Periodicity

In a nutshell

The periodic table puts elements in order of atomic (proton) number. The same pattern of outer electrons repeats in each period, so physical and chemical properties repeat too. That repeating pattern is periodicity.

This subtopic uses that idea to explain two trends you must be able to explain in an exam: first ionisation energy and melting point across Periods 2 and 3. It also uses successive ionisation energies as evidence for electron shells.

Assumed knowledge: Electron structure (sub-shells, orbitals, electron configurations) and Bonding and structure (covalent bonding, intermolecular forces).

Core content

The structure of the periodic table

The periodic table arranges the elements:

• by increasing atomic (proton) number
• in periods (rows) that show repeating trends in physical and chemical properties (this repetition is periodicity)
• in groups (columns) of elements that have similar chemical properties, because they have the same number of outer-shell electrons in the same type of sub-shell

How the table was developed

• Early chemists, most famously Mendeleev, ordered elements by atomic mass and grouped elements with similar properties together.
• Mendeleev left gaps for elements not yet discovered and predicted their properties. When elements such as germanium were later found with properties close to his predictions, the scientific community accepted his Periodic Law.
• He also swapped a few elements out of mass order (for example tellurium and iodine) so they sat in the group whose chemistry they matched. Ordering by atomic number, introduced later, explained why this worked.
• The table is still being extended. New, very heavy elements are made in particle accelerators, and an element is only added and named once other laboratories have confirmed the discovery.

Periodic trend in electron configuration and ionisation energy

Electron configurations across Periods 2 and 3

Across each period, electrons fill the s sub-shell, then the p sub-shell of the same outer shell. Period 3 repeats the Period 2 pattern one shell further out:

Only the outer shell is shown. Elements in the same group have the same outer-shell configuration, which is why their chemistry is similar.

s-, p- and d-blocks

An element's block is named after the sub-shell that holds its highest-energy electron.

• s-block: Groups 1 and 2 (highest-energy electron in an s sub-shell). Hydrogen (1s¹) and helium (1s²) are s-block by configuration, even though helium is placed with the noble gases.
• p-block: Groups 13 to 18 (highest-energy electron in a p sub-shell).
• d-block: the central block, from scandium to zinc in Period 4 (highest-energy electron in a d sub-shell).

First ionisation energy

The 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.

Put simply, it is the removal of 1 mol of electrons from 1 mol of gaseous atoms. The equation always has the same shape, with (g) on every species:

• sodium: Na(g) → Na⁺(g) + e^-
• chlorine: Cl(g) → Cl⁺(g) + e^-

Write it like this even for an element that is not a gas at room temperature, and even for one that never normally forms a 1+ ion.

What decides the size of an ionisation energy

Ionisation energy measures how strongly the nucleus attracts the outer electron. Three factors control that attraction:

1. Nuclear charge: more protons means a greater attraction.
2. Atomic radius: an outer electron further from the nucleus is attracted less.
3. Shielding: inner shells of electrons repel the outer electron and reduce the nuclear attraction it feels.

The same shape repeats in Period 2 (Li to Ne, Z = 3 to 10) and Period 3 (Na to Ar, Z = 11 to 18). That repeat is periodicity, and each feature has an explanation you need.

Across a period: general increase

1. Nuclear charge increases (one more proton each time).
2. Electrons are added to the same shell, so shielding is similar.
3. So the atomic radius decreases and the nuclear attraction on the outer electrons increases.
4. More energy is needed to remove an outer electron, so first ionisation energy increases.

Down a group: decrease

1. The atomic radius increases (more shells), so the outer electron is further from the nucleus.
2. There is more shielding from extra inner shells.
3. So the nuclear attraction on the outer electron decreases, even though the nuclear charge has increased.
4. Less energy is needed to remove it, so first ionisation energy decreases.

This also explains the big drop from a noble gas to the next alkali metal (Ne to Na, Ar to K): the outer electron is now in a new shell, further out and more shielded.

The two small decreases across a period

These are the details examiners test most often.

Be to B (and Mg to Al): a drop as the p sub-shell starts.

• Boron's outer electron is in the 2p sub-shell, which is higher in energy than the 2s sub-shell.
• So the 2p electron is easier to remove, and boron's first ionisation energy is lower than beryllium's.
• The same happens from Mg (3s²) to Al (3s²3p¹).

N to O (and P to S): a drop as the first p-orbital pairs up.

