Periodic Trends – Interactive Study Tool

Why Study Periodic Trends?

They explain the variations in the properties of the elements

The big idea

The periodic table is not just a list. Elements are arranged so that their properties repeat in a predictable pattern — periodically. Once you know the four trends on this page, you can predict which of two atoms is bigger, which holds its electrons more tightly, and which end of a bond the shared electrons will sit closest to — without looking up a single number.

Everything here is explained by exactly two ideas: the effective nuclear charge (Z_eff) and the number of energy levels (the shielding effect). Learn to reach for the right one and every trend explanation writes itself.

The four definitions

Trend	Definition	Where it matters
Atomic radius	Half the distance between the nuclei of two bonded atoms — i.e. the size of the atom.	Size comparisons; underlies every other trend
Ionization energy	The energy required to remove an electron from an atom in the gas phase (endothermic).	Ionic bonding in metals — the loss of electrons
Electron affinity	The energy released when a negative ion forms from a gaseous atom (exothermic).	Ionic bonding in non-metals — the gain of electrons
Electronegativity	The tendency of an atom to attract electron density towards itself when it is covalently bonded.	Covalent bonding — bond polarity

Don't mix up EA and electronegativity. Electron affinity is a measured energy for a lone gaseous atom grabbing a free electron (ionic bonding). Electronegativity is a unitless tendency to pull on electrons that are already being shared in a covalent bond. They trend the same way, but they are not the same quantity.

What you already know — energy levels increase down a group

Going down a group, each element has more electrons, and those extra electrons have to occupy new, higher energy levels. Hydrogen uses 1 shell, lithium 2, sodium 3, potassium 4.

Going across a period the number of energy levels does not change — every element in period 3 uses exactly 3 shells. The new electrons all pile into the same outer shell.

Show:

What you need next — Coulomb's Law

F = k Q₁Q₂ r²

Coulombic attraction is the force of attraction between positive and negative charges. In an atom that means the pull between the protons in the nucleus and the electrons around it.

Q₁More protons ⇒ stronger attraction. The force is directly proportional to the charges.

Q₂More electrons being attracted ⇒ stronger total attraction, for the same reason.

r²Greater distance ⇒ much weaker attraction. The force is inversely proportional to r², so doubling the distance cuts the force to a quarter.

✓Every trend on this site is one of these two effects: more protons pulling harder, or electrons sitting further away.

Where to next

Head to Z_eff & the Master Rule — that single tab is the one that makes every other trend on this page explainable.

Effective Nuclear Charge (Z_eff)

The net pull the nucleus has on the valence electrons

Shielding — why the valence electrons don't feel the full nuclear charge

An outer electron is shielded from the pull of the protons by all the electrons sitting between it and the nucleus. Those core (inner) electrons repel the outer electron and partly cancel the attraction — so the valence electron feels a smaller, effective nuclear charge.

Effective nuclear charge = (actual nuclear charge) − (shielding effect)

Z_eff = Z − S

Z = the number of protons in the nucleus.

S = the number of core (inner) electrons doing the shielding.

Z_eff is a Coulombic attraction — the net attractive force between the protons in the nucleus and the valence electrons. The bigger the Z_eff, the tighter the atom holds its outer electrons.

Z_eff Calculator

Pick an element, or type an electron configuration by shell (e.g. 2,8,5). Core electrons are shown in blue, valence electrons in rose.

Or type a configuration

Protons (Z)

Protons, Z

15

Core electrons, S

10

Z_eff

5

Try:

Handy check. For a main-group element, Z_eff comes out equal to the group number (using the old 1–8 numbering). Phosphorus is in group 5 and its Z_eff is 5. If your answer doesn't match the group number, you have probably counted the valence electrons as core electrons.

How Z_eff behaves

increases across a period

Why? The number of protons increases, so the nuclear pull gets stronger. Meanwhile the number of energy levels stays the same, so no extra shielding is added to cancel that pull out. Na (Z_eff = 1) → Cl (Z_eff = 7).

stays constant down a group

Why? Z − S stays roughly the same. The increase in nuclear charge (more protons) is cancelled out by the increase in the number of core/shielding electrons. Li, Na, K and Cs all have Z_eff = 1.

Show:

The master rule — for every trend explanation

Rule Pick your explanation by direction

ACROSS
a period →

Explain it with the effective nuclear charge (Z_eff). More protons pulling on electrons in the same energy level.

DOWN
a group ↓

Explain it with the increased number of energy levels (the shielding effect). Valence electrons sit at a greater average distance from the nucleus.

Never explain a trend using something that is constant in that direction. Z_eff is constant down a group — so "Z_eff increases down the group" is always wrong. The number of energy levels is constant across a period — so "there are more energy levels across the period" is always wrong. If it doesn't change, it can't be the cause.

This one rule is worth more marks than any single fact in this unit. Every "why" answer Miss Peters is looking for starts by identifying the direction first, then reaching for the matching cause.

Interactive Periodic Trend Map

Pick a property — see the direction it moves across and down

Choose a trend

Brighter cells = larger values. Click any element for its value. The arrows show the direction the property increases.

across a period

down a group

low

high ▪ picometres (pm) ▪ grey = no meaningful value

Click an element to see its value.

