What fraction of Fe exists as Fe (III) in Fe_ 0.96 O ? (Consider Fe_ 0.96 to be made up of Fe (II) and Fe (III) only)

Correct answer: A. (1)/(12) Let's assume Fe in Fe^ 2+ =x Let's assume Fe in Fe^ 3+ (0.96-x) Using the charge conservation formula for Fe_ 0.96 O_ 1.00 We get: (x)(+2)+(0.96-x)(+3)-2=0 aligned & =2 x+2.88-3 x-2=0 & x=0.88 aligned So, number of ions of Fe^ 3+ =0.96-x=0.96-0.88=0.08 Fraction of Fe^ 3+ ion =(0.08)/(0.96) =(1)/(12)

NEET 2022

Exam: NEET
Year: 2022
Subject: Chemistry
Chapter: Solid State

NEET 2022 · Chemistry · Solid State

What fraction of Fe exists as Fe (III) in Fe_ 0.96 O ? (Consider...

What fraction of

\mathrm{Fe}

exists as

\mathrm{Fe}

(III) in

\mathrm{Fe}_{0.96} \mathrm{O}

? (Consider

\mathrm{Fe}_{0.96}

to be made up of

\mathrm{Fe}

(II)

and \mathrm{Fe}

(III) only)

Options

A $\frac{1}{12}$
B0.08
C $\frac{1}{16}$
D $\frac{1}{20}$

Correct answer

A. $\frac{1}{12}$

Step-by-step solution

Let's assume

\mathrm{Fe}

\mathrm{Fe}^{2+}=\mathrm{x}

Let's assume

\mathrm{Fe}

\mathrm{Fe}^{3+}(0.96-x)

Using the charge conservation formula for

\mathrm{Fe}_{0.96} \mathrm{O}_{1.00}

We get:

(x)(+2)+(0.96-x)(+3)-2=0

\begin{aligned} & =2 x+2.88-3 x-2=0 \\ & x=0.88 \end{aligned}

So, number of ions of

\mathrm{Fe}^{3+}=0.96-x=0.96-0.88=0.08 \text{Fraction of } \mathrm{Fe}^{3+} \text{ ion } =\frac{0.08}{0.96} =\frac{1}{12}$

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