JEE Chemistry 2026 Question with Solution

Given: A mixture of common salt, $K_2Cr_2O_7$ and concentrated $H_2SO_4$ is heated.

Find: The formula of the gas evolved and the oxidation state of the central metal atom in that gas.

This is the Chromyl Chloride Test, used to detect the presence of chloride ions.

The reaction is:

$$K_2Cr_2O_7 + 4NaCl + 6H_2SO_4 \xrightarrow{\Delta} 2KHSO_4 + 4NaHSO_4 + 2CrO_2Cl_2 \uparrow + 3H_2O$$

So, the gas evolved is $CrO_2Cl_2$.

Now let the oxidation state of chromium in $CrO_2Cl_2$ be $x$.

$$x + 2(-2) + 2(-1) = 0$$ $$x - 4 - 2 = 0$$ $$x = +6$$

Therefore, the gas is $CrO_2Cl_2$ and the oxidation state of chromium is $+6$. The correct option is A.

Given: The evolved gas is identified through the chromyl chloride test.

Find: Oxidation state of the central chromium atom.

In $CrO_2Cl_2$:

• Oxygen has oxidation state $-2$ each.
• Chlorine has oxidation state $-1$ each.

Using overall neutrality,

$$x + 2(-2) + 2(-1) = 0$$ $$x - 6 = 0$$ $$x = +6$$

Hence chromium is in the $+6$ oxidation state.