Question

According to molecular orbital theory, which is correct statement regarding $O_2$?

• A. Paramagnetic and bond order$ < O^{-}_{2}$
• B. Paramagnetic and bond order $ > O^{-}_{2}$
• C. Diamagnetic and bond order $ < O^{-}_{2}$
• D. Diamagnetic and bond order $ > O^{-}_{2}$

Answer 1

Answer: (B)

$O_{2}(16)=\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{z}^{2}, \pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2}, \pi^{*} 2 p_{x}^{1} \approx \pi^{*} 2 p_{y}^{1}$
Bond order $=\frac{10-6}{2}=2$ Due to the presence of unpaired electrons, $O_{2}$ is paramagnetic.
$O_{2}^{-}(17)=\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{z}^{2}, \pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2}, \pi^{*} 2 p_{x}^{2} \approx \pi^{*} 2 p_{y}^{1}$
Bond order $=\frac{10-7}{2}=1.5$
$\therefore $ B.O. of $O_{2}>O_{2}^{-}$ paramagnetic.