The atomic number of cerium `(Ce)` is 58. The correct electronic confguration of `Ce^(3+)` ions is :

Options

A. `[Xe]4f^(1)`
B. `[Kr]4f^(1)`
C. `[Xe]4f^(13)`
D. `[Kr]4d^(1)`

Explanation

`Z = 54` is `[Xe]=Z= 56` is `Ba = [ Xe] 6s^(2)`
`Z =57` ( La) is `[Xe]5d^(1)` ( electron 5d in place of expected 4f)
But in `Z= 58` ( Ce ) , electron enters 4f and `5d^(1)` also shifts to 4f. Hence, electronic configuration of `Ce( 58) ` is `[Xe] 4f^(2) 5d^(0) 6s^(2)`.
`:. Ce^(3+) = [Xe]4f^(1)`