Question

Iron oxide has a formula $F{e_{0.96}}{O_{1.00}}$. What fraction of $Fe$ exist as $F{e^{2 + }}$ and $F{e^{3 + }}$ ions?

Answer 1

Hint: The electronic configuration of iron is $[Ar]3{d^6}4{s^2}$, so it exists in its two oxidation states which are $ + 3$ and $ + 2$. The iron in its $ + 3$ oxidation state is relatively more stable than the $ + 2$ oxidation state because of achieving a stable half-filled d-orbital. For the given question, find the number of molecules of each ion using the charge conservation formula for a compound and then the fraction of each ion can be evaluated by dividing it with the total number of iron molecules present in the sample.

Complete answer:
Let us assume that the x molecules of $F{e^{2 + }}$ ions and y molecules of $F{e^{3 + }}$ ions are present in iron oxide. So, the formula of the sample can be represented as ${\left( {F{e^{2 + }}} \right)_x}{\left( {F{e^{3 + }}} \right)_y}O$.
As per question, the formula of iron oxide is given as $F{e_{0.96}}{O_{1.00}}$. So, we can say that the ions of iron cannot exceed than $0.96$. Hence, we can relate x and y as per following expression:
$x + y = 0.96\;\;\;\;\;\;...(1)$
Also, we know that for a neutral molecule, the charge on the electropositive ions counterbalances the charge on the electronegative element. So, in the given formula the total positive charge on ferrous and ferric ions must balance the two units of negative charge on the oxygen atom. Thus, x and y can also be related by another expression which is as follows:
$2x + 3y = 2$
$ \Rightarrow x + \dfrac{3}{2}y = 1\;\;\;\;\;\;\;\;\;...(2)$
On subtracting equation (1) from equation (3), the value of y will be as follows:
$\dfrac{3}{2}y - y = 1 - 0.96$
$ \Rightarrow y = 0.08$
Substituting the value of y in equation (1), the value of x will be as follows:
$x + 0.08 = 0.96$
$ \Rightarrow x = 0.88$
Fraction of $F{e^{2 + }}$ in the given sample $ = \dfrac{{0.88}}{{0.96}} \Rightarrow 0.92$
Fraction of $F{e^{3 + }}$ in the given sample $ = \dfrac{{0.08}}{{0.96}} \Rightarrow 0.083$
Hence, in the given sample fraction of $Fe$ exist as $F{e^{2 + }}$ and $F{e^{3 + }}$ ions are $0.92$ and $0.083$ respectively.

Note:
It is important to note that this question can alternatively be solved without involving the second variable and considering y as $0.96 - x$ and directly substituting it into the charge conservation formula. It will reduce the calculations for solving the linear equations. Also, if the given method is applied, then remember to check the coefficients of x and y before simplifying the linear equations.