$$SnCl_2 + 2HgCl_2 \rightarrow Hg_2Cl_2 + SnCl_4$$

In the given reaction:

A

$$HgCl_2$$ is oxidised

B

$$SnCl_2$$ is oxidised

C

$$Hg_2Cl_2$$ is oxidised

D

$$SnCl_2$$ is reduced