Lesson 5 · 25 min

Periodic trends: size, ionization energy and electronegativity

Predict, then check the data: explain atomic size, ionization energy, electron affinity and electronegativity with effective nuclear charge, and count valence electrons from successive ionization energies.

In this lesson

Use shielding and the effective nuclear charge $Z_\mathrm{eff}$ to explain why atoms shrink across a period and grow down a group
Describe the periodic trends in first ionization energy, electron affinity and electronegativity, and explain “exceptions” such as Be→B and N→O
Read the number of valence electrons (and the group) from the jump in successive ionization energies
Connect the trends to chemistry: metals lose electrons, nonmetals gain them

Sodium reacts violently with water and releases hydrogen; chlorine gas makes iron wire burn; argon reacts with almost nothing. The three sit close together in the periodic table yet behave utterly differently. The answer hides in one question: how tightly does the nucleus hold the outermost electrons?

Shielding and effective nuclear charge

Outer electrons do not “see” the whole nuclear charge: inner electrons sit in between and cancel part of it — shielding. The charge an outer electron actually feels is the effective nuclear charge $Z_\mathrm{eff}$ . In 1930 Slater proposed simple rules to estimate it:

each other electron in the same group ( $n$ s, $n$ p) shields 0.35 (0.30 within 1s);
each electron in the next shell in ( $n-1$ ) shields 0.85;
each electron deeper still shields 1.00.

Across a period each extra proton comes with just one poorly shielding electron in the same shell, so $Z_\mathrm{eff}$ rises by about 0.65 per step; down a group the valence electrons move to a new, larger shell while $Z_\mathrm{eff}$ barely changes.

Z_\mathrm{eff} = Z - S

The larger

Z_\mathrm{eff}

and the closer the electron, the more tightly it is held.

$Z$: nuclear charge (proton number)
$S$: shielding constant: the charge screened by other electrons
$Z_\mathrm{eff}$: the effective nuclear charge that electron feels

Worked example

Compare the effective nuclear charge in sodium and chlorine

Use Slater’s rules to estimate $Z_\mathrm{eff}$ for the 3s electron of Na and a 3p electron of Cl, and explain the difference in their radii and ionization energies.

Predict first

Going from Na to Cl along period 3, atoms gain more and more electrons. What happens to their radius?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

It grows: more electrons, bigger atom It shrinks It stays about the same: the shell does not change It grows, then shrinks

Four trends, one cause

Atomic radius: half the distance between the nuclei of two identical bonded atoms (the covalent radius). It decreases across a period and increases down a group.
First ionization energy $IE_1$ : the energy to remove one electron from a gaseous atom. It generally rises across a period and falls down a group.
Electron affinity: the energy released when a gaseous atom gains an electron. It is largest for nonmetals on the right; noble gases, N, Be and Mg form no stable anion.
Electronegativity: an atom’s pull on shared electrons in a bond (Pauling scale, fluorine highest at 3.98). It rises across a period and falls down a group.

All four are faces of one fact: how far the valence electrons are from the nucleus, and how large a $Z_\mathrm{eff}$ they feel.

\ce{X(g) -> X+(g) + e-}\quad \Delta E = IE_1 \qquad \ce{X(g) + e- -> X-(g)}\quad \Delta E = -E_\mathrm{ea}

This book takes the electron affinity

E_\mathrm{ea}

as the energy released (positive = exothermic); some books use the opposite sign.

$IE_1$: first ionization energy (always positive: removing an electron costs energy)
$E_\mathrm{ea}$: electron affinity

Predict · check the data · explain

Periodic trend explorer

Along

Along Period 2 (Li → Ne), how does the first ionization energy change?

Make a prediction to reveal the data.

Predict first, then reveal. The first ionization energies of period 2 step backwards twice — find both and read why. Then switch to electron affinity down group 17: why does fluorine come after chlorine?

