The classification of an element as s, p, d or f block according to its outer electron configuration. Trends in atomic radius and first ionisation energy across Period 3 and down a group, explained by nuclear charge, shielding and atomic radius. The trend in melting point across Period 2 and Period 3, explained by the structure and bonding of the elements.

What this dot point is asking

AQA wants you to classify elements into the s, p, d and f blocks from their outer electron configuration, describe and explain the trends in atomic radius and first ionisation energy across Period 3 and down a group, and explain the melting point trend across Periods 2 and 3 in terms of structure and bonding.

The four blocks

For example sodium ($1s^2 2s^2 2p^6 3s^1$) is s block, and chlorine ($1s^2 2s^2 2p^6 3s^2 3p^5$) is p block.

Atomic radius across Period 3

Across Period 3 the atomic radius decreases. Each successive element has one more proton (greater nuclear charge), but the added electron goes into the same outer shell, so shielding is roughly constant. The increasing nuclear charge pulls the outer shell in more strongly, so the radius shrinks.

Down a group the radius increases, because each element down the group has an extra electron shell and more shielding, outweighing the rise in nuclear charge.

First ionisation energy across Period 3

The general trend across Period 3 is an increase, because nuclear charge rises while shielding is about constant, so the outer electron is held more tightly. There are two dips:

• Mg to Al: the electron is removed from a 3p orbital (Al) rather than a 3s orbital (Mg). The 3p is higher in energy and slightly shielded by 3s, so it is easier to remove.
• P to S: in sulfur the 3p sub-shell now has a paired electron ($3p^4$), and the repulsion between the paired electrons makes one easier to remove than the equivalent electron in phosphorus ($3p^3$, all unpaired).

Melting point across a period

Try this

Q1. State which block manganese ($[Ar]3d^5 4s^2$) is in and why. [1 mark]

• Cue. d block; its highest-energy electrons are in the 3d sub-shell.

Q2. Explain why atomic radius decreases across Period 3. [2 marks]

• Cue. Nuclear charge increases, shielding stays roughly constant, so the outer shell is pulled in more strongly.

Q3. Explain why silicon has the highest melting point in Period 3. [2 marks]

• Cue. Giant covalent structure; many strong covalent bonds must be broken, needing a lot of energy.