# An organic compound 'A' is oxidized with $Na_{2}O_{2}$ followed by boiling with $HNO_{3}$. 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 : Option 1)SulphurOption 2)NitrogenOption 3)Phosphorus Option 4)Fluorine

According to question :

$A+Na_{2}O_{2}\rightarrow y\left ( Resultant\: \: solution \right )$

$Y+HNO_{3}\overset{\left ( NH_{4} \right )_{2}M_{o}0_{4}}{\rightarrow}yellow\: \: ppt.$

$\left ( NH_{4} \right )_{2}MoO_{4}$ is used for detection of phosphorus. This yellow coloured precipitate helps us to conclude that Phosphorus is present in the organic compound.

Option 1)

Sulphur

Option 2)

Nitrogen

Option 3)

Phosphorus

Option 4)

Fluorine

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