Periodic trends
Trends in atomic radius, ionisation energy, electronegativity, metallic character, reactivity and melting point across periods and down groups, explained with nuclear charge, shielding and electron configuration, plus successive ionisation energies and ionic radii.
Drop a pea-sized piece of lithium into water and it fizzes gently across the surface. Sodium melts into a ball and skates around. Potassium bursts into a lilac flame, and caesium explodes violently enough to shatter the container. Same group, same single outer electron, wildly different behaviour. One idea, the pull of the nucleus on its outermost electrons, explains this trend and nearly every other one in the periodic table.
The key idea: nuclear charge and shielding
An outer electron is attracted by the positive nucleus but repelled by all the electrons between it and the nucleus. Inner (core) electrons shield the outer electrons from part of the nuclear charge. The net pull felt by an outer electron is called the effective nuclear charge, roughly the nuclear charge minus the number of core electrons.
- Across a period (left to right): protons are added to the nucleus, but the extra electrons go into the same outer shell, which shields very little. Effective nuclear charge rises, so outer electrons are pulled closer and held more tightly.
- Down a group: each element has one more shell. The outer electrons are further from the nucleus and shielded by more inner shells, so they are held less tightly, even though the nuclear charge is larger.
Every explanation of a trend should use these two ideas: the size of the attraction (nuclear charge versus shielding) and the distance of the outer electrons from the nucleus.
Atomic radius
Atoms have no sharp edge, so the atomic radius is usually taken as half the distance between the nuclei of two identical atoms bonded together (the covalent radius).
- Decreases across a period: increasing effective nuclear charge pulls the same shell closer (Na 166 pm to Cl 102 pm).
- Increases down a group: each extra shell puts the outer electrons further out (Li 128 pm, Na 166 pm, K 203 pm).
Ions. A cation is smaller than its atom: losing the outer electrons often removes a whole shell, and the remaining electrons are pulled in more tightly. An anion is larger: added electrons repel each other and the same nuclear charge holds more electrons. In an isoelectronic series, more protons means a smaller ion: O²⁻ (140 pm) > F⁻ (133 pm) > Na⁺ (102 pm) > Mg²⁺ (72 pm), all with 10 electrons.
Ionisation energy
The first ionisation energy is the energy needed to remove one mole of electrons from one mole of gaseous atoms:
$$\text{Na}(g) \rightarrow \text{Na}^+(g) + e^- \qquad \Delta H = +496\text{ kJ mol}^{-1}$$
- Increases across a period: higher effective nuclear charge and smaller radius hold the outer electron more tightly.
- Decreases down a group: the outer electron is further away and better shielded.
- Noble gases sit at the peaks; group 1 metals at the troughs.
Two small dips reveal subshells. Boron (801) is lower than beryllium (900) because its outer electron is in 2p, slightly higher in energy and better shielded than 2s. Oxygen (1314) is lower than nitrogen (1402) because oxygen's fourth 2p electron has to pair up in an orbital that already holds one electron, and the repulsion between the pair makes it easier to remove. The same dips appear at aluminium and sulfur in period 3.
The first ionisation energies of Na, Mg and Al are 496, 738 and 578 kJ mol⁻¹. Explain the pattern.
Na to Mg: one more proton, and the extra electron is in the same 3s subshell, so effective nuclear charge rises and the 3s electron is held more tightly. IE increases.
Mg to Al: aluminium's outer electron is in 3p, which is higher in energy than 3s and partly shielded by the 3s² electrons. Less energy is needed to remove it, so IE falls even though nuclear charge has increased.
Successive ionisation energies
Electrons can be removed one at a time: the first, second, third ionisation energies and so on. Each is larger than the one before, because each electron is removed from an increasingly positive ion. But the increase is not steady. When all the valence electrons have gone, the next electron must come from a full inner shell that is much closer to the nucleus and far less shielded, and the ionisation energy jumps. Counting the electrons removed before the big jump gives the number of valence electrons, and so the group.
The first five ionisation energies of element X are 578, 1817, 2745, 11 577 and 14 842 kJ mol⁻¹. Which group is X in?
Ratios: 1817/578 = 3.1, 2745/1817 = 1.5, 11 577/2745 = 4.2, 14 842/11 577 = 1.3.
The largest jump is between the 3rd and 4th electrons, so X has 3 valence electrons: group 13. (It is aluminium.)
- Start on sodium. The jump between IE1 and IE2 is ×9.2. Explain why in terms of shells.
- Slide to magnesium (Z = 12). Where is the big jump now, and how large is it? (You should see ×5.3 between IE2 and IE3.)
- Turn off the log scale. Why is a log scale more useful for reading these graphs?
