Brown ring is made for
A) \[NO_{3}^{-}\]
B) \[C{{l}^{-}}\]
C) \[{{I}^{-}}\]
D) \[B{{r}^{-}}\]
Correct Answer: A
Solution :
Ring test for nitrate ions is\[FeS{{O}_{4}}(aq)+NO(g)+5{{H}_{2}}O(l)\xrightarrow{{}}\]\[\underset{\text{dark}\,\text{brown}\,\text{ring}}{\mathop{[Fe{{({{H}_{2}}O)}_{5}}NO]}}\,S{{O}_{4}}\] At the junction of \[{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\]and (\[\text{FeS}{{\text{O}}_{4}}+\]nitrate solution)
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