Question

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 :

• A. Na$_3$PO$_4$.12MoO$_3$
• B. (NH$_4$)$_3$PO$_4$.12(NH$_4$)$_2$MoO$_4$
• C. (NH$_4$)$_3$PO$_4$.12MoO$_3$
• D. MoPO$_4$.21NH$_4$NO$_3$

Answer 1

The Correct Option is C

To qualitatively detect phosphorus, the compound is first heated with an oxidizing agent to facilitate its analysis. The test proceeds as follows:

1. Oxidation of the compound to phosphoric acid (H₃PO₄).
2. Treatment of the oxidized product with nitric acid (HNO₃) to ensure phosphorus is in its +5 oxidation state.
3. Addition of ammonium molybdate ((NH₄)₂MoO₄) under acidic conditions.

This reaction yields a yellow precipitate, ammonium phosphomolybdate, with the formula (NH₄)₃PO₄.12MoO₃. The characteristic yellow color and this specific composition confirm the presence of phosphorus.

Evaluating the given options:

• \(Na_3PO_4.12MoO_3\): Sodium-based compounds are not typically used in this qualitative test.
• (NH₄)₃PO₄.12(NH₄)₂MoO₄: This formula does not represent the correct ammonium phosphomolybdate complex.
• (NH₄)₃PO₄.12MoO₃: This matches the standard formulation of the ammonium phosphomolybdate precipitate. This is the correct identification.
• \(MoPO_4.21NH_4NO_3\): This compound is not relevant to the standard phosphorus analysis test.

Consequently, the formation of (NH₄)₃PO₄.12MoO₃ is the correct indicator for the presence of phosphorus.