The non stiochiometric compound ${F e}_{0.94} O$ is formed when $x$ % of ${F e}^{2 +}$ ions are replaced by as many $\frac{2}{3} {F e}^{3 +}$ ions, $x$ is

18

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12

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15

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6

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Solution

The correct option is B $18$
$x$ % of ${F e}^{2 +}$ ions are replaced by as many $\frac{2}{3} {F e}^{3 +}$ ions.

The number of ${F e}^{3 +}$ ions replacing $x {F e}^{2 +}$ ions $= \frac{2 x}{3}$

The vacancies of cations $= x - \frac{2 x}{3} = x / 3$
But the non stiochiometric compound has formula ${F e}_{0.94} O$ .
$x / 3 = 1 - 0.94 = 0.06$

$x = 0.06 \times 3 = 0.18 = 18 %$

Hence, the correct option is $A$

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