In qualitative test for identification of presence of phosphorous, the compound is heated with an oxidising agent.…

Step 1: Understand the Test for Phosphorus Detection

Phosphorus detection involves these steps:

1. The organic compound is oxidised with an oxidising agent (e.g., $\ce{Na2O2}$ or fuming $\ce{HNO3}$)
2. Phosphorus is converted to phosphate: $\ce{P -> H3PO4}$
3. The solution is treated with dilute $\ce{HNO3}$ (to acidify)
4. Ammonium molybdate $\ce{(NH4)_2MoO4}$ is added

Step 2: The Reaction

$\ce{H3PO4 + 12(NH4)_2MoO4 + HNO3 -> (NH4)_3PO4.12MoO3 + NH4NO3 + H2O}$

The yellow precipitate formed is ammonium phosphomolybdate: $\ce{(NH4)3PO4.12MoO3}$

Step 3: Identify the Correct Formula

(a) $\ce{Na3PO4.12MoO3}$ ❌ — Sodium is not present in this test; the reagent is ammonium molybdate.

(b) $\ce{(NH4)3PO4.12(NH4)2MoO4}$ ❌ — The Mo is present as $\ce{MoO3}$ (not as $\ce{MoO4^{2-}}$) in the precipitate.

(c) $\ce{(NH4)3PO4.12MoO3}$ ✅ — This is the correct formula for ammonium phosphomolybdate, the characteristic yellow precipitate.

(d) $\ce{MoPO4.21NH4NO3}$ ❌ — Incorrect formula; not the product of this reaction.

Step 4: Conclusion

Answer: (c) $\ce{(NH4)3PO4.12MoO3}$

Key Points to Remember:

• Phosphorus test: compound → oxidise → $\ce{H3PO4}$ → + $\ce{HNO3}$ + $\ce{(NH4)_2MoO4}$ → yellow precipitate
• Yellow precipitate = ammonium phosphomolybdate $\ce{(NH4)_3PO4.12MoO3}$
• This is a qualitative (not quantitative) test for phosphorus in organic compounds
• Compare with: nitrogen (Prussian blue), sulphur (violet), halogens (AgX precipitate)