- Identify Mystery X, Y and Z by choosing a group. Mystery Y has two smaller steps before its big jump; explain them using subshells.
Electronegativity
Electronegativity is the ability of an atom in a covalent bond to attract the shared electrons towards itself. The most common scale is Linus Pauling's, on which fluorine has the highest value (3.98) and caesium and francium the lowest (about 0.8). Electronegativity increases across a period (greater effective nuclear charge, smaller atom, so shared electrons are closer to a stronger pull) and decreases down a group (bonding electrons further from the nucleus and better shielded). Noble gases are usually given no value because helium, neon and argon form no stable compounds. Electronegativity is central to bonding in lessons 1G to 1I.
Metallic character, reactivity and state
Metals tend to lose electrons; non-metals tend to gain or share them. Metallic character decreases across a period and increases down a group, following ionisation energy in reverse: the easier it is to lose an outer electron, the more metallic the element. A diagonal band of metalloids (B, Si, Ge, As, Sb, Te) separates metals from non-metals.
- Group 1 and 2 metals with water: reactivity increases down the group because the outer electron is lost more easily. 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g). Group 2 metals are less reactive than group 1 in the same period because they must lose two electrons, held by a larger nuclear charge: magnesium barely reacts with cold water but burns in steam.
- Halogens: reactivity decreases down group 17, because a halogen reacts by gaining an electron, and the incoming electron is attracted less strongly by larger atoms. Fluorine is the most reactive non-metal. A more reactive halogen displaces a less reactive one from its salts: Cl₂(aq) + 2KBr(aq) → 2KCl(aq) + Br₂(aq).
State and melting point across period 3 depend on structure, not just on the atoms: Na, Mg and Al are metallic solids with rising melting points (more delocalised electrons and smaller cations); silicon is a covalent network with the highest melting point (1414 °C); P₄, S₈, Cl₂ and Ar are molecular, held only by dispersion forces, so their melting points are low (S₈, the largest molecule, is highest of these).
- With Atomic radius, highlight Period 3. Check that radius falls from Na (166 pm) to Cl (102 pm). Why does Ar's bar look odd?
- Switch to First ionisation energy and highlight Group 1. Then highlight Group 18. What do the two patterns have in common?
- On Electronegativity, find the highest value. Which elements have no bar, and why?
- On Melting point, highlight Period 3. Explain the peak at silicon (1414 °C) and the drop to phosphorus (44 °C).
- Look at elements 21 to 30 on Atomic radius. Why does the radius barely change across the transition metals?
Test yourself
Which list is in order of increasing atomic radius?
Radius decreases across period 2 (N > O > F) and P, in period 3, is largest.
The first ionisation energy of sulfur (1000 kJ mol⁻¹) is lower than that of phosphorus (1012 kJ mol⁻¹) because
P is 3p³ (all single); S is 3p⁴, so one orbital holds a pair.
Successive IEs of an element (kJ mol⁻¹): 738, 1451, 7733, 10 543. The element is most likely in group
The big jump is after the second electron, so there are 2 valence electrons.
Which ion is the smallest?
All have 18 electrons; Ca²⁺ has the most protons (20), so it pulls them in most tightly.
Which statement about group 17 is correct going down the group?
Each element has an extra shell; EN, reactivity and IE all decrease.
Complete the summary of periodic trends.
Order these elements from lowest to highest first ionisation energy.
The net attraction felt by an outer electron is called the . Inner electrons reduce this attraction by . The element with the highest electronegativity is . Across a period, atomic radius . Metallic character increases a group.
Explain the trends in atomic radius and first ionisation energy (a) across period 3 from sodium to chlorine and (b) down group 1 from lithium to potassium. Use data to support your answer: radius (pm) Na 166, Cl 102, Li 128, K 203; first IE (kJ mol⁻¹) Na 496, Cl 1251, Li 520, K 419.
(a) From Na to Cl the number of protons increases from 11 to 17, but every added electron goes into the third shell. Electrons in the same shell shield each other poorly, so the effective nuclear charge on the outer electrons rises from about +1 to about +7. The outer shell is pulled closer, so the radius falls from 166 pm to 102 pm, and the outer electrons are held more tightly, so the first IE rises from 496 to 1251 kJ mol⁻¹.
(b) From Li to K each element has an extra occupied shell. The single outer electron is further from the nucleus and is shielded by more inner electrons. Although the nuclear charge increases, the effective nuclear charge stays about +1 while the distance increases, so the attraction is weaker. Radius increases from 128 to 203 pm and IE decreases from 520 to 419 kJ mol⁻¹, because a more distant, more shielded electron is easier to remove.
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