Question

What is the total number of electrons in a superoxide ion, \[O_2^ - \]?

Answer 1

Hint: Electronic configuration of oxygen,\[O\] atom is
\[O\]-\[1{s^2}2{s^2}2{p^4}\]
Therefore, total number of electrons in an oxygen atom is \[2 + 2 + 4 = 8\]
Superoxide ion is a free radical that exhibits paramagnetism due to the presence of an unpaired electron.
Superoxides are compounds in which the oxidation number of oxygen is \[ - \dfrac{1}{2}\].

Complete answer:
Total number of electrons in an oxygen atom= \[8\]
Total number of electrons in \[2\] oxygen atoms = \[2 \times 8 = 16\]
We know, \[1\] negative charge is equal to \[1\] electrons.
Superoxide ion contains \[1\] negative charge on its molecule.
 Total number of electrons on superoxide ion,\[O_2^ - \]=( number of electrons in \[2\] oxygen atoms + number of electron in negative charge)
\[\;Total{\text{ }}number{\text{ }}of{\text{ }}electrons{\text{ }}on{\text{ }}superoxide{\text{ }}ion,O_2^ - = 16 + 1 = 17\]
Hence, total number of electrons on superoxide ion, \[O_2^ - \]= \[17\]

Note:
Superoxide ion contains \[2\] oxygen atoms linked together by a single bond and one of the oxygen atoms having an extra electron thus forming \[O_2^ - \]. In case of peroxide ion, it contains \[2\] extra electrons, \[1\] on each oxygen atom which results in \[O_2^{2 - }\]. In case of oxide ion, it contains only \[1\] oxygen atom and carries \[2\]extra electrons and hence is present as \[O_{}^{2 - }\].