Among KO_2,KAlO_2,CaO_2 and NO_2^+,unpaired e | vedclass

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(C) To determine the presence of an unpaired electron,we analyze the electronic structure of each species:$1$. $KO_2$ (Potassium superoxide): It consi

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$KO_2$ only

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(C) To determine the presence of an unpaired electron,we analyze the electronic structure of each species:$1$. $KO_2$ (Potassium superoxide): It consists of $K^+$ and $O_2^-$ ions. The superoxide ion $(O_2^-)$ has $17$ valence electrons. According to Molecular Orbital Theory,the configuration is $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^2 (\pi^* 2p_y)^1$. It contains one unpaired electron and is paramagnetic.$2$. $KAlO_2$: This is an ionic compound containing $K^+$ and $AlO_2^-$ ions. The $AlO_2^-$ ion has an even number of electrons and all electrons are paired.$3$. $CaO_2$ (Calcium peroxide): It consists of $Ca^{2+}$ and $O_2^{2-}$ (peroxide ion). The peroxide ion has $18$ electrons,all of which are paired.$4$. $NO_2^+$ (Nitronium ion): It has $16$ valence electrons $(5 + 6 	imes 2 - 1 = 16)$. All electrons are paired.Therefore,only $KO_2$ contains an unpaired electron.Hence,option $C$ is correct.

Among $KO_2$,$KAlO_2$,$CaO_2$ and $NO_2^+$,unpaired electron is present in

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