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An aqueous solution of $X$ on addition of hydrogen peroxide in ice cold conditions gives blue colour to the ethereal layer. Then, $X$ can be
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The correct answer is:
acidified potassium chromate
Both chromate and dichromate radicals on reaction with acidified $\mathrm{H}_{2} \mathrm{O}_{2}$ give blue solution which turns green on standing. In the presence of organic solvents such as diethyl ether, amyl alcohol or pyridine, permanent blue colour obtained.
$$
\begin{aligned}
&\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}_{2} \stackrel{\text { Ether }}{\longrightarrow} \mathrm{CrO}_{5}+\mathrm{K}_{2} \mathrm{SO}_{4} \\
&+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+4 \mathrm{H}_{2} \mathrm{SO}_{4}+4 \mathrm{H}_{2} \mathrm{O}_{2} \stackrel{\text { Ether }}{\longrightarrow} 2 \mathrm{CrO}_{5}+\mathrm{K}_{2} \mathrm{SO}_{4} \\
&\text { (Blue } \\
&\text { colour) } \\
&+5 \mathrm{H}_{2} \mathrm{O}
\end{aligned}
$$
$$
\begin{aligned}
&\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}_{2} \stackrel{\text { Ether }}{\longrightarrow} \mathrm{CrO}_{5}+\mathrm{K}_{2} \mathrm{SO}_{4} \\
&+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+4 \mathrm{H}_{2} \mathrm{SO}_{4}+4 \mathrm{H}_{2} \mathrm{O}_{2} \stackrel{\text { Ether }}{\longrightarrow} 2 \mathrm{CrO}_{5}+\mathrm{K}_{2} \mathrm{SO}_{4} \\
&\text { (Blue } \\
&\text { colour) } \\
&+5 \mathrm{H}_{2} \mathrm{O}
\end{aligned}
$$
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