Answer:
expected configuration for Cr is
1s2 2s2 2p6 3s2 3p6 4s2 3d4
but in real, it is
1s2 2s2 2p6 3s2 3p6 4s1 3d5
electron from 4s orbital jumps to 3d orbital to get stable configuration.
so the last electron comes in 3d orbital as filling of 3d takes place after filling of 4s orbital.
Hence,quantum numbers for last electron in Cr is :-
n = 3
l = 2
m = +2
s = +1/2
Step-by-step explanation: