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4.Chemical Bonding and Molecular Structure
medium
Bond order of ${O_2}$ is
A
$2$
B
$1.5$
C
$3$
D
$3.5$
Solution
(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$.
Standard 11
Chemistry
