$\underline{\mathrm{A}} \xrightarrow{4 \mathrm{KOH}, \mathrm{O}_{2}} \underset{\text { (Green) }}{2 \mathrm{~B}}+2 \mathrm{H}_{2} \mathrm{O}$ $3 \mathrm{~B} \xrightarrow{4 \mathrm{HCl}} \underset{\text { (Purple) }}{2 \mathrm{C}}+\mathrm{MnO}_{2}+2 \mathrm{H}_{2} \mathrm{O}$ $2 \underline{\mathrm{C}} \xrightarrow{\mathrm{H}_{2} \mathrm{O}, \mathrm{KI}} 2 \underline{\mathrm{~A}}+2 \mathrm{KOH}+\underline{\mathrm{D}}$ In the above sequence of reactions, $\underline{A}$ and $\underline{D}$, respectively are:

(B) $\mathrm{MnO}_{2}$ and $\mathrm{KIO}_{3}$

(A) Ki and $\mathrm{KMnO}_{4}$

(C) $\mathrm{KIO}_{3}$ and $\mathrm{MnO}_{2}$

(D) KI and $\mathrm{K}_{2} \mathrm{MnO}_{4}$

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · d and f Block Elements

\underline{A} 4 KOH, O_{2} (Green) 2 B + 2 H_{2} O

3 B 4 HCl (Purple) 2 C + MnO_{2} + 2 H_{2} O

2 \underline{C} H_{2} O, KI 2 \underline{A} + 2 KOH + \underline{D}

In the above sequence of reactions,

\underline{A}

and

\underline{D}

, respectively are:

Answer: (B)

Following the sequence:

MnO X_{2}

(A) heated with KOH and

O X_{2}

gives the green manganate

K X_{2} MnO X_{4}

(B); in acid this disproportionates to the purple permanganate

KMnO X_{4}

MnO X_{2}

; and

KMnO X_{4}

oxidising KI in water is itself reduced back to

MnO X_{2}

(A) while iodide is oxidised to iodate

KIO X_{3}

(D). So A is

MnO X_{2}

and D is

KIO X_{3}

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option