JEE MAIN - Chemistry (2025 - 23rd January Evening Shift - No. 19)
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 :
$\mathrm{K}_2 \mathrm{Cr}(\mathrm{OH})_6$ and $\mathrm{Cr}_2 \mathrm{O}_3$
$\mathrm{K}_2 \mathrm{CrO}_4$ and $\mathrm{Cr}_2 \mathrm{O}_3$
$\mathrm{K}_2 \mathrm{CrO}_4$ and $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$
$\mathrm{K}_2 \mathrm{CrO}_4$ and $\mathrm{CrO}$
Explanation
$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 \xrightarrow[-\mathrm{H}_2 \mathrm{O}]{\mathrm{KOH}} \underset{\text { [A] }}{\mathrm{K}_2 \mathrm{CrO}_4} \xrightarrow[-\mathrm{H}_2 \mathrm{O}]{\mathrm{H}_2 \mathrm{~S}_4} \underset{\text { [B] }}{\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7}+\mathrm{K}_2 \mathrm{SO}_4$
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