Na and Ar electron configuration

Determine the electron configuration ($1s^2,…$) for sodium ($Z=11$) and for the rare gas argon ($Z=18$). Discuss the radii if you compare Na and Ar.  Compare the ionization potentials between Na and Ar.

For Z=11 the electron shell configuration is $Na= 1s^2,2s^2,2p^6,3s^1$

For Z=18 the electron shell configuration is $Ar= 1s^2,2s^2,2p^6,3s^2,3p^6$

For many electron atoms, the radius of the atom is given by the mean radius of the last energy level, in this case the third energy level. From quantum mechanics one can write this radius as

$<r_n>=<\psi_{n l m}|r|\psi_{n l m}>=<R_{n l}(r)|r|R_{n l}(r)>=\int {R_{n l}^2*r*(r^2 d r)}$

Where $r^2*d r$ comes from the element of volume written in spherical coordinates. If one makes the computations for the case of $Na(n=3, l=0)$ and $Ar(n=3,l=1)$ the value of the mean radius will depend just on the principal quantum number. A simpler explanation is the following: the radius of the atom is about the radius of the last electron energy shell $(n=3)$ derived from the Bohr


$r_n=(n^2*\hbar^2)/(Z m k e^2)$

Since Ar has more electrons (and thus nuclear positive charge $Z$) than Na it will have a lower radius. All the elements in the 3rd group that have the electrons on the last energy level will have about the same radius corresponding to $n=3$ energy level, but at the same time the radius will decrease with increasing the $Z$ number (the nuclear charge). An intuitive view is that this increasing positive charge is attracting more the exterior electrons (from n=3 level) towards the nucleus. The same discussion is valid for the ionization energies.  The energy of level n can be approximated from the Bohr model with

$E_n=-(Z^2*E_1)/n^2 =-(Z^2*13.6 eV)/n^2$

The ionization energy is the absolute value of this energy so is about the same for all elements in group 3 $(n=3)$ but increasing from Na to Ar. Intuitively the positive increasing nucleus charge $(Z)$ is keeping more tightly the outer electrons (from the last shell) bonded in the atom, so one

will need more energy to ionize the Ar atom.