A common nitrate test, known as the brown ring test can be performed by adding iron(II) sulfate to a solution of a nitrate, then slowly adding concentrated sulfuric acid such that the acid forms a layer below the aqueous solution. A brown ring will form at the junction of the two layers, indicating the presence of the nitrate ion.

I know for sure that certain oxides of nitrogen, such as $\ce{NO}$ and $\ce{NO2}$ test positively in the brown ring test for nitrates, but I had never found a reliable source stating the same for the other oxides of nitrogen. But one of the questions in a class test I took was:

Which one of the following does not give the Brown Ring test?

  1. $\ce{N2O}$
  2. $\ce{NO}$
  3. $\ce{N2O3}$
  4. $\ce{N2O5}$

The answer was given as $\ce{N2O}$ and I am unable to identify the logic behind the answer. Could it be because the nitrogen compound is reduced in the process of forming the brown complex with iron, while $\ce{N2O}$ already has too small an oxidation state to be reduced to $\ce{NO+}$?

  • 3
    $\begingroup$ You may put it this way, but that's a post factum explanation. Just remember that N2O is relatively inert and doesn't react like that. $\endgroup$ May 7, 2019 at 14:26

1 Answer 1


The brown ring nitrate test is based on forming ferrous nitrosyl cation, with nitrous oxide formed from nitrates by iron oxidation in acidic solution:

$$\begin{align} \ce{3 [Fe(H2O)6]^2+ + NO3- + 4 H+ &->3 [Fe(H2O)6]^3+ + NO + 2 H2O} \\ \ce{[Fe(H2O)6]^2+ + NO &->3 [Fe(H2O)(NO)]^2+} + H2O\\ \end{align}$$

Positive results are given also by nitrites and nitrogen oxides able to form nitric or nitrous acid, directly by hydration, or indirectly by disproportionation.

$$\begin{align} \ce{ [Fe(H2O)6]^2+ + NO2- + 2 H+ &-> [Fe(H2O)6]^3+ + NO + H2O} \\ \ce{N2O3 + H2O &-> 2 HNO2}\\ \ce{2 NO2 + H2O &-> HNO2 + HNO3}\\ \ce{N2O5 + H2O &-> 2 HNO3}\\ \end{align}$$

Therefore, $\ce{NO}$ and all higher nitrogen oxides oxidizing $\ce{Fe^2+}$, forming $\ce{NO}$, react positively in the brown ring test.

$\ce{N2O}$ cannot do that, because it cannot form $\ce{NO}$ at given conditions.


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