In the Cr3+ion, how many electrons have ms=+1/2?

1 Answer
Mar 20, 2018

I get 12 spin-up electrons in "Cr"^(3+). How many spin-down electrons are there in "Cr"^(3+)? Did you come up with 9? What orbitals are they in?


"Cr" has the electron configuration:

[Ar]3d^5 4s^1

Take out one 4s and two 3d electrons to get "Cr"^(3+):

[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 m_s = +1/2, so... we have one electron for EVERY atomic orbital occupied.

With l =0,1,2 for s,p,d orbitals, there are then...

  • overbrace(2(0) + 1)^("1s orbital(s)") = bb1xx1s electron of m_s = +1/2
  • overbrace(2(0) + 1)^("2s orbital(s)") = bb1xx2s electron of m_s = +1/2
  • overbrace(2(1) + 1)^("2p orbitals") = bb3xx2p electrons of m_s = +1/2
  • overbrace(2(0) + 1)^("3s orbital(s)") = bb1xx3s electron of m_s = +1/2
  • overbrace(2(1) + 1)^("3p orbitals") = bb3xx3p electrons of m_s = +1/2
  • overbrace([2(2) + 1])^"Total 3d orbitals" - overbrace(2)^"2 empty 3d orbitals" = bb3xx3d electrons of m_s = +1/2

Thus, 1+1+3+1+3+3 = color(blue)(bb12) electrons in "Cr"^(3+) that are spin-up.