Question

Question 32

Using molecular orbital theory, compare the bond energy and magnetic character of $O_2^{+}$ and $O_2^{+}$ species.

Answer 1

According to molecular orbital theory electronic configurations of $O_2^{+}$ and $O_2^{+}$ species are as follows

$O_2^{+}:(\sigma 1s{)}^2,({\sigma }^{\ast }1s{)}^2,(\sigma 2s{)}^2,({\sigma }^{\ast }2s{)}^2,(\sigma 2p_z{)}^2,(\pi 2p_x^2,\pi 2p_y^2),({\pi }^{\ast }2p_x^1)$

Bond order of $O_2^{+}=\frac{10-5}{2}=\frac{5}{2}=2.5$

$O_2^{-}=(\sigma 1s{)}^2,({\sigma }^{\ast }1s{)}^2,(\sigma 2s{)}^2,({\sigma }^{\ast }2s{)}^2,(\sigma 2p_z{)}^2,(\pi 2p_x^2,\pi 2p_y^2),({\pi }^{\ast }2p_x^2,{\pi }^{\ast }2p_y^1)$

Bond order of $O_2^{-}=\frac{10-7}{2}=\frac{3}{2}=1.5$

Higher bond order of $O_2^{+}$ shows that it is more stable than $O_2^{-}$ . Both the species have unpaired electrons. So, both are paramagnetic in nature.