JEE MAIN - Chemistry (2024 - 30th January Evening Shift - No. 7)
A and B formed in the following reactions are:
$$\begin{aligned} & \mathrm{CrO}_2 \mathrm{Cl}_2+4 \mathrm{NaOH} \rightarrow \mathrm{A}+2 \mathrm{NaCl}+2 \mathrm{H}_2 \mathrm{O}, \\ & \mathrm{A}+2 \mathrm{HCl}+2 \mathrm{H}_2 \mathrm{O}_2 \rightarrow \mathrm{B}+3 \mathrm{H}_2 \mathrm{O} \end{aligned}$$
$$\mathrm{A}=\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7, \mathrm{~B}=\mathrm{CrO}_5$$
$$\mathrm{A}=\mathrm{Na}_2 \mathrm{CrO}_4, \mathrm{~B}=\mathrm{CrO}_5$$
$$\mathrm{A}=\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_4, \mathrm{~B}=\mathrm{CrO}_4$$
$$\mathrm{A}=\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7, \mathrm{~B}=\mathrm{CrO}_3$$
Explanation
$$\mathrm{Cr{O_2} + 4NaOH \to \mathop {N{a_2}Cr{O_4}}\limits_{(A)} + 2NaCl + 2{H_2}O}$$
$$\mathrm{N{a_2}Cr{O_4} + 2{H_2}{O_2} + 2HCl \to \mathop {Cr{O_5}}\limits_{(B)} + \mathop {2NaCl}\limits_{(Mis\sin g\,from\,balanced\,equation)} + 3{H_2}O}$$
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