Chemical Bonding - Result Question 51
####26. Among $KO _2, AlO _2^{-}, BaO _2$ and $NO _2^{+}$, unpaired electron is present in
(1997 C, 1M)
(a) $NO _2^{+}$and $BaO _2$
(b) $KO _2$ and $AlO _2^{-}$
(c) Only $KO _2$
(d) Only $BaO _2$
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Solution:
- Molecular orbital electronic configuration are
$KO _2\left(O _2^{-}\right): \sigma 1 s^{2} \stackrel{}{\sigma} 1 s^{2} \sigma 2 s^{2} \stackrel{}{\sigma} 2 s^{2} \sigma 2 p _x^{2}\left|\begin{array}{l|l}\pi 2 p _y^{2} & \stackrel{}{\pi} 2 p _y^{2} \ \pi 2 p _z^{2} & * \ \pi 2 p _z^{1}\end{array}\right| \stackrel{}{\sigma} 2 p _x^{0}$
Has one unpaired electron in $\stackrel{*}{\pi} 2 p$ orbital.
$AlO _2^{-}$has both oxygen in $O^{2-}$ state, therefore, no unpaired electron is present.
Has no unpaired electron.
$NO _2^{+}$has $[O=\stackrel{+}{N}=O]$ bonding, hence no unpaired electron.
