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 energy varies considerably
for different elements but it is influenced by 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, this will obviously increase the amount of energy needed to remove them.
The further away the electrons
are from the nucleus then the easier they will be to remove since
the electrostatic force of attraction from the positively charged protons
in the
nucleus and the negatively charged electrons in the shells will decrease with distance.
The last factor to consider is the amount of shielding. The
electrons in the valence shell (outer shell) will not feel
the full electrostatic attractive force from the positively charged nucleus because the
inner or coreelectrons
will effective shield or
screen the nucleus from them.
Shielding
This screening or shielding effect by the inner or core electrons partly cancel out some of the positive charge from the nucleus that the outer valence electrons feel, this shielding works simply because the inner core electrons are closer to the nucleus|.
This shielding effect
will reduce the size of the electrostatic attractive force
from
the nucleus that the outer electrons feel; this means that it will require less energy
to remove these shieldedelectrons. A phrase you may hear when discussing shielding is effective nuclear charge
effective nuclear charge (Zeff), this is simply the net or overall positive charge experienced by an electron, taking into account both the actual nuclear charge and the shielding effect of the inner or core electrons.
Calculating the shielding in an atom of sodium
As a simple example of shielding consider an atom of sodium, with an electron configuration of: 1s22s22p63s1, the 10 core electrons occupy the 1s, 2s and 2p sub-levels and it is these core electrons which provide the shielding; so a simple calculation to find the effective nuclear charge (Zeff) that the 3s1electron feels would be:
Zeff = Z-S
where:
Z= nuclear charge = +11
S= number of core electrons = 10
so Zeff = 11-10= +1
The low value for Zeff explains why sodium has a relatively low first ionisation energy; however you should be aware that the actual effective nuclear charge ( Zeff) felt by the 3s electron will be larger than the +1 value calculated above simply because the inner core electrons don't completely cancel the nuclear charge.
Understanding Shielding and Effective Nuclear Charge — Group 2 (Be, Mg, Ca)
As a simple example consider the group 2 elements Be, Mg and Ca; now each of these elements has 2 outer valence electrons and the remaining electrons present can be considered as core electrons which will shield the nuclear charge from the outer valence electrons. This means that in each case the outer valence electrons will experience an effective nuclear charge (Zeff) of 2+; this is outlined in the activity below:
Number of protons in the nucleus (equals electrons in a neutral atom).
S (core electrons shown)
Grey dots representing inner/core electrons that shield the outer electrons from the full nuclear charge.
For A-level here we take: Be → S=2 (1s² core); Mg → S=10 (Ne core); Ca → S=18 (Ar core).
Zeff (effective nuclear charge)
The pull of the nucleus that outer electrons actually feel after shielding. We display Z − S.
In Group 2 this gives Be=2, Mg=2, Ca=2 (roughly constant down the group).
Zeff and shielding
It should be mentioned that the picture of shielding described above is a little simplistic but it can act as a useful starting point and is fine for A-level chemistry, if you wish to obtain more accurate values for the shielding effects of electrons within each principal energy level and sub-level a quick internet or YouTube search on Slater values maybe useful, but it is also worth mentioning that knowledge of these values or how to calculate them then are not covered in the A-level specification, but they are relatively easy to use and learn and may provide a better understanding of the shielding effects of electrons.
What is shielding?
The reduction in the attraction between the nucleus and an outer electron, caused by the presence of inner-shell electrons.
How does shielding change across period 3 in the periodic table?
It remains fairly contant because as you cross period 3 most of the electrons are being added to the same sub-level (3p), these electrons on average are at the same distance from the nucleus, so they don’t shield each other effectively.
What is meant by the phrase: effective nuclear charge?
The effective nuclear charge is the net positive charge experienced by an electron in an atom, taking into account the actual nuclear charge and the shielding effect of inner core electrons.
It is also worth noting that electrons in the same principal energy level or shell shield each other very poorly. So for example as you cross a period in the periodic table the strength of the shielding effect does not vary by much- this is outlined in the two images shown above in the image slider.
Trends in the ionisation energy for the group2 elements
The values for 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 first 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 principal energy level or 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 chargeincreases
as we descend group 2 but the increasing nuclear charge is offset
by the fact
that the electrons in the valence
ns sub-level are further from the nucleus and shielded from the nuclear charge.
We have also seen that due to shielding effects the effective nuclear charge (Zeff) that each valence electron feels is 2+ but the atoms are obviously getting larger as we descend group 2.
Calculating the effective nuclear charge for group 2 elements
We can carry out a very rough and simplistic calculation similar to the one above for sodium to get an idea of the actual effective nuclear charge
that
the valence shell electrons in the group 2 elements will feel by simply subtracting
the number of electrons in the lower electron shells; the core electrons from
the nuclear charge e.g.
beryllium (Be) has 4 protons, so its nuclear charge is 4+, now beryllium also has 4 electrons in total with an electron configuration of 1s22s2.
There are 2 valence electrons in the 2s sub-shell and with these electrons being in
the same sub-shell they will shield each other
weakly, so we will assume all the shielding
comes from the inner
1s2electrons. These 2 electrons can shield 2 protons. This means that the outer electrons
in theory 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+
From the information in the table we can see that the valence electrons in all the 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 principal energy levels or
electron energy levels and sub-levels or sub-shells within atoms. So far
we have only considered the enthalpy changes for the firstionisation 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
secondionisation 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 ionincreases with each additional electron removed.
As an example consider the trend in the ionisation energies for the group III metal aluminium; atomic number 13 and with an
electronic configuration: 1s22s22p63s23p1.
