QJEEM08Ch08b9: JEE Main 2008 Chemistry - Class 11 JEEM08Ch08b9 Solution: Which of the following pair of species have the same bond order?

Number of electron in NO⁺ = number of electron in CN^– = 14 electrons.

As both have same number of electrons so their bond order is equal.

Moleculer orbital configuration of NO⁺ (14 electrons) is

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^0}^ * = \,\,\pi _{2p_y^0}^ *$

$\therefore$ B.O = ${1 \over 2}\left[ {10 - 4} \right]$ = 3

Moleculer orbital configuration of CN^- (14 electrons) is

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$

$\therefore$ B.O = ${1 \over 2}\left[ {10 - 4} \right]$ = 3