Can lanthanum (Z = 57) exhibit + 4 oxidation state ?

Explanation

Electronic configuration of `La : [Xe]5d^(1)6s^(2)`
Electronic configuration of `La^(3+):[Xe]5d^(0)6s^(0)`
In + 3 oxidation state lanthanum loses 3 electrons to attain the stable electronic configuration of the nearest noble gas, xenon. However, removal of one more electron from the highly stable `La^(3+)` ion to attain the + 4 oxidation state is very difficult. Hence, `La^(4+)` does not exist. In other words, lanthanum does not exhibit + 4 oxidation state.