In the scheme given below, X and Y, respectively, are

The correct option is MnO₄⁻ and Cl₂.

Let us analyze the reaction sequence shown in the scheme step-by-step:

1. Reaction of Metal Halide with aqueous NaOH:

The image shows a reaction scheme starting with a Metal halide reacting with aq. NaOH to form a White precipitate (P) and a Filtrate (Q).

Here, the metal halide is manganese chloride, ${\text{MnCl}}_2$. Reaction with $\text{NaOH}$ gives manganese(II) hydroxide, ${\text{Mn(OH)}}_2$, which forms a white precipitate, and sodium chloride, $\text{NaCl}$, in the filtrate.

${\text{MnCl}}_2+2\text{NaOH}\to {\text{Mn(OH)}}_2\downarrow \text{ (White precipitate, P)}+2\text{NaCl (Filtrate, Q)}$

2. Identification of X:

In the second reaction step from the image, P [${\text{Mn(OH)}}_2$] reacts with aq. H₂SO₄ and excess PbO₂ under heating to produce X (a coloured species in solution).

Lead dioxide (${\text{PbO}}_2$) in an acidic medium is a strong oxidizing agent (bismuthate test / lead peroxide test for ${\text{Mn}}^{2+}$ ions). It oxidizes ${\text{Mn}}^{2+}$ ions to purple permanganate ions, MnO₄⁻.

$2{\text{Mn(OH)}}_2+5{\text{PbO}}_2+3{\text{H}}_2{\text{SO}}_4\to 2{\text{MnO}}_4^{-}\text{ (Purple species, X)}+5{\text{Pb}}^{2+}+3{\text{SO}}_4^{2-}+5{\text{H}}_2\text{O}$

3. Identification of Y:

In the third reaction step, Q [containing chloride ions, ${\text{Cl}}^{-}$] is warmed with MnO(OH)₂ and Conc. H₂SO₄ to evolve gas Y (gives blue-coloration with KI-starch paper).

Hydrated manganese dioxide oxidizes chloride ions to chlorine gas (${\text{Cl}}_2$):

$\text{MnO}{\text{(OH)}}_2+2{\text{Cl}}^{-}+3{\text{H}}^{+}\to {\text{Mn}}^{2+}+{\text{Cl}}_2\uparrow \text{ (Gas Y)}+3{\text{H}}_2\text{O}$

The evolved chlorine gas, Cl₂ (Y), oxidizes iodide ions in KI-starch paper to iodine (${\text{I}}_2$), which turns starch blue.

Therefore, X is MnO₄⁻ and Y is Cl₂.