Potassium dichromate acts as a strong oxidising agent in the acidic medium. When heated with a metal chloride and conc. sulphuric acid, it gives orange red fumes. On passing these fumes through sodium hydroxide, it gives a yellow solution (X) and this acidified solution gives yellow precipitate (Y) with lead acetate. On heating potassium dichromate with KOH, it forms a yellow solution (Z) which on acidification changes to orange colour. Acidified solution of dichromate ion gives deep blue colour with `H_2O_2` due to the formation of (P)
The orange red fumes obtained on heating potassium dichromate with a metal chloride and conc. `H_2SO_4` is due to the formation of

Options

A. `CrO_3`
B. `CrO_5`
C. `CrO_2Cl_2`
D. `CrOCl_2`

Explanation

To solve the question step by step, we will analyze the reactions involving potassium dichromate and its interactions with other chemicals as described in the problem.

### Step 1: Understanding the Reaction
Potassium dichromate (K2Cr2O7) acts as a strong oxidizing agent in acidic medium. When it is heated with a metal chloride (like NaCl) and concentrated sulfuric acid (H2SO4), it produces orange-red fumes.

### Step 2: Identifying the Product of the Reaction
The orange-red fumes produced in this reaction are due to the formation of chromyl chloride (CrO2Cl2). The balanced equation for the reaction can be represented as:
\[ K_2Cr_2O_7 + 6 H_2SO_4 + 4 NaCl \rightarrow 2 CrO_2Cl_2 + 2 KHSO_4 + 4 NaHSO_4 + 3 H_2O \]

### Step 3: Passing Fumes Through Sodium Hydroxide
When these orange-red fumes (CrO2Cl2) are passed through sodium hydroxide (NaOH), a yellow solution (X) is formed. The reaction can be represented as:
\[ CrO_2Cl_2 + 4 NaOH \rightarrow Na_2CrO_4 + 2 NaCl + 2 H_2O \]
Here, sodium chromate (Na2CrO4) is the yellow solution (X).

### Step 4: Formation of Yellow Precipitate
When the yellow solution (X) is acidified, it reacts with lead acetate (Pb(CH3COO)2) to form a yellow precipitate (Y), which is lead chromate (PbCrO4):
\[ Na_2CrO_4 + Pb(CH_3COO)_2 \rightarrow PbCrO_4 \downarrow + 2 NaCH_3COO \]

### Step 5: Heating Potassium Dichromate with KOH
When potassium dichromate is heated with potassium hydroxide (KOH), it forms a yellow solution (Z), which is also sodium chromate (K2CrO4):
\[ K_2Cr_2O_7 + 2 KOH \rightarrow 2 K_2CrO_4 + H_2O \]
This yellow solution (Z) changes to orange color upon acidification, reverting back to potassium dichromate (K2Cr2O7).

### Step 6: Reaction with Hydrogen Peroxide
The acidified solution of dichromate ions (Cr2O7^2-) gives a deep blue color with hydrogen peroxide (H2O2) due to the formation of chromium (III) ions (Cr^3+):
\[ Cr_2O_7^{2-} + H_2O_2 + 14 H^+ \rightarrow 2 Cr^{3+} + 7 H_2O \]
The deep blue color is due to the formation of the complex ion with chromium.

### Conclusion
The orange-red fumes obtained on heating potassium dichromate with a metal chloride and concentrated sulfuric acid are due to the formation of **chromyl chloride (CrO2Cl2)**.

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