JEE MAIN - Chemistry (2025 - 3rd April Morning Shift - No. 22)
Consider the following reactions
$$ \begin{aligned} & \mathrm{A}+\underset{\substack{ \text { Little } \\ \text { amount }}}{\mathrm{NaCl}}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CrO}_2 \mathrm{Cl}_2+\text { Side Products } \\ & \mathrm{CrO}_2 \mathrm{Cl}_{2 \text { (Vapour) }}+\mathrm{NaOH} \rightarrow \mathrm{~B}+\mathrm{NaCl}+\mathrm{H}_2 \mathrm{O} \\ & \mathrm{~B}+\mathrm{H}^{+} \rightarrow \mathrm{C}+\mathrm{H}_2 \mathrm{O} \end{aligned} $$
The number of terminal ' $O$ ' present in the compound ' C ' is__________
Explanation
$$\begin{aligned} & \mathrm{Cr}_2 \mathrm{O}_7^{2-}+\mathrm{NaCl}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CrO}_2 \mathrm{Cl}_2 \\ & \mathrm{CrO}_2 \mathrm{Cl}_2 \text { (Vapour) }+\mathrm{NaOH} \rightarrow \\ & \mathrm{Na}_2 \mathrm{CrO}_4+\mathrm{NaCl}+\mathrm{H}_2 \mathrm{O} \\ & \mathrm{Na}_2 \mathrm{CrO}_4+\mathrm{H}^{\oplus} \rightarrow \mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{H}_2 \mathrm{O} \\ & (\mathrm{C}) \\ & \mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7 \rightarrow 2 \mathrm{Na}^{\oplus}+\mathrm{Cr}_2 \mathrm{O}_7^{2-} \end{aligned}$$
_3rd_April_Morning_Shift_en_22_1.png)
No of terminal "O" = 6
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