Question

In the Cr3+ ion,how many electrons have n+1=4 and ms=-1/2?

Answer 1

Well, I count four, but if that's not your question, you'll have to specify...

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Are you asking how many electrons in `"Cr"^(3+)` in atomic orbitals with principal quantum number `n+1 = 4` have `m_s = -1/2`?

If so...

`n = 3`

corresponds to the `3d`, `3p`, and `3s` orbitals, and `m_s = -1/2` designates the spin-down electrons.

`"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...

`3d`: ZERO electrons with `m_s = -1/2`

`underbrace(ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr)))_(3d)`

`3p`: THREE electrons with `m_s = -1/2` (and three with `+1/2`)

`underbrace(ul(uarr color(blue)(darr))" "ul(uarr color(blue)(darr))" "ul(uarr color(blue)(darr)))_(3p)`

`3s`: ONE electron with `m_s = -1/2` (and one with `+1/2`)

`underbrace(ul(uarr color(blue)(darr)))_(3s)`

So there are `bb4` total...