The atomic number of cerium (Ce) is 58. The correct electronic configuration of C{{e}^{3+}} isa.) [Xe]4{{f}^{1}} b.) [Kr]4{{f}^{1}}c.) [Xe]4{{f}^{13}}d.) [Kr]4{{d}^{1}}

The atomic number of cerium (Ce) is 58, which means it has 58 electrons in its neutral state. When cerium loses three electrons to form the Ce^3+ ion, it has 55 electrons remaining.

To determine the correct electronic configuration of Ce^3+, we need to follow the Aufbau principle, Hund's rule, and the Pauli exclusion principle. The electron configuration for Ce^3+ can be determined as follows:

Ce (neutral) electron configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 4f² 5s² 5p⁶ 5d¹ 6s²

Ce^3+ electron configuration: Remove 3 electrons from the outermost shell (5s² 5p⁶) and 1 electron from the 4f subshell.

So, the correct electron configuration of Ce^3+ is:
[Xe]4f¹

The correct answer is (a) [Xe]4f^1.