Question

Oxidation number of Iron in $F{e_{0.94}}O$is:
(A) $2\,x\,0.94$
(B) $\dfrac{{200}}{{94}}$
(C) $\dfrac{{94}}{{200}}$
(D) $0.94$

Answer 1

As we know that the oxygen atom is an electronegative atom and has $- 2$ charge due to only six electrons in its outermost orbital. The iron element is a transition element and can have variable valency.

Complete step by step answer:
The oxidation number of Iron can be calculated as-
Suppose y is the oxidation number of Iron in iron oxide, it is a neural oxide so the charge on $F{e_{0.94}}O$is zero and the charge on oxygen is $- 2$.

0.94 \times y + 1( - 2) = 0\\\ 0.94y = 2\\\ y = \dfrac{2}{{0.94}}\\\ = \dfrac{{200}}{{94}} \end{array}$$ **Therefore, the correct option is option (B).** **Additional information** -The iron is a transition element and exist in variable valency such as $$F{e^{ + 2}}$$and $$F{e^{ + 3}}$$so, in the crystal $$F{e_{0.94}}O$$ we can also know the total number of $$F{e^{ + 2}}$$and $$F{e^{ + 3}}$$as- The formula $$F{e_{0.94}}O$$shows that the ratio between iron and oxygen is$$0.94:1$$. Thus, if there are $$100$$oxygen atoms then the number of total iron atom is $$94$$ Charge on$$100$$oxygen is $$ - 200$$ Suppose there are $$x$$ number of $$F{e^{ + 2}}$$in the crystal Then the number of $$F{e^{ + 3}}$$$$ = (94 - x)$$ The total charge on $$F{e^{ + 2}}$$and $$F{e^{ + 3}}$$$$ = (282 - x)$$ As the iron oxide is neutral, the total charge on cation is the same as the total charge on anion. $$\begin{array}{c} \,(282 - x) = 200\\\ x = 82 \end{array}$$ Therefore, the total number of $$F{e^{ + 2}}$$is $$82$$ and the total number of $$F{e^{ + 3}}$$is $$200$$. **Note:** A metal with higher valency replaces more than one metal with lower valency, thereby creating vacancies.