A) lead nitrate
B) nitric acid
C) ferrous sulphate
D) sodium hydroxide
E) carbon disulphide
Correct Answer: C
Solution :
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. \[FeS{{O}_{4}}+NO\xrightarrow{{}}\underset{brown\,ring}{\mathop{FeS{{O}_{4}}NO}}\,\] \[(6FeS{{O}_{4}}+2HN{{O}_{3}}+2{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}\] \[3F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+4{{H}_{2}}O+2NO\uparrow )\]
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