When $$Fe_{0.93}O$$ is heated in presence of oxygen, it converts to $$Fe_2O_3$$. The number of correct statement/s from the following is ______.
A. The equivalent weight of $$Fe_{0.93}O$$ is $$\frac{\text{Molecular weight}}{0.79}$$
B. The number of moles of $$Fe^{2+}$$ and $$Fe^{3+}$$ in 1 mole of $$Fe_{0.93}O$$ is 0.79 and 0.14 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 $$Fe_{0.93}O$$ is 85% and 15% respectively.

We need to analyze each statement about $$Fe_{0.93}O$$.

In $$Fe_{0.93}O$$, let the number of $$Fe^{2+}$$ ions be $$x$$ and $$Fe^{3+}$$ ions be $$y$$ per formula unit.

$$x + y = 0.93$$ ... (i)

For charge neutrality with one $$O^{2-}$$: $$2x + 3y = 2$$ ... (ii)

From (i): $$x = 0.93 - y$$. Substituting in (ii):

$$2(0.93 - y) + 3y = 2$$

$$1.86 - 2y + 3y = 2$$

$$y = 0.14$$, so $$x = 0.79$$

Statement A: When $$Fe_{0.93}O$$ is oxidized to $$Fe_2O_3$$, each $$Fe^{2+}$$ loses 1 electron. The n-factor (number of electrons lost per formula unit) = number of $$Fe^{2+}$$ ions = 0.79. So equivalent weight = Molecular weight / 0.79. This is correct.

Statement B: From our calculation, moles of $$Fe^{2+} = 0.79$$ and $$Fe^{3+} = 0.14$$ per mole of $$Fe_{0.93}O$$. This is correct.

Statement C: $$Fe_{0.93}O$$ has a structure similar to FeO (rock salt type) where $$O^{2-}$$ ions form a cubic close packed (FCC) arrangement. It is a metal-deficient defect (some Fe sites are vacant or have Fe³⁺ instead of Fe²⁺). This is correct.

Statement D: % of $$Fe^{2+} = \frac{0.79}{0.93} \times 100 = 84.95\% \approx 85\%$$

% of $$Fe^{3+} = \frac{0.14}{0.93} \times 100 = 15.05\% \approx 15\%$$

This is correct.

All four statements are correct.

The answer is 4.