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Edexcel IAL Chemistry · Topic 1.5
The Periodic Table
Big idea: as you move across a period, atoms get harder to ionise but easier to bond tightly in a metal lattice — and both trends come down to one thing: how strongly the nucleus pulls on the outer electrons.
Summary — the whole chapter in one scan
First ionisation energy generally increases across a period (left → right) because nuclear charge goes up while shielding stays about the same, so the atom shrinks and holds its electrons tighter.
First ionisation energy decreases down a group because extra electron shells add distance and shielding, which outweighs the bigger nuclear charge.
There are two predictable "dips" in the ionisation energy graph across Period 2 and Period 3 — one from a change in subshell (Be→B, Mg→Al) and one from electron-electron spin-pair repulsion (N→O, P→S).
Melting points across Period 3 rise steeply from Na to Si, then crash from Si to Ar — because the bonding and structure change completely partway through the period.
Na, Mg, Al are giant metallic structures; Si is a giant covalent (molecular) structure; P, S, Cl, Ar are simple molecular structures held together by weak intermolecular forces.
1. Ionisation Energy Across a Period
First, let's be precise about what "first ionisation energy" (IE₁) actually means, because half of getting this topic right is knowing the definition cold:
Definition
First ionisation energy is the energy required to remove one electron from each atom in one mole of gaseous atoms, to form one mole of gaseous 1+ ions.
X(g) → X⁺(g) + e⁻
The periodic table isn't just a random grid — elements are lined up in order of increasing atomic number, arranged into groups (columns) and periods (rows). Because electron configuration repeats in a pattern as atomic number increases, properties like ionisation energy repeat too. That repeating pattern is called periodicity — and it's the reason a graph of IE₁ against atomic number looks like a jagged sawtooth that resets at the start of every new period.
FIRST IONISATION ENERGY (kJ/mol) — Period 2 example
2500 | He
| *
2000 |
|
1500 | N F Ne
| * \ * *
1200 | C * O
| * \ /
900 | Be *
| * \ /
600 | / B
| / *
300 | H Li
| * *
0 +--------------------------------------------- Na
1 2 3 4 5 6 7 8 9 10 11 (atomic number)
Notice: overall trend rises across the period, but dips at
Be→B and N→O. Then it CRASHES back down at Na — a new
period, a new outer shell, much easier to remove an electron.
Think of it like this: imagine the nucleus is a magnet and the outer electrons are paperclips stuck to it. As you move across a period, you're adding protons to the nucleus (stronger magnet) but the paperclips are still in the same shell — same rough distance away. A stronger magnet pulling on paperclips at the same distance means it's harder to pull one off. That's the whole trend in one sentence.
Why ionisation energy increases across a period
Four linked facts, each one causing the next:
Nuclear charge increases — each element across the period has one more proton than the last.
Shielding stays roughly constant — you're adding electrons to the same outer shell, so the number of inner, shielding shells doesn't change.
Atomic radius decreases — with more positive charge pulling on the same shell, and no extra shielding to counteract it, the outer electrons get tugged in closer to the nucleus.
Result: the outer electron is held more tightly (closer + stronger pull), so you need more energy to remove it. IE₁ rises.
Analogy
Picture a tug-of-war. The nucleus is pulling the electron in; distance and shielding are on the electron's side, helping it "escape." Across a period, the nucleus's team gets stronger (more protons) while the electron's team stays the same size (same shell, same shielding) — so the nucleus keeps winning by a bigger margin each time. That's why it gets progressively harder to remove an electron.
Practice Question
Explain why the first ionisation energy of magnesium is higher than that of sodium.
The two dips in Period 2 (and why they matter)
If the trend were perfectly smooth, IE₁ would rise in a straight staircase from Li to Ne. It doesn't — there are two small dips, and examiners love asking you to explain them because they test whether you actually understand electron configuration, not just "more protons = harder to remove."
