A) NO
B) CO
C) \[{{N}_{2}}\]
D) \[N{{H}_{2}}\]
Correct Answer: A
Solution :
\[3FeS{{O}_{4}}+N{{O}_{2}}+3{{H}_{2}}S{{O}_{4}}\to \]\[F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+\underset{(\text{Black}\,\,\,\text{brown})}{\mathop{FeS{{O}_{4}}.NO}}\,+{{H}_{2}}O\]
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