The bond order of O_2^ + is the same as in | vedclass

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Question

(a) $O_2^ + (15{e^ - }) = K{K^*}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}{(\sigma 2{p_x})^2}$

${(\pi 2{p_y})^2}{(\pi 2{p_z})^2}{({\pi ^*}2{p_y})^1}{({\pi

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$N_2^ + $

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(a) $O_2^ + (15{e^ - }) = K{K^*}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}{(\sigma 2{p_x})^2}$

${(\pi 2{p_y})^2}{(\pi 2{p_z})^2}{({\pi ^*}2{p_y})^1}{({\pi ^*}2{p_z})^0}$

Hence, bond order $ = \frac{1}{2}(10 - 5) = 2.5$$N_2^ + (13{e^ - }) = K{K^*}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}{(\sigma 2{p_x})^2}$

${(\pi \,2{p_y})^2}{(\pi \,2{p_z})^1}$

Hence, bond order $ = \frac{1}{2}(9 - 4) = 2.5$.

The bond order of $O_2^ + $ is the same as in

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