The ionisation energy required to remove the first 7
electrons from aluminium are 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 since it is the highest
in energy and will require
the least amount of energy
to be removed as it is furthest from the positively charged nucleus and will be shielded
by all the inner core electrons in the inner electron shells and sub-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 secondionisation energy of aluminium
is 1820 kJ/mol, this is 1242KJ/mol more energy than the first
ionisation energy.
The third ionisation energy of aluminium requires 2750 kJ/mol of energy. At this stage, the remaining electron is a 3s electron, which is being removed from an ion with a 2+ charge. The increased ionisation energy, 930 kJ/mol higher than the second ionisation energy, can be partly put down to the higher positive charge of the ion. The 2+ ion has a larger effective nuclear charge, which means the remaining electrons are more strongly attracted to the positively charged nucleus.
Additionally, as the electrons are removed, the radius of the ion decreases due to reduced electron-electron repulsion. The smaller size of the ion means the remaining electrons are closer to the nucleus and more tightly bound. Therefore, more energy is required to remove these electrons.
The fourthionisation energy of
aluminium shows a dramatic and massive
increase in energy over the third ionisation energy.
A massive 8820kJ/mol of extra
energy is required to remove
the 4thelectron 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 reactions will lose 3 electrons, these three electrons will be the
valence electrons in the third electron shell, once these three electrons have been removed then we are left with an ion with a
noble gas configuration. Removing more
electrons will require removing the inner core electrons
and this will require much more energy.
More evidence for shells and sub-shells
As a further example consider the alkali metal sodium, which has an electron configuration of 1s22s22p63s1 .
Sodium has 1 valence electronin the 3s sub-shell. Once this electron
is removed the sodium ion (Na+) formed will have a noble
gas (np6) electron configuration, in this case the noble gas
will be neon. Removal
of a further electron will mean removing an electron
from the
second electron shell, one of the electrons in the 2p sub-level or sub-shell would be removed.
These inner or core electrons are much closer
to the nucleus and will
be much more tightly held by the electrostatic attraction to the positively charged nucleus;
this coupled with the fact that we will be
removing an electron from a smaller positively chargedsodium ion means much more energy will be
needed.
The firstionisation energy of sodium is 496 kJ/mol while the
secondionisation energy is 4560 kJ/mol, quite an increase! This large increase in energy would be
good evidence for the existence of electron shells within atoms.
A similar
pattern is found with
the group II metal magnesium (Mg), which has the electron configuration: 1s2222p63s2. Magnesium has
2 valence electrons
in the outer 3s sub-shell. You should
be able to predict that removing the first two electrons, that is the valence electrons
that would normally be lost
in a chemical reaction will require energy. Once these
two electrons are lost then magnesium
will have a noble gas electron configuration (the same as Neon).
However to remove a thirdelectron would involve removing one of the electrons 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 remove 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
Flashcards self-check
Use the flashcards below to review and test your understanding of the key terms on ionisation energies.
State whether the first ionisation process is endothermic or exothermic, and explain why.
Endothermic – energy must be supplied to overcome the electrostatic attraction between the nucleus and the electron.
List the three main factors that affect the size of an atom's ionisation energy
1. Nuclear charge
2. Distance of the electron from the nucleus
3. Amount of shielding by inner electrons
What is meant by shielding?
The reduction in the attraction between the nucleus and an outer electron due to repulsion by inner electrons.
Why does shielding remain fairly constant across Period 3?
Because electrons are added to the same principal energy level and do not shield each other effectively.
Define effective nuclear charge (Zeff).
The net positive charge experienced by an electron, taking into account both nuclear charge and shielding by inner electrons.
Using sodium as an example, explain why its first ionisation energy is low.
Sodium has +11 nuclear charge and 10 inner electrons that shield the nucleus. The outer 3s electron feels an effective nuclear charge of about +1.
Describe the trend in first ionisation energies down Group 2.
They decrease because although nuclear charge increases, outer electrons are further from the nucleus and shielding increases, reducing the attraction.
The ion becomes more positively charged after electrons removed, radius decreases, remaining electrons feel a stronger effective nuclear charge, more energy is needed.
How do successive ionisation energies provide evidence for the existence of electron shells?
Large jumps in successive ionisation energies show when an electron is being removed from an inner shell closer to the nucleus
Explain how the ionisation energies of magnesium support the existence of subshells.
The 1st and 2nd ionisation energies are low and close together, electrons are from the 3s subshell, the 3rd ionisation energy is high, electron lost from the inner 2p subshell.
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 formed are also decreasing in size
and with the effective nuclear charge increasing it will require more energy to remove the
electrons from the increasing strong attractive pull of the nucleus.
This shielding effect
will reduce the size of the electrostatic attractive force
from
the nucleus that the outer electrons feel; this means that it will require less energy
to remove these shielded electrons. A phrase you may hear when discussing shielding is effective nuclear charge
effective nuclear charge (Zeff), this is simply the net or overall positive charge experienced by an electron, taking into account both the actual nuclear charge and the shielding effect of the inner or core electrons.
As a further example consider the alkali metal sodium, which has an electron configuration of 1s22s22p63s1 .
Sodium has 1 valence electron in the 3s sub-shell. Once this electron
is removed the sodium ion (Na+) formed will have a noble
gas (np6) electron configuration, in this case the noble gas
will be neon. Removal
of a further electron will mean removing an electron
from the
second electron shell, one of the electrons in the 2p sub-level or sub-shell would be removed.
These inner or core electrons are much closer
to the nucleus and will
be much more tightly held by the electrostatic attraction to the positively charged nucleus;
this coupled with the fact that we will be
removing an electron from a smaller positively charged
sodium ion means much more energy will be
needed.
The first ionisation energy of sodium is 496 kJ/mol while the
second ionisation energy is 4560 kJ/mol, quite an increase! This large increase in energy would be
good evidence for the existence of electron shells within atoms.
