The non stoichiometric compound Fe_ 0.94 O is formed when x \% of Fe^ 2+ ions are replaced by as many (2)/(3) Fe^ 3+ ions, x is

Correct answer: A. 18 The number of Fe^ 3+ ions replacing x Fe^ 2+ ions = 2 x 3 vacancies of cations array l = x-(2 x)/(3)=x / 3 But x / 3=1-0.94=0.06, x=0.06 ×3 =0.18=18 \% array

BITSAT 2016

Exam: BITSAT
Year: 2016
Subject: Chemistry
Chapter: Solid State

BITSAT 2016 · Chemistry · Solid State

The non stoichiometric compound Fe_ 0.94 O is formed when x \% of...

The non stoichiometric compound

\mathrm{Fe}_{0.94} \mathrm{O}

is formed when

\mathrm{x} \%

\mathrm{Fe}^{2+}

ions are replaced by as many

\frac{2}{3} \mathrm{Fe}^{3+}

ions,

\mathrm{x}

Options

Correct answer

A. 18

Step-by-step solution

The number of

\mathrm{Fe}^{3+}

ions replacing

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

ions

=\frac{2 \mathrm{x}}{3}

vacancies of cations

\begin{array}{l} = x-\frac{2 x}{3}=x / 3 \\ \text{ But } x / 3=1-0.94=0.06, x=0.06 \times3 \\ =0.18=18 \% \end{array}

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