Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Chemistry
The oxidation state of K in KO2 is
Q. The oxidation state of
K
in
K
O
2
is
1807
180
The s-Block Elements
Report Error
A
−
1
B
+
1
C
0
D
+
2
Solution:
The oxidation number of oxygen in superoxide ion
(
O
2
−
)
is
−
1
; since the compound is neutral, therefore, the oxidation state of potassium is
+
1
.