Animated Solution for Chemistry - Organic Chemistry: An organic compound A is oxidised with ${\text{Na}}_2{\text{O}}_2$ followed by boiling with ${\text{HNO}}_3$. 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

Imagine you are a detective in a chemistry laboratory, and you are handed a mysterious organic compound. Your mission is to find out which elements are hiding inside its carbon-based structure. This problem takes us through a classic, textbook procedure used in qualitative analysis to unmask a very specific element: Phosphorus.

The problem describes a sequence of three distinct chemical treatments applied to an unknown organic compound 'A': 1. Oxidation with sodium peroxide ($N{a}_2{O}_2$). 2. Boiling with concentrated nitric acid ($HN{O}_3$). 3. Treatment with ammonium molybdate.

The final clue is the formation of a bright yellow precipitate. If you have studied the qualitative analysis of organic compounds, this sequence should immediately ring a bell. It is the definitive test for the presence of phosphorus.

Oxidation

In an organic compound, phosphorus is covalently bonded within the carbon framework. To test for it, we first need to break it out of this organic cage and convert it into a simple, inorganic ion.

This is achieved by heating the compound with sodium peroxide ($N{a}_2{O}_2$), a powerful oxidizing agent. The intense oxidation destroys the organic matter and converts any phosphorus present into sodium phosphate ($N{a}_{3}P{O}_4$).

Now, the phosphorus is in the form of the phosphate ion ($P{O}_4^{3-}$), which is ready to react.

Acidification

Before we can add our final testing reagent, we must prepare the environment. The aqueous extract containing the sodium phosphate is boiled with concentrated nitric acid ($HN{O}_3$).

This step serves a dual purpose. First, it converts the sodium phosphate into phosphoric acid ($H_{3}P{O}_4$). Second, it ensures the solution is highly acidic, which is a strict requirement for the final precipitation reaction to occur successfully.

The Yellow Precipitate

With the stage set, we introduce the star reagent: ammonium molybdate ($(N{H}_4{)}_{2}Mo{O}_4$). When added to the acidic solution containing phosphoric acid, a complex and beautiful reaction takes place.

The product of this reaction is ammonium phosphomolybdate ($(N{H}_4{)}_{3}P{O}_4\cdot 12Mo{O}_3$). This massive coordination complex is highly insoluble in acidic conditions and immediately crashes out of the solution, forming a distinct, canary-yellow precipitate.

The appearance of this yellow solid is the ultimate confirmation. It proves beyond a shadow of a doubt that the original organic compound contained phosphorus. Therefore, the correct answer is phosphorus.