On adding AgNO₃ solution to KI solution, a negatively charged colloidal sol will be formed in which of the following conditions ? 

In order to form a negatively charged colloidal sol on adding AgNO₃​ solution into KI solution, KI should be in excess.

From options,

(A) For 100 mL of 0.1 M AgNO₃ , no. of millimoles= $100\times 0.1=10$

       For 100 mL of 0.1 M KI, no. of millimoles= $100\times 0.1=10$

       both have same no. of moles, hence it is incorrect

(B) For 100 mL of 0.1 M AgNO₃ , no. of millimoles=$100\times 0.1=10$

       For  50 mL of 0.2 M KI = $50\times 0.2=10$

      both have same no. of moles, hence it is incorrect

(C) For 100 mL of 0.2 M AgNO₃ , no. of millimoles= $100\times 0.2=20$

      For 100 mL of 0.1 M KI, no. of millimoles= $100\times 0.1=10$

       excess of AgNO₃ is present, hence it is incorrect

(D) For 100 mL of 0.1 M AgNO₃ , no. of millimoles= $100\times 0.1=10$

      For 100 mL of 0.15 M KI, no. of millimoles=$100\times 0.15=15$

      Excess of KI is present, hence it is the correct answer.