In hydrogen all orbitals in the same shell have the same energy:
s
3
2
1
p
d
but once some orbitals contain electrons the energies of other
orbitals are affected due to repulsion
so the p and d orbitals have higher energy than the s orbitals of the same shell:
s
3
p
d
2
You need to copy this diagram
1
Hydrogen
1s1
Helium
1s2
Lithium
1s2 2s1
Beryllium
1s2 2s2
Boron
1s2 2s2 2p1
Carbon
1s2 2s2 2p2
Nitrogen
1s2 2s2 2p3
Oxygen
1s2 2s2 2p4
Try to write down the electronic configurations of :
Neon (z = 10)
1s2 2s2 2p6
Sodium (z = 11)
1s2 2s2 2p6 3s1
Silicon (z = 14)
1s2 2s2 2p6 3s2 3p 2
Potassium (z = 19)
1s2 2s2 2p6 3s2 3p 6 4s 1
You may have expected the final electron in potassium to be in the
3d sub-shell
1s2 2s2 2p6 3s2 3p 6 3d 1
but potassium is so similar to sodium that the electronic structure
should also be similar with a outer s electron
Na
K
2
1s
2
2s
6
2p
1
3s
1s2 2s2 2p6 3s2 3p 6 4s 1
The explanation is that the increase in energy of the 3d sub-shell
puts it higher than the 4s.
4f
d
4
3
2
p
s
1
NB 3d sub-shell higher than the 4s.
Potassium
d
4
3
2
p
s
1
1s2 2s2 2p6 3s2 3p6 4s1
2
8
8
1
This accounts for the odd series of lengths of periods in the
Periodic Table :
2, 8, 8, 18
when the maximum number of electrons per shell is
2, 8, 18, 36
Bromine for example is the 17th element across the PT
but has only 7 outer electrons.
Bromine
d
4
3
2
p
s
1
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5
2
8
18
7