Ionisation energy
The first ionisation energy is the
amount of energy required to remove 1 mole of electrons from an isolated atom in the
gaseous state. It can be represented by the equation:
X(g) → X+(g) + e
This process will obviously be an endothermic one, energy will have to be provided to
remove the electron from
the attractive force it feels from the positively charged nucleus.
The first ionisation energies varies considerable
for different elements. The three factors
that you must consider when discussing ionisation energy
are:
- the size of the nuclear charge, the larger the number of
positively charged protons present in the nucleus
then the greater will be the attraction for the electrons.
- The further away the electrons
are from the nucleus then the easier they will be to remove since
the force of attraction from the positvely charged protons
in the
nucleus will decrease with distance.
- The last factor to consider is shielding. The
electrons in the valence shell (outer shell) will not feel
the full affect of the positively charged nucleus because the
inner or core electrons
will effective shield or
screen the nucleus from them. This shielding effect
will reduce the size of the attractive force
from
the nucleus that the electrons feel and so it will require less energy
to remove them.
Trends in the ionisation energy for the group2 elements
The first ionisation energies for the elements
in group 2 of the periodic table are shown in the bar chart below.
The general trend is fairly obvious, as we go down group 2 from the elements
beryllium to barium the
ionisation energy drops.
To help explain this trend in the ionisation energies
we need to consider the electronic configuration for the
group 2 elements, these are shown in the table below:
| element |
atomic number |
electron configuration |
| beryllium |
4 |
1s22s2 |
| magnesium |
12 |
1s22s22p63s2 |
| calcium |
20 |
1s22s22p63s23p64s2 |
| strontium |
38 |
1s22s22p63s23p64s23d104p65s2 |
| barium |
56 |
1s22s22p63s23p64s23d104p65s24d105p66s2 |
As we move down group 2 from one element to the next a
new electron shell is added and the
electrons in the last shell or
valence
shell are in a higher principal energy level and so will be
further from the nucleus. The
size of the nuclear charge increases
as we descend group 2 but the increasing nuclear charge is offset
by the fact
that the electrons in the valence
ns sub-shell are further from the nucleus.
We can carry out a rough calculation to get an idea of the actual effective nuclear charge
that
the valence shell electrons will feel by simply subtracting
the number of electrons in the lower electron shells from
the nuclear charge e.g.
Be has 4 protons, so its nuclear charge is 4+.
It has 4 electrons in total, 1s22s2.
There are 2 valence electrons in the 2s sub-shell, these being in
the same shell will shield each other very
weakly, so we will assume all the shielding
comes from the inner
1s2 electrons. These 2 electrons
will shield 2 protons. This means that the outer electrons
will feel an effective nuclear charge of only 2+.
We can carry out a similar calculation for all the group 2 elements, as shown in the table below:
| element |
atomic number (nuclear charge) |
number of inner screening electrons |
number of valence electrons |
effective nuclear charge felt by valence electrons |
| beryllium |
4 |
2 |
2 |
2+ |
| magnesium |
12 |
10 |
2 |
2+ |
| calcium |
20 |
18 |
2 |
2+ |
| strontium |
38 |
36 |
2 |
2+ |
| barium |
56 |
54 |
2 |
2+ |
This means that the valence electrons in all group 2 elements
will feel an effective nuclear charge
of 2+, but
of course as we descend the group the distance from the
nucleus to the valence electrons increases greatly,
so much less energy will be required to separate the
outer valence electrons when they are further
from the
positively charged nucleus, which means that the
ionisation energy will get lower
as the atoms in group get larger. We can show this simply as:
Successive ionisation energies
Ionisation energies are a good source of evidence for the presence of
electron shells and sub-shells. So far
we have only considered the enthalpy changes for the first ionisation energy of an element:
X(g) → X+(g) + e
However there is no reason to stop at removing just one electron, we can continue and remove more. The
second ionisation energy of an element can be represented by the change:
X+(g) → X2+(g) + e
This is the enthalpy change (amount of heat energy required) to remove 1 mole
of electrons from 1 mole of gaseous ions.
The third ionisation energy would be:
X2+(g) → X3+(g) + e
As you might expect the ionisation energy required to remove successive
electrons from an increasingly positively charged ions
increases with each additional electron removed.
As As an example consider the ionisation energies for aluminium, atomic number 13, with an
electronic configuration: 1s22s22p63s231.
The ionisation energy required to remove the first 7
electrons from aluminium is shown in the table below.
