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Metal deficiency defect is shown by ${Fe}_{0.93} O$ . In the crystal, some ${Fe}^{2 +}$ cations are missing and loss of positive charge is compensated by the presence of ${Fe}^{3 +}$ ions. The percentage of ${Fe}^{2 +}$ ions in the ${Fe}_{0.93} O$ crystals is (Nearest integer)

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Detailed Solution

In ${Fe}_{0.93} O$ for every 93 Fe ions 14 are ${Fe}^{+ 3}$ and (93-14)=79 are ${Fe}^{+ 2}$ ions
$∴ % {Fe}^{+ 2} = \frac{79}{93} \times 100 = 84.9 %$

${nearest integer} = 85 %$

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