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)

$O_2(16)=\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2\approx \pi 2p_y^2,{\pi }^{\ast }2p_x^1\approx {\pi }^{\ast }2p_y^1$
Bond order $=\frac{10-6}{2}=2$ Due to the presence of unpaired electrons, $O_2$ is paramagnetic.
$O_2^{-}(17)=\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2\approx \pi 2p_y^2,{\pi }^{\ast }2p_x^2\approx {\pi }^{\ast }2p_y^1$
Bond order $=\frac{10-7}{2}=1.5$
$\therefore$ B.O. of $O_2>O_2^{-}$ paramagnetic.