.....#1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(1)3d^(5).........#
Ordinarily, the #4s# orbital would be filled, however, #"Hund's rule of maximum multiplicity"# operates in this scenario. You should check with your teacher whether you are expected to know this (I suspect that you are not). Upon oxidation, the electronic configuration of the ions is less ambiguous:
#Cr^(2+):# #1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(4)#
#Cr^(3+):# #1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(3)#
The metal ions respectively (and necessarily) have 22, and 21 electrons. Why #"necessarily"#?
