Consider the following reactions $$\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \xrightarrow[-\mathrm{H}_{2} \mathrm{O}]{\mathrm{KOH}}[\mathrm{~A}] \xrightarrow[-\mathrm{H}_{2} \mathrm{O}]{\mathrm{H}_{2} \mathrm{SO}_{4}}[\mathrm{~B}]+\mathrm{K}_{2} \mathrm{SO}_{4}$$ The products [A] and [B], respectively are :

(C) $\mathrm{K}_{2} \mathrm{CrO}_{4}$ and $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$

(A) $\mathrm{K}_{2} \mathrm{Cr}(\mathrm{OH})_{6}$ and $\mathrm{Cr}_{2} \mathrm{O}_{3}$

(B) $\mathrm{K}_{2} \mathrm{CrO}_{4}$ and $\mathrm{Cr}_{2} \mathrm{O}_{3}$

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

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · d and f Block Elements

Consider the following reactions

K_{2} Cr_{2} O_{7} KOH - H_{2} O [A] H_{2} SO_{4} - H_{2} O [B] + K_{2} SO_{4}

The products [A] and [B], respectively are :

Answer: (C)

Treating orange potassium dichromate with KOH converts dichromate to the yellow chromate ion, since the two interconvert with pH:

Cr X_{2} O X_{7} X^{2 -} + 2 OH X^{-} 2 CrO X_{4} X^{2 -} + H X_{2} O

, giving [A] =

K X_{2} CrO X_{4}

. Acidifying the chromate with

H X_{2} SO X_{4}

reverses this equilibrium, converting the yellow chromate back to orange dichromate:

2 CrO X_{4} X^{2 -} + 2 H X^{+} Cr X_{2} O X_{7} X^{2 -} + H X_{2} O

, giving [B] =

K X_{2} Cr X_{2} O X_{7}

, with

K X_{2} SO X_{4}

as the accompanying salt. The two products are

K X_{2} CrO X_{4}

and

K X_{2} Cr X_{2} O X_{7}

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option