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JEE Main 201912 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
12 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Practical Chemistry

JEE Main 2019 · 12 April · Morning Shift · Chemistry · Practical Chemistry

An organic compound ‘A’ is oxidized with N a _ 2 O _ 2...

An organic compound ‘A’ is oxidized with Na2O2N{a}_{2}{O}_{2} followed by boiling with HNO3.HN{O}_{3} . The resultant solution is then treated with ammonium molybdate to yield a yellow percipitate. Based on a above observation, the element present in the given compound is:An organic compound ‘A’ is oxidized with N a _ 2 O _ 2 followed by boiling with HN O _ 3 . The resultant

Options

  1. AFluorine
  2. BPhosphorus
  3. CNitrogen
  4. DSulphur

Correct answer

B. Phosphorus

Step-by-step solution

The phosphorus containing organic compounds are detected by 'Lassaigne's test' by heated with an oxidizing agent (sodium peroxide). The phosphorus present in the compound is oxidised to phosphate.

The solution is boiled with nitric acid and then treated with ammonium molybdate to produced canary yellow precipitate.

Phosphorus gets oxidised when it is treated with fuming conc. HNO3.HN{O}_{3} .
Na3PO4+3HNO3→H3PO4+3NaNO3{Na}_{3}{PO}_{4} + 3HN{O}_{3} \to {H}_{3}P{O}_{4} + 3{NaNO}_{3}An organic compound ‘A’ is oxidized with N a _ 2 O _ 2 followed by boiling with HN O _ 3 . The resultant

Phosphoric acid is used along with molybdenum compound for the identification of Magnesium.

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