JEE Main 2023 · 1 February · morning · Single correct · +4 / −1
A solution of FeCl₃ when treated with K₄[Fe(CN)₆] gives a prussium blue precipitate due to the formation of :
- (a) Fe[Fe(CN)₆]
- (b) Fe₄[Fe(CN)₆]₃
- (c) Fe₃[Fe(CN)₆]₂
- (d) K[Fe₂(CN)₆]
Answer and solution
Answer (b) Fe₄[Fe(CN)₆]₃
Solution
Chemistry ma’am · 3 steps
Write the two ions with their charges, then balance them: Fe³⁺ meets [Fe(CN)₆]⁴⁻.
- Identify the ions. FeCl₃ supplies Fe³⁺. K₄[Fe(CN)₆] is potassium ferrocyanide, and it supplies the complex ion [Fe(CN)₆]⁴⁻, with iron in the +2 state inside it.
- Balance the charges. The precipitate must be neutral. The lowest common multiple of 3 and 4 is 12, so you need 4 Fe³⁺ (+12) for every 3 [Fe(CN)₆]⁴⁻ (-12).
4Fe³⁺ + 3[Fe(CN)₆]⁴⁻ → Fe₄[Fe(CN)₆]₃ v
- So. The Prussian blue precipitate is ferric ferrocyanide, Fe₄[Fe(CN)₆]₃, which is option (b).
The idea to keep. For any iron and cyanide-complex precipitate, first check which reagent is ferro (4-) and which is ferri (3-), then balance the charges. Fe³⁺ with ferrocyanide gives Prussian blue.
Answer from the NTA final answer key. Solution by PrepWiser.