Among KO_(2), AlO_(2)^(-), BeO_(2) and NO_(2)^(+), unpaired electron is present in

Among KO_(2), AlO_(2)^(-), BeO_(2) and NO_(2)^(+), unpaired electron i

1.	Among KO_(2), AlO_(2)^(-), BeO_(2) and NO_(2)^(+), unpaired electron is present in
Answer» `NO_(2)^(+) and BaO_(2)` `KO_(2) and AlO_(2)^(-)` `KO_(2)` only `BaO_(2)`only Solution :ELECTRONS present : `NO_(2)^(+) = 7 + 16 - 1 = 22 , BaO_(2) = 56 + 16 = 72 ` `AlO_(2)^(-) = 13 + 16 + 1 = 30 ` and `KO_(2) = 19 + 16 = 35` Thus , OLNY `KO_(2)` is odd electron species.