In the scheme given below, $\mathbf{X}$ and $\mathbf{Y}$, respectively, are Metal halide $\xrightarrow{\text { aq. } \mathrm{NaOH}}$ White precipitate (P) + Filtrate (Q) $$\begin{aligned} & \mathbf{P} \xrightarrow[\text { heat }]{\substack{\text { aq. } \mathrm{H}_{2} \mathrm{SO}_{4} \\ \mathrm{PbO}_{2} \text { excess }}} \mathbf{X} \text { (a coloured species in solution) } \\ & \mathbf{Q} \xrightarrow[\text { warm }]{\substack{\mathrm{MnO}(\mathrm{OH})_{2}, \\ \text { Conc. } \mathrm{H}_{2} \mathrm{SO}_{4}}} \mathbf{Y} \text { (gives blue-coloration with KI-starch paper) } \end{aligned}$$

(C) $\mathrm{MnO}_{4}{ }^{-}$and $\mathrm{Cl}_{2}$

(A) $\mathrm{CrO}_{4}{ }^{2-}$ and $\mathrm{Br}_{2}$

(B) $\mathrm{MnO}_{4}{ }^{2-}$ and $\mathrm{Cl}_{2}$

(D) $\mathrm{MnSO}_{4}$ and HOCl

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · d and f Block Elements

In the scheme given below,

X

and

Y

, respectively, are

Metal halide

aq. NaOH

White precipitate (P) + Filtrate (Q)

P aq. H_{2} SO_{4} PbO_{2} excess 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)

Answer: (C)

The metal halide is

MnCl X_{2}

.

Step 1:

MnCl X_{2} + NaOH Mn (OH) X_{2} ↓ + NaCl

White precipitate P is

Mn (OH) X_{2}

, filtrate Q contains

Cl X^{-}

.

Step 2:

Mn (OH) X_{2}

is oxidized by

PbO X_{2}

in acidic medium (

H X_{2} SO X_{4}

) with heat:

Mn (OH) X_{2} PbO X_{2}, H X^{+}, heat MnO X_{4} X^{-} + Pb X^{2 +}

X is

MnO X_{4} X^{-}

(permanganate ion), which is purple/colored.

Step 3: Filtrate Q (containing

Cl X^{-}

) is treated with concentrated

H X_{2} SO X_{4}

and warmed:

Cl X^{-} conc \cdot H X_{2} SO X_{4}, warm Cl X_{2}

Y is

Cl X_{2}

, which gives blue coloration with KI-starch paper (liberates

I X_{2}

from KI).

Answer: (C)

MnO X_{4} X^{-}

and

Cl X_{2}

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Single correct option