As you can see the more electrons that are removed
the more energy is required, however it is not a stepwise
or steady increase.
| ionisation energy |
1st |
2nd |
3rd |
4th |
5th |
6th |
7th |
| energy required/kJmol-1 |
578 |
1820 |
2750 |
11 570 |
14 840 |
18 375 |
23 299 |
- The first ionisation energy of aluminium is:
Al(g) → Al+(g) + e
Here the electron in the 3p sub-shell will be removed first. It is the highest
in energy and will require
the least amount of energy
to be removed as it is furthest from the nucleus and will be screened
by all the electrons in the inner shells (1s22s22p63s2).
- The next electron to be removed will be in the 3s sub-shell.
Electrons in the 3s sub-shell will
be closer to the nucleus than those in the 3p sub-shell so will require more energy
to remove them.
We are also trying to remove a negatively charged electron from a
positive charged ion, this will also
increase the amount of energy required
due to electrostatic attraction between the positively charged
ion and the electron. The equation below represents the second
ionisation energy of aluminium:
Al+(g) → Al2+(g) + e
The second ionisation energy of aluminium
is 1820 kJ/mol, this is 1242KJ/mol more energy than the first
ionisation energy.
- The third ionisation energy requires 2750kJ/mol of energy, this time the
remaining 3s electron will
be removed, but it is being removed from an ion with a 2+ charge. This
would partly explain
the increase in energy of 930kJ/mol over the second
ionisation energy. However you also
need to consider the fact that as we remove electrons
the ions will decrease in size, since
there will be less electron-electron repulsion, as we have less electrons present. The
smaller ions mean that the remaining electrons
will be more tightly held by the nucleus.
This means more energy will be required to remove them.
- The fourth ionisation energy of
aluminium shows a quite dramatic and massive
increase in energy over the 3rd ionisation energy.
A massive 8820kJ/mol of extra
energy is required to remove
the 4th electron compared to the third.
Al3+(g)→ Al4+(g) + e
This huge increase in energy is due to
the fact that we are removing an electron
from the 2p sub-shell which is much closer to the nucleus than
the 3s sub-shell. Aluminium
in its chemical reaction will lose 3 electrons, these would be the
valence electrons in the
3rd electron shell, principal quantum number 3,
once we have
remove these 3 electrons then we are left with an ion with a
noble gas configuration. Removing more
electron will require removing the core inner electron
and this will require much more energy.
As a further example consider sodium, Na: 1s22s22p631 .
Sodium has 1 valence electron,
the electron in the 3s sub-shell. Once this electron
is removed the ion formed will have a noble
gas (np6) electron arrangement, in this case the noble gas
will be neon. Removal
of further electron will mean removing an electron
from the
second electron shell, the 2p electrons.
These inner or core electrons are much closer
to the nucleus and will
be much more tightly held by the electrostatic attraction to the nucleus
this coupled with the fact that we will be
removing an electron from a samller positively charged
sodium ion means more energy will be
needed. The 1st
ionisation energy of sodium is 496k/mol the
second is 4560kJ/mol, quite an increase! This would be
good evidence for the existence of electron shells within atoms. A similar
thing is found with
Mg, 1s2222p63s2. Magnesium has
2 valence electrons
in the last shell, the 3s shell. You should
be able to predict that removing the first two electrons, the valence electrons
that would be lost
in the normal chemical reactions of magnesium will require some energy but
not too much. Once these
two electrons are lost then magnesium
will have a noble gas configuration (the same as Neon).
However to remove a third electron would mean removing one from the second principal energy level,
this will require a large increase in energy. The table below give
the values for the first three ionisation energies of magnesium. This data provides clear evidence for electron shells.
Here we have 2 electrons which are relatively
easy to removed followed by a
third which requires a huge increase in energy to remove:
| ionisation energy |
1st |
2nd |
3rd |
| energy required/kJmol-1 |
740 |
1819 |
7737 |
Key Points
- The factors that affect the ionisation energy of a particular atom are:
- The size of the nuclear charge.
- The distance between the electron being removed and the nucleus.
- The amount of shielding.
- The ionisation energies drop as we descend any group in the periodic table
- Successive ionisation energies are always larger simply because there is
less shielding of the nucleus, so the effective nuclear charge is increasing. The ions are also decreasing in size
and with the effective nuclear charge increasing it will require more energy to remove the
electrons from the increasing attractive pull of the nucleus.
Practice questions
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