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?
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?
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Ammonium Molybdate + Yellow Precipitate always signals the presence of Phosphorus in qualitative organic analysis.
Updated On: Jun 3, 2026
- S
- N
- P
- I
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The Correct Option is C
Solution and Explanation
Step 1: Concept
The identification of extra elements (like N, S, P, and halogens) in organic chemistry is handled via targeted qualitative analysis pathways. Phosphorus is specifically detected using the ammonium molybdate confirmation sequence.
Step 2: Meaning
Oxidizing phosphorus with sodium peroxide converts it into inorganic phosphate ($PO_{4}^{3-}$). Boiling it with nitric acid prepares the chemical environment for complexation.
Step 3: Analysis
When an organic molecule holding phosphorus is heated with an oxidizing agent like sodium peroxide ($Na_{2}O_{2}$), the phosphorus changes into sodium phosphate ($Na_{3}PO_{4}$). Treating this with concentrated $HNO_{3}$ and adding ammonium molybdate yields a distinct canary-yellow precipitate of ammonium phosphomolybdate: $Na_{3}PO_{4} + 3HNO_{3} \longrightarrow H_{3}PO_{4} + 3NaNO_{3}$ $H_{3}PO_{4} + 12(NH_{4})_{2}MoO_{4} + 21HNO_{3} \longrightarrow (NH_{4})_{3}PO_{4}\cdot12MoO_{3}\downarrow + 21NH_{4}NO_{3} + 12H_{2}O$
Step 4: Conclusion
The generation of a yellow color with ammonium molybdate uniquely confirms the presence of Phosphorus (P).
Final Answer: (C)
The identification of extra elements (like N, S, P, and halogens) in organic chemistry is handled via targeted qualitative analysis pathways. Phosphorus is specifically detected using the ammonium molybdate confirmation sequence.
Step 2: Meaning
Oxidizing phosphorus with sodium peroxide converts it into inorganic phosphate ($PO_{4}^{3-}$). Boiling it with nitric acid prepares the chemical environment for complexation.
Step 3: Analysis
When an organic molecule holding phosphorus is heated with an oxidizing agent like sodium peroxide ($Na_{2}O_{2}$), the phosphorus changes into sodium phosphate ($Na_{3}PO_{4}$). Treating this with concentrated $HNO_{3}$ and adding ammonium molybdate yields a distinct canary-yellow precipitate of ammonium phosphomolybdate: $Na_{3}PO_{4} + 3HNO_{3} \longrightarrow H_{3}PO_{4} + 3NaNO_{3}$ $H_{3}PO_{4} + 12(NH_{4})_{2}MoO_{4} + 21HNO_{3} \longrightarrow (NH_{4})_{3}PO_{4}\cdot12MoO_{3}\downarrow + 21NH_{4}NO_{3} + 12H_{2}O$
Step 4: Conclusion
The generation of a yellow color with ammonium molybdate uniquely confirms the presence of Phosphorus (P).
Final Answer: (C)
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