A) \[{{N}_{2}}O\]
B) \[NO\]
C) \[N{{O}_{2}}\]
D) \[{{N}_{2}}{{O}_{5}}\]
Correct Answer: B
Solution :
\[NO_{3}^{-}+{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}HSO_{4}^{-}+HN{{O}_{3}}\] \[6FeS{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}+2HN{{O}_{3}}\xrightarrow[{}]{{}}\] \[3F{{e}_{2}}{{(S{{O}_{4}})}_{3}}4{{H}_{2}}O+\underset{\begin{smallmatrix} nitric \\ oxide \end{smallmatrix}}{\mathop{2NO}}\,\] \[FeS{{O}_{4}}+NO\xrightarrow{{}}\underset{\begin{smallmatrix} ferrous\,nitrososulphate\, \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,(brown\,ring) \end{smallmatrix}}{\mathop{Fe(NO)S{{O}_{4}}}}\,\] Hence, nitric oxide (NO) reacts with \[FeS{{O}_{4}}\]to give brown compound.
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