A hydrated mettalic salt(A). Light green in colour , on careful heating gives a white anhydrates residue (B),(B) is soluble in water and its aqueous gives a dark blue precipitate ( C ) with potassium hexacyanidoferrate (III) . (B ) on strong heating gives a brown residue(D) and a mixture of two gases(E ) and (F ). The gaseous mixture when passed through acidified potassium dichromate , produce green colour solution and when passed through lead acetate solution gave a white precipitate. Out of two gase( E) can act as both reducing as well as oxidising agent.
Identify (A) ,(B) ,(C ),(D) , and (F) and gives the reaction involved.
Explain why salt(A) become white on heating .

Answer

To solve the problem step by step, let's analyze the information provided and identify the compounds and reactions involved.

### Step 1: Identify the Hydrated Metallic Salt (A)
The hydrated metallic salt (A) is described as light green in color. The only common hydrated ferrous salt that fits this description is **Ferrous Sulfate Heptahydrate (FeSO4·7H2O)**.

### Step 2: Heating of (A) to form (B)
When Ferrous Sulfate Heptahydrate is carefully heated, it loses its water of hydration and forms an anhydrous residue (B), which is **Anhydrous Ferrous Sulfate (FeSO4)**. This compound is white in color.

### Step 3: Solubility of (B) and Formation of (C)
Anhydrous Ferrous Sulfate (B) is soluble in water. When dissolved, it dissociates into Fe²⁺ ions and sulfate ions. The presence of Fe²⁺ ions in the solution reacts with potassium hexacyanidoferrate(III) (K3[Fe(CN)6]) to form a dark blue precipitate (C), known as **Turnbull's Blue (K4[Fe(CN)6]·Fe)**.

### Step 4: Strong Heating of (B) to form (D) and Gases (E) and (F)
On strong heating, Anhydrous Ferrous Sulfate (B) decomposes to form a brown residue (D), which is **Ferric Oxide (Fe2O3)**, along with a mixture of two gases (E) and (F). The gases produced are **Sulfur Dioxide (SO2)** and **Sulfur Trioxide (SO3)**.

### Step 5: Gaseous Mixture Reaction with Acidified Potassium Dichromate
The gaseous mixture (E and F) when passed through acidified potassium dichromate (K2Cr2O7) produces a green color solution. This indicates that **SO2** is oxidized to **SO4²⁻** ions, which is a characteristic reaction of sulfur dioxide with a strong oxidizing agent.

### Step 6: Gaseous Mixture Reaction with Lead Acetate
When the gaseous mixture is passed through lead acetate solution, it gives a white precipitate. This white precipitate is **Lead Sulfate (PbSO4)**, which forms when SO2 or SO3 reacts with lead acetate.

### Step 7: Identify Gases (E) and (F)
From the reactions, we identify:
- (E) = **Sulfur Dioxide (SO2)** (which can act as both a reducing and oxidizing agent)
- (F) = **Sulfur Trioxide (SO3)**

### Summary of Identifications:
- (A) = **Ferrous Sulfate Heptahydrate (FeSO4·7H2O)**
- (B) = **Anhydrous Ferrous Sulfate (FeSO4)**
- (C) = **Turnbull's Blue (K4[Fe(CN)6]·Fe)**
- (D) = **Ferric Oxide (Fe2O3)**
- (E) = **Sulfur Dioxide (SO2)**
- (F) = **Sulfur Trioxide (SO3)**

### Reactions Involved:
1. **Heating Reaction:**
\[ \text{FeSO}_4 \cdot 7\text{H}_2\text{O} \xrightarrow{\text{heat}} \text{FeSO}_4 + 7\text{H}_2\text{O} \]

2. **Formation of Precipitate (C):**
\[ \text{Fe}^{2+} + \text{K}_3[\text{Fe(CN)}_6] \rightarrow \text{Turnbull's Blue (C)} \]

3. **Decomposition Reaction:**
\[ 2\text{FeSO}_4 \xrightarrow{\text{heat}} \text{Fe}_2\text{O}_3 + \text{SO}_2 + \text{SO}_3 \]

4. **Reaction with Acidified Potassium Dichromate:**
\[ \text{SO}_2 + \text{K}_2\text{Cr}_2\text{O}_7 \rightarrow \text{Cr}^{3+} + \text{SO}_4^{2-} \quad (\text{green solution}) \]

5. **Reaction with Lead Acetate:**
\[ \text{SO}_2 + \text{Pb(OAc)}_2 \rightarrow \text{PbSO}_4 \downarrow \quad (\text{white precipitate}) \]

### Explanation of Color Change of Salt (A) on Heating
The salt (A), Ferrous Sulfate Heptahydrate, is light green due to the presence of Fe²⁺ ions. Upon heating, it loses water and becomes anhydrous, which changes its structure and oxidation state, resulting in a white color. The white color of the anhydrous salt is due to the absence of water molecules that contribute to the green color of the hydrated form.