In qualitative test for identification of presence of phosphorous, the compound is heated with an oxidising agent. Which is further treated with nitric acid and ammonium molybdate respectively. The yellow coloured precipitate obtained is :

Solution

Chemistry ma’am · 3 steps

Turn phosphorus into phosphate first; then ammonium molybdate locks it into yellow ammonium phosphomolybdate.

1. Oxidise to phosphate. Heating the compound with an oxidising agent such as Na₂O₂ converts all the phosphorus into phosphate, Na₃PO₄. Boiling with nitric acid then gives phosphoric acid, H₃PO₄.
2. Add ammonium molybdate. In the hot nitric acid solution, phosphate reacts with ammonium molybdate. Each phosphate group binds 12 MoO₃ units, and three NH₄⁺ ions balance its charge.

   H₃PO₄ + 12(NH₄)₂MoO₄ + 21HNO₃ → (NH₄)₃PO₄·12MoO₃ v + 21NH₄NO₃ + 12H₂O

3. So. The yellow precipitate is ammonium phosphomolybdate, (NH₄)₃PO₄·12MoO₃, which is option (d).

The idea to keep. Yellow precipitate with ammonium molybdate in HNO₃ means phosphate. The precipitate is ammonium phosphomolybdate, with three NH₄⁺ ions and 12 MoO₃ units for every phosphate.

Answer from the NTA final answer key. Solution by PrepWiser.