Ammonium molybdate is used to test the radical :

P O_{4}^{3 -}

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A s O_{4}^{3 -}

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C u^{2 +}

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A g^{+}

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Solution

The correct options are
A $P O_{4}^{3 -}$
B $A s O_{4}^{3 -}$
Option (A),(B) are correct.
Both $P O_{4}^{3 -}$ and $A s O_{4}^{3 -}$ give yellow ppt. with ammonium molybdate.

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Similar questions

{H N O}_{3}

{A s O}_{4}^{3 -}

{P O}_{4}^{3 -}

X + H N O_{3} \to Y + N O_{2} + H_{2} O + S

\to

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