JEE Main 2021 · 16 March · evening · Single correct · +4 / −1
Fex₂ and Fey₃ are known when x and y are :
- (a) x = F, Cl, Br, I and y = F, Cl, Br, I
- (b) x = F, Cl, Br and y = F, Cl, Br, I
- (c) x = F, Cl, Br, I and y = F, Cl, Br
- (d) x = Cl, Br, I and y = F, Cl, Br, I
Answer and solution
Answer (c) x = F, Cl, Br, I and y = F, Cl, Br
Solution
Chemistry ma’am · 4 steps
Fe³⁺ is a strong enough oxidant to pull an electron off iodide, so FeI₃ cannot survive.
- Iron(II) halides. In FeX₂ iron is at +2, a modest oxidation state. No halide ion is oxidised by Fe²⁺, so FeF₂, FeCl₂, FeBr₂ and FeI₂ all exist.
- Iron(III) halides. In FeY₃ iron is at +3, which is an oxidising state. F⁻, Cl⁻ and Br⁻ hold their electrons too tightly to be oxidised, so FeF₃, FeCl₃ and FeBr₃ are stable.
2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂
- Why not FeI3. Iodide is the strongest reducing agent among the halides. It reduces Fe³⁺ to Fe²⁺ and is itself oxidised to I₂, so FeI₃ breaks down into FeI₂ and iodine.
- So. x can be F, Cl, Br or I, but y can only be F, Cl or Br. That is option (c).
The idea to keep. A compound cannot exist if the metal in its higher oxidation state would oxidise its own anion. Fe³⁺ and I⁻ is the classic clash, just like Cu²⁺ and I⁻ giving CuI.
Answer from the NTA final answer key. Solution by PrepWiser.