Brown ring in the test of $N O_{3}^{-}$ is formed due to the formation of :

F e S O_{4} \cdot N O

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[F e {(S O_{4})}_{2} \cdot H O] \cdot H_{2} O

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F e_{2} {(S O_{4})}_{3} \cdot N O

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none of these

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Solution

The correct option is B $F e S O_{4} \cdot N O$
Option (A) is correct.
Brown ring in the test of $N O_{3}^{-}$ is formed due to the formation of $F e S O_{4} \cdot N O$ .
Brown ring test is the test to find out the presence of nitrate radical. For this equal volume of the freshly prepared $F e S O_{4}$ is mixed with solution to be analysed. To this concentrated $H_{2} S O_{4}$ is added along the sides of the test tube. A brown ring will forms at the junction of the two liquids. This is Nitroso ferrous sulphate $(F e S O_{4} . N O)$

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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}

N O_{2}^{-}

N O_{3}^{-}

[{F e (H_{2} O)}_{5} N o]^{2 +}

{N O}_{3}^{+}

F e

F e S O_{4}

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