Of the following outer electronic configurations of atoms, the highest oxidation state is achieved by which one?

Transition metals can achieve high oxidation states by losing/sharing both $ns$ and $(n-1)d$ electrons.$(n-1)d^{5}n{s}^2$ has a total of $7$ valence electrons (like Manganese) and can achieve an oxidation state of $+7$.$(n-1)d^{8}n{s}^2$ ($10$ valence electrons) does not usually reach $+10$.$(n-1)d^{3}n{s}^2$ reaches $+5$.$(n-1)d^{5}n{s}^1$ reaches $+6$ (like Chromium).
Highest is achieved by $(n-1)d^{5}n{s}^2$.