Step 1: Recall features of actinides relevant to oxidation states.
Actinides (Z = 90 to 103) are $f$-block elements. The energies of $5f$, $6d$, and $7s$ subshells are very close to each other. Electrons from all three subshells can participate in bonding, allowing actinides to show a very wide range of oxidation states.
Step 2: Identify the elements in the question.
The four elements are: Uranium (U, Z = 92), Neptunium (Np, Z = 93), Americium (Am, Z = 95), and Protactinium (Pa, Z = 91).
Step 3: List the maximum oxidation states.
Pa: maximum $+5$. U: maximum $+6$ (as in $UF_6$). Am: maximum $+6$. Np: maximum $+7$.
Step 4: Compare the oxidation states.
\[ Np\ (+7) > U\ (+6) = Am\ (+6) > Pa\ (+5) \] Neptunium is the only element among the four to reach the $+7$ oxidation state.
Step 5: Explain why Np reaches $+7$.
Neptunium (Z = 93) has configuration $[Rn]\,5f^4\,6d^1\,7s^2$. The closeness of $5f$, $6d$, and $7s$ energies allows removal of up to 7 electrons under strongly oxidizing conditions, enabling the $+7$ state.
Step 6: State the final answer.
\[ \boxed{Np} \]