When a mixture of \(NaCl\) and \(K_2Cr_2O_7\) is gently warmed with conc.\(H_2SO_4\)?

(A) Deep red vapours are evolved

(B) The vapours when passed through \(NaOH\) solution give yellow solution of \(Na_2CrO_4\)

(C) Chlorine gas is evolved

(D) Chromyl chloride is formed

Choose the correct answer from the options given below:

A, B and D only

The correct answer is option 1. A, B and D only.

When a mixture of NaCl and K₂Cr₂O₇ is gently warmed with concentrated H₂SO₄, the following reactions occur:

1. Oxidation of Chloride (Cl⁻) to Chlorine (Cl₂):

H₂SO₄ acts as a strong oxidizing agent and oxidizes Cl⁻ ions from NaCl to Cl₂ gas. This explains option (C), chlorine gas is evolved.

2. Formation of Chromyl Chloride (CrO₂Cl₂):

K₂Cr₂O₇ (potassium dichromate) acts as an oxidizing agent and reacts with NaCl and H₂SO₄ to form CrO₂Cl₂ (chromyl chloride). This explains option (D), chromyl chloride is formed.

3. Chromyl Chloride reacts with water to form chromate (CrO₄²⁻):

CrO₂Cl₂ reacts with water to form H₂CrO₄ (chromic acid) and HCl. H₂CrO₄ dissociates in water to form CrO₄²⁻ ions.

4. Reaction with NaOH:

When the vapors containing CrO₂Cl₂ and HCl are passed through an NaOH solution, CrO₂Cl₂ reacts with NaOH to form Na₂CrO₄ (sodium chromate), giving a yellow solution. This explains option (B), the vapors when passed through NaOH solution give a yellow solution of Na₂CrO₄.

Therefore, options A, B, and D are all correct, while option C alone is not accurate as the chlorine gas formed does not contribute to the yellow solution observed with NaOH.

Remember, strong oxidizing agents like concentrated H₂SO₄ can readily oxidize chloride ions to chlorine gas. Additionally, CrO₂Cl₂ is a characteristic reddish-brown fuming compound formed in this reaction.