In the scheme given below, $\mathbf{X}$ and $\mathbf{Y}$, respectively, are

$\text{Metal halide} \stackrel{ \mathrm{aq}. \mathrm{NaOH}}{\to } \text{ White precipitate} \left(\mathrm{P}\right)+ \text{Filtrate} \left(\mathrm{Q}\right)$

$\mathrm{P}\mathbf{ }\mathbf{ }\mathbf{ }\underset{ \mathrm{heat} }{\stackrel{ \mathrm{aq}. {\mathrm{H}}_2{\mathrm{SO}}_4}{\stackrel{{\mathrm{PbO}}_2 (\mathrm{excess}) }{⟶}\mathbf{ }\mathbf{ }}}\mathrm{X} \left(\text{a coloured species in solution}\right)$

$\mathrm{Q} \underset{ \mathrm{warm} }{\stackrel{\mathrm{MnO}(\mathrm{OH}{)}_2, \mathrm{Conc}.{\mathrm{H}}_2{\mathrm{SO}}_4}{⟶}} \mathrm{Y}\left(\text{gives blue-coloration with} \mathrm{KI}-\text{starch paper}\right)$

The metal halide can be manganese dichloride. It gives white precipitate of manganese hydroxide.

${\mathrm{MnCl}}_2\stackrel{\mathrm{aq}.\mathrm{NaOH}}{\to }\underset{\left(\mathrm{P}\right)}{\mathrm{Mn}{\left(\mathrm{OH}\right)}_2}+\underset{\left(\mathrm{Q}\right)}{\mathrm{NaCl}}$

The white precipitate manganese hydroxide on reaction with lead dioxide and concentrated sulphuric acid gives purple coloured solution permanganate.

$\mathrm{Mn}{\left(\mathrm{OH}\right)}_2\underset{\mathrm{Conc}.{\mathrm{H}}_2{\mathrm{SO}}_4,\mathrm{heat}}{\overset{{\mathrm{PbO}}_2\left(\mathrm{excess}\right)}{\to }}{\mathrm{MnO}}_4^{-}\left(\mathrm{purple} \mathrm{solution}\right)$

The sodium chloride gives ${\mathrm{Cl}}_2$ on reaction with $\mathrm{MnO}{\left(\mathrm{OH}\right)}_2, \mathrm{conc}.{\mathrm{H}}_2{\mathrm{SO}}_4$. The chlorine formed can oxidise iodide to iodine which gives blue colouration with starch paper.