JEE MAIN - Chemistry (2023 - 24th January Morning Shift - No. 15)
When $$\mathrm{Fe_{0.93}O}$$ is heated in presence of oxygen, it converts to $$\mathrm{Fe_2O_3}$$. The number of correct statement/s from the following is ________
A. The equivalent weight of $$\mathrm{Fe_{0.93}O}$$ is $${{\mathrm{Molecular\,weight}} \over {0.79}}$$
B. The number of moles of Fe$$^{2+}$$ and Fe$$^{3+}$$ in 1 mole of $$\mathrm{Fe_{0.93}O}$$ is 0.79 and 0.14 respectively
C. $$\mathrm{Fe_{0.93}O}$$ is metal deficient with lattice comprising of cubic closed packed arrangement of O$$^{2-}$$ ions
D. The % composition of Fe$$^{2+}$$ and Fe$$^{3+}$$ in $$\mathrm{Fe_{0.93}O}$$ is 85% and 15% respectively
Answer
4
Explanation
Let the number of $\mathrm{Fe}^{2+}$ ions be $x$ and number of $\mathrm{Fe}^{3+}$ ions be $0.93-x$.
According to the charge neutrality,
$$ \begin{aligned} & 2 x+3(0.93-x)=2 \Rightarrow 2 x-3 x+2.79=2 \Rightarrow x=0.79 \\\\ & \mathrm{Fe}^{2+} \text { ion }=x=0.79 \\\\ & \mathrm{Fe}^{3+} \text { ion }=0.93-x=0.93-0.79=0.14 \\\\ & \% \text { of } \mathrm{Fe}^{2+} \text { ion }=\frac{0.79}{0.93} \times 100=84.95 \% \approx 85 \% \\\\ & \% \text { of } \mathrm{Fe}^{3+} \text { ion }=\frac{0.14}{0.93} \times 100=15.05 \% \end{aligned} $$
As it is a metal deficiency defect,
so, 1 mol of $\mathrm{Fe}_{0.93} \mathrm{O}$ contains $1 \mathrm{~mol}$ of oxide ions and $0.93 \mathrm{~mol}$ of iron ions as $\mathrm{Fe}^{2+}$ and $\mathrm{Fe}^{3+}$.
So, moles of $\mathrm{Fe}^{2+}=0.79$
and moles of $\mathrm{Fe}^{3+}=0.14$
$$ \text { The equivalent weight }=\frac{\text { Molecular weight }}{n \text {-factor }} $$
$$F{e^{2 + }} \to F{e^{3 + }} + {e^ - }$$
For one $\mathrm{Fe}^{2+}, n$-factor $=1$
$\therefore $ For $0.79 \,\mathrm{Fe}^{2+}, n$-factor $=0.79$
Out of $0.93 \mathrm{~mol}$, there are $0.79 \mathrm{~mol} \,\mathrm{Fe}^{2+}$ ions are present.
According to the charge neutrality,
$$ \begin{aligned} & 2 x+3(0.93-x)=2 \Rightarrow 2 x-3 x+2.79=2 \Rightarrow x=0.79 \\\\ & \mathrm{Fe}^{2+} \text { ion }=x=0.79 \\\\ & \mathrm{Fe}^{3+} \text { ion }=0.93-x=0.93-0.79=0.14 \\\\ & \% \text { of } \mathrm{Fe}^{2+} \text { ion }=\frac{0.79}{0.93} \times 100=84.95 \% \approx 85 \% \\\\ & \% \text { of } \mathrm{Fe}^{3+} \text { ion }=\frac{0.14}{0.93} \times 100=15.05 \% \end{aligned} $$
As it is a metal deficiency defect,
so, 1 mol of $\mathrm{Fe}_{0.93} \mathrm{O}$ contains $1 \mathrm{~mol}$ of oxide ions and $0.93 \mathrm{~mol}$ of iron ions as $\mathrm{Fe}^{2+}$ and $\mathrm{Fe}^{3+}$.
So, moles of $\mathrm{Fe}^{2+}=0.79$
and moles of $\mathrm{Fe}^{3+}=0.14$
$$ \text { The equivalent weight }=\frac{\text { Molecular weight }}{n \text {-factor }} $$
$$F{e^{2 + }} \to F{e^{3 + }} + {e^ - }$$
For one $\mathrm{Fe}^{2+}, n$-factor $=1$
$\therefore $ For $0.79 \,\mathrm{Fe}^{2+}, n$-factor $=0.79$
Out of $0.93 \mathrm{~mol}$, there are $0.79 \mathrm{~mol} \,\mathrm{Fe}^{2+}$ ions are present.
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