Today we’ll be going through:
- What is Ionisation Energy?
- Factors affecting Ei
- Trends in Ei
- down groups
- across periods
- Deducing the electronic configurations
- Position of an element in the Periodic Table
What is Ionisation Energy?
The energy needed to remove electrons from gaseous atoms at their ground state (thus IONISING the atom).
More specific definition:
The energy needed to remove 1 mole of electrons from 1 mole of gaseous atoms at their ground state.
It is commonly denoted as Ei or ΔHi, where
1st Ionisation energy = ΔHi1
2nd Ionisation energy = ΔHi2
Measuring Ionisation Energy
- An element is prepared in gaseous state.
- High-speed electrons are fired at an atom.
- The electrons transfer ENERGY & knock out electrons from the atom, one by one.
- The energy supplied to knock out each electron is the IONISATION ENERGY.
When the 1st electron is knocked out, the (initially neutral) atom becomes a +1 ion.
The energy used here is the 1st ionisation energy of the atom.
More energy will be needed to knock out another electron, making the ion have a charge of +2.
The energy used here is the 2nd ionisation energy of the atom, & so on.
Definition of 1st Ionisation Energy
The energy needed to remove 1 mole of electrons from 1 mole of gaseous atoms of an element in their ground state to form 1 mole of gaseous +1 ions.
The first electrons to get knocked out are from the outer shell.
When all of those have been removed, the electrons from the shell below it are the ones to be removed, & so on.
When knocking out all the electrons from the atom until only the nucleus is left, you can measure the energy needed to remove each one.
With this data, a graph of SUCCESSIVE IONISATION ENERGIES of an element can be plotted.
Successive Ionisation Energies of an Element
Remember that you remove the OUTER electrons first.
Notice a few things:
- The energy needed to remove electrons INCREASES as the number of electrons removed INCREASES.
It is more difficult to remove electrons as there are fewer electrons in an atom.
- There is a large jump in energy needed between the 1st & the 2nd
This indicates a difference in SHELL. The 2nd electron was in a shell CLOSER to the nucleus than the 1st electron, thus more energy is required.
- The energies are plotted on a LOGARITHMIC scale.
This means that each number on the scale represents a new POWER of 10.
This is because the difference in ionisation energy between electrons is quite large.
Identifying Elements by Successive Ionisation Energies
The pattern that has large jumps between the shells makes it easy for scientists to identify the electron arrangements in atoms.
Thus, you can use this diagram to identify a certain atom.
All you have to do is to write out the electron arrangement, from outside to inside.
In this case, you can see that the atom has an arrangement of 2.8.1.
1 valence electron, 8 in the shell below it, & 2 in the innermost shell.
Thus, it can be concluded that this element is sodium.
3 Factors Affecting Ionisation Energy
- Size of Nuclear Charge
As the proton number of an atom increases, the positive nuclear charge increases.
The larger this positive charge, the stronger the attraction between the electrons & protons*.
Thus, higher energy would be required to remove the electrons.
- Distance from Nucleus
The further the shell from the nucleus, the weaker the forces of attraction between the electrons & the nucleus.
Thus, lower energy would be required to remove electrons.
- Shielding Effect
Electrons are negatively charged & repel each other.
The electrons in lower shells will repel the electrons in outer shells.
The more electrons in lower shells, the greater this shielding effect.
Thus, the lower the attractive forces between the nucleus & outer electrons.
*However, another thing must be taken into consideration before choosing this as a factor in any specific case:
Effective Nuclear Charge
The DIFFERENCE in charge between the positive PROTONS & the negative ELECTRONS in the INNER shells of an atom.
The negative charges of the inner electrons cancel out most of the positive charge of the protons.
Thus, the valence electron will only feel the NET attraction of the protons.
For example: the nuclear charge of Magnesium is +12.
Its effective nuclear charge is +(12 – 2 – 8) = +2.
When comparing the ionisation energies of different elements, it is important to check if they have different EFFECTIVE nuclear charges.
For example: there won’t be an increase in the 1st IE between magnesium & calcium.
Although Ca (+20) has a larger NUCLEAR CHARGE than Mg (+12), they BOTH have the SAME EFFECTIVE nuclear charge (+2).
In fact, IE DECREASES between Mg & Ca, due to the higher shielding effect.
However, there IS an increase in the 1st IE between Magnesium & Aluminium.
Because Al has BOTH a larger nuclear charge (13) & EFFECTIVE nuclear charge (+3) than Mg.
As you can see, the EFFECTIVE nuclear charge is determined by the number of shells of an element.
Trends of FIRST Ionisation Energies of Elements
Let’s plot the 1st Ionisation Energies of each element in group 1:
You can see a trend: the ionisation energies DECREASE going down groups.
- More shells are being filled
- Valence electron further away from the nucleus
(Atomic radius increases)
- More inner shells being filled
- Greater shielding effect
- Weaker attraction from the nucleus to an electron in the outer shell
- Although Nuclear Charge INCREASES, the EFFECTIVE Nuclear charge is THE SAME
Let’s plot the 1st Ionisation Energies of each element in period 3:
You can see a trend: the ionisation energies INCREASE going across periods.
There are DROPS in certain elements (in this case: Mg to Al & P to S).
Why this trend?
- Nuclear Charge INCREASES
- Number of protons increases
- Force of attraction between valence electron & nucleus increases
- Atomic radius DECREASES slightly
- Higher number of electrons in the (same) outer shell
- Valence electrons are attracted closer to the nucleus
- Same number of shells
- Same number of electrons in inner shells
- Reasonably equal shielding effect
The DROPS (anomalies) are due to:
- Electronic configuration (p-orbitals vs s-orbitals)
- Spin-pair repulsion (p1 vs p2)
Let’s explain these factors:
Anomalies in IE due to Electronic Configuration
If you recall the explanations on orbitals, you should remember that:
A p-orbital is on a HIGHER sub-level than an s-orbital, if they are on the SAME PRINCIPAL energy level.
Thus, an electron in a p-orbital of a certain shell will require LESS energy to remove as the p-orbital has a slightly higher energy level than an s-orbital of the same shell.
In the case of Mg & Al,
|Mg has an electron configuration of1s2 2s2 2p6 3s2||Al has an electron configuration of1s2 2s2 2p6 3s2 3p1|
It is this electron in the 3p orbital of Al that has lower ionisation energy than the 1st electron in the 3s orbital of Mg.
Differences in IE due to Spin-Pair Repulsion
You should remember that in one orbital, there can be a maximum of 2 electrons.
If there are 2 electrons in the same orbital (a PAIR), they will have a slight repulsion towards each other (due to like negative charges).
This means that it is actually EASIER to remove a PAIRED electron than an UNPAIRED one.
Let’s look at an example in terms of orbitals:
|P has an electron configuration of 1s2 2s2 2p6 3s2 3p3||S has an electron configuration of 1s2 2s2 2p6 3s2 3p4|
|Since the p-subshell consists of 3 different p-orbitals (x,y,z), let’s separate each one.
1s2 2s2 2p6 3s2 3px1 3py1 3pz1
|Hund’s Rule says that each of the 3 electrons in the p-subshell will occupy a separate p-orbital.||Since the p-subshell consists of 3 different p-orbitals (x,y,z), let’s separate each one.
1s2 2s2 2p6 3s2 3px2 3py1 3pz1
|There are 4 electrons in the p-subshell.||The ones in 2 of the orbitals are alone, but 1 orbital must have a PAIR.|
|All of these electrons are UNPAIRED, & thus are more DIFFICULT to remove.||2 of the electrons are PAIRED, & thus are EASIER to remove.|
Trend across Periodic Table