The objective is to identify which element among the provided ones exhibits a +4 oxidation state, correlating it with its electron configuration.
Step 1: Electron Configuration Analysis. We examine the listed elements:
Cerium (Ce): Atomic number 58, electron configuration \( [Xe] 4f^1 5d^1 6s^2 \). In a +4 oxidation state, cerium achieves a \( [Xe] \) configuration by losing its 4f and 5d electrons.
Terbium (Tb): Atomic number 65, electron configuration \( [Xe] 4f^9 6s^2 \). While a +4 state would involve electron loss from 4f and 6s orbitals, Tb predominantly displays a +3 oxidation state.
Europium (Eu): Atomic number 63, electron configuration \( [Xe] 4f^7 6s^2 \). Europium commonly exists in +2 and +3 oxidation states, not typically +4.
Lutetium (Lu): Atomic number 71, electron configuration \( [Xe] 4f^{14} 5d^1 6s^2 \). Lutetium typically forms a +3 state by losing its 5d and 6s electrons, and a +4 state is not observed.
Step 2: Conclusion. Based on the electron configuration analysis, Cerium (Ce) is the sole element capable of achieving a +4 oxidation state by shedding its 4f and 5d electrons.
Consequently, the correct answer is \({\text{Ce}} \).