Which halogen is known to cause the reaction given below :
2 Cu²⁺+4 X⁻ → Cu₂ X₂(s)+X₂

Solution

Chemistry ma’am · 4 steps

Ask which halide is a strong enough reducing agent to hand an electron to Cu²⁺. Only iodide is.

1. What happens. In this reaction Cu²⁺ is reduced to Cu⁺, which separates out as the solid Cu₂X₂. At the same time X⁻ is oxidised to X₂. So the halide must be a good enough reducing agent to give up its electron to Cu²⁺.
2. Reducing power. A halide gives up its electron more easily when its E°(X₂/X⁻) is lower. Down the group this value falls, so reducing power rises from F⁻ to I⁻.

   Halide	E°(X₂/X⁻)	Does Cu²⁺ oxidise it?
   F⁻	2.87 V	No, stable CuF₂ forms
   Cl⁻	1.36 V	No, stable CuCl₂ forms
   Br⁻	1.09 V	No, CuBr₂ forms
   I⁻	0.54 V	Yes, Cu₂I₂ and I₂ form

3. Why iodide works. CuI is very insoluble, so it keeps leaving the solution as it forms. This pulls the reaction forward and lets Cu²⁺ oxidise I⁻. That is why CuI₂ does not exist: 2Cu²⁺ + 4I⁻ → Cu₂I₂(s) + I₂.
4. So. Only iodide is a strong enough reducing agent to cause this reaction, so the answer is (b), only iodine.

The idea to keep. Iodide is the strongest reducing halide. Any oxidising cation, such as Cu²⁺ or Fe³⁺, can turn it into I₂. That is why CuI₂ and FeI₃ do not exist as stable compounds.

Answer from the NTA final answer key. Solution by PrepWiser.