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JEE Advanced Chemistry d and f Block Elements 2023 JEE Advanced 2023 (Paper 1)

JEE Advanced Chemistry Question (2023) — Solution

Question

In the scheme given below, X and Y , respectively, are Metal halide         →   aq .  NaOH          White precipitate   P +   Filtrate   Q P       ⟶ PbO 2   ( excess )        aq .  H 2 SO 4   heat   X   a coloured species in solution   Q       ⟶ MnO ( OH ) 2 ,   Conc . H 2 SO 4   warm       Y gives blue-coloration with   KI - starch paper  

Options

  1. A. CrO 4 2 -  and  Br 2
  2. B. MnO 4 2 -   and       Cl 2
  3. C. MnO 4 -     and       Cl 2
  4. D. MnSO 4     and     HOCl

Answer

C. MnO 4 -     and       Cl 2

Step-by-step solution

The metal halide can be manganese dichloride. It gives white precipitate of manganese hydroxide. MnCl 2 → aq . NaOH Mn OH 2 P + NaCl Q The white precipitate manganese hydroxide on reaction with lead dioxide and concentrated sulphuric acid gives purple coloured solution permanganate. Mn OH 2 → Conc . H 2 SO 4 , heat PbO 2 excess MnO 4 - purple   solution The sodium chloride gives  Cl 2  on reaction with  MnO OH 2 ,   conc . H 2 SO 4 . The chlorine formed can oxidise iodide to iodine which gives blue colouration with starch paper.

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