Among KO_2,AlO_2^-,BaO_2,and NO_2^+,in which | vedclass

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(C) To determine the presence of unpaired electrons,we analyze the electronic configuration or total electron count of each species:$1$. $NO_2^+$: Tot

Vedclass · Accepted Answer

$KO_2$ only

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(C) To determine the presence of unpaired electrons,we analyze the electronic configuration or total electron count of each species:$1$. $NO_2^+$: Total electrons = $7 + (2 	imes 8) - 1 = 22$ (even,all paired).$2$. $BaO_2$: Contains the peroxide ion $O_2^{2-}$. Total electrons in $O_2^{2-} = 16 + 2 = 18$ (even,all paired).$3$. $AlO_2^-$: Total electrons = $13 + (2 	imes 8) + 1 = 30$ (even,all paired).$4$. $KO_2$: Contains the superoxide ion $O_2^-$. Total electrons in $O_2^- = 16 + 1 = 17$ (odd). An odd number of electrons implies at least one unpaired electron.Therefore,only $KO_2$ contains an unpaired electron.

Column $-I$	Column $-II$
$(a)$. Super oxide	$(p)$. $Na_2O_2$
$(b)$. Peroxide	$(q)$. $CO_2$
$(c)$. Dioxide	$(r)$. $C_3O_2$
$(d)$. Sub oxide	$(s)$. $CsO_2$

Among $KO_2$,$AlO_2^-$,$BaO_2$,and $NO_2^+$,in which species is an unpaired electron present?

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Column $-I$ Column $-II$

$(a)$. Super oxide $(p)$. $Na_2O_2$

$(b)$. Peroxide $(q)$. $CO_2$

$(c)$. Dioxide $(r)$. $C_3O_2$

$(d)$. Sub oxide $(s)$. $CsO_2$

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