MCQ
Bond order of ${O_2}$ is
- ✓$2$
- B$1.5$
- C$3$
- D$3.5$
✓
Answer
Correct option: A.
$2$
(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$.
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