According to molecular orbital theory, antibonding molecular orbitals of $\mathrm{O}_{2}$ contain
- $4$ electrons
- $6$ electrons
- $10$ electrons
- $8$ electrons
Solution
$\left(\begin{array}{l}\sigma^{*} 2 \mathrm{~s}=2 \mathrm{e}^{-} \\ \pi * 2 \mathrm{p}_{\mathrm{x}}=1 \mathrm{e}^{-} \\ \frac{\pi^{*} 2 \mathrm{p}_{\mathrm{y}}=1 \mathrm{e}^{-}}{\text {Total }=4 \text { electrons }}\end{array}\right)$Asked in: MHT CET 2020 (15 Oct Shift 1)
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