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 blue colour of compound P fades due to the formation of

Options

A. `Cr^(3+)` and `O_2`
B. `CrO_4^(2-)` and `Cr^(3+)`
C. `Cr_2O_7^(2-)` and `Cr^(3+)`
D. `CrO_3`

Explanation

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

### Step 1: Understanding the Reaction of Potassium Dichromate
Potassium dichromate (K₂Cr₂O₇) acts as a strong oxidizing agent in acidic medium. When it is heated with a metal chloride and concentrated sulfuric acid, it produces orange-red fumes. This is due to the formation of chromium(VI) oxide (CrO₃) or similar chromium species.

**Hint:** Remember that potassium dichromate is a source of Cr(VI), which is responsible for the orange-red color in the fumes.

### Step 2: Reaction with Sodium Hydroxide
When the orange-red fumes are passed through sodium hydroxide (NaOH), a yellow solution (X) is formed. This yellow solution is likely potassium chromate (K₂CrO₄), which is yellow in color.

**Hint:** The color change from orange-red to yellow indicates a change in the oxidation state of chromium from +6 to +6 but in a different chemical form (chromate).

### Step 3: Precipitation with Lead Acetate
When the yellow solution (X) is acidified with lead acetate (Pb(C₂H₃O₂)₂), it forms a yellow precipitate (Y). This precipitate is lead chromate (PbCrO₄), which is a well-known yellow precipitate.

**Hint:** Lead chromate is formed from the reaction of chromate ions with lead(II) ions in an acidic medium.

### Step 4: Heating with Potassium Hydroxide
Next, when potassium dichromate is heated with potassium hydroxide (KOH), it forms another yellow solution (Z), which is also potassium chromate (K₂CrO₄). Upon acidification, this solution changes color to orange, indicating the formation of potassium dichromate again.

**Hint:** The heating with KOH leads to the formation of chromate ions, which are yellow, and acidification converts them back to dichromate ions, which are orange.

### Step 5: Reaction with Hydrogen Peroxide
The acidified solution of dichromate ions reacts with hydrogen peroxide (H₂O₂) to produce a deep blue color due to the formation of a compound (P), which is likely chromium(VI) peroxide (CrO₅).

**Hint:** The deep blue color indicates the formation of a complex or a specific oxidation state of chromium that is not stable.

### Step 6: Decomposition of Compound P
The blue color of compound P fades due to its decomposition into chromium(III) ions (Cr³⁺) and oxygen (O₂). The decomposition reaction can be summarized as:
\[ \text{CrO}_5 \rightarrow \text{Cr}^{3+} + \text{O}_2 \]

**Hint:** The fading of color indicates that the compound is unstable and breaks down into simpler species.

### Final Answer
The final answer to the question is that the blue color of compound P fades due to the formation of chromium(III) ions and oxygen gas.

**Summary of the Steps:**
1. Potassium dichromate produces orange-red fumes in acidic medium.
2. Passing fumes through NaOH gives yellow solution (K₂CrO₄).
3. Acidifying with lead acetate gives yellow precipitate (PbCrO₄).
4. Heating with KOH forms yellow solution (K₂CrO₄), which turns orange upon acidification.
5. Acidified dichromate reacts with H₂O₂ to form deep blue compound (CrO₅).
6. Blue color fades due to decomposition into Cr³⁺ and O₂.