Chemistry - CHEMICAL BONDING Question with Solution | TestHub

$O_2^{-}=\sigma 1s^2<\stackrel{*}{\sigma }1s^2<\sigma 2s^2<\stackrel{*}{\sigma }2s^2<\sigma 2p_x^2=\pi 2p_y^2=\pi 2p_z^2<\stackrel{*}{\pi }2p_y^2=\stackrel{*}{\pi }2p_z^1$

Bond Order$=\frac{6-3}{2}=1.5$

$O_2=\sigma 1s^2<\stackrel{*}{\sigma }1s^2<\sigma 2s^2<\stackrel{*}{\sigma }2s^2<\sigma 2p_x^2<\pi 2p_y^2=\pi 2p_z^2<\stackrel{*}{\pi }2p_y^1=\stackrel{*}{\pi }2p_z^1$

Bond order$=\frac{6-2}{2}=2$

$N_2=\sigma 1s^2<\stackrel{*}{\sigma }1s^2<\sigma 2s^2<\stackrel{*}{\sigma }2s^2<\pi 2p_z^2=\pi 2p_z^2<\sigma 2p_x^2$

Bond order$=\frac{6-0}{2}=3$

According to$MOT$; electronic configuration of$N_2^{+}$is:

$N_2^{+}=\sigma 1s^2<\stackrel{*}{\sigma }1s^2<\sigma 2s^2<\stackrel{*}{\sigma }2s^2<\pi 2p_x^2=\pi 2p_y^2<\sigma 2p_z^1$

Bond order$=\frac{1}{2}\left(9-4\right)=2.5$

i.e.,$2\pi$bonds and$0.5\sigma$bonds.