Alkanline earth metal nitrate `(A)` on heating decompose, leaving a solid residue `(B)` which goes into solution with dilute `HCI`. The solution of `(B)` gives a white precipitate `(C )` is dissolved in dilute `HCl` and the solution is treated with potassium chromae to get yellow precipitate `(D)`. The solution `(B)` with dilute `H_(2)SO_(4)` also gives a white precipitate `(E)` insoluble in diute `HCl` and nitric acid. the precipitate `(E)` is a part of a white pigment lithopone.
The nitrate `(A)` can be confirmed by flame test. The colour imparted by the salt to the Bunsen flame is

Options

A. Yellow
B. Green
C. Blue
D. Red

Explanation

To solve the problem, we will break down the steps involved in the decomposition of alkaline earth metal nitrate `(A)` and the subsequent reactions that lead to the formation of various compounds `(B)`, `(C)`, `(D)`, and `(E)`.

### Step-by-Step Solution:

1. **Identify the Alkaline Earth Metal Nitrate `(A)`**:
- The alkaline earth metal nitrate that fits the description is **Barium Nitrate (Ba(NO3)2)**. This is because it decomposes upon heating to form a solid residue and gases.

2. **Decomposition of `(A)`**:
- When Barium Nitrate is heated, it decomposes according to the reaction:
\[ 2 \text{Ba(NO}_3\text{)}_2 \rightarrow 2 \text{BaO} + 4 \text{NO}_2 + \text{O}_2 \]
- The solid residue `(B)` left after heating is **Barium Oxide (BaO)**.

3. **Dissolving `(B)` in Dilute HCl**:
- Barium Oxide reacts with dilute hydrochloric acid (HCl):
\[ \text{BaO} + 2 \text{HCl} \rightarrow \text{BaCl}_2 + \text{H}_2\text{O} \]
- The solution now contains **Barium Chloride (BaCl2)**.

4. **Formation of White Precipitate `(C)`**:
- When Barium Chloride is treated with a solution that contains sulfate ions (like from sulfuric acid), it forms a white precipitate:
\[ \text{BaCl}_2 + \text{Na}_2\text{SO}_4 \rightarrow \text{BaSO}_4 \downarrow + 2 \text{NaCl} \]
- The white precipitate `(C)` is **Barium Sulfate (BaSO4)**.

5. **Treatment with Potassium Chromate**:
- When the solution of Barium Chloride is treated with potassium chromate (K2CrO4), it forms a yellow precipitate:
\[ \text{BaCl}_2 + \text{K}_2\text{CrO}_4 \rightarrow \text{BaCrO}_4 \downarrow + 2 \text{KCl} \]
- The yellow precipitate `(D)` is **Barium Chromate (BaCrO4)**.

6. **Formation of White Precipitate `(E)` with Dilute H2SO4**:
- When Barium Chloride is treated with dilute sulfuric acid (H2SO4), it forms another white precipitate:
\[ \text{BaCl}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4 \downarrow + 2 \text{HCl} \]
- The precipitate `(E)` is again **Barium Sulfate (BaSO4)**, which is insoluble in dilute HCl and nitric acid.

7. **Flame Test for Confirmation of `(A)`**:
- The color imparted to the Bunsen flame by Barium Nitrate is **apple green**.

### Final Answer:
The color imparted by the salt (Barium Nitrate) to the Bunsen flame is **apple green**.