A) \[HgC{{l}_{2}}\]
B) \[SnC{{l}_{2}}\]
C) \[SnC{{l}_{4}}\]
D) \[Hg\]
Correct Answer: B
Solution :
Mercury (a more electropositive element) is added as \[HgC{{l}_{2}}-\] Therefore, it is reduced. Chlorine (a more electronegative element) is added as \[SnC{{l}_{2}}-\] Therefore, it is oxidized.You need to login to perform this action.
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