Qualitative analysis of organic compounds is performed by Lassaigne's test by fusion with metallic sodium, by which the covalent compounds are converted into ionic compounds. Extra elementsline `N,S,P` andhalogensaredetected by their usual tests.
Q. Yellow precipitate in the detection of phosphorous when an organic compound is heated with `Na_2O_2` and then boiled with conc. `HNO_3` followed by the addition of ammonium molybdate is duw to the formation of :

Options

A. `(NH_4)_3.PO_4.12MoO_3`
B. `(NH_4)_3.PO_4.6MoO_3`
C. `(NH_4)_3.PO_4.12MoO_2`
D. `(NH_4)_3PO_4.6MoO_2`

Answer

To solve the question regarding the yellow precipitate formed during the detection of phosphorus in an organic compound, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Process**: The qualitative analysis of organic compounds involves detecting elements such as phosphorus (P) using specific reagents. In this case, we are using sodium peroxide (Na2O2) and ammonium molybdate ((NH4)6Mo7O24).

2. **Fusion with Sodium Peroxide**: When the organic compound is heated with sodium peroxide, it leads to the formation of sodium phosphate (Na3PO4). This reaction is crucial as it converts the phosphorus in the organic compound into a more detectable form.

**Reaction**:
\[ \text{Organic Compound} + \text{Na}_2\text{O}_2 \rightarrow \text{Na}_3\text{PO}_4 + \text{other products} \]

3. **Extraction and Boiling with HNO3**: The fused mass containing sodium phosphate is then extracted with water. The aqueous solution is boiled with concentrated nitric acid (HNO3). This step helps in further purifying the phosphate.

**Reaction**:
\[ \text{Na}_3\text{PO}_4 + 21 \text{HNO}_3 \rightarrow \text{Products} \]

4. **Addition of Ammonium Molybdate**: After boiling with HNO3, ammonium molybdate is added to the solution. This reagent is specifically used to detect phosphorus.

**Reaction**:
\[ \text{Na}_3\text{PO}_4 + \text{(NH}_4)_6\text{Mo}_7\text{O}_{24} \rightarrow \text{NH}_4\text{PO}_4 \cdot 12\text{MoO}_3 + \text{other products} \]

5. **Formation of Yellow Precipitate**: The reaction between ammonium molybdate and the phosphate leads to the formation of a yellow precipitate known as ammonium phosphomolybdenate.

**Final Compound**:
\[ \text{Ammonium Phosphomolybdenate} \quad (\text{NH}_4)_3\text{PO}_4 \cdot 12\text{MoO}_3 \]

6. **Conclusion**: The yellow precipitate observed is due to the formation of ammonium phosphomolybdenate, which confirms the presence of phosphorus in the organic compound.

### Final Answer:
The yellow precipitate is due to the formation of **ammonium phosphomolybdenate** \((\text{NH}_4)_3\text{PO}_4 \cdot 12\text{MoO}_3\).

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