Q.
The solubility product of Hg2I2 is equal to
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a
Hg22+I−
b
Hg2+I−
c
Hg22+I−2
d
Hg2+I−2
answer is C.
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Detailed Solution
Hg2I2⇌Hg22++2I− Ksp=Hg22+I−2mercurous ion is in dimeric form which exist as Hg2+2 not simply as Hg+
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