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Q.

The solubility product of Hg2I2 is equal to

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a

Hg22+I 

b

Hg2+I

c

Hg22+I2

d

Hg2+I2

answer is C.

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Detailed Solution

Hg2I2Hg22++2I Ksp=Hg22+I2mercurous ion is in dimeric form which exist as Hg2+2 not simply as Hg+

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