From H To B

 
 

Hydrogen:
z  =  1
 

	n	s	p
L	2
K	1	Ý		H

One electron (1s), the ground state is ²S_½
The binding energy (Ionization energy) is 13,6 eV

Helium:
z  =  2
 

	n	s	p
L	2
K	1	Ýß		He

2 electrons (1s) (1s), ground state 1s²

Since both electrons have the same main quantum numbers then the ground state should be ¹S₀ (total spin S = ­ + ¯ = 0). If we excite an electron into n = 2, i.e. 2s state (1s) (2s) then both electrons have different main quantum numbers n and total spin can be S = 1. Here we show a rough scheme of Helium energy levels and possible transitions:

Lithium:
z = 3
 

	n	s	p
L	2	Ý
K	1	Ýß		Li

According to Pauli there is no 1s³ configuration but
1s² 2s  º  ²S_½.
­- fulfilled shell with L = 0 and S = 0, J = 0.

If the nucleus has been covered fully by two 1s electrons then the ionization energy would be ^{13,6 eV}/₄ = 3,4 eV. Since the wavefunction of 2s electron penetrates deeper into the nucleus we therefore have a slightly higher ionization energy: 5,4 eV or Z_eff = 1,3.

Since the 1s² shell is fully occupied, we have a noble gas configuration. Therefore, we sometimes write He(2s) for the Li-ground state configuration. Another way is to use capital letters K, L, M, N.... for fully occupied shells, or K(1s) for Li.

Beryllium:
z = 4
 

	n	s	p
L	2	Ýß
K	1	Ýß		Be

(1s²) (2s²) ≡¹S₀

The binding energy would be ^{24,6 eV}/₄ = 6,2 eV (24,6 eV ≡ He). Because of wavefunctions' penetration into the nucleus we have 9,3 eV. Thus Beryllium isn't as chemically inert as He.
 

Energy states of Beryllium

Boron:
z = 5
 

	n	s	p
L	2	Ýß	Ý
K	1	Ýß		B

(1s²) (2s²) (2p) ≡²P_½

²P_3/2 is also principally possible as the ground state. Nevertheless ²P_½-state lies a bit lower energetically (Hunds rule (2)); this can be described as follows:

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