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Stereoisomerism

FeSO4 forms b...

Question

$F e S O_{4}$ forms brown ring with:

N O_{3}

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N O

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N O_{2}

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N_{2} O_{3}

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Solution

The correct option is B $N O$
The brown ring test is used to detect the presence of $N O$ (nitrate) ion. This is a three-step reaction at the end of which $F e$ and $N O$ form a deep brown coloured complex called iron nitrososulphate $(F e S O_{4} . N O)$ .

The reactions involved in the brown ring test are as follows:

1. $N a N O_{3} + H_{2} S O_{4} \to N a H S O_{4} + H N O_{3}$

2. $2 H N O_{3} + 6 F e S O_{4} + 3 H_{2} S O_{4} \to 3 F e_{2} (S O_{4})_{3} + 2 N O + 4 H_{2} O$

3. $F e S O_{4} + N O \to F e S O_{4} . N O$

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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}^{-}

C l^{-}

N O_{2}^{-}

N O_{3}^{-}

F e {S O}_{4}

F e {S O}_{4}

{N O}_{2}^{-}

{N O}_{3}^{-}

H_{2} {S O}_{4}

(I)

(I I)

N O

(I I I)

{F e}_{2} {({S O}_{4})}_{3}

N O_{3}^{-}

F e S O_{4}

H_{2} S O_{4}

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