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

Hint: 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 \[ - 2\].
\[\begin{array}{c}
0.94 \times y + 1( - 2) = 0\\
0.94y = 2\\
y = \dfrac{2}{{0.94}}\\
 = \dfrac{{200}}{{94}}
\end{array}\]
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.