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The ionisation energy of an element is the energy required to remove the one of the outermost electrons from an atom of the element in its gaseous state.
This is quite a difficult process to measure the energy requirements for however, so it is usually scaled up to the energy required or enthalpy change to remove one mole of electrons from one mole of atoms of an element in the gaseous state, which allows us to measure it in J mol-1 (Energy released per mole). Although, it is much more usual to see kJ mol-1 used, as working in J mol-1 means using some unnecessarily large numbers.
The first ionisation energy is the energy requirement for a single electron to be removed from the uncharged elemental electronic configuration (i.e. for lithium this would be 1s2 2s1) to give a cation with a +1 charge.
This could also be written as a chemical formula: X(g) -> X+(g) + e-, where ‘X’ is a particular element.
If another electron was removed from this cation, to give a new cation with a +2 charge, the process would be referred to as the second ionisation energy and so on.
The second ionisation energy could be represented by: X+(g) -> X2+(g) + e-see more
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 3rd 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:
1st 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, 1st IE decreases down the group.see more