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.