Ionization Energies
The quantity of energy required to remove an electron form an atom or ion in the gaseous state is called the
ionization energy. In atoms containing more than one electron this energy requirement depends on two
factors, the effective charge of the nucleus and the the average distance of the electron from the nucleus.
The Rydberg equation, originally derived from the Bohr model can be used to describe the energy for a one
electron system such as H, He+ or Li2+, all of which have only one electron.
E = (-2.178 x 10-18 J) (Z2/n2) (Equation 1)
where Z is the effective charge of the nucleus and n refers to the quantum number in which the electron
resides. Usually energies are expressed per mole of atoms, so the Rydberg equation multiplied by
Avogadro’s number becomes:
E = (6.022 x 1023 )(-2.178 x 10-18 J) (Z2/n2) = -1312 (Z2/n2) (Equation 2)
The Rydberg equation calculates the energy released when an electron returns to its ground state. Since the
ionization energy is the energy required to remove an electron from an atom it has a positive sign rather than
a negative sign.
The shielding effect
For atoms with more than one electron the ionization energy does not conform exactly with the energy
calculated from the formula above.The effective ionization energy of the 3s electron of sodium has been
measured experimentally and found to be 4.95 x 102 kJ/mol. If this value is substituted into equation 2 above
and the resulting expression is solved for Z, the effective charge of the particle is +1.84 rather than +11, the
number of protons in the nucleus.
In a sodium atom the 3s electron is separated from the nucleus by 10 intervening electrons. These interior
electrons repel the external 3s electron, reducing the attraction of the nucleus on this external electron. This
phenomena is called shielding.
Suppose that another electron were added to the 3 s orbital as in a magnesium atom (atomic number =12). In
general the electron shielding effect is very small for electrons sharing the same orbital; since they also repel
each other. Therefore they tend to stay as far away from each other as they can. In general core electrons
closer to the nucleus shield the outer electrons from the electron attracting effect of the nucleus. The addition
of an energy level to an atom means that the outermost electrons are further from the nucleus that the
corresponding electron of the element of the previous period. Hence metals, especially the alkali metals and
the alkaline earth metals increase in reactivity as on progresses down a column in the periodic table.
The Penetration Effect
Since there are 10 electrons between the nucleus and the outer electron, we might expect the effective charge
to be 1.00 rather than 1.84. While the 3s electron does spend most of its time on the outside of the atom, there
is some probability that at some point in time it will also be close to the nucleus. In other words the external
3s electron, occasionally penetrates the kernel electrons reducing the effect of shielding and increasing the
effective charge.
Because of this so called penetration effect electrons prefer the s orbital over a p orbital in a given energy
level. In term s of orbital energies an s orbital is always lower than a p orbital, which is lower than a d
orbital, which in turn is lower energy than an f orbital.
The penetration effect can be best understood if we consider the atom from the point of view of a probability
model rather than a simple Bohr model. Although the 3s electron of a sodium atom for example, spends most
of its time far from the nucleus there is a still a small but never the less significant probability that it will be
close to the nucleus. In other words, this ability to penetrate the core electrons reduces the effect of the
shielding and thus increases the effective charge of the 3s electron in a sodium atom. This effect is known as
the penetration effect.
Because the 4s orbital allows for significant penetration, the 4 s orbital is filled in the Aufbau process before
the 3d orbital. The most probable distance of the electron in a 3d orbital is still less than a 4s, but the 4s
orbital allows for a greater penetration effect and hence it is preferred over the 3d orbital.
Analysis of Ionization Energies
The ionization energy is the energy required to remove an electron from a gaseous atom or ion.
X(g) ---------> X+ (g) + eThe atom or ion is assumed to be in the ground state.
Each successive electron that is removed from an atom requires additional energy. The ionization energies
for aluminum for example are given below:
Al (g) -------> Al+ (g) + e- I1 = 580 kJ mol-1
Al+ (g) -------> Al2+ (g) + e- I2 = 1815 kJ mol-1
Al 2+ (g) -------> Al 3+ (g) + e- I3 = 2740 kJ mol-1
Al 3+ (g) -------> Al4+ (g) + e- I4 = 11,600 kJ mol-1
The highest energy electron is always removed first. The energy required to remove the first electron is
called the first ionization energy. The energy required to remove the second electron is called the second
ionization energy. It is always larger than the first ionization energy. After the removal of the first electron
there is in effect one additional proton attracting the electrons, making them more tightly held. This is also
true for the third, fourth, and successive ionization energies.
The valence electrons, those in the highest energy level are more loosely held than the core electrons.
Therefore there is a very large increase when one attempts to remove a core electron. In the above example,
the fourth ionization energy of aluminum is about 4 times greater than the third, but the third is only about
1.5 times greater than the second.
Then electron configuration is 1s2 2s2 2p6 3s2 3p1
The fourth ionization involves the removal of a core 2 p electron. The core electrons are much more tightly
held than the valence electrons.
There is also a smaller but significantly larger difference between the first and second ionization energies for
aluminum than there is between the second and third ionization energies. the first ionization energy
represents the removal of a 3p electron. The second and third ionization energies represent the removal of the
two 3s electrons. Since the 3p electron is in a higher energy sublevel than the 3s electrons it is less tightly
held and therefore easier to remove. There is a very large increase in the ionization energy with the removal
of a core electron. There is a smaller but still significantly large increase in the ionization energy when one
switches from p to s electrons in the valence energy level. This helps to explain why first ionization energy of
magnesium is higher than that of aluminum even though aluminum has 1 more proton.
Finally within the same sublevel paired electrons are easier to remove than unpaired electrons. As a result the
first ionization energy of sulfur is lower than that of phosphorous.even though both involve the removal of a
p electron and sulfur has one more proton. The p electron removed from sulfur is a paired electron, whereas
in phosphorous all of the 3p electrons are unpaired.
The Ionization Energies for Elements in Period 3.
All values are in kJ per mole
Element I1
I2
Na
495
4560
Mg
735
1445
I3
7730
I4
I5
I6
I7
Al
580
1815
2740
11600
Si
780
1575
3220
4350
16100
P
1060
1890
2905
4950
6270
21200
S
1005
2260
3375
4565
6950
8490
27000
Cl
1255
2295
3850
5160
9360
11000
Ar
1527
2665
3945
5770
8780
12000
6560
7230