The correct option is C HgI2−4,I−3
In a solution containing HgCl2, I2 and I−, both HgCl2 and I2 compete for I−.
Since the formation constant of [HgI4]2− is 1.9×1030
which is very large as compared to I−3(Kf=700), I− will preferentially combine with HgCl2.
HgCl2+2I−→HgI2Redppt↓+2Cl−
HgI2+2I−→[HgI4]2−soluble