In the Cr3+ion, how many electrons have ms=+1/2?
1 Answer
Mar 20, 2018
I get
[Ar]3d^5 4s^1
Take out one
[Ar] 3d^3 = 1s^2 2s^2 2p^6 3s^2 3p^6 3d^3 .
BY CONVENTION, the electron in a singly-occupied orbital is spin-up, i.e. has
With
overbrace(2(0) + 1)^("1s orbital(s)") = bb1xx1s electron ofm_s = +1/2 overbrace(2(0) + 1)^("2s orbital(s)") = bb1xx2s electron ofm_s = +1/2 overbrace(2(1) + 1)^("2p orbitals") = bb3xx2p electrons ofm_s = +1/2 overbrace(2(0) + 1)^("3s orbital(s)") = bb1xx3s electron ofm_s = +1/2 overbrace(2(1) + 1)^("3p orbitals") = bb3xx3p electrons ofm_s = +1/2 overbrace([2(2) + 1])^"Total 3d orbitals" - overbrace(2)^"2 empty 3d orbitals" = bb3xx3d electrons ofm_s = +1/2
Thus,