Bond order of {O_2} is | vedclass

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Question

(a) Electronic configuration of ${O_2}$ is

${O_2} = {(\sigma 1s)^2}{({\sigma ^*}1s)^2}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}{({\sigma ^*}2p_z)^2}$

$

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(a) Electronic configuration of ${O_2}$ is

${O_2} = {(\sigma 1s)^2}{({\sigma ^*}1s)^2}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}{({\sigma ^*}2p_z)^2}$

$(\pi 2p_x^2 \equiv \pi 2p_y^2)\;({\pi ^*}2p_x^1 \equiv {\pi ^*}2p_y^1)$

Hence bond order $ = \frac{1}{2}\left[ {{N_b} - {N_a}} \right]$ $ = \frac{1}{2}[10 - 6] = 2$.

Bond order of ${O_2}$ is

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