Ionisation Energy

The definition, and the equation it has to generate

First ionisation energy is the energy needed to remove one electron from each atom in one mole of gaseous atoms, forming one mole of gaseous 1+ ions.

The trend across Period 3

First ionisation energy increases across the period: more protons, an increased nuclear charge, electrons in the same shell, no extra shielding, so a stronger attraction between the nucleus and the outer electron. The word outer has to be there; a decreasing atomic radius is not the reason.

A whole period is three matched stages, each with its own justification: the general rise, the s to p dip, the pairing dip. Both dips run in Period 2, and each must be named in the period the question names.

The aluminium dip, Mg to Al

Al, aluminium, has the lower first ionisation energy. Its outer electron is in a 3p sub-shell, and 3p is higher in energy than 3s, so the electron is easier to remove: write both halves. Sub-shell or orbital, never energy level or shell, and name the element the dip runs to, Al.

The sulfur dip, P to S

S is lower than P because the outer electrons in the 3p sub-shell begin to pair, and paired electrons repel, so less energy is needed to remove one. The word pair carries it: p⁴ against p³ explains nothing, and the repulsion sits in the p orbital, never the s. Where a plot is asked for, put S below P before explaining why.

The trend down Group 2

First ionisation energy decreases down the group: more shells, so the outer electron is further from the nucleus and more shielded, giving a weaker attraction between the nucleus and the outer electron. Write decrease first; any other direction voids the reasoning under it.

At the top of the group, Be has the highest first ionisation energy in Group 2: the smallest atom, the least shielding, the strongest attraction between the nucleus and the outer electron.

• Ba has the lowest second ionisation energy in Group 2.
• Mg has the lowest second ionisation energy in Period 3.

Comparisons have to be comparative

A comparative answer needs a comparative ending: further from the nucleus, higher energy orbital, more shielding. You can write a superlative or a converse argument instead. Naming one orbital and stopping is not a comparison, so name both and join them with and: Ca⁺ loses an electron from a 4s orbital and K⁺ loses an electron from a 3p orbital. It covers the dips, Group 2 and the successive energies.

Why the second ionisation energy is larger than the first

The electron is removed from a positive ion, so the attraction holding it is stronger.

The big jump in successive ionisation energies

1. Name the element, or the group.
2. Give where the jump is: a large increase from the 5th to the 6th ionisation energy.
3. Give why, using two of these and not one: removed from a lower energy level, closer to the nucleus, less shielding.

The group is the number of electrons removed before the jump.

Second against third of one element keeps the shape: removed from 1s rather than 2s, lower in energy, less shielding, a stronger attraction between the nucleus and the outer electron.

Reading a log10 plot

The axis is log10 of the ionisation energy, so raise 10 to the power of the reading and write the result in kJ mol⁻¹. For the reading marked on the plot, any value from 2500 to 3200 is the right answer.