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Question

(A) Mercuric chloride $(HgCl_2)$ reacts with stannous chloride $(SnCl_2)$ in two steps.First,$HgCl_2$ is reduced to mercurous chloride $(Hg_2Cl_2)$:$2

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Liquid $Hg, SnCl_4$

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(A) Mercuric chloride $(HgCl_2)$ reacts with stannous chloride $(SnCl_2)$ in two steps.First,$HgCl_2$ is reduced to mercurous chloride $(Hg_2Cl_2)$:$2HgCl_2 + SnCl_2 \rightarrow Hg_2Cl_2 + SnCl_4$When $SnCl_2$ is in excess,the mercurous chloride $(Hg_2Cl_2)$ is further reduced to metallic mercury $(Hg)$:$Hg_2Cl_2 + SnCl_2 \rightarrow 2Hg + SnCl_4$Thus,the final products are liquid $Hg$ and $SnCl_4$.

When mercury $(II)$ chloride is treated with excess of $SnCl_2$,the products obtained are

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