• In oxygen, two electrons are paired in the same 2p orbital.
• These paired electrons repel each other.
• So one of them is easier to remove, and oxygen's first ionisation energy is lower than nitrogen's.
• The same happens from P (3p³) to S (3p⁴).

Still don't get it? · the small dips at boron and oxygen

Think of climbing a staircase where each step is a bit higher than the last. That is the general rise across a period: each new proton makes the electrons a little harder to pull off.

Now picture two places where a step is slightly lower than you expect.

1. Boron. Up to beryllium, every electron has gone into s sub-shells. Boron's fifth electron is the first to go into a p sub-shell. A p sub-shell sits slightly higher in energy than the s sub-shell of the same shell, so that electron starts "higher up" and needs a little less energy to remove. The step dips.
2. Nitrogen to oxygen. Nitrogen's three 2p electrons each have an orbital to themselves. Oxygen's fourth 2p electron has no empty orbital left, so it has to share one. Two electrons in one orbital push against each other (they are both negative). That extra repulsion makes one of them easier to remove. The step dips again.
3. After each dip the general rise carries on, because the nuclear charge keeps going up.

In exam wording: for B, the outer electron is in a 2p sub-shell, which is higher in energy than 2s; for O, there is repulsion between the paired electrons in a 2p orbital. Do not swap the two reasons.

Successive ionisation energies

Successive ionisation energies remove electrons one at a time from the same element. Each step starts from the ion made in the step before, and every species is gaseous:

• second ionisation of sodium: Na⁺(g) → Na²⁺(g) + e^-
• third ionisation of aluminium: Al²⁺(g) → Al³⁺(g) + e^-

Each successive ionisation energy is larger than the one before. The electron is removed from an increasingly positive ion, with the same nuclear charge holding fewer electrons, so the ion is smaller and the remaining electrons are attracted more strongly.

A large jump appears whenever the next electron comes from a new shell closer to the nucleus: it is much nearer the nucleus and much less shielded.

The y-axis is a log scale, so the values (from about 500 to about 160 000 kJ mol^-1) fit on one graph. Read it from left to right:

• 1 electron is easy to remove: the outer shell (shell 3) holds 1.
• Then a big jump, and 8 electrons with similar, steadily rising values: shell 2 holds 8.
• Then another big jump, and 2 electrons that are hardest of all: shell 1 holds 2.

So sodium is 2, 8, 1: Group 1, Period 3.

To find the group, count the electrons removed before the first big jump. That number is the number of outer-shell electrons.

These jumps were early evidence that electrons sit in shells, supporting the Bohr model of the atom.

Still don't get it? · reading successive ionisation energies

Imagine peeling an onion where each layer is stuck on more tightly than the last.

1. The outer layer comes off fairly easily, a few pieces at a time, with each piece slightly harder than the one before (because there is less onion left pulling them apart, and the core holds on harder).
2. Then you reach a new, tighter layer. The first piece of that layer is suddenly much harder to pull off than the last piece of the layer before. That is the big jump.
3. Count how many pieces came off before that first big jump: that is how many were in the outer layer.

For electrons: each electron removed leaves a smaller, more positive ion, so each ionisation energy is a bit bigger than the last. When the next electron has to come from a shell closer to the nucleus, with less shielding, the ionisation energy leaps up.

In exam wording: a large increase shows the electron is being removed from a new shell, closer to the nucleus, with less shielding, so it is more strongly attracted. The number of electrons removed before the first large increase is the number of outer-shell electrons, which gives the group (1 or 2 electrons: Group 1 or 2; 3 to 8 electrons: Group 13 to 18).

Periodic trend in structure and melting point

Metallic bonding

Metallic bonding is the strong electrostatic attraction between cations (positive ions) and delocalised electrons.

• Each metal atom gives up its outer-shell electrons, which become delocalised: free to move through the whole structure.
• The cations sit in a regular arrangement called a giant metallic lattice. All metals have this structure.

Properties of giant metallic lattices

Giant covalent lattices

A giant covalent lattice is a network of atoms bonded by strong covalent bonds. On the spec: carbon as diamond, graphite and graphene, and silicon.

Properties of giant covalent lattices

• Very high melting and boiling points. To melt the solid, many strong covalent bonds must be broken, which takes a lot of energy.
• Insoluble in water and other solvents. The covalent bonds are too strong to be broken by interaction with a solvent.
• Do not conduct electricity, except graphite and graphene, which have delocalised electrons that can move.

Graphene: why it matters

• Very strong for its mass, because of its network of strong covalent bonds.
• Conducts electricity very well, because of its delocalised electrons.
• It is also very light, thin and flexible, and almost transparent.

