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

Jun 23, 2017

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, $\text{W}$, 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 $4 s$ and $4 p$ electrons, $n = 4$.

Two examples:

$\text{Ca}$, with configuration $\left[A r\right] \textcolor{b l u e}{4 {s}^{2}}$. This has $\boldsymbol{2}$ valence electrons.

$\text{P}$, with configuration $\left[N e\right] \textcolor{b l u e}{3 {s}^{2} 3 {p}^{3}}$. This has $\boldsymbol{5}$ valence electrons.

EXCEPTIONS BELOW!

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

Three examples:

$\text{Sc}$ (scandium), with configuration $\left[A r\right] \textcolor{b l u e}{3 {d}^{1} 4 {s}^{2}}$. This is why $\text{Sc}$ can have a maximum oxidation state of $+ 3$ (e.g. in ${\text{ScCl}}_{3}$); it has $\boldsymbol{3}$ valence electrons.

$\text{W}$ (tungsten), with configuration $\left[X e\right] \textcolor{red}{4 {f}^{14}} \textcolor{b l u e}{5 {d}^{4} 6 {s}^{2}}$ --- note that the $4 f$ electrons are hardly used, even though they are listed after "$\left[X e\right]$", the noble gas core.
$\text{W}$ is commonly going to have a $+ 6$ maximum oxidation state (e.g. in ${\text{WO}}_{3}$), which means it probably has $\boldsymbol{6}$ valence electrons most of the time.

$\text{Os}$ (osmium), with configuration $\left[X e\right] \textcolor{red}{4 {f}^{14}} \textcolor{b l u e}{5 {d}^{6} 6 {s}^{2}}$. It has up to $\boldsymbol{8}$ valence electrons, e.g. in ${\text{OsO}}_{4}$.

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

Three examples:

$\text{Pa}$ (protactinium), with configuration $\left[R n\right] \textcolor{b l u e}{5 {f}^{2} 6 {d}^{1} 7 {s}^{2}}$. This is why $\text{Pa}$ usually has a $+ 5$ oxidation state in its compounds (such as ${\text{Pa"_2"O}}_{5}$); it has $\boldsymbol{5}$ valence electrons.

$\text{Bk}$ (berkelium), with configuration $\left[R n\right] \textcolor{red}{5 {f}^{9}} \textcolor{b l u e}{7 {s}^{2}}$ --- note that $11$ valence electrons would be insane. In berkelium, only some of the $5 f$ electrons are considered valence, depending on context.
Since $+ 4$ is the highest known easily-accessible oxidation state (e.g. in ${\text{BkO}}_{2}$), $\text{Bk}$ has around $\boldsymbol{4}$ valence electrons (not $2$!).

$\text{Gd}$ (gadolinium), with configuration $\left[X e\right] \textcolor{red}{4 {f}^{7}} \textcolor{b l u e}{5 {d}^{1} 6 {s}^{2}}$. It most reasonably has $\boldsymbol{3}$ valence electrons, and indeed, its highest oxidation state is usually $+ 3$ (such as in ${\text{Gd"_2"O}}_{3}$).