JEE Advanced2023Chemistryd and f Block ElementsActual
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 
Options
- ACrO 4 2 - and Br 2
- BMnO 4 2 -   and       Cl 2
- CMnO 4 -     and       Cl 2
- DMnSO 4     and     HOCl
Correct 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 bl