Q. The bond order of $O_{2}^{+}$ is the same as in

A) $N_{2}^{+}$
B) $CN^{-}$
C) CO
D) $NO^{+}$
Explanation

(a)
$O_{2}^{+}\ \left( 15\ e^{-} \right) = KK^{*}\ (\sigma 2s)^{2}\left( \ _{\sigma}^{*}2s \right)^{2}\left( \sigma 2p_{x} \right)^{2},\left( \pi 2p_{y} \right)^{2}$
${= \left( \pi 2p_{z} \right)}^{2}\left( \ _{\pi}^{*}2p_{y} \right)^{1} = \left( \ _{\pi}^{*}2p_{z} \right)^{0}$
Hence, bond order $= \frac{1}{2}(10 - 5) = 2.5$
$N_{2}^{+}\ \left( 13\ e^{-} \right) = KK^{*}\ (\sigma 2s)^{2}\left( \ _{\sigma}^{*}2s \right)^{2}\left( \pi 2p_{y} \right)^{2}$
${= \left( \pi 2p_{z} \right)}^{2}{,\left( \sigma 2p_{x} \right)}^{1}$
Hence, Bond order $= \frac{1}{2}(9 - 4) = 2.5$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.