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4.Chemical Bonding and Molecular Structure
easy
The molecular orbital configuration of a diatomic molecule is
$\sigma \,\,1{s^2}\,\,{\sigma ^*}\,\,1{s^2}\,\sigma \,\,2{s^2}\,{\sigma ^*}\,2{s^2}\,\,\sigma \,2p_x^2\,\left\{ {{}_{\pi \,2p_z^2}^{\pi \,2p_y^2}} \right.$
Its bond order is
A
$3$
B
$2.5$
C
$2$
D
$1$
Solution
(a) $B.O.$ $ = \frac{{{N_b} – {N_a}}}{2} = \frac{{10 – 4}}{2} = 3$.
Standard 11
Chemistry
