Question

An organic compound ‘A’ is oxidized with $\text{N}{{\text{a}}_{2}}{{\text{O}}_{2}}$ followed by boiling with $\text{HN}{{\text{O}}_{3}}$. 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). Sulfur
(b). Nitrogen
(c). Fluorine
(d). Phosphorus

Answer 1

Hint: This element is used in the production of fertilizers. The most common allotrope of this element is red and white in color. The red one is used in the side of matchboxes which is used to strike safety matches against it to light them.

Complete step by step answer:
The question given is actually the qualitative test for phosphorus. It is given in the form of a question. The test is done for the detection of phosphorus elements in the organic compound.
There is a test named Lassaigne’s test for the detection of special elements other than carbon like nitrogen, sulphur, halogens etc.
Preparation of Lassaigne’s extract: Nitrogen, Sulfur, Phosphorus and Halogens are detected by Lassaigne’s test. The elements present in organic compounds are converted to ionic form by fusing with Na-metal. The reactions involved are-
$\text{Na+ C+N}\to \text{NaCN}$
$\text{2Na+ S}\to \text{N}{{\text{a}}_{2}}\text{S}$
$\text{Na+ X}\to \text{NaX}$
Now, Let us discuss the main reaction involved in the detection of phosphorus. Phosphorus is detected by fusing the organic compound or Lassaigne’s extract with an oxidizing agent like $\text{N}{{\text{a}}_{2}}{{\text{O}}_{2}}$. Phosphorus in the compound is oxidized to $\text{P}{{\text{O}}_{4}}^{3-}$ which is then extracted with water. The solution is boiled with $\text{HN}{{\text{O}}_{3}}$ and then treated with ammonium molybdate. A yellow colored precipitate indicates the presence of phosphorus.
$\text{N}{{\text{a}}_{3}}\text{P}{{\text{O}}_{4}}+\text{HN}{{\text{O}}_{3}}\to {{\text{H}}_{3}}\text{P}{{\text{O}}_{4}}$
$\text{N}{{\text{a}}_{2}}{{\text{O}}_{2}}+\text{P}\to \text{2N}{{\text{a}}_{3}}\text{P}{{\text{O}}_{4}}+2\text{N}{{\text{a}}_{2}}\text{O}$
${{\text{H}}_{3}}\text{P}{{\text{O}}_{4}}+12{{(\text{N}{{\text{H}}_{4}}\text{)}}_{2}}\text{Mo}{{\text{O}}_{4}}+21\text{HN}{{\text{O}}_{3}}\to {{(\text{N}{{\text{H}}_{4}})}_{3}}.12\text{Mo}{{\text{O}}_{3}}+21\text{N}{{\text{H}}_{4}}\text{N}{{\text{O}}_{3}}+12{{\text{H}}_{2}}\text{O}$

Note: The yellow precipitate that is formed in the reaction is a double salt which means its constituent particles can retain its identity in the solution even after breaking down.