$\mathrm{Fex}_{2}$ and $\mathrm{Fey}_{3}$ are known when x and y are:

(A) $x=\mathrm{F}, \mathrm{Cl}, \mathrm{Br}, \mathrm{I}$ and $\mathrm{y}=\mathrm{F}, \mathrm{Cl}, \mathrm{Br}$

(B) $x=\mathrm{Cl}, \mathrm{Br}, \mathrm{I}$ and $y=\mathrm{F}, \mathrm{Cl}, \mathrm{Br}, \mathrm{I}$

(C) $x=\mathrm{F}, \mathrm{Cl}, \mathrm{Br}$, I and $y=\mathrm{F}, \mathrm{Cl}, \mathrm{Br}, \mathrm{I}$

(D) $x=\mathrm{F}, \mathrm{Cl}, \mathrm{Br}$ and $y=\mathrm{F}, \mathrm{Cl}, \mathrm{Br}, \mathrm{I}$

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · d and f Block Elements

Fex_{2}

and

Fey_{3}

are known when x and y are:

Answer: (A)

Iron forms the dihalide with all four halogens, because

F e^{2 +}

is too weak an oxidant to attack any halide ion. The trihalide is another matter:

F e^{3 +}

is a moderate oxidising agent and iodide is the most easily oxidised of the halides, so the two react together as

2 F e^{3 +} + 2 I^{-} \to 2 F e^{2 +} + I_{2}

and the solid iron(III) iodide cannot be isolated. Fluoride, chloride and bromide are held far more firmly and their iron(III) salts are stable. So

Fe X_{2}

exists for fluorine, chlorine, bromine and iodine, while

Fe Y_{3}

exists only for fluorine, chlorine and bromine.

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Single correct option