2
$\begingroup$

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 $\ce{NO+}$?

$\endgroup$
  • 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. $\endgroup$ – Ivan Neretin May 7 at 14:26
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.