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 compound `(E)` is

Options

A. `BaSO_(4)`
B. `MgSO_(4)`
C. `CaSO_(4)`
D. `Na_(2)SO_(4)`

Explanation

To solve the question step by step, we will analyze the decomposition of alkaline earth metal nitrate and the subsequent reactions leading to the identification of compound E.

### Step 1: Identify the Alkaline Earth Metal Nitrate (A)
The question states that an alkaline earth metal nitrate (A) decomposes upon heating. The common alkaline earth metal nitrates are barium nitrate, calcium nitrate, and magnesium nitrate. For this problem, we will consider **barium nitrate (Ba(NO₃)₂)** as it is known to decompose to form barium oxide.

### Step 2: Decomposition of Barium Nitrate
When barium nitrate is heated, it decomposes as follows:
\[ \text{2 Ba(NO}_3\text{)}_2 \rightarrow \text{2 BaO} + \text{4 NO}_2 + \text{O}_2 \]
This gives us a solid residue (B), which is **barium oxide (BaO)**.

### Step 3: Solubility of Barium Oxide in Dilute HCl
The solid residue (B), which is barium oxide (BaO), is soluble in dilute hydrochloric acid (HCl):
\[ \text{BaO} + 2 \text{HCl} \rightarrow \text{BaCl}_2 + \text{H}_2\text{O} \]
This results in a solution of barium chloride (BaCl₂).

### Step 4: Formation of White Precipitate (C)
The solution of barium chloride (BaCl₂) can react with ammonium carbonate ((NH₄)₂CO₃) to form a white precipitate (C):
\[ \text{BaCl}_2 + (NH_4)_2\text{CO}_3 \rightarrow \text{BaCO}_3 \downarrow + 2 \text{NH}_4\text{Cl} \]
Here, the white precipitate (C) is **barium carbonate (BaCO₃)**.

### Step 5: Solubility of Barium Carbonate in Dilute HCl
The white precipitate (C), barium carbonate (BaCO₃), is soluble in dilute HCl:
\[ \text{BaCO}_3 + 2 \text{HCl} \rightarrow \text{BaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \]
This gives us a solution of barium chloride (BaCl₂) again.

### Step 6: Formation of Yellow Precipitate (D)
When the solution of barium chloride (BaCl₂) is treated with potassium chromate (K₂CrO₄), a yellow precipitate (D) is formed:
\[ \text{BaCl}_2 + \text{K}_2\text{CrO}_4 \rightarrow \text{BaCrO}_4 \downarrow + 2 \text{KCl} \]
The yellow precipitate (D) is **barium chromate (BaCrO₄)**.

### Step 7: Formation of White Precipitate (E) with Dilute H₂SO₄
The solution of barium chloride (BaCl₂) when treated with dilute sulfuric acid (H₂SO₄) gives a white precipitate (E):
\[ \text{BaCl}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4 \downarrow + 2 \text{HCl} \]
This white precipitate (E) is **barium sulfate (BaSO₄)**.

### Step 8: Properties of Precipitate (E)
The precipitate (E), barium sulfate (BaSO₄), is known to be insoluble in dilute HCl and nitric acid, confirming its identity. Additionally, it is a component of the white pigment lithopone, as mentioned in the question.

### Conclusion
The compound (E) is **barium sulfate (BaSO₄)**.

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