$250 \mathrm{mL}$ of $1\mathrm{M}, {\mathrm{I}}_2$ molar is separately reacted with $0.5 \mathrm{M} {\mathrm{Cu}}_2\mathrm{S}$ solution, $0.5 \mathrm{M} \mathrm{CuS}$ solution and $0.5 \mathrm{M} {\mathrm{S}}_2{{\mathrm{O}}_3}^{2-}$ solution, causing production of ${\mathrm{Cu}}^{+2}$ and ${{\mathrm{SO}}_4}^{2-}$ in the first two and ${\mathrm{S}}_4{{\mathrm{O}}_6}^{2-}$ in the last case along with iodide ions. Which of the following options is /are correct assuming $100\%$ completion of the reaction?

$$\begin{gathered} \underset{({\mathrm{n}}_{\mathrm{f}}=10)}{{\mathrm{Cu}}_2\mathrm{S} } + \underset{({\mathrm{n}}_{\mathrm{f}}=2)}{ {\mathrm{I}}_2} \to {\mathrm{Cu}}^{+2} + {{\mathrm{SO}}_4}^{-2} + {\mathrm{I}}^{-} .....\left(1\right) \\ \underset{({\mathrm{n}}_{\mathrm{f}}=8)}{\mathrm{CuS} } + \underset{({\mathrm{n}}_{\mathrm{f}}=2)}{ {\mathrm{I}}_2} \to {\mathrm{Cu}}^{+2} + {{\mathrm{SO}}_4}^{-2} + {\mathrm{I}}^{-} .....\left(2\right) \\ \underset{({\mathrm{n}}_{\mathrm{f}}=2)}{{\mathrm{I}}_2 } + \underset{({\mathrm{n}}_{\mathrm{f}}=1)}{ {\mathrm{S}}_2{{\mathrm{O}}_3}^{-2}} \to {\mathrm{I}}^{-} + \underset{({\mathrm{n}}_{\mathrm{f}}=2)}{{{\mathrm{S}}_4{\mathrm{O}}_6}^{-2} } .....\left(3\right) \end{gathered}$$

In reaction (1):

$$\begin{gathered} \mathrm{Meq} \mathrm{of} {\mathrm{Cu}}_2\mathrm{S} = \mathrm{Meq} \mathrm{of} {\mathrm{I}}_2 \\ 0.5 \times 10 \times \mathrm{V} = 250 \times 1 \times 2 \\ \mathrm{V} = 100 \mathrm{ml} (\mathrm{of} {\mathrm{Cu}}_2\mathrm{S}) \end{gathered}$$

In reaction (2):

$$\begin{gathered} \mathrm{Meq} \mathrm{of} \mathrm{CuS} = \mathrm{Meq} \mathrm{of} {\mathrm{I}}_2 \\ 0.5 \times 8 \times {\mathrm{V}}_{\mathrm{CuS}} = 250 \times 1 \times 2 \\ {\mathrm{V}}_{\mathrm{CuS}} = 125 \mathrm{ml} \\ \mathrm{eQ}.\mathrm{wt} \mathrm{of} {\mathrm{I}}_2 = \frac{\mathrm{mol}.\mathrm{wt}}{{\mathrm{n}}_{\mathrm{f}}} = \frac{254}{2} = 127 \end{gathered}$$