Ce (Z-58) and Yb (Z=70) exhibit stable +4 and +2 oxidation states, respectively. This is because

Options

A. `Ce^(4+)` and `Yb^(2+)` acquire `f^7` configurations
B. `Ce^(4+)` and `Yb^(2+)` acquire `f^0` configurations
C. `Ce^(4+) and Yb^(2+)` aceuire ` f^0` and ` f^(14)` configuations
D. `Ce^(4+) and Yb^(2+)` acquire `f^7` and `f^(14)` configurations

Answer

C

Explanation

both acquire stable configurations that is `, f^0 and f^(14)`