Q. Oxidation number of iron in $F e_{0.94} O$ is

2447 237 Redox Reactions Report Error

A

+2

20%

B

200/94

60%

C

94/200

14%

D

+1

6%

Solution:

O.N. of Fe in $F e_{0.94} O$ is
$x \times 0.94 + 2 x (- 1) = 0$ or
$94 x = 200$ or $x = 200/94$ .