A) \[{{N}_{2}}O\]
B) \[NO\]
C) \[{{N}_{2}}{{O}_{3}}\]
D) \[{{N}_{2}}{{O}_{5}}\]
Correct Answer: B
Solution :
\[FeS{{O}_{4}}+NO\,\to \ \underset{\text{(Brown)}}{\mathop{FeS{{O}_{4}}.NO}}\,\]
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