The brown ring in the detection of $N O_{3}^{-}$ ion is due to the formation of

$N O_{2}$

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$N_{2} O_{3}$

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$N_{2} O_{5}$

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$F e S O_{4} . N O$

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Solution

The correct option is D
$F e S O_{4} . N O$

During the brown ring test $F e S O_{4} . N O$ is formed from brown colored ion $F e (H_{2} O)_{5} . N O]^{+ 2}$

$N O_{3}^{-} + 3 F e^{+ 2} + 4 H^{⨁} \to N O + 3 F e^{+ 3} + 2 H_{2} O$

$[F e (H_{2} O)_{6}]^{+ 2} + N O \to [F e (H_{2} O)_{5} . N O]^{+ 2} N i t r o s o F e r r o u s I o n (b r o w n) + H_{2} O$

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Similar questions

The brown ring in the detection of $N O_{3}^{-}$ ion is due to the formation of

N O_{3}

N O_{3}^{-} i o n

N O_{3}^{-}

N O_{3}^{-}

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