MCQ
The bond order of $O_2^ + $ is the same as in
- ✓$N_2^ + $
- B$C{N^ - }$
- C$CO$
- D$N{O^ + }$
✓
Answer
Correct option: A.
$N_2^ + $
(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$.
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