Question

The values of Quantum numbers for the 5th electron of boron and last entering electron of potassium?

Answer 1

Elements on the second row of the periodic table occupy `n = 2` orbitals, and boron has three valence electrons, occupying `2s` and `2p` orbital(s). Potassium on the other hand, is on the fourth row (occupies only an `s` orbital with `n = 4`) and has one valence electron.

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Boron only has 5 electrons, so two of them go into the `1s`, two into the `2s`, and one into a `2p` orbital.

`underbrace(ul(uarr color(white)(darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr)))_(2p)`

`underbrace(ul(uarr darr))_(2s)`

`" "`
`" "`
`" "`
`" "`
`" "`
`" "`

`underbrace(ul(uarr darr))_(1s)`

• Any `color(blue)(2)p` orbital has `color(blue)(n = 2)` (given in the name).
• All `p` atomic orbitals have angular momenta of `color(blue)(l = 1)`, seeing as `l = 0, color(blue)(1), 2, 3 harr s, color(blue)(p), d, f`.
• The z-projection of the angular momentum can only take on values `{-l, -l+1, . . . , l-1, l}`, so `color(blue)(m_l = {-1,0,+1})` are your options.
• The electron can be spin up or down, and being a fermion, it is a spin-half particle. So we have `color(blue)(m_s = +1/2)` (by convention).

When you figure out how to define the `4s` orbital of potassium, you should say what you think the quantum numbers are for the unpaired electron...