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Q.
Which of the following species is diamagnetic?
AMUAMU 2001
Solution:
If a species does not have any unpaired electron, then it is called diamagnetic and if a species has unpaired electron, then it is called paramagnetic.
(a) Oxygen molecule $\left( O _{2}\right)$ has following electronic configuration.
$O _{2} K K$
$(\sigma 2 s)^{2}(\overset{*}{\sigma} 2 s)^{2}\left(\sigma 2 p_{z}\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2}$
$\left(\overset{*}{\pi} 2 p_{x}\right)^{1}\left(\overset{*}{\pi} 2 p_{y}\right)^{1}$
This is paramagnetic due to presence of two unpaired electrons.
(b) Peroxide ion $\left( O _{2}^{2-}\right)$ has following electronic configuration.
$O _{2}^{2-} K K$
$\left.(\sigma 2 s)^{2}\left(\overset{*}{\sigma}2 s\right)^{2}\left(\sigma 2 p_{z}\right)^{2}\left(\pi 2 p_{x}\right)^{2}\right)\left(\pi 2 p_{y}\right)^{2}$
$\left(\overset{*}{\pi} 2 p_{x}\right)^{2}\left(\overset{*}{\pi} 2 p_{y}\right)^{2}$
There is no unpaired electron in $O _{2}^{2-}$ ion.
So, it is diamagnetic.
(c) Superoxide ion $\left( O _{2}^{-}\right)$has following electronic configuration.
$O _{2}^{-} K K$
$(\sigma 2 s)^{2}\left(\overset{*}{\sigma} 2 s\right)^{2}\left(\sigma 2 p_{z}\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2}$
$\left(\overset{*}{\pi} 2 p_{x}\right)^{2}\left(\overset{*}{\pi} 2 p_{y}\right)^{1}$
There is one unpaired electron in $O _{2}^{-} .$So, it is paramagnetic.
(d) Oxygen molecule ion $\left( O _{2}^{+}\right)$has following electronic configuration.
$O _{2}^{+} K K$
$(\sigma 2 s)^{2}(\overset{*}{\sigma} 2 s)^{2}\left(\sigma 2 p_{z}\right)^{2}\left(\pi 2 p_{x}\right)^{2} \left( \pi 2p_{y}\right)^2 \left(\overset{*}{\pi} 2p_{x}\right)^1 $
There is one unpaired electron in $O _{2}^{+}$. So, it is paramagnetic.