Question

An organic compound containing an extra element E on reaction with $Na_{2}O_{2}$ followed by boiling with $HNO_{3}$ gives a compound. This on treatment with ammonium molybdate solution gives yellow precipitate. What is E?

• A. S
• B. N
• C. P
• D. I

Hint:

Ammonium Molybdate + Yellow Precipitate always signals the presence of Phosphorus in qualitative organic analysis.

Answer 1

The Correct Option is C

Step 1: Understand the test.
We are detecting an extra element E in an organic compound. The clue is that the final test uses ammonium molybdate and gives a yellow precipitate.

Step 2: Recall the yellow precipitate test.
A yellow precipitate with ammonium molybdate is the classic confirming test for phosphorus, present as phosphate.

Step 3: Follow the first reaction.
Heating the organic compound with sodium peroxide oxidises the phosphorus into phosphate ions, which become sodium phosphate.

Step 4: Follow the acid step.
Boiling with nitric acid turns the phosphate into phosphoric acid in an acidic medium, ready for the next test.

Step 5: Form the yellow precipitate.
Phosphoric acid then reacts with ammonium molybdate to give a canary-yellow precipitate of ammonium phosphomolybdate. \[ H_3PO_4 + 12(NH_4)_2MoO_4 + 21HNO_3 \longrightarrow (NH_4)_3PO_4{\cdot}12MoO_3 \downarrow + \ldots \]

Step 6: Identify E.
This yellow precipitate confirms the element is phosphorus. \[ \boxed{E = P} \]