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QJEEM23Ch8d9c: JEE Main 2023 Chemistry - Class 11 JEEM23Ch8d9c Solution: When is heated in presence of oxygen, it converts to . The…

When Fe0.93O\mathrm{Fe_{0.93}O} is heated in presence of oxygen, it converts to Fe2O3\mathrm{Fe_2O_3} . The number of correct statement/s from the following is ________

A. The equivalent weight of Fe0.93O\mathrm{Fe_{0.93}O} is Molecular weight0.79{{\mathrm{Molecular\,weight}} \over {0.79}}

B. The number of moles of Fe 2+^{2+} and Fe 3+^{3+} in 1 mole of Fe0.93O\mathrm{Fe_{0.93}O} is 0.79 and 0.14 respectively

C. Fe0.93O\mathrm{Fe_{0.93}O} is metal deficient with lattice comprising of cubic closed packed arrangement of O 2−^{2-} ions

D. The % composition of Fe 2+^{2+} and Fe 3+^{3+} in Fe0.93O\mathrm{Fe_{0.93}O} is 85% and 15% respectively

Step-by-Step Verified Solution

Let the number of Fe2+\mathrm{Fe}^{2+} ions be xx and number of Fe3+\mathrm{Fe}^{3+} ions be 0.93−x0.93-x .

According to the charge neutrality,

amp;2x+3(0.93−x)=2⇒2x−3x+2.79=2⇒x=0.79amp;Fe2+ ion =x=0.79amp;Fe3+ ion =0.93−x=0.93−0.79=0.14amp;% of Fe2+ ion =0.790.93×100=84.95%≈85%amp;% of Fe3+ ion =0.140.93×100=15.05%\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 Fe0.93O\mathrm{Fe}_{0.93} \mathrm{O} contains 1 mol1 \mathrm{~mol} of oxide ions and 0.93 mol0.93 \mathrm{~mol} of iron ions as Fe2+\mathrm{Fe}^{2+} and Fe3+\mathrm{Fe}^{3+} .

So, moles of Fe2+=0.79\mathrm{Fe}^{2+}=0.79

and moles of Fe3+=0.14\mathrm{Fe}^{3+}=0.14

 The equivalent weight = Molecular weight n-factor \text { The equivalent weight }=\frac{\text { Molecular weight }}{n \text {-factor }}

Fe2+→Fe3++e−F{e^{2 + }} \to F{e^{3 + }} + {e^ - }

For one Fe2+,n\mathrm{Fe}^{2+}, n -factor =1=1

∴\therefore For 0.79 Fe2+,n0.79 \,\mathrm{Fe}^{2+}, n -factor =0.79=0.79

Out of 0.93 mol0.93 \mathrm{~mol} , there are 0.79 mol Fe2+0.79 \mathrm{~mol} \,\mathrm{Fe}^{2+} ions are present.

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