In the scheme given below, $X$ and $Y$, respectively, are $\begin{aligned} \text{Metal halide}…
In the scheme given below, $X$ and $Y$, respectively, are
$\begin{aligned}
\text{Metal halide} &\xrightarrow[\text{NaOH}]{\text{aq}} \text{White precipitate} + \text{Filtrate} \\
P &\xrightarrow[\text{H}_2\text{SO}_4]{\text{aq, heat}} X \quad (\text{a coloured species in solution}) \\
Q &\xrightarrow[\text{Conc. H}_2\text{SO}_4]{\text{warm}} Y \quad (\text{gives blue-coloration with KI-starch paper})
\end{aligned}$
and
Solution
The metal halide can be manganese dichloride. It gives white precipitate of manganese hydroxide.
$MnCl_{2} \xrightarrow{aq. NaOH} Mn(OH)_{2} \downarrow + NaCl$
The white precipitate manganese hydroxide on reaction with lead dioxide and concentrated sulphuric acid gives purple coloured solution permanganate.
$Mn(OH)_{2} \xrightarrow[excess PbO_{2}]{Conc. H_{2}SO_{4}, heat} MnO_{4}^{-}$ (purple solution)
The sodium chloride gives $Cl_{2}$ on reaction with $Mn(OH)_{2}, Conc. H_{2}SO_{4}$. The chlorine formed can oxidise iodide to iodine which gives blue colouration with starch paper.