Dip 1 — Beryllium → Boron (subshell change)
Element
Electron configuration
IE₁ (kJ/mol)
Beryllium (Be)
1s² 2s²
900
Boron (B)
1s² 2s² 2pₓ¹
800
Why IE₁ drops
In beryllium, the electron being removed comes from the 2s subshell. In boron, the fifth electron sits in the 2p subshell instead. The 2p subshell has slightly higher energy and is, on average, further from the nucleus than the 2s subshell (it's also shielded slightly by the 2s electrons). A further-out, higher-energy electron is easier to remove — so despite boron having one more proton than beryllium, its IE₁ is actually lower.
Dip 2 — Nitrogen → Oxygen (spin-pair repulsion)
Element
Electron configuration
IE₁ (kJ/mol)
Nitrogen (N)
1s² 2s² 2pₓ¹ 2p_y¹ 2p_z¹
1400
Oxygen (O)
1s² 2s² 2pₓ² 2p_y¹ 2p_z¹
1310
Why IE₁ drops
In nitrogen, all three 2p electrons sit in separate orbitals, each unpaired (this follows Hund's rule — electrons spread out before pairing). In oxygen, the fourth 2p electron has to pair up with one already in the 2pₓ orbital. Two electrons crammed into the same orbital repel each other (spin-pair repulsion), which pushes them apart and makes it easier to knock one of that pair off. So even though oxygen has more protons than nitrogen, its IE₁ is slightly lower.
Don't mix these two dips up
Be→B dip = a subshell change (2s → 2p, further from nucleus).
N→O dip = electron pairing within the same subshell (repulsion between two electrons sharing an orbital).
Examiners can tell instantly if you've memorised the "shape" of the graph without understanding why each dip happens — always name the correct mechanism for the correct pair.
The same two dips repeat in Period 3
Element
Electron configuration
IE₁ (kJ/mol)
Magnesium (Mg)
1s² 2s² 2p⁶ 3s²
738
Aluminium (Al)
1s² 2s² 2p⁶ 3s² 3pₓ¹
578
Phosphorus (P)
1s² 2s² 2p⁶ 3s² 3pₓ¹ 3p_y¹ 3p_z¹
1012
Sulfur (S)
1s² 2s² 2p⁶ 3s² 3pₓ² 3p_y¹ 3p_z¹
1000
Same logic: Mg→Al is a 3s→3p subshell jump (further from nucleus, easier to remove). P→S is spin-pair repulsion in the 3pₓ orbital (two electrons forced into one orbital repel and destabilise each other).
Practice Question
Sulfur has a lower first ionisation energy than phosphorus, even though sulfur has a greater nuclear charge. Explain why.
2. Ionisation Energy Down a Group
Now flip direction. Going down a group (e.g. Li → Na → K), IE₁ decreases — the opposite trend to across a period. It's tempting to think "more protons = always harder to remove an electron," but that's exactly the trap this section exists to catch.
Here's the chain of reasoning:
Nuclear charge increases — more protons as you go down the group. (True, but it's not the deciding factor here.)
Extra electron shells are added — each element down the group has one more full shell of electrons than the one above it.
Atomic radius increases — more shells means a physically bigger atom, so the outer electron sits much further from the nucleus.
Shielding increases — more inner shells of electrons sit between the nucleus and the outer electron, blocking (shielding) some of the nuclear attraction.
Result: the increase in distance and shielding outweighs the increase in nuclear charge, so the outer electron is actually held less tightly. It's easier to remove → IE₁ falls.
The key exam phrase
"The increased shielding and increased distance from the nucleus outweigh the increased nuclear charge." — this exact idea of one factor outweighing another is what separates a full-mark answer from a partial one.
Analogy
Imagine shouting to get someone's attention. Across a period, you're shouting louder (more nuclear charge) at someone standing at the same distance — so they hear you more clearly. Down a group, you're shouting louder too, but the person keeps walking further away and more people keep stepping in between you (shielding) — so despite shouting louder, they hear you less clearly overall.
