Question

An Organic compound ‘A’ is oxidizing with Na₂ O₂ followed by boiling with HNO₃. The resultant solution is then treated with ammonium molybdate to yield a yellow precipitate. Based on above observation, the element present in the given compound is:
1. Nitrogen
2. Phosphorus
3. Fluorine
4. Sulphur

Answer 1

Correct Answer - Option 2 : Phosphorus

Concept:

Phosphorus is detected in the form of yellow ppt (precipitate) of ammonium phosphate molybdate on reaction with ammonium molybdate.

This is qualitative test for phosphorous

\(p + N{a_2}{O_3} \to N{a_3}P{O_4}\mathop \to \limits^{3HN{O_3}} {H_3}P{O_4} + 3NaN{O_3}\)      ----(1)

H₂PO₄ + 12(NH₄)MoO₄ + 21HNO₃ → (NH₄)PO₄.12MoO₃ + 21NH₄NO₃ + 12H₂O      ----(2)

Ammonium molybdate yellow ppt    

Based on the above observation, an organic compound ‘A’ is Phosphate. From the reaction (1), the phosphate is oxidizing with Na₂ O₃ followed by boiling with HNO₃. The resultant solution is then treated with ammonium molybdate ((NH₄)MoO₄) to yield the yellow precipitate.