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

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.