Which mixture of the solutions will lead to the formation of negatively charged colloidal ${\text{ [AgI]I }}^{-}\text{sol? }$

If colloidal sol of AgI is prepared by adding KI solution to ${\mathrm{AgNO}}_3$ till KI is in slightly  excess, iodide ion $\left({\mathrm{I}}^{-}\right)$ will be adsorbed on the surface of AgI thereby giving a negative charge to the sol.In second option $50 \mathrm{X} {10}^{-3} \mathrm{moles} \mathrm{of} {\mathrm{AgNO}}_{3 } \mathrm{is} \mathrm{mixed} \mathrm{with} 75 \mathrm{X} {10}^{-3 } \mathrm{moles} \mathrm{of} \mathrm{KI}$. KI is excess 

In third option $50 \mathrm{X} {10}^{-3} \mathrm{moles} \mathrm{of} {\mathrm{AgNO}}_{3 } \mathrm{is} \mathrm{mixed} \mathrm{with} 100 \mathrm{X} {10}^{-3 } \mathrm{moles} \mathrm{of} \mathrm{KI}$ ..KI is excess 

$$\begin{gathered} \mathrm{AgI} + {\mathrm{I}}^{-} ⟶ \mathrm{AgI}:{\mathrm{I}}^{-} \\ (\mathrm{from} \mathrm{KI}) \mathrm{Negative} \mathrm{Sol} \end{gathered}$$