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Q.
Bond order of $ O_{2} $ is:
Delhi UMET/DPMTDelhi UMET/DPMT 2004
Solution:
Bond order is calculated by using following formula after writing electronic configuration of molecule according to molecular orbital theory.
$O_{2}=8+8=16=\sigma 1 s^{2}, \overset{*}{\sigma} 1 s^{2}, \sigma 2 s^{2}, \overset{*}{\sigma} 2 s^{2}$,
$\sigma 2 p_{z}^{2}, \pi 2 p_{x}^{2}, \pi 2 p_{y}^{2}, \overset{*}{\pi} 2 p_{x}^{1}, \overset{*}{\pi} p_{y}^{1}$
Bond order $=\frac{N_{b}-N_{a}}{2}$
Where, $N_{b}=$ no. of electron in bonding molecular orbitals
$Na =$ no. of electron in antibonding molecular orbitals.
Bond order $=\frac{10-6}{2}=2$