Question

oxidation number of Fe in Fe0.94O is
1..200 2 .200/94
3. 94/200 3 .none
I want descriptive answer.

Answer 1

Usually, Fe in oxides will be in either +2 oxidation state or in +3 oxidation state. Fe0.94O is a non-stoichiometric compound. A non-stoichiometric compound is that in which the cation/anion ratio is not simple.

Let us suppose that the oxidation number of iron is x.

We know that the Oxidation number of O is −2.

Therefore, Oxidation number of iron can be calculated as:

0.94x − 2 = 0

x = 2/0.94 = 200/94 = 2.12

Therefore, the Oxidation number of Fe in Fe0.94O is = 200/94 = 2.12

Answer 2

first of all just supposed that the oxidation number of Fe is X.
As we know that he oxygen have oxidation number of -2
So in the FeO.94O iron oxidation number will be 0.94X-2 =O
Just solve the equation simply    0.94X=2
So furthur simplifying it X=2/0.94
we can also write the equation in the form of X= 2*100/94 = 200/94 
So the oxidation number for the iron will be 200/94  which is option 2 in the question.