JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Practical and Analytical Chemistry

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 :

(A)Phosphorus✓ Correct
(B)Sulphur
(C)Nitrogen
(D)Fluorine

Answer: (A)

Solution

Fusing the compound with

Na X_{2} O X_{2}

and then boiling with

HNO X_{3}

oxidises any phosphorus present all the way to phosphate,

PO X_{4} X^{3 -}

. Adding ammonium molybdate then throws down the characteristic canary-yellow precipitate of ammonium phosphomolybdate,

(NH X_{4}) X_{3} [PO X_{4} (MoO X_{3}) X_{12}]

. This test is diagnostic of phosphorus in the organic compound.

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option

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