The aqueous solution/liquid of ferrous sulphate $ (FeS{{O}_{4}}) $ absorbs nitric oxide to a considerable extent to give brown coloured
$ FeS{{O}_{4}}.NO $ .
This is the ring test for identifying the
$ NO_{3}^{-} $ radical. $ FeS{{O}_{4}}+NO\xrightarrow{{}}\underset{brown\,ring}{\mathop{FeS{{O}_{4}}NO}}\, $
$ (6FeS{{O}_{4}}+2HN{{O}_{3}}+2{{H}_{2}}S{{O}_{4}}\xrightarrow{{}} $ $ 3F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+4{{H}_{2}}O+2NO\uparrow ) $