HgCl2andI2 both when dissolved in water containing I− ions the pair of species formed is :
A
HgI2−4,I−3
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B
Hg2I2,I−
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C
HgI2,I−
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D
HgCl2,I−3
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Solution
The correct option is AHgI2−4,I−3 I2 reacts with I− and maintains the following equillibrium. I2+I−⇌I−3 Hg2+ gives precipitate of HgI2 on reaction with I− but HgI2 is soluble in solution of I− Hg2++2I−→HgI2 HgI2+2I−⇌[HgI4]2−
Option (c) is correct