An aqueous solution `FeSO_(4).Al_(2)(SO_(4))_(3)` and chrome alum is heated with excess of `Na_(2)O_(2)` and fltered. 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

Answer

To solve the problem, we need to analyze the reaction of the given compounds when heated with sodium peroxide (Na₂O₂) and determine the resulting filtrate and residue.

### Step-by-Step Solution:

1. **Identify the Compounds:**
- The compounds in the solution are:
- Iron(II) sulfate (FeSO₄)
- Aluminum sulfate (Al₂(SO₄)₃)
- Chrome alum (KCr(SO₄)₂·12H₂O)

2. **Understand the Role of Sodium Peroxide:**
- Sodium peroxide (Na₂O₂) acts as an oxidizing agent. It contains peroxide ions (O₂²⁻) which can oxidize certain metal ions.

3. **Oxidation of Iron(II) Ions:**
- In the solution, iron is present in the +2 oxidation state (Fe²⁺).
- When Fe²⁺ is treated with Na₂O₂, it gets oxidized to Fe³⁺ (ferric ion).
- The oxidation of Fe²⁺ to Fe³⁺ results in the formation of brown-colored ferric compounds, which are typically insoluble.

4. **Oxidation of Chromium(III) Ions:**
- Chrome alum contains chromium in the +3 oxidation state (Cr³⁺).
- The chromium ions (Cr³⁺) will also be oxidized by the peroxide ions to form chromate ions (CrO₄²⁻), where chromium is in the +6 oxidation state.
- The chromate ions are typically yellow in color and are soluble in water.

5. **Filtration Process:**
- After heating the mixture with excess Na₂O₂, the insoluble brown ferric compounds will remain as residue when the mixture is filtered.
- The soluble chromate ions will pass through the filter paper, resulting in a yellow filtrate.

6. **Conclusion:**
- The materials obtained after filtration are:
- **Filtrate:** Yellow (due to chromate ions, CrO₄²⁻)
- **Residue:** Brown (due to ferric ions, Fe³⁺)

Based on the analysis, the correct answer is **Option C: a yellow filtrate and a brown residue.**