Species Changing Orange Color of Acidic Potassium Dichromate

Potassium dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$) in an acidic medium acts as a powerful oxidizing agent. The dichromate ion ($\text{Cr}_2\text{O}_7^{2-}$), which gives the solution its characteristic orange color, is reduced to the chromium(III) ion ($\text{Cr}^{3+}$), which is green in color:

$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \longrightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$

For a chemical species to change the orange color of acidic $\text{K}_2\text{Cr}_2\text{O}_7$ to green, it must act as a reducing agent and undergo oxidation.

Let's analyze the species given in the options:

1. $\text{Fe}^{2+}$ (Ferrous ion): Can be oxidized to ferric ion ($\text{Fe}^{3+}$): $\text{Fe}^{2+} \longrightarrow \text{Fe}^{3+} + e^-$

2. $\text{Sn}^{2+}$ (Stannous ion): Can be oxidized to stannic ion ($\text{Sn}^{4+}$): $\text{Sn}^{2+} \longrightarrow \text{Sn}^{4+} + 2e^-$

3. $\text{I}^-$ (Iodide ion): Can be oxidized to iodine ($\text{I}_2$): $2\text{I}^- \longrightarrow \text{I}_2 + 2e^-$

4. $\text{S}^{2-}$ (Sulfide ion): Can be oxidized to elemental sulfur ($\text{S}$): $\text{S}^{2-} \longrightarrow \text{S} + 2e^-$

5. Non-reducing species present in other options:

   • $\text{Fe}^{3+}$, $\text{Sn}^{4+}$, and $\text{SO}_4^{2-}$ are already in their highest common oxidation states ($+3$, $+4$, and $+6$ for sulfur, respectively) and cannot be oxidized further under these conditions.

Thus, all species in the set $\{\text{Fe}^{2+}, \text{Sn}^{2+}, \text{I}^-, \text{S}^{2-}\}$ are reducing agents capable of reducing orange dichromate ($\text{Cr}_2\text{O}_7^{2-}$) to green chromium(III) ions ($\text{Cr}^{3+}$).

Therefore, the correct option is A.