The electron configuration of a neutral Samarium (Sm) atom is:
\[ \text{[Xe]} 6s^2 4f^6 \]
Samarium has an atomic number of 62. When it forms a +3 ion (Sm\(^{3+}\)), it loses three electrons. The electrons are typically removed from the outermost energy levels first, which in this case means removing the 6s electrons and then two of the 4f electrons.
So, the removal of three electrons would proceed as follows:
- Remove 2 electrons from the 6s subshell: \(6s^2\) → \(6s^0\) (removing 2 electrons).
- Remove 1 electron from the 4f subshell: \(4f^6\) → \(4f^5\) (removing 1 electron).
Thus, the electron configuration for the Sm\(^{3+}\) ion is:
\[ \text{[Xe]} 4f^5 \]
Therefore, the answer is:
\[ \text{[Xe]} 4f^5 \]