Consider the following reactions $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 \xrightarrow[-\mathrm{H}_2…
$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 \xrightarrow[-\mathrm{H}_2 \mathrm{O}]{\mathrm{KOH}}[\mathrm{A}] \xrightarrow[-\mathrm{H}_2 \mathrm{O}]{\mathrm{H}_2 \mathrm{SO}_4}[\mathrm{~B}]+\mathrm{K}_2 \mathrm{SO}_4$
The products $[A]$ and $[B]$, respectively are :
- $\mathrm{K}_2 \mathrm{CrO}_4$ and CrO
- $\mathrm{K}_2 \mathrm{CrO}_4$ and $\mathrm{Cr}_2 \mathrm{O}_3$
- $\mathrm{K}_2 \mathrm{CrO}_4$ and $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$
- $\mathrm{K}_2 \mathrm{Cr}(\mathrm{OH})_6$ and $\mathrm{Cr}_2 \mathrm{O}_3$
Solution
[A] $\mathrm{K}_2 \mathrm{CrO}_4$
[B] $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$
Asked in: JEE Main 2025 (23 Jan Shift 2)