Ionisation Energy and Period 3 Trends
Definition of First Ionisation Energy
The energy required to remove one electron from each atom in a mole of gaseous atoms.
Example of first ionisation energy: Example of second ionisation energy:
Mg (g) -> Mg+ (g) + e- Mg+ (g) -> Mg2+ (g) + e-
Trends in Ionisation energy
Successive ionisation energy for an element
Successive
ionisation energy provides evidence that electrons are arranged in shells.
As the number of electrons removed increases more ionisation energy
is required because the shells become increasingly closer to the influence
of the nucleus.
Notice
the large difference in ionisation energy between the 1st
and 2nd electron. This is because electrons are arranged
in shells.
The
first electron removed is the outer electron in the n=3 shell. The second
electron is in the n=2 shell. The n=2 shell is lower in energy as it
is closer to the influence of the nucleus. This means there is decreased
shielding from inner electrons and stronger nuclear attraction.
Trends in first ionisation
energy across the period
General increase across the
period because:
·
The nuclear charge increases;
·
Electrons are removed from the same energy level;
·
Therefore there’s stronger nuclear attraction of the
outer electron so more energy is needed to remove the outer electron.
1. Decrease in ionisation energy between Be and
B because:
- When you look at the elements
subshells Be is 1s12s2 and B 1s22s22p1
- The s subshell is lower
in energy (closer to the nucleus) than the p subshell;
- Therefore less energy
is required to remove the outer electron in Boron despite an increased
nuclear charge.
2. Decrease in ionisation energy between N and
O because:
- When you look at the orbitals
the arrangement is:
N O
- The extra repulsion from
the paired electron in the outer subshell of Oxygen lowers the energy
required to remove the outer electron despite an increased nuclear
charge.
3. Large decrease in ionisation energy between
Ne and Na because:
- The outer electron in
Ne is in the n=2 subshell but the outer electron in Na is in the n=3
subshell;
- There is more shielding
of inner electrons in the Na atom;
- Therefore less energy
is needed to remove the outer electron despite an increased nuclear
charge.
Trends in first ionisation
energy down the group
General
decrease because:
- The outer electrons are
in different energy shells (e.g. n=2, n=3)
- Therefore there is increased
shielding from inner electrons and weaker attraction to outer electrons;
- This means less energy
is needed to remove the outer electron despite and increased nuclear
charge.
Period 3 Trends
Atomic radius
As
you move across the period the atomic radius decreases because the nuclear
charge increases, however the outer electrons are in the same shell
(n=3) and therefore shielding from inner electrons is the same so the
outer electrons are attracted closer to the nucleus.
Na |
Mg |
Al |
Si |
P |
S |
Cl |
Ar |
2,8,1 |
2,8,2 |
2,8,3 |
2,8,4 |
2,8,5 |
2,8,6 |
2,8,7 |
2,8,8 |
Conductivity
Between Na and Al there is increased electrical conductivity as there are more delocalised electrons per cation and therefore there are more electrons to carry the charge.
Melting and boiling points
There
is an increase in melting and boiling points from Na to Al as there
are a greater number of delocalised electrons per atom therefore the
strength of the metallic bonding increases.
Silicon
has a high melting and boiling point because it forms 4 covalent bonds
which need a large amount of energy to break.
Phosphorus,
Why is the boiling point of
Al above Si?
This
is because once Silicon has melted the strong covalent structure has
broken down and therefore it has weaker bonds than molten Aluminium
so has a lower boiling point than ‘expected’.