Trends in the ionisation energy for the period 3 elements
Ionisation energy
The first ionisation energy is the
amount of energy required to remove 1 mole of electrons from 1 mole of isolated atoms in the
gaseous state. It can be represented by the equation:
X(g) → X+(g) + e(g)
This process will obviously be an endothermic one since 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 vary 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 positively 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. 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 energies for the period 3 elements
The ionisation energies for the period 3 elements Na-Ar are shown in the graph below. There are a few
observations worth making:
- The general trend in ionisation energy
across period 3 elements is that it increases with increasing nuclear
charge. There are 2 "dips" in this trend; these are with the elements
aluminium and sulfur.
- The 2 dips in the ionisation energies, one as we go from Mg to Al and the other as we go from the
element P to S. The drop in ionisation energy from Mg to Al is due to the fact that the
electron being removed is from a 2s sub-shell in the case of Mg and a
p sub-shell in the case of Al. Since the 3p sub-shell is further from the nucleus it will be higher in
energy and also the electron in the 3p sub-shell is shielded more than an electron in the 3s sub-level, so it requires less energy to remove. That is the 3p electron in aluminium will experience a smaller
effective nuclear charge than the 3s electron in magnesium because it will be
shielded more effectively by the inner core electrons.
We can show this as:
Mg: 1s22s22p63s2 → Mg+: 1s22s22p63s1
Al: 1s22s22p63s23p1 → Al+: 1s22s22p63s2
Since the 3p electron is further from the
nucleus than the 3s electron and it is also shielded
more by
the inner core electrons and so it will require less energy to remove it.
Trends in the ionisation energies of the period 3 elements
The graph opposite shows the trend in the ionisation energies across period 3.
There is also a drop in the ionisation energy
as we go from the element phosphorus to sulfur. If we consider
the electronic configuration of these two elements
then we can easily offer an explanation as to why this drop happens:
P: 1s22s22p63s23p3 → P+: 1s22s22p63s2
S: 1s22s22p63s23p4 → S+: 1s22s22p63s3
In phosphorus the 3p electrons all occupy separate p-orbitals, however in sulfur the
electrons begin to pair up in the p-orbitals.
This pairing up of electrons will introduce some repulsion
between the paired electrons, this means that a filled orbital will be
slightly higher
in energy than a half-filled orbital so less
energy will be needed to remove this one electron.
Now recall Hund's rule of
maximum multiplicity, this rule will require the three p-electrons which remains in the p-orbitals to all have parallel spins and occupy separate orbitals, so the one electron
in the sulfur p-orbitals which has a spin in the opposite direction to the other 3 electrons, will be the one
which is removed. This is shown in the diagram below:
In the sulfur atom the electrons in the p-orbitals begin to pair up. This pairing up will introduce some repulsion between the two electrons in this orbital. This means that when we ionise the sulfur atom and remove one of the electrons in the 3p-orbitals then less energy than expected will be required to remove it
because the other electron in the p-orbital will give it a bit of “a push” and help it leave due to this replusion between them.
Key Points
- The general trend in ionisation energies
across a period is that it increases due to
increasing nuclear charge.
- The size of atoms across a period decreases, the main reason
for this is increasing nuclear charge and the fact that additional electrons are being added to the same principal energy level,
so as we cross a period the atoms also get smaller and with
increasing nuclear charge the ionisation energies
will generally increase.
- There a 2 dips in the first ionisation in period 3, one when we leave the
s-block and enter the p-block, this happens in the move from magnesium to aluminium, here the outer valence electron moves from a 3s-orbital to 3p-orbital which is higher in energy and so requires less energy to remove the electron since there is an increase in nuclear charge and an increase in shielding.
- The second dip in the first ionisation energy in period 3
occurs when we go from a 3p3 to a 3p4
electronic
configuration, this happens when we move from phosphorus to sulfur. The addition pairing
up of
the electrons in the p-sub shell raises its energy
and so makes it easier to remove an electron.
Practice questions
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