Question

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

• A. $NO_2^+$ and $BaO_2$
• B. $KO_2$, and $AlO_2^-$
• C. Only $KO_2$
• D. Only $BaO_2$

Answer 1

Answer: (C)

Only $KO_2$

Answer 2

The correct option is(C): KO2.

Number of electrons has to be calculated:

NO2+: 5+6(2)-1 =16 electrons

AlO2- = 3+6(2) +1 = 16 electrons

BaO2 = 2+6(2)= 14 electrons

KO2 =1 +6(2) = 13 electrons

All the compounds have paired electrons except KO2. Therefore, the answer is KO2