Answer
The electron configuration of Fe is $[\mathrm{Ar}] 3 d^{6} 4 \mathrm{s}^{2}$ and for $\mathrm{Fe}^{3+},[\mathrm{Ar}] 3 d^{5} .$ The $3 d^{5}$ subshell, being half-filled, provides stability, which favors the oxidation state of $+3$. Cobalt $\left([\mathrm{Ar}] 3 d^{7} 4 s^{2}\right)$ and nickel $\left([\mathrm{Ar}] 3 d^{8} 4 s^{2}\right)$ must lose four and five electrons, respectively, to achieve a half-filled $3d$ subshell. However, they both lose the 2 electrons in $4S$ subshell (the higher energy subshell in transition metals), yielding the $+2$ oxidation state.
