When $\mathrm{Fe}_{0.93} \mathrm{O}$ is heated in presence of oxygen, it converts to $\mathrm{Fe}_{2} \mathrm{O}_{3}$. The number of correct statement/s from the following is ______. A. The equivalent weight of $\mathrm{Fe}_{0.93} \mathrm{O}$ is $\frac{\text { Molecular weight }}{0.79}$. B. The number of moles of $\mathrm{Fe}^{2+}$ and $\mathrm{Fe}^{3+}$ in 1 mole of $\mathrm{Fe}_{0.93} \mathrm{O}$ is 0.79 and 0.14 respectively. C. $\mathrm{Fe}_{0.93} \mathrm{O}$ is metal deficient with lattice comprising of cubic closed packed arrangement of $\mathrm{O}^{2-}$ ions. D. The % composition of $\mathrm{Fe}^{2+}$ and $\mathrm{Fe}^{3+}$ in $\mathrm{Fe}_{0.93} \mathrm{O}$ is $85 \%$ and $15 \%$ respectively.

JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Solid State

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_{0.93} O

\frac{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 +}

Fe_{0.93} O

85%

and

15%

respectively.

Answer: 4

Charge balance in

F e_{0.93} O

requires the iron charges to total

+ 2

per oxide ion: with

x

mole of

F e^{2 +}

and

(0.93 - x)

mole of

F e^{3 +}

2 x + 3 (0.93 - x) = 2

gives

x = 0.79

, so there are

0.79

mol

F e^{2 +}

and

0.14

mol

F e^{3 +}

. These correspond to

\frac{0.79}{0.93} \times 100 = 85%

and

\frac{0.14}{0.93} \times 100 = 15%

. Oxidising all the iron to

F e^{3 +}

F e_{2} O_{3}

changes the charge by

0.79

per formula unit, so the equivalent weight is the molecular weight divided by

0.79

, and the solid is a metal-deficient oxide with cubic close-packed oxide ions. All four statements are correct.

JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Integer answer