Answer:
Option C
Explanation:
The bond order of B2 molecule is 1. The ground state electron configuration of B is 1s2 2s22p1.So, B2 molecule has a total of ten electrons, which are arranged in MO s.
$(\sigma1s)^{2}(\sigma^{*}1s)^{2}(\sigma2s)^{2}(\sigma^{*}2s)^{2}(\pi_{2p_y^1}=\pi_{2p_y^1})$
Bond order = $\frac{1}{2}$ (Number of bonding electrons - Number of anti-bonding electrons )
=$\frac{1}{2}(6-4)=1$