These properties make it a candidate for flexible electronics and touchscreens, and for strengthening lightweight composite materials.

Melting points across Period 3

Across Periods 2 and 3, structure changes from giant metallic, to giant covalent, to simple molecular.

• Na to Al: across the metals, the cation charge increases (Na⁺, Mg²⁺, Al³⁺), there are more delocalised electrons per atom, and the cations get smaller. So the electrostatic attraction between cations and delocalised electrons gets stronger, and more energy is needed to overcome it.
• Si: giant covalent. Many strong covalent bonds must be broken, so silicon has the highest melting point in the period.
• P₄, S₈, Cl₂, Ar: only weak London forces between molecules (or atoms) are overcome; the covalent bonds inside the molecules do not break. Little energy is needed, so melting points are low.
• Order within the simple molecular elements: S₈ has 128 electrons, P₄ has 60, Cl₂ has 34 and Ar has 18. More electrons means stronger London forces, so melting point falls S₈ > P₄ > Cl₂ > Ar.

Melting points across Period 2

Period 2 follows the same structural pattern:

Worked examples

Example 1: Deduce the group of an element from its successive ionisation energies.

The first six successive ionisation energies of element X are:

Look for the largest increase. Comparing each value with the one before makes the jump obvious:

$\frac{IE_2}{IE_1} = \frac{1577}{787} = 2.00$

$\frac{IE_3}{IE_2} = \frac{3232}{1577} = 2.05$

$\frac{IE_4}{IE_3} = \frac{4356}{3232} = 1.35$

$\frac{IE_5}{IE_4} = \frac{16\,091}{4356} = 3.69$

$\frac{IE_6}{IE_5} = \frac{19\,805}{16\,091} = 1.23$

The big jump is between the 4th and 5th ionisation energies.

1. Four electrons are removed before the big jump, so X has 4 outer-shell electrons.
2. The 5th electron comes from a new shell closer to the nucleus, with less shielding, so it is much more strongly attracted.
3. So X is in Group 14. (These are the values for silicon.)

Where marks go: answering "Group 15" because the jump appears at the 5th value. Count the electrons removed before the jump, not the position of the jump.

Example 2: Use the successive ionisation energy graph for sodium (in Core content) to deduce the number of electrons in each shell.

Count the points in each group between the big jumps, from left to right:

$1 \text{ (outer shell)} + 8 + 2 \text{ (innermost shell)} = 11 = \text{atomic number of Na}$

Written from the nucleus outwards, the shells hold 2, 8, 1.

Where marks go: writing the shells in the order you read them off the graph (1, 8, 2). The first electrons removed are the outermost.

Example 3: Write equations for the first and third ionisation energies of aluminium.

• First: Al(g) → Al⁺(g) + e^-
• Third: Al²⁺(g) → Al³⁺(g) + e^-

Where marks go: writing Al(g) → Al³⁺(g) + 3e^- for the third ionisation energy (that is the sum of the first three), or leaving off any (g).

Example 4: Explain why the first ionisation energy of oxygen is less than that of nitrogen.

The model answer, one creditable point per line:

1. N is 1s² 2s² 2p³; O is 1s² 2s² 2p⁴.
2. In O, two electrons are paired in one 2p orbital.
3. The paired electrons repel each other.
4. So less energy is needed to remove one of them from O.

Where marks go: explaining it with "nitrogen has a stable half-filled sub-shell" and never mentioning repulsion between paired electrons, which is the reason the spec names.

Example 5: Explain why the first ionisation energy decreases from magnesium to calcium (down Group 2).

1. Calcium's outer electron is in a shell further from the nucleus (larger atomic radius).
2. There is more shielding from extra inner shells.
3. So the nuclear attraction on the outer electron is weaker, even though Ca has more protons.
4. Less energy is needed to remove it.

Where marks go: writing "more protons, so a higher ionisation energy". The increase in nuclear charge is outweighed by the greater distance and shielding.

Example 6: Explain the difference in melting point between silicon (1687 K) and phosphorus (317 K).

1. Silicon has a giant covalent lattice.
2. Melting silicon breaks many strong covalent bonds, which needs a lot of energy.
3. Phosphorus has a simple molecular lattice of P₄ molecules.
4. Melting phosphorus only overcomes weak London forces between molecules, which needs much less energy.

Where marks go: saying the covalent bonds in P₄ break when it melts, or using "intermolecular forces" for silicon. Silicon has no separate molecules.

