An aqueous solution of `FeSO_4.Al_2(SO_4)_3` and chrome alum is heated with excess of `Na_2O_2` and filtered. The materials obtained are

Options

A. A colourless filtrate and a green residue
B. A yellow filtrate and a green residue
C. A yellow filtrate and a brown residue
D. A green filtrate and a brown residue

Explanation

To solve the problem, we need to analyze the reaction of the given compounds when heated with excess sodium peroxide (Na₂O₂) and then determine the materials obtained after filtration.

### Step-by-Step Solution:

1. **Identify the Compounds**:
The compounds involved are `FeSO₄` (ferrous sulfate), `Al₂(SO₄)₃` (aluminum sulfate), and chrome alum, which can be represented as `KCr(SO₄)₂·12H₂O`.

2. **Heating with Sodium Peroxide**:
When the aqueous solution containing these compounds is heated with excess sodium peroxide (Na₂O₂), it acts as an oxidizing agent.

3. **Formation of Hydroxides**:
- The `Fe²⁺` ions from `FeSO₄` will react with hydroxide ions (OH⁻) formed during the reaction to precipitate as `Fe(OH)₂` (ferrous hydroxide).
- The `Al³⁺` ions from `Al₂(SO₄)₃` will form `Al(OH)₄⁻` (aluminate ions) in the presence of excess hydroxide ions.
- The `Cr³⁺` ions from chrome alum will form `CrO₄²⁻` (chromate ions) in an alkaline medium.

4. **Filtration**:
After the reaction, the mixture is filtered. The soluble species will pass through the filter paper, while the precipitates will remain on the filter.

5. **Color of the Filtrate**:
The filtrate will contain the yellow chromate ions (`CrO₄²⁻`), resulting in a yellow solution.

6. **Residue on the Filter**:
The residue will primarily consist of `Fe(OH)₃` (ferric hydroxide) which is a brown precipitate formed from the oxidation of `Fe(OH)₂` in the presence of air.

### Conclusion:
After filtration, the materials obtained are:
- **Yellow filtrate** (due to chromate ions)
- **Brown residue** (due to ferric hydroxide)

### Final Answer:
The correct answer is: **A yellow filtrate and a brown residue.**