Filling up electronic orbitals

The inert elements in periodic table are noble gases with number of electrons \( el=2,10,18,36,54,86,118 \). Here I present how to obtain \( el \). In this case we need two quantum numbers \( n,\ell \)   (principal, angular). \( n=1,2,3,\ldots ;\ell =0,1,2,\ldots ,n-1 \). In periodic table, practically we use \( \ell \)   up to 3, i.e. \( \ell =0,1,2,3=s,p,d,f \)  .

Step 1 : calculate \( f(n,\ell )=n+\ell \)   for \( n=1,2,\ldots ,8 \) and \( \ell =0,1,\ldots ,n-1\) ; \( \ell _{m a x}=3 \), sort by \( n \) then \( \ell \)

Step 2 : sort by \( f(n,\ell ) \), then \( n \) then \( \ell \)

Step 3 : calculate maximum number of electron \( el=2 (2 l+1) \)

Step 4 : summarise \(el\), and we get the noble gases electron numbers.

If an orbital is half-filled or less, then the electrons tend to have their spin aligned. However, there are some anomalous electron configurations, the electron spin aligns accoss orbitals^[1], e.g.

 

	should be	actual
Cr	\( [Ar]4s^23d^4 \)	\( [Ar]4s^13d^5 \)
Cu	\( [Ar]4s^23d^9 \)	\( [Ar]4s^13d^{10} \)
Mo	\([Kr]5s^24d^4\)	\([Kr]5s^14d^5\)
Ru	\([Kr]5s^24d^6\)	\([Kr]5s^14d^7\)
Rh	\([Kr]5s^24d^7\)	\([Kr]5s^14d^8\)
Pd	\([Kr]5s^24d^8\)	\([Kr]4d^{10}\)
Ag	\([K r]5s^24d^9\)	\([K r]5s^14d^{10}\)
Pt	\([Xe]6s^24f^{14}5d^8\)	\([Xe]6s^24f^{14}5d^9\)

 

^[1] TL Brown et al, Chemistry the Central Science, 12^th ed, Prentice Hall, 2012, p. 236.