Why does the First Ionisation Energy decrease down a group in the Periodic Table?

Two major factors control how tightly held the outermost electron is and therefore how much energy is required to remove it, which gives us the size of the Ionisation energy. The first of these factors is shielding. Electrons in orbitals/shells other than the orbital occupied by the outermost electron repel it slightly, reducing the amount of energy required to ‘pull’ away this outer electron. For example, if we consider Magnesium, which has an electronic arrangement of 1s² 2s² 2p⁶ 3s² our outermost electron is either of the two from the 3s sub-shell (these are both equivalent to each other) and the two full shells below the 3^rd shell will act to shield the outermost electron from the attraction of the nucleus.

The second factor is the effective nuclear charge: the attraction to the positively charged nucleus felt by the negatively charged outermost electron. As the number of protons within the nucleus (i.e. the atomic number) increases, so does the effective nuclear charge.

It may also be worth mentioning the atomic radius (distance between the nucleus and the outermost electron) as this can often be worth a mark on some exam boards. Increased shielding will increase atomic radii, whereas an increase in effective nuclear charge will decrease atomic radii.

We can use these three properties to explain the trend in first ionisation energy:

1^st IE decreases down the group: this is because the number of filled shells increases down the group, increasing shielding and the distance between the nucleus and the outermost electrons, for very similar effective nuclear charge. This means the outermost electron is more loosely held down the group and so less energy is required to remove it. Thus, 1^st IE decreases down the group.