`KCI + "conc" H_(2)SO_(4) + K_(2)Cr_(2)O_(7) overset(Delta) to (X) overset(NaOH) to (Y),(X) ` is reddish brown coloured gas soluble in `NaOH` forming `(Y),(X) and (Y) `are

Options

A. `Cr_(2)OCI_(2)Na_(2)CrO_(3)`
B. `Cr_(2)O_(2)CI_(2)Na_(2)CrO_(3)`
C. `CrO_(2)CI_(2)Na_(2)CrO_(6)`
D. `CrO_(2)CI_(2)Na_(2)CrO_(4)`

Answer

To solve the problem step by step, we will analyze the reaction and identify the compounds involved.

### Step 1: Identify the Reactants
The reactants given in the question are:
- Potassium Chloride (KCl)
- Concentrated Sulfuric Acid (H₂SO₄)
- Potassium Dichromate (K₂Cr₂O₇)

### Step 2: Understand the Reaction
When KCl, concentrated H₂SO₄, and K₂Cr₂O₇ are heated together, they undergo a reaction that produces a reddish-brown gas. This gas is known as chromyl chloride (CrO₂Cl₂).

### Step 3: Identify Product X
The reddish-brown gas produced from the reaction is:
- **X = Chromyl Chloride (CrO₂Cl₂)**

### Step 4: Reaction of X with NaOH
Next, we need to determine what happens when the chromyl chloride (X) reacts with sodium hydroxide (NaOH). The reaction of chromyl chloride with NaOH produces sodium chromate (Na₂CrO₄) and sodium chloride (NaCl), along with water (H₂O).

### Step 5: Identify Product Y
From the reaction of X with NaOH, we find that:
- **Y = Sodium Chromate (Na₂CrO₄)**

### Step 6: Final Answer
Thus, the final answer is:
- **X = CrO₂Cl₂ (Chromyl Chloride)**
- **Y = Na₂CrO₄ (Sodium Chromate)**

### Summary
The products of the reaction are:
- **X = CrO₂Cl₂**
- **Y = Na₂CrO₄**