First ionisation energy
The first ionisation energy of an element is the energy required to remove one electron from each of a mole of free gaseous atoms of that element.
How is it written as an equation?
M(g) --> M+(g) + e-
You must include state symbols (always a gas) and the charges of the ion and the electron.
Trends in first ionisation energies
The greater the number of protons, the greater the attraction of the electrons to the nucleus and the harder it is to remove the electrons. The nuber of protons in the nucleus is known as the nuclear charge.
This added nucleur charge is partly cancelled out by the other electrons int he ato. Each inner shell and inner sub-sell electron canels out one unit of charge rom the nucleus. This is known as shielding.
The outermost electrons in the ato only feel the residual positive charge after all inner shell and inner sub-shell electrons have cancelled out much of the nuclear charge. This residual positive charge is known as the effective nuclear charge.
Electrons repel each other. The degree of repulsion between the outmost electrons affects the ease with which electrons can be removed.
Trends in the periodic table
Trends in the first few elements
Hydrogen is a very small atom, and the single electron is close to the nucleus and is therefore strongly attracted to it. There are no electrons shielding it from the nucleus, so it has a higher first ionisation energy. Helium has a higher ionisation energy than hydrogen because it has a higher nuclear charge due to having an extra proton. It has no more shielding than hydrogen does, so it requires more energy than hydrogen to remove its outermost electron. Lithiums outer electron is in the second energy level (2s1) so it is much more distant form the nucleus. The electron is shielded by the electrons in the first energy level, so it has a lower first ionisation energy.
Trends in periods 2 and 3
In general, ionisation energies increase across a period.
In the whole of period 2, the outer electrons are in 2-level orbitals (2s or 2p).
These are all the same distances from the nucleus, and are shielded by the same number of electrons.
The difference is the increasing number of protons across the period. It causes a greater attraction between the nucleus and the electrons and so it increases the ionisation energies.
In period 3, the trend remains: the ionisation energy increases across the period. All the electrons being removed are in the third level and are shielded by the same number of electrons. They is still an increasing nuclear charge though, and this is what causes the higher ionisation energies.
Drops in ionisation energies
The ionisation energies drops between groups 2 and 3.
Boron has one more proton than beryllium, but boron's outer electron is in a 2p orbital, not a 2s. 2p orbitals have higher energy than the 2s orbital, and the electron is found further from the nucleus.
The increased distance results in a reduced attraction and so a reduced ionisation energy.
The 2p orbital is shielded by the 1s2 , but also by the electrons 2s2 electrons, to some extent. This reduces the pull from the nucleus and also reduces the ionisation energy.
More drops in ionisation energies
There is an increase in proton number from group 5 to group 6 - so why does the ionisation energy decrease?
The shielding is the same and the electron is being removed from the same orbital. What is different is that the electron is being removed from the 2px2 pair (in the case of oxygen), so the repulsion between the two electrons in this orbital results in the electron being easier to remove than it would otherwise be.
Trends in the transition series
Except for zinc, all of the transition elements have very similar ionisation energies. Each of these elements lose the electron in the 4s orbital.
The number of protons in the nucleus increases, but so does the number of 3d electrons. The 3d electrons shield the elements, and the extra proton and extra shielding cancel each other out.
Zinc has an extra proton, so the attraction is greater than in all other transition elements. There is a degree of repulsion between the paired up electrons in the 4s orbital, but it is not enough outweight the effect of the extra proton in this case.