When a mixture of \(K_2Cr_2O_7\) and \(NaCl\) is warmed concentrated \(H_2SO_4\) red vapours of \((A)\) are evolved, which turn \(NaOH\) yellowish due to formation of compound \((B)\). \(A\) and \(B\) respectively are :

\(A: CrO_2Cl_2 \text{ and } B: Na_2CrO_4\)

The correct answer is (3)\(A: CrO_2Cl_2\) and \(B: Na_2CrO_4\).
The reaction involves the mixture of \(K_2Cr_2O_7\) (potassium dichromate) and \(NaCl\) (sodium chloride) being warmed with concentrated \(H_2SO_4\) (sulfuric acid). The chemical equation is as follows:
\[ 2K_2Cr_2O_7 + 8NaCl + 6H_2SO_4 \rightarrow 2CrO_2Cl_2 + 2K_2SO_4 + 8NaHSO_4 + 3H_2O \]
Here's a step-by-step explanation:
1. Formation of Chromyl Chloride (\(CrO_2Cl_2\)):
\(K_2Cr_2O_7\) is a powerful oxidizing agent. When it reacts with \(H_2SO_4\), it gets reduced. The chromium in the \(K_2Cr_2O_7\) changes its oxidation state from +6 to +4, forming \(CrO_2Cl_2\), which is a red-orange volatile compound known as chromyl chloride.

2. Reaction Overview:
\(2K_2Cr_2O_7 + 8NaCl + 6H_2SO_4 \rightarrow 2CrO_2Cl_2 + 2K_2SO_4 + 8NaHSO_4 + 3H_2O\)

3. Chromyl Chloride (\(CrO_2Cl_2\)) in the Presence of Moist \(NaOH\):
Chromyl chloride reacts with moist \(NaOH\) to form yellow sodium chromate (\(Na_2CrO_4\)):
\[ CrO_2Cl_2 + 2NaOH \rightarrow Na_2CrO_4 + 2HCl \]
The yellow color of sodium chromate imparts a yellowish tint to the solution.

In summary, the red vapors of \(CrO_2Cl_2\) are evolved during the reaction with \(K_2Cr_2O_7\) and \(NaCl\) in the presence of concentrated \(H_2SO_4\). When these vapors react with moist \(NaOH\), they form yellowish sodium chromate (\(Na_2CrO_4\)). Therefore, \(A\) is \(CrO_2Cl_2\) (chromyl chloride), and \(B\) is \(Na_2CrO_4\) (sodium chromate).

So, the correct answer is: 3. \(A: CrO_2Cl_2\) and \(B: Na_2CrO_4\)