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 )

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

TestHub · Accepted Answer

The metal halide can be manganese dichloride. It gives white precipitate of manganese hydroxide.
${MnCl}_{2} \overset{aq . NaOH}{\to} \underset{(P)}{Mn {(OH)}_{2}} + \underset{(Q)}{NaCl}$
The white precipitate manganese hydroxide on reaction with lead dioxide and concentrated sulphuric acid gives purple coloured solution permanganate.
$Mn {(OH)}_{2} \to_{Conc . H_{2} {SO}_{4}, heat}^{{PbO}_{2} (excess)} {MnO}_{4}^{-} (purple solution)$
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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