The non-stoichiometric compound $F e_{0.94} O$ is formed when x% of $F e^{2 +}$ ions are replaced by $F e^{3 +}$ ions then x is

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Solution

The correct option is D
18

Number of $F e^{3 +}$ ions replacing $x F e^{2 +} = \frac{2 x}{3}$ ; Vacancies of cations

$= x - \frac{2 x}{3} = \frac{x}{3}$
But $\frac{x}{3} = 1 - 0.94 = 0.06$ or x = 0.06*3 = 0.18 = 18%

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