MCQ
In the reaction $SnCl_2(excess)+HgCl_2\ \rightarrow\ A+SnCl_4$, '$A$' is
- A$Hg_2Cl_2$
- ✓$Hg$
- C$HgCl$
- D$HgCl_3$
✓
Answer
Correct option: B.
$Hg$
b
$SnCl _2+ HgCl _2 \rightarrow Hg + SnCl _4$
$SnCl _2+ HgCl _2 \rightarrow Hg + SnCl _4$
Here $A$ is $Hg$.
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