Define first ionization energy or ionization energy:
The first ionization energy is the energy required to remove one mole of
electrons from one mole of gaseous atoms to form one mole of gaseous positive
ions.
General Equation:
X(g) → X+(g) + e−
Where:
X(g) is a gaseous atom.
X⁺(g) is a gaseous ion with +1 charge.
e⁻ is the electron removed.
Important points:
The atom must be in the gaseous state.
The process involves removing one electron from each atom.
It is measured in kilojoules per mole (kJ/mol).
For example, first ionization energy of calcium in kJ mol⁻¹.
Ca (g) → Ca⁺ (g) + e⁻; ΔHi1 = +590 kJ mol⁻¹
If a second electron is removed from the gaseous 1+ ions, it is the 2nd ionization
energy, ΔHi2.
Ca⁺ (g) → Ca²⁺ (g) + e⁻; ΔHi2 = +1150 kJ mol⁻¹
The 2nd ionization energy, ΔHi2 is the energy needed to remove one electron from
each gaseous 1+ ion in one mole of the ions to form one mole of gaseous 2+ ion.
The continuous removal of electrons until the nucleus is left only will result in
successive ionization energies.
Factors affecting the ionization energy
1) Charge on the nucleus (Number of proton) : The greater the number of proton
in the nucleus, the greater the amount of positive charge. The greater the positive
charge, the greater the attractive force between the nucleus and outer electrons. More
energy is needed to overcome the attractive force. So, the ionization energy is
higher. - The greater the nuclear charge, the higher the ionization energy.
2) Distance between nucleus and outer electrons (Size of atom/ion) : The larger
the size of the atom, the greater the distance between the nucleus and the outer
electrons. - The greater the distance between the nucleus and the outer electrons, the
weaker the attractive force between nucleus and outer electrons. - - Furthermore, the
outer electrons experience greater shielding effect by the inner electrons. Less
energy is required to overcome the attractive force. So, the ionization energy is
lower. - The greater the distance between nucleus and outer electrons, the lower
the ionization energy.
3) Shielding effect by the inner electrons: All electrons are negatively-charged, so
they repel each other. Electrons in full inner shells will repel the outer electrons and
so prevent the full nuclear charge being felt by the outer electrons. This is called
shielding or screening. - - The greater the shielding effect, the weaker the attractive
force between the nucleus and outer electrons. Less energy is required to overcome
the attractive force. So, the ionization energy is lower. - The greater the shielding
effect, the lower the ionization energy.
Trends in Ionization Energies
1. Across a Period (→ left to right):
Trend: Ionization energy increases across a period.
Explanation:
Nuclear charge increases (more protons are added).
Electrons are added to the same energy level, so shielding stays roughly
the same.
The atomic radius decreases → electrons are held more tightly.
Therefore, more energy is needed to remove an electron.
2. Down a Group (↓ top to bottom):
Trend: Ionization energy decreases down a group.
Explanation:
Atomic radius increases (more electron shells).
Shielding increases (inner electrons block the outer ones from the nucleus).
Although nuclear charge increases, the effect is outweighed by shielding and
distance.
So, less energy is required to remove the outermost electron.
Exceptions:
Some small dips occur in the trend across a period:
1. The drop between (Be-B) and (Mg-Al) :
There is a slight decrease in first ionization energy between beryllium-born and
magnesium-aluminium because the fifth electron in boron is located in the 2p sub-
shell, which is slightly further away from the nucleus. The outer electron in boron is
shielded by the 1s² as well as 2s² electrons.
Be: 1s²2s² B: 1s²2s²2p¹
The decrease in first ionization energy between magnesium and aluminium has the
same reason, except that everything is happening at the third energy level.
Mg: 1s²2s²2p⁶3s² Al: 1s²2s²2p⁶3s²3p¹
2. The drop between (N-O) and (P-S)
There is a slight decrease in first ionization energy between nitrogen-oxygen and
phosphorus-sulphur because the electron being removed in oxygen is from the
orbital which contains a pair of electrons. The extra repulsion between the pair of
electrons results in less energy needed to remove the electron. This is called spin-
pair repulsion.
N: 1s²2s²2px¹2py¹2pz¹ O: 1s²2s²2px²2py¹2pz¹
The decrease in first ionization energy between phosphorus and sulphur has the same
reason, except that everything is happening at the third energy level.
P: 1s²2s²2p⁶3s²3px¹3py¹3pz¹ S: 1s²2s²2p⁶3s²3px²3py¹3pz¹
Successive ionization energies can be used to determine the electronic
configuration of an element by analyzing the jumps in energy required to remove
successive electrons. A large jump in ionization energy indicates that the next
electron is being removed from a lower energy level (closer to the nucleus),
revealing the number of valence electrons and the element's group number.
How it works:
1. Identify the jumps:
Look for significant increases in ionization energy between successive
values. These jumps indicate a change in energy level as you move from removing
valence electrons (outermost shell) to core electrons (inner shells).
2. Determine the number of valence electrons:
The number of electrons removed before the first large jump corresponds to the
number of valence electrons in the atom.
3. Deduce the group number:
For elements in the s and p blocks, the number of valence electrons directly
corresponds to the group number in the periodic table.
4. Infer the electronic configuration:
Knowing the number of valence electrons and the element's period (from the
successive ionization energies), you can deduce the electronic configuration,
including the arrangement of electrons in different energy levels and sublevels.
Example:
If a large jump in ionization energy occurs after removing the second electron, it
means the element has two valence electrons, placing it in Group 2.
Another example using data:
Let's say the first five ionization energies for an element.
Ionization energy KJ/mol
First 577
Second 1820
Third 2740
Fourth 11600
Fifth 14800
A large jump occurs between the 3rd and 4th ionization energies, meaning the
element has 3 valence electrons. This element is likely in Group III.
Therefore, by analyzing the trends in successive ionization energies, one can deduce
the electronic configuration and group number of an element.
Fig: Successive ionization energy
Another Example: Chlorine
The electronic configuration of chlorine is 1s²2s²2p⁶3s²3px²3py²3pz¹.
Between the second and third ionization energy, there is a slight increase in
difference in ionization energy because the first two electrons being removed come
from the orbitals which contain a paired electrons. The extra repulsion between the
electrons result in the ionization energy being lower. There is also a slight increase
in difference in ionization energy between the fifth and sixth electron being
removed because the sixth electron being removed comes from the 3s sub-shell,
which is slightly closer to the nucleus. The drastic increase in ionization energy
between the seventh and eighth electrons suggests that the eighth electron comes
from a principal quantum shell closer to the nucleus.
