From H To B

Hydrogen:
z = 1

| n | s | p
L | 2 |  |  |  |
K | 1 | Ý |  | H |

One electron (1s), the ground state is 2S½
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 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 2S½.
- 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 therefore we have a bit higher ionization energy: 5,4 eV or Zeff = 1,3.

Since 1s2 shell is fully occupied, that corresponds to the noble gas therefore it's sometimes written 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 area we have 9,3 eV here. Beryllium isn't so 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) ≡2P½

2P3/2 is also principally possible as the ground state. Nevertheless 2P½-state lays a bit deeper energetically (Hunds rule (2)); that can be described as follows:

L→ ·S®≡ ½ · cSL(J(J + 1) - L(L + 1) - S(S + 1))