Excess aqueous $\mathrm{HgCl}{}_2$ is added to a small amount of $\mathrm{SnCl}{}_2$ solution. The white precipitate formed is $\mathrm{Hg}{}_2\mathrm{Cl}{}_2$. Which of the following describes the oxidation-state changes?

Tin(II) is a good reductant because the inert pair effect is weaker for tin than for lead, so $+4$ is the more stable state. It reduces Hg(II) to Hg(I) in $\mathrm{Hg}{}_2\mathrm{Cl}{}_2$ (white) when $\mathrm{HgCl}{}_2$ is in excess; with excess $\mathrm{SnCl}{}_2$ the reduction would continue to grey metallic Hg. So Sn goes $+2\to +4$ and Hg goes $+2\to +1$.