JEE Advanced 2023 · Paper 1 · Chemistry Q4

← JEE Advanced 2023 · Paper 1

ChemistrySingle correct+3 / -1Question 38 of 51

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

Metal halide → [aq. NaOH] → white precipitate (P) + filtrate (Q). P → [aq. H2SO4, PbO2 (excess), heat] → X (a coloured species in solution). Q → [MnO(OH)2, conc. H2SO4, warm] → Y (gives blue coloration with KI-starch paper).

\ce{CrO4^2-}

and

\ce{Br2}

\ce{MnO4^2-}

and

\ce{Cl2}

\ce{MnO4-}

and

\ce{Cl2}

\ce{MnSO4}

and

\ce{HOCl}

Answer and worked solution

Official answer: C

Identifying the salt. The halide is $\ce{MnCl2}$ .

The precipitate (P). $\ce{Mn^2+ + 2OH- -> Mn(OH)2}$ is a white precipitate.

X. Heating with excess $\ce{PbO2}$ and acid oxidises Mn(II) to purple permanganate:
$\ce{2Mn^2+ + 5PbO2 + 4H+ -> 2MnO4- + 5Pb^2+ + 2H2O}.$

Y. The filtrate (Q) contains $\ce{Cl-}$ . Warming it with $\ce{MnO(OH)2}$ (hydrated $\ce{MnO2}$ ) and concentrated $\ce{H2SO4}$ releases $\ce{Cl2}$ . Chlorine oxidises $\ce{I-}$ to $\ce{I2}$ , which turns starch blue.

X = $\ce{MnO4-}$ and Y = $\ce{Cl2}$ .

Topics:Cation analysis Anion analysis Important compounds (KMnO4, K2Cr2O7)

← Previous question Next question →

JEE Advanced 2023 · Paper 1, Chemistry question 4