Tin(II) chloride reduces mercury(II) chloride in two stages: 2HgCl2 + SnCl2 -> Hg2Cl2 + SnCl4 (white silky), then Hg2Cl2 + SnCl2 -> 2Hg + SnCl4 (grey). The total moles of SnCl2 needed to convert 1 mol of HgCl2 completely to metallic mercury are

JEE Advanced levelSingle correct optionDifficulty 3 of 3qc181

Tin(II) chloride reduces mercury(II) chloride in two stages: $2 HgCl X_{2} + SnCl X_{2} Hg X_{2} Cl X_{2} + SnCl X_{4}$ (white silky), then $Hg X_{2} Cl X_{2} + SnCl X_{2} 2 Hg + SnCl X_{4}$ (grey). The total moles of $SnCl X_{2}$ needed to convert 1 mol of $HgCl X_{2}$ completely to metallic mercury are

Stage 1 uses 0.5 mol

SnCl X_{2}

per mol

HgCl X_{2}

and gives 0.5 mol

Hg X_{2} Cl X_{2}

. Stage 2 uses 1 mol

SnCl X_{2}

per mol

Hg X_{2} Cl X_{2}

, i.e. 0.5 mol. Total

0.5 + 0.5 = 1.0

(Hg(II) to Hg(0) needs 2 electrons, Sn(II) gives 2).