An organic compound $'A'$ is oxidized with ${Na2O2}$ 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 :
- Sulphur
- Nitrogen
- Fluorine
- Phosphorus
The Correct Option is D
Solution and Explanation
The phosphorus present in the compound in oxidised to phosphate.
The solution is boiled with nitric acid and then treated with ammonium molybdate to produced canary yellow precipitate.
${ Na3PO4 + 3HNO3 -> H3PO4 + 3NaNO3}$ ${H3PO4 + $\underset{\text{ (Ammonium molybdate)}}{{12 (NH4)2MoO4 }}$ + 21 HNO3 -> (NH4)3PO4.12MoO3 v + 21 NH4NO3 + 12 H2O }$
(Ammonium phosphomolybdate)
(canary yellow precipitate)
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Concepts Used:
Group 15 Elements
Group 15 elements are also known as nitrogen group elements or nitrogen family.
It consists of:
- Nitrogen (N)
- Phosphorus (P)
- Arsenic (As)
- Antimony (Sb)
- Bismuth (Bi)
- Moscovium (Mc)
They are present on the right side of the periodic table. All the elements are arranged based on their atomic weight. The elements with similar properties reside in a column called the group.
All Group 15 elements have the electron configuration ns2np3 in their outer shell, where n is the principal quantum number.
Periodic Trends in Group 15 Elements
