Question

An organic compound ${}^{\prime }{A}^{\prime }$ is oxidized with $N{a}_2{O}_2$ followed by boiling with $HN{O}_3$. 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:

• A. Flurine
• B. Phosphorus
• C. Nitrogen
• D. Sulphur

Answer 1

The correct option is B Phosphorus

The phosphorus containing organic compound are detected by 'Lassaigne's test' by heated with an oxidizing agent (sodium peroxide).
The phosphorus present in the compound is oxidised to phosphate.

The solution is boiled with nitric acid and then treated with ammonium molybdate to produced canary yellow precipitate.

$N{a}_{3}P{O}_4+3HN{O}_3\to H_{3}P{O}_4+3NaN{O}_3$

$H_{3}P{O}_4+\underset{\text{Ammonium molybdate}}{12(N{H}_4{)}_{2}Mo{O}_4}+21HN{O}_3⟶\underset{\text{Ammonium phosphomolybdate (canary yellow ppt.)}}{(N{H}_4{)}_{3}P{O}_4.12Mo{O}_3}\downarrow +21N{H}_{4}N{O}_3+12H_{2}O$

Phosphorus is detected in the form of canary yellow $ppt$ on reaction with ammonium molybdate.

Thus, option (c) is correct.