⚽ Goal: calculate the ground state configuration of each atom
👑 Properties:
The elements are in order of atomic number (number of protons in the nucleus) $Z$
The rows are called periods
The first of the longer rows (from scandium to zinc) are called transition metals (although they are not a group they have similar properties_
The columns are called groups and generally have similar chemical properties
The last column are called noble or inert gases
The $\rm Li^{+}$ ion is helium-like, its ground state is like parahelium with two electrons in an $S=0$ antisymmetric spin state and both in the $1s$ state.
We use the notation
$$ \mathrm{He}=(1s)^2 $$
to indicate the structure of helium is two $1s$ electrons
Due to the Pauli exclusion principle the $1s$ orbital is now full and the third electron that is added must settle into one of the $n=2$ states $2s$ or $2p$.
The $1s$ state is spherically symmetric it has a mean $r$ value $\langle r\rangle\sim a_0/Z=a_0/3$ thus as the third electron approaches a spherically-symmetric electric field from a total charge $(Z-2)e$.
Thus when the third electron settles into the $2s$ or $2p$ orbital it is a good approximation to say that it is the orbital of a hydrogen atom.
The $n=2$ orbitals have $\langle r\rangle\sim4a_0$ a factor of $12$ times larger than the $1s$ orbitals. There is therefore almost no overlap between the two levels.
🗒️ Note: This can be called the “shell” model: the $n=2$ electron is in a completely different shell far away from the two $1s$ electrons
The hydrogen wavefunctions go like $\psi\sim r^\ell$ for small $r$. Thus, while both orbitals have the electron probability concentrated around $r\sim4a_0$, in the $2s$ orbital, the electron spends a higher fraction of its time at small radii “inside the shell” of the $1s$ electron.
💎 Conclusion: the energy of $2s$ state is slightly lowered by the small fraction of time the electron experiences the full $Z=3$ charge rather than the shielded charge $Z-2$
The first electron occupies the $2s$ orbital
The ground state thus is
$$ \mathrm{Li} = (1s)^2(2s) $$
or
$$ \mathrm{Li} = (\mathrm{He})(2s) $$
the helium structure plus one $2s$ electron
For beryllium $(Z=4)$ the first two electrons fill the $1s$ orbital the next two can both reach the lowest energy by occupying the $2s$ orbital so we have $(1s)^2(2s)^2$
Continuing the pattern we get
$$ \begin{aligned} \mathrm{Be} &= (\mathrm{He})(2s)^2 \\ \mathrm{B} &= (\mathrm{He})(2s)^2(2p) \\ \mathrm{C} &= (\mathrm{He})(2s)^2(2p)^2 \\ \mathrm{N} &= (\mathrm{He})(2s)^2(2p)^3 \\ \mathrm{O} &= (\mathrm{He})(2s)^2(2p)^4 \\ \mathrm{F} &= (\mathrm{He})(2s)^2(2p)^5 \\ \mathrm{Ne} &= (\mathrm{He})(2s)^2(2p)^6\end{aligned} $$
$$ \sum_m\Bigl|Y^m_{\ell}(\theta,\phi)\Bigr|^2=\frac1{4\pi} \qquad\text{for all $\ell$} $$
This means that a full orbital is spherically symmetric. Thus although one electron in a $2p$ state is not spherically symmetric, once all six states are occupied, the total probability distribution of the six electrons is spherically symmetric
💎 Conclusion: thus sodium’s eleventh electron experiences a total charge of $(Z-10)e=1e$ since the two $n=1$ electrons are at $\langle r\rangle\sim a_0/11\sim0.1a_0$ the eight $n=2$ electrons are at $\langle r\rangle\sim 4a_0/9\sim0.4a_0$ and the $n=3$ orbitals have $\langle r\rangle\sim 9a_0$
The next eight atoms are
$$ \begin{aligned} \mathrm{Na} &= (\mathrm{Ne})(3s) \\ \mathrm{Mg} &= (\mathrm{Ne})(3s)^2 \\ \mathrm{Al} &= (\mathrm{Ne})(3s)^2(3p) \\ \mathrm{Si} &= (\mathrm{Ne})(3s)^2(3p)^2 \\ \mathrm{P} &= (\mathrm{Ne})(3s)^2(3p)^3 \\ \mathrm{S} &= (\mathrm{Ne})(3s)^2(3p)^4 \\ \mathrm{Cl} &= (\mathrm{Ne})(3s)^2(3p)^5 \\ \mathrm{Ar} &= (\mathrm{Ne})(3s)^2(3p)^6\end{aligned} $$
🗒️ Note: elements in the same group have similar chemical properties. The inner core of electrons is very small and tightly bound and only the outer shell is involved in interactions with other atoms.