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 :

(NH_4)_3 PO_4 \cdot 12(NH_4)_2 MoO_4

P Block Elements PYQ — JEE Main 2024 · 8 Apr · Shift 2 · Q20 with Solution

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Correct answer: D

Principle of the test for phosphorus

In qualitative analysis, a compound suspected to contain phosphorus is first heated with an oxidising agent so that phosphorus is converted into phosphate ion, $\mathrm{PO_4^{3-}}$ .
Treatment with nitric acid and ammonium molybdate

The solution is then acidified with $\mathrm{HNO_3}$ and treated with ammonium molybdate. In presence of phosphate, a canary yellow precipitate is obtained.
Nature of the yellow precipitate

This precipitate is ammonium phosphomolybdate, commonly represented as:
$(\mathrm{NH_4})_3\mathrm{PO_4}\cdot 12\mathrm{MoO_3}$
Checking the options
- A: $\mathrm{Na_3PO_4\cdot 12MoO_3}$
  Incorrect, because sodium is not part of the characteristic precipitate.
- B: $(\mathrm{NH_4})_3\mathrm{PO_4}\cdot 12(\mathrm{NH_4})_2\mathrm{MoO_4}$
  Incorrect representation of the precipitate.
- C: $\mathrm{MoPO_4\cdot 21NH_4NO_3}$
  Incorrect.
- D: $(\mathrm{NH_4})_3\mathrm{PO_4}\cdot 12\mathrm{MoO_3}$
  Correct. This is ammonium phosphomolybdate.
Final answer

Therefore, the yellow precipitate obtained is:
$(\mathrm{NH_4})_3\mathrm{PO_4}\cdot 12\mathrm{MoO_3}$

View written solutionFree

Correct answer: D

Principle of the test for phosphorus

In qualitative analysis, a compound suspected to contain phosphorus is first heated with an oxidising agent so that phosphorus is converted into phosphate ion, $\mathrm{PO_4^{3-}}$ .
Treatment with nitric acid and ammonium molybdate

The solution is then acidified with $\mathrm{HNO_3}$ and treated with ammonium molybdate. In presence of phosphate, a canary yellow precipitate is obtained.
Nature of the yellow precipitate

This precipitate is ammonium phosphomolybdate, commonly represented as:
$(\mathrm{NH_4})_3\mathrm{PO_4}\cdot 12\mathrm{MoO_3}$
Checking the options
- A: $\mathrm{Na_3PO_4\cdot 12MoO_3}$
  Incorrect, because sodium is not part of the characteristic precipitate.
- B: $(\mathrm{NH_4})_3\mathrm{PO_4}\cdot 12(\mathrm{NH_4})_2\mathrm{MoO_4}$
  Incorrect representation of the precipitate.
- C: $\mathrm{MoPO_4\cdot 21NH_4NO_3}$
  Incorrect.
- D: $(\mathrm{NH_4})_3\mathrm{PO_4}\cdot 12\mathrm{MoO_3}$
  Correct. This is ammonium phosphomolybdate.
Final answer

Therefore, the yellow precipitate obtained is:
$(\mathrm{NH_4})_3\mathrm{PO_4}\cdot 12\mathrm{MoO_3}$