Practice Question
Explain why the first ionisation energy of potassium is lower than that of sodium, even though potassium has a greater nuclear charge.
Across a period vs down a group — side by side
Factor
Across a Period (IE₁ ↑)
Down a Group (IE₁ ↓)
Nuclear charge
Increases
Increases
Shell number
Same
Increases
Distance to nucleus
Decreases
Increases
Shielding
Roughly constant
Increases
Atomic radius
Decreases
Increases
Outer electron held...
More tightly → harder to remove
Less tightly → easier to remove
3. Thermal Trends — Melting Points Across Period 3
Now we switch from ionisation energy to a completely different property — melting point — but the underlying skill is the same: link a macroscopic property (does it melt easily or not?) to what's happening at the level of bonding and structure.
Element
Na
Mg
Al
Si
P
S
Cl
Ar
Melting point (K)
371
923
932
1683
317
392
172
84
MELTING POINT ACROSS PERIOD 3 (K)
1800 | Si
| ****
1500 | ****
| ****
1200 | ****
| ****
900 | Mg Al ****
| **** **** ****
600 | **** **** ****
| **** **** **** S
300 | Na **** **** **** P ****
|**** **** **** **** **** **** Cl
0 +----------------------------------------****----Ar
Na Mg Al Si P S Cl Ar
Rises steadily Na→Si (giant structures, strong bonds getting stronger),
then CRASHES Si→Ar (switch to weak, simple molecular structures).
Period 2 follows exactly the same shape for the same reasons, so once you understand Period 3, Period 2 is free marks.
These three are metals. A metal is a giant lattice of positive ions held together by a "sea" of delocalised electrons that can move freely throughout the whole structure — not attached to any one atom.
METALLIC BONDING — "sea of electrons" model
+ - + - + + = positive metal ion
- + - + - - = delocalised electron
+ - + - + (free to move anywhere
- + - + - throughout the lattice)
+ - + - +
The electrostatic attraction between the + ions and the
"sea" of - electrons is what holds the whole lattice together.
Why melting point rises from Na → Mg → Al
Na donates 1 electron per atom into the sea, Mg donates 2, and Al donates 3. Going from Na to Al: (1) the positive ion charge increases (Na⁺ → Mg²⁺ → Al³⁺), and (2) the number of delocalised electrons in the sea increases. Both effects make the electrostatic attraction between the ions and the sea of electrons stronger. Stronger metallic bonding = more energy needed to break the lattice apart = higher melting point.
Exam trap — drawing the lattice
If you're asked to draw a metallic lattice, examiners want to see: a regular arrangement of positive ions in neat rows/columns, and the ions tightly packed (touching/close together — remember, metals are solids at room temperature!). You will lose marks for large gaps between ions (looks like a gas) or a random/scattered arrangement (looks like a liquid). You do not need to draw every delocalised electron individually — but the ion arrangement itself must clearly look like a solid.
Why Si has the highest melting point of all
Silicon does not form metallic bonds — instead, every Si atom is joined to its neighbouring Si atoms by strong covalent bonds that extend throughout the entire structure (a "giant covalent" or "giant molecular" lattice, just like diamond). Melting silicon means breaking huge numbers of strong covalent bonds throughout the whole 3D network — that takes an enormous amount of energy, which is why Si's melting point (1683 K) towers over everything else in the period.
After silicon, the bonding changes completely. P, S, Cl and Ar are non-metals that exist as small, discrete, simple molecules: P₄, S₈, Cl₂, and Ar as single free atoms.
Why melting points crash after silicon
Inside each molecule (e.g. within a single P₄ molecule), the covalent bonds are strong. But when a substance like this melts, you are not breaking those covalent bonds — you're only overcoming the weak instantaneous dipole–induced dipole forces (a type of van der Waals force) that hold separate molecules to each other. These intermolecular forces are far weaker than covalent or metallic bonds, so it takes very little energy to melt these elements — hence the sudden drop in melting point.
