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In a solution containing HgCl2, I2, and Γ, both HgCl2 and I2 compete for Γ. Since formation constant of [HgI4]2− is very large (1.9 × 10^30) as compared with Γ3 (Kf = 700), Γ will preferentially combine with HgCl2.

1. HgCl2 + 2Γ → [HgΓ4]2− + 2Cl−
2. HgI2 + 2Γ → [HgΓ4]2−
3. HgCl2 + 2Γ → [HgΓ4]2− + 2I−
4. HgI2 + 2Γ → [HgΓ4]2− + 2Cl−

Correct answer: HgI2 + 2Γ → [HgΓ4]2−

Solution

The reaction that occurs is between HgI2 and Γ to form [HgΓ4]2−, as HgI2 is the intermediate formed when HgCl2 reacts with Γ. This aligns with the high formation constant of [HgI4]2−.

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