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 yellow precipitate `(D)` is

Options

A. `PbCrO_(4)`
B. `BaCrO_(4)`
C. `CaCrO_(4)`
D. none of these

Explanation

To solve the problem, we will break it down step by step based on the information provided in the question and the video transcript.

### Step 1: Identify the Alkaline Earth Metal Nitrate (A)
We start with an alkaline earth metal nitrate, which we will denote as (A). The most common alkaline earth metal nitrates are those of barium, strontium, calcium, magnesium, and radium.

### Step 2: Decomposition of (A)
Upon heating, alkaline earth metal nitrates decompose to form metal oxides, nitrogen dioxide, and oxygen. For example, barium nitrate decomposes as follows:
\[ \text{Ba(NO}_3\text{)}_2 \rightarrow \text{BaO} + 2 \text{NO}_2 + \frac{1}{2} \text{O}_2 \]
Here, the solid residue (B) left after decomposition is barium oxide (BaO).

### Step 3: Solubility of (B) 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 gives us a solution of barium chloride (BaCl₂).

### Step 4: Formation of White Precipitate (C)
When the solution of (B) (BaCl₂) is treated with ammonium carbonate \((\text{NH}_4\text{CO}_3)\), a white precipitate (C) of barium carbonate (BaCO₃) is formed:
\[ \text{BaCl}_2 + \text{(NH}_4\text{)}_2\text{CO}_3 \rightarrow \text{BaCO}_3 \downarrow + 2 \text{NH}_4\text{Cl} \]

### Step 5: Dissolving (C) in Dilute HCl
The white precipitate (C), which is barium carbonate (BaCO₃), is soluble in dilute HCl:
\[ \text{BaCO}_3 + 2 \text{HCl} \rightarrow \text{BaCl}_2 + \text{CO}_2 + \text{H}_2\text{O} \]

### Step 6: Treatment with Potassium Chromate
When the solution of barium chloride (from the previous step) is treated with potassium chromate (K₂Cr₂O₇), a yellow precipitate (D) of barium chromate (BaCrO₄) is formed:
\[ \text{BaCl}_2 + \text{K}_2\text{Cr}_2\text{O}_7 \rightarrow \text{BaCrO}_4 \downarrow + 2 \text{KCl} \]

### Step 7: Formation of White Precipitate (E) with Dilute H₂SO₄
When the solution of (B) (BaCl₂) is treated with dilute sulfuric acid (H₂SO₄), a white precipitate (E) of barium sulfate (BaSO₄) is formed:
\[ \text{BaCl}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4 \downarrow + 2 \text{HCl} \]
Barium sulfate (BaSO₄) is insoluble in dilute HCl and nitric acid, which confirms its identity.

### Step 8: Conclusion
The yellow precipitate (D) formed when barium chloride is treated with potassium chromate is:
\[ \text{BaCrO}_4 \]
Thus, the answer to the question is that the yellow precipitate (D) is **barium chromate (BaCrO₄)**.

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