 p _ { x } ^ { 2 } {\begin{matrix} \pi * 2 p_y^{1} \\ \pi * 2 p_z^{0} \end{matrix} .. $
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 $=N_2^{2-}<N_2^{-}<N_2$
