Idea involved

For transition elements, the highest oxidation state is generally related to the total number of valence electrons available in the outer $ns$ and inner $(n-1)d$ orbitals.

So for a configuration of the form $(n-1)d^x ns^y$, the maximum oxidation state is often close to $x+y$ for early transition elements.

Check each option

Option A: $(n-1)d^3 ns^2$

Total valence electrons $=3+2=5$. This corresponds to elements like V-group, which can show a highest oxidation state of $+5$.

Option B: $(n-1)d^5 ns^1$

Total valence electrons $=5+1=6$. This is like chromium, whose highest oxidation state is $+6$.

Option C: $(n-1)d^8 ns^2$

Total valence electrons $=8+2=10$. But in later transition metals, the highest oxidation state does not keep increasing with $d$ count because $d$ electrons become less available for bonding. Such elements typically do not achieve oxidation state $+10$ in normal chemistry. This configuration resembles Ni, whose common highest oxidation state is much lower.

Option D: $(n-1)d^5 ns^2$

Total valence electrons $=5+2=7$. This corresponds to Mn-group elements, and manganese attains the very high oxidation state of $+7$ (for example in $\mathrm{MnO_4^-}$).

Compare the highest oxidation states

• A $\to +5$
• B $\to +6$
• C $\to$ much lower than $+7$
• D $\to +7$

Therefore, the highest oxidation state among the given configurations is achieved by $(n-1)d^5 ns^2$

Final answer

The correct option is D.