Successive ionization energies: counting valence electrons

Remove electrons one at a time and each step is harder than the last (the ion left behind is ever more positive). But the rise is not smooth: once the valence electrons are gone, the next one comes from a filled inner shell and the cost suddenly jumps several-fold. The jump counts the valence electrons for you. Magnesium’s first two ionization energies are 7.6 eV and 15.0 eV, but the third is 80.1 eV — which is why magnesium appears in compounds as $\ce{Mg^2+}$ and never as $\ce{Mg^3+}$ .

Strip the electrons off one by one

Successive ionization energies

The first 2 electrons are valence electrons and come off fairly easily; electron 3 comes from the core, and the cost jumps 5.3-fold. That is why Mg appears in compounds as Mg²⁺: no ordinary chemical reaction can pay for one more electron.

k	From	IEₖ / eV	kJ/mol	Ratio to previous
1	3s	7.65	738	—
2	3s	15.04	1,451	×1.97
3	2p	80.14	7,733	×5.33
4	2p	109.27	10,543	×1.36
5	2p	141.33	13,636	×1.29
6	2p	186.76	18,020	×1.32
7	2p	225.02	21,711	×1.20
8	2p	265.92	25,658	×1.18
9	2s	327.99	31,646	×1.23
10	2s	367.49	35,457	×1.12
11	1s	1761.80	169,988	×4.79
12	1s	1962.66	189,368	×1.11

Switch from Mg to Na, Al, Si and Cl: after which electron does the first big jump appear, and how does that relate to the group number? Try the linear axis to see why a logarithmic one is needed.

Concept check

An element’s successive ionization energies (kJ/mol) are 578, 1817, 2745, 11577, 14842. Which group is it most likely in?

Group 1 Group 2 Group 13 Group 14

The chemistry: metals lose electrons, nonmetals gain them

Metals on the left have low ionization energies and few valence electrons, so in reactions they tend to lose electrons, forming cations such as $\ce{Na+}$ , $\ce{Mg^2+}$ and $\ce{Al^3+}$ ; nonmetals on the right have high electronegativity and electron affinity, so they tend to gain electrons, forming anions such as $\ce{Cl-}$ and $\ce{O^2-}$ , or to share electrons. Metallic character therefore falls across a period and rises down a group, and nonmetallic character does the opposite. The larger the electronegativity difference between two atoms, the more ionic the bond between them — the starting point for understanding chemical bonds.

Try the calculation

Magnesium’s second ionization energy is 15.035 eV. What is it in kJ/mol? (1 eV per atom = 96.485 kJ/mol)

Try the calculation

Use Slater’s rules to estimate $Z_\mathrm{eff}$ for a 2p electron in oxygen ( $1s^2\,2s^2\,2p^4$ ).

Key takeaways

Effective nuclear charge $Z_\mathrm{eff} = Z - S$ rises by about 0.65 per step across a period and stays nearly constant down a group, where shells are added.
Across a period radius shrinks while first ionization energy, electron affinity and electronegativity grow; down a group the reverse, broadly.
The ionization-energy dips Be→B and N→O (and Mg→Al, P→S) and electron affinity F < Cl are evidence of subshell energies and electron pairing.
Successive ionization energies jump once the valence electrons are gone, revealing the group; metals lose electrons to form cations, nonmetals gain or share them.

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Data and references

J. C. Slater, “Atomic Shielding Constants”, Phys. Rev. 36, 57 (1930).
PubChem Periodic Table of Elements (first ionization energies, electron affinities, Pauling electronegativities), National Center for Biotechnology Information, accessed 2026-09.
B. Cordero et al., “Covalent radii revisited”, Dalton Trans. 2008, 2832.
A. Kramida, Yu. Ralchenko, J. Reader and NIST ASD Team, NIST Atomic Spectra Database, Ionization Energies (ver. 5.11), https://physics.nist.gov/asd.
L. Pauling, “The Nature of the Chemical Bond. IV.”, J. Am. Chem. Soc. 54, 3570 (1932).
P. Atkins, T. Overton, J. Rourke, M. Weller, F. Armstrong, Shriver & Atkins’ Inorganic Chemistry, 5th ed., Oxford (2010), Ch. 1 (atomic properties and the Born–Haber cycle for NaCl).