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Q.

$(A) :$ Addition of $A g N O_{3} (a q)$ to excess of $K I (a q)$ gives $- v e$ sol whereas addition of $K I (a q)$ to excess of $A g N O_{3}$ gives $+ v e$ sol
$(R) :$ The sol particles adsorbs the common ions present in solution and acquire their charge

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a

A is correct but R is wrong

b

A is wrong but R is correct

c

Both A and R are correct and R is correct explanation of A

d

Both A and R are correct and R is not correct explanation of A

answer is B.

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Detailed Solution

$A g N O_{3} + \underset{(e x c e s s)}{K I} \to [A g I] I^{-}$

$K I + \underset{(e x c e s s)}{A g N O_{3}} \to [A g I] A g^{+}$

The preferential adsorption of common ion on $A g I$ particles provides $- v e$ and $+ v e$ charge.

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(A): Addition of AgNO3(aq) to excess of KI(aq) gives −ve sol whereas addition of KI(aq) to excess of AgNO3 gives +ve sol(R): The sol particles adsorbs the common ions present in solution and acquire their charge