JEE Chemistry 2024 Question with Solution

Given: NaCl reacts with concentrated $H_2SO_4$ and $K_2Cr_2O_7$ to form reddish fumes $(B)$, and $(B)$ reacts with NaOH to give yellow solution $(C)$.

Find: The identities of $(B)$ and $(C)$ and hence the correct option.

NaCl with concentrated $H_2SO_4$ and $K_2Cr_2O_7$ gives chromyl chloride, which is reddish fuming in nature. Therefore, $(B) = CrO_2Cl_2$.

On reaction with NaOH, chromyl chloride gives sodium chromate, which is yellow:

$$CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 + 2NaCl + 2H_2O$$

Thus, $(C) = Na_2CrO_4$.

Therefore, $(B)$ and $(C)$ respectively are $CrO_2Cl_2$ and $Na_2CrO_4$. The correct option is A.

Given: The reddish fumes formed from NaCl, concentrated $H_2SO_4$ and $K_2Cr_2O_7$ must be identified first.

Find: The products $(B)$ and $(C)$.

The reddish vapours produced in the chromyl chloride test are $CrO_2Cl_2$.

The subsequent reaction with NaOH forms yellow sodium chromate:

$$CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 + 2NaCl + 2H_2O$$

So the yellow solution is $Na_2CrO_4$.

Hence the pair is $CrO_2Cl_2, Na_2CrO_4$.