Common exam mistakes

• Ionisation equations without state symbols. Every species in an ionisation energy equation needs (g), including the atom: Cl(g), not Cl₂(g) or Cl(s).
• Definition missing a condition. The definition needs one electron removed from each atom, one mole, and gaseous atoms forming gaseous 1+ ions.
• Wrong second or third ionisation equation: starting from the atom and removing several electrons at once. Each successive ionisation starts from the ion made in the previous step and removes one electron.
• "Bigger nucleus" or "more mass" instead of greater nuclear charge (more protons).
• "Shielding increases across a period". Electrons are added to the same shell, so shielding is similar; the nuclear charge is what increases.
• Down a group, citing only the extra protons and concluding the ionisation energy rises. The larger radius and extra shielding outweigh it.
• Swapping the two dips. Be to B is the p sub-shell being higher in energy than s; N to O is repulsion between paired electrons in a p-orbital.
• Successive ionisation energies: using the position of the jump as the group. A jump between the 4th and 5th values means 4 outer electrons, so Group 14.
• Metallic bonding without the key words. The mark needs attraction between positive ions (cations) and delocalised electrons. "Attraction between atoms and electrons", "negative ions" or electrons that are just "free" without being delocalised all lose the mark.
• Mixing structure words: "intermolecular forces" or "molecules" for silicon or diamond, or "ions" in a giant covalent lattice.
• Saying covalent bonds break when P₄, S₈ or Cl₂ melt. Only the London forces between molecules are overcome.
• Explaining S₈ > P₄ > Cl₂ with mass or M_r. The cause is more electrons, giving stronger London forces.
• Graphite conducting because of "free ions" or because the layers slide. It conducts because of delocalised electrons that can move.
• Diamond and graphite: wrong number of bonds. Diamond carbon forms 4 covalent bonds; graphite and graphene carbon form 3.

Key definitions

• Periodicity: a repeating trend in physical and chemical properties across the periods of the periodic table.
• First ionisation energy: the energy required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous 1+ ions.
• Successive ionisation energies: the energies needed to remove each further electron, one at a time, from the gaseous ion formed in the previous step (a formal definition is not required).
• s-, p- and d-block elements: elements whose highest-energy electron is in an s, p or d sub-shell respectively.
• Metallic bonding: the strong electrostatic attraction between cations (positive ions) and delocalised electrons.
• Giant metallic lattice: a regular arrangement of cations surrounded by delocalised electrons (all metals).
• Giant covalent lattice: a network of atoms bonded by strong covalent bonds (diamond, graphite, graphene, silicon).
• Delocalised electrons: electrons that are not attached to one atom or bond and are free to move through the structure.

Specification

• I can describe the periodic table as elements arranged by increasing atomic (proton) number, in periods showing repeating trends (periodicity), and in groups with similar chemical properties. (3.1.1(a))
• I can outline how the Periodic Law was developed and accepted, and how new elements are discovered and confirmed. (3.1.1(a), HSW)
• I can describe the trend in electron configurations across Periods 2 and 3. (3.1.1(b)(i))
• I can classify elements into the s-, p- and d-blocks. (3.1.1(b)(ii))
• I can define first ionisation energy and write equations for first and successive ionisation energies with state symbols. (3.1.1(c))
• I can explain the trend in first ionisation energies across Periods 2 and 3, including the small decreases from Be to B and from N to O, and down a group, in terms of attraction, nuclear charge and atomic radius. (3.1.1(c)(i))
• I can use successive ionisation energies to predict the number of electrons in each shell and the group of an element. (3.1.1(c)(ii))
• I can explain metallic bonding as strong electrostatic attraction between cations and delocalised electrons, and describe a giant metallic lattice. (3.1.1(d))
• I can explain the giant covalent lattices of diamond, graphite, graphene and silicon as networks of atoms bonded by strong covalent bonds, and the properties and uses of graphene. (3.1.1(e))
• I can explain the melting and boiling points, solubility and electrical conductivity of giant metallic and giant covalent lattices in terms of particles, forces and mobility. (3.1.1(f))
• I can explain the variation in melting points across Periods 2 and 3, from giant metallic to giant covalent to simple molecular. (3.1.1(g))

Related notes

• Electron structure: sub-shells, orbitals and electron configurations
• Bonding and structure: covalent bonding, London forces and simple molecular lattices
• Atomic structure and isotopes: atomic number and protons
• Group 2: how ionisation energy explains reactivity down the group
• The halogens: trends down Group 17