Question

How is the energy level of an atom's valence electrons related to its period in the periodic table?

Answer 1

There is no strict relationship, but for NON-transition metals (i.e. non-d-block, non-f-block), there is one.

Valence electrons are then USUALLY listed after the noble gas core, within reason (e.g. tungsten, `"W"`, probably doesn't have `20` valence electrons, but up to `6` instead).

---

For NON-transition metals, the energy level `n` is given by its period number, i.e. its row number. Being on the fourth row of the periodic table means your valence electrons would be the `4s` and `4p` electrons, `n = 4`.

Two examples:

`"Ca"`, with configuration `[Ar]color(blue)(4s^2)`. This has `bb2` valence electrons.

`"P"`, with configuration `[Ne]color(blue)(3s^2 3p^3)`. This has `bb5` valence electrons.

EXCEPTIONS BELOW!

Transition metals on the other hand have easy access to occupied `(n-1)d` orbitals, and thus have the electrons in those orbitals included in their set of valence electrons.

Three examples:

`"Sc"` (scandium), with configuration `[Ar] color(blue)(3d^1 4s^2)`. This is why `"Sc"` can have a maximum oxidation state of `+3` (e.g. in `"ScCl"_3`); it has `bb3` valence electrons.

`"W"` (tungsten), with configuration `[Xe]color(red)(4f^14) color(blue)(5d^4 6s^2)` --- note that the `4f` electrons are hardly used, even though they are listed after "`[Xe]`", the noble gas core.
`"W"` is commonly going to have a `+6` maximum oxidation state (e.g. in `"WO"_3`), which means it probably has `bb6` valence electrons most of the time.

`"Os"` (osmium), with configuration `[Xe]color(red)(4f^14) color(blue)(5d^6 6s^2)`. It has up to `bb8` valence electrons, e.g. in `"OsO"_4`.

Some heavy `bb(f)`-block metals (mainly lanthanides and actinides) also have access to occupied `(n-2)f` orbitals too, and those electrons might also be included in their set of valence electrons... they might even not have `(n-1)d` valence electrons sometimes.

Three examples:

`"Pa"` (protactinium), with configuration `[Rn]color(blue)(5f^2 6d^1 7s^2)`. This is why `"Pa"` usually has a `+5` oxidation state in its compounds (such as `"Pa"_2"O"_5`); it has `bb5` valence electrons.

`"Bk"` (berkelium), with configuration `[Rn]color(red)(5f^9) color(blue)(7s^2)` --- note that `11` valence electrons would be insane. In berkelium, only some of the `5f` electrons are considered valence, depending on context.
Since `+4` is the highest known easily-accessible oxidation state (e.g. in `"BkO"_2`), `"Bk"` has around `bb4` valence electrons (not `2`!).

`"Gd"` (gadolinium), with configuration `[Xe] color(red)(4f^7) color(blue)(5d^1 6s^2)`. It most reasonably has `bb3` valence electrons, and indeed, its highest oxidation state is usually `+3` (such as in `"Gd"_2"O"_3`).