Question

When $Fe _{0.93} O$ is heated in presence of oxygen, it converts to $Fe _2 O _3$ The number of correct statement/s from the following is
A. The equivalent weight of $Fe _{093} O$ is Molecular wejght $0.79$
B. The number of moles of $Fe ^{2+}$ and $Fe ^{3+}$ in 1 mole of $Fe _{0.93} O$ is $0.79$ and $014$ respectively
C. $Fe _{0.93} O$ is metal deficient with lattice comprising of cubic closed packed arrangement of $O ^{2-}$ ions
D. The $\%$ composition of $Fe ^{2+}$ and $Fe ^{3+}$ in $Fe _{093} O$ is $85 \%$ and $15 \%$ respectively

Answer 1

Correct Answer: 4

Step 1: Calculating \( n_f \) (number of Fe atoms per formula unit). For \( \text{Fe}_{0.93}\text{O} \): \[ n_f = 3 \left( 1 - \frac{200}{93} \right) \times 0.93 \] Simplifying gives: \[ n_f = 0.79 \]
Step 2: Mole ratio of \( \text{Fe}^{2+} \) and \( \text{Fe}^{3+} \). From the above, - Moles of \( \text{Fe}^{2+} \): \( 0.79 \) - Moles of \( \text{Fe}^{3+} \): \( 1 - 0.79 = 0.14 \).
Step 3: Verifying statements. A: The equivalent weight of \( \text{Fe}_{0.93}\text{O} \) is indeed \( \frac{\text{Molecular weight}}{n_f} \), where \( n_f = 0.79 \). Conclusion: True.
B: The calculated values of \( \text{Fe}^{2+} \) and \( \text{Fe}^{3+} \) are 0.79 and 0.14, respectively. Conclusion: True.
C: \( \text{Fe}_{0.93}\text{O} \) is a metal-deficient oxide with a lattice structure of cubic close-packed \( \text{O}^{2-} \) ions. Conclusion: True.
D: The percentage of \( \text{Fe}^{2+} \) and \( \text{Fe}^{3+} \) is calculated as: \[ \text{Fe}^{2+}: \frac{0.79}{0.93} \times 100 = 85\%, \quad \text{Fe}^{3+}: \frac{0.14}{0.93} \times 100 = 15\%. \] Conclusion: True.