Classic mistake
Students often say "the covalent bonds in P₄ break when phosphorus melts" — wrong. Melting a simple molecular substance only overcomes the weak forces between molecules. The strong covalent bonds within each molecule stay completely intact. Always be precise about what type of bond/force is actually being broken.
Why is S's melting point higher than P's, if both are "simple molecular"?
Even within the simple molecular block, size matters. Sulfur exists as S₈ molecules — larger than phosphorus's P₄ molecules. Bigger molecules have more electrons and a greater surface area of contact, which means stronger instantaneous dipole–induced dipole forces between them. Stronger intermolecular forces need more energy to overcome, so S (392 K) melts at a higher temperature than P (317 K) — even though S comes later in the period and "should" in theory be a smaller atom.
Practice Question
Explain, in terms of structure and bonding, why the melting point of silicon is much higher than that of phosphorus.
Practice Question
Explain why the melting point of aluminium is higher than that of sodium.
What to Memorise
First ionisation energy: energy to remove 1 electron from each atom in 1 mole of gaseous atoms, forming 1 mole of gaseous 1+ ions. X(g) → X⁺(g) + e⁻
Periodicity: the repeating pattern of physical/chemical properties across periods, caused by the repeating pattern of electron configurations.
Across a period: IE₁ increases — nuclear charge ↑, shielding ≈ constant, atomic radius ↓, outer electron harder to remove.
Down a group: IE₁ decreases — extra shells → distance ↑ and shielding ↑, which outweighs the increased nuclear charge.
Be→B dip: caused by a subshell change (2s → 2p); the 2p electron is further from the nucleus and easier to remove.
N→O dip: caused by spin-pair repulsion — two electrons forced into the same p orbital repel each other, making one easier to remove.
Mg→Al and P→S dips: the exact same two mechanisms repeat in Period 3 (3s→3p subshell change; spin-pair repulsion in 3pₓ).
Metallic bonding: electrostatic attraction between a giant lattice of positive ions and a "sea" of delocalised electrons.
Na → Al melting point rise: increasing ionic charge (1+, 2+, 3+) and increasing number of delocalised electrons → stronger metallic bonds.
Silicon: giant covalent/molecular structure — every atom joined by strong covalent bonds throughout the lattice → highest melting point in the period.
P, S, Cl, Ar: simple molecular structures (P₄, S₈, Cl₂, Ar). Melting only overcomes weak instantaneous dipole–induced dipole forces between molecules — covalent bonds within molecules stay intact.
S vs P melting point: S₈ molecules are bigger than P₄ molecules, so S has stronger intermolecular forces and a higher melting point than P.
Concepts Checklist
Exam Tips & Common Mistakes
"Outweighs" is a mark-scheme keyword
For the down-a-group ionisation energy trend, simply listing "shielding increases, distance increases, nuclear charge increases" without saying which factor wins often loses the final mark. Always finish with: "...and this outweighs the increased nuclear charge."
Name the correct dip mechanism
Never just say "there's a dip because of electron configuration" — that's too vague for full marks. State explicitly whether it's a subshell change (Be→B, Mg→Al) or spin-pair repulsion (N→O, P→S), and explain *why* that specific mechanism makes removal easier.
Melting simple molecular substances
Never say covalent bonds break when P₄, S₈, or Cl₂ melt. Only the weak intermolecular forces (instantaneous dipole–induced dipole forces) between molecules are overcome. This single mix-up is one of the most common ways to lose marks in this topic.
Drawing metallic lattices
Positive ions must be shown in a regular, tightly-packed arrangement (rows and columns, touching). A random scatter or large gaps between ions will be marked as showing a liquid or gas, not a solid metal — even if you've correctly drawn the delocalised electrons.
"More protons" is not a full answer on its own
Increased nuclear charge is only half the story. Always pair it with what's happening to shielding and distance (constant across a period; both increasing down a group) — examiners want the whole causal chain, not just the starting fact.