A) \[N{{O}_{2}}\]
B) \[HCl\]
C) \[N{{H}_{3}}\]
D) \[S{{O}_{2}}\]
Correct Answer: D
Solution :
\[S{{O}_{2}}\] gas is a reducing agent, hence turns dichromate paper green due to formation of chromium sulphate. Sulphite radical \[(SO_{3}^{2-})\] is identified by this. \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}+3S{{O}_{2}}+{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}\] \[\underset{\text{green}}{\mathop{C{{r}_{2}}{{(S{{O}_{4}})}_{2}}}}\,+{{K}_{2}}S{{O}_{4}}+{{H}_{2}}O\]
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