Why does bromine react with iron(II) but not iodine?
All four of the ions are reducing agents. But usually halogen gases' tendency to undergo reduction to become halide ions decreases down the group (along with their reactivity). Out of the $4$ ions, only $\ce{Fe^{2+}}$ has a mild enough reducing power. 

$$\ce{2Fe^{2+} +Br2 -&gt; 2Br^- +2Fe^{3+}}$$

But $\ce{Fe^{2+}}$ is not strong enough to reduce $\ce{I2}$ as it is relatively stable.

In fact, $\ce{SO3^{2-}}$ has the strongest reducing power among the four options, as it can reduce both $\ce{I2}$ and $\ce{Br2}$.

$\ce{OH^{-}}$ is a mildly strong reducing agent, it can react with both $\ce{I2}$ and $\ce{Br2}$, and its typical reaction is

$$\ce{4OH^- -&gt; 2H2O +O2 + 4e^-}$$

$\ce{Mn^{2+}}$ is the least reducing, because it is the product of a strong oxidizing agent that has been reduced:

$$\ce{MnO4^{-} + 8H+ +5e^- -&gt; Mn^{2+} +4H2O}$$

Here, permanganate ion $\left(\ce{MnO4^-}\right)$ is a very strong oxidizing agent; the product thus would be weakly reducing.

Actually, a similar idea can be seen with $\ce{SO3^{2-}}$ (sulphite) ions. Since it is a strong reducing agent, we can expect the product formed after it is oxidized is a weak oxidizing agent.

$$\ce{SO3^{2-} + H2O -&gt; SO4^{2-} + 2H+ + 2e^-}$$

Indeed, sulphate ions alone are weak oxidizers, and stable.

I suggest you refer to E.C.S. (electrochemical series) to know whether a reaction would occur. It is easy to find online; e.g., at Wikipedia.