The summary diagram from class

Property	Across a period →	Down a group ↓
Atomic / ionic radius	decreases	increases
Ionization energy	increases	decreases
Electron affinity	increases	decreases
Electronegativity	increases	decreases
Z_eff	increases	constant

Atomic size is the odd one out. It goes the opposite way to the other three. That makes sense: IE, EA and electronegativity are all measures of how strongly an atom pulls on electrons, and an atom pulls hardest when its valence electrons are closest to the nucleus — i.e. when it is small.

So there is really only one thing to remember: the smaller the atom, the harder it holds and attracts electrons. Fluorine — tiny, top right — is the champion at all three. Francium and caesium — huge, bottom left — are the worst at all three.

Trend: Atomic Radius

Half the distance between the nuclei of two bonded atoms

The definition

You cannot measure a single atom with a ruler — an atom has no hard edge. Instead you measure the distance between the nuclei of two bonded atoms and take half of it. That half-distance is the atomic radius, and it is the closest thing we have to "the size of the atom".

Trend & explanation

decreases across a period

Why? Z_eff increases. Going across a period there is an increasing number of protons and an increasing number of valence electrons being added to the SAME energy level. That increased attraction pulls the energy levels in closer to the nucleus, shrinking the atom.

increases down a group

Why? There is an increase in the number of energy levels. Valence electrons are being added to increasingly higher energy levels at a greater average distance from the nucleus. This increases the size of the atom.

Show:

Class examples

Which atom is larger: Na or Cl?

1Na and Cl are both in period 3 — this is an across a period comparison, so use Z_eff.

2Cl has more protons (17 vs 11) with the same 3 energy levels, so Cl has the larger Z_eff (7 vs 1).

3The stronger pull draws Cl's valence electrons in tighter, so chlorine is the smaller atom.

✓Na is larger (190 pm vs 79 pm).

Which atom is larger: Li or Cs?

1Li and Cs are both in group 1 — this is a down a group comparison, so use the number of energy levels.

2Li uses 2 energy levels; Cs uses 6. Cs's valence electron is far further from the nucleus.

3Do not say "Cs has a bigger Z_eff" — Z_eff is constant down a group (both are 1).

✓Cs is larger (298 pm vs 167 pm) — in fact it is one of the largest atoms there is.

Trend: Ionic Radius

What happens to size when an atom becomes an ion

Cations are smaller than the parent atom

Why? To form a cation an atom must lose its outer electrons. In some cases that means losing an entire energy level (Na 2,8,1 → Na⁺ 2,8). In other cases it means minimising electron–electron repulsion, letting the remaining electrons be pulled in tighter. Either way, the ion is smaller.

There is a second effect: the ion still has all its protons but fewer electrons, so each remaining electron gets a bigger share of the nuclear pull.

Anions are larger than the parent atom

Why? To form an anion an atom must gain electrons. The additional electrons cause electron–electron repulsion, which pushes the electrons away from each other — they spread out more, making the ion larger.

Same protons, more electrons — so the nuclear pull is now shared among more electrons and each one is held a little more loosely.

Atom ⇄ Ion size visualiser

Pick a species below, then press play to watch the atom change size as it becomes an ion. Circles are drawn to scale.

Atom radius

—

Ion radius

—

Cations — metals losing electrons

Anions — non-metals gaining electrons

Note: going across a period and down a group, ionic radius follows the same trend direction as atomic radius — bigger down a group, smaller across a period. Compare Li⁺ → Rb⁺ (0.60 → 1.48 Å) in the cation row, and F⁻ → I⁻ (1.36 → 2.16 Å) in the anion row.

About the numbers. These radii are quoted in ångströms (Å) from the class slides (1 Å = 100 pm) and come from a different data set than the picometre radii used elsewhere on this site. Compare atoms with atoms and ions with ions within a single table — never mix the two sets.

Class examples

Which is larger: K or K⁺?

1K is 2, 8, 8, 1. To become K⁺ it loses its single 4th-shell electron.

2That removes an entire energy level — K⁺ is 2, 8, 8, using only 3 shells.

3The 19 protons now pull on only 18 electrons, tightening the remaining shells further.

✓K is larger (2.31 Å vs 1.33 Å) — the cation is little more than half the size.

Which is larger: Br or Br⁻?

1Br has 7 valence electrons and gains one to complete its octet, forming Br⁻.

2No new energy level is added — but there are now more electrons repelling each other in the same shell.

3That extra electron–electron repulsion pushes the electron cloud outward.

✓Br⁻ is larger (1.85 Å vs 1.14 Å).

Trend: Ionization Energy

Ionic bonding in metals — the loss of electrons

The definition

X_(g) + heat → X⁺_(g) + e⁻

Ionization energy (IE) is the energy required to remove an electron from an atom in the gas phase. Because energy must be put in, the process is endothermic — which is why "+ heat" sits on the reactant side of the equation.

The value tells you how strongly an atom's nucleus holds on to its valence electrons. IE is specific to metals undergoing a loss of electrons in ionic bonding.

Trend & explanation

increases across a period

Why? Z_eff increases. Going across a period there is an increasing number of protons and an increasing number of valence electrons being added to the SAME energy level. The stronger attraction between the nucleus and the valence electrons makes it more difficult to remove an electron.

decreases down a group

Why? There is an increase in the number of energy levels. Valence electrons sit at a greater average distance from the nucleus, so the attraction is weaker (the shielding effect) and it becomes easier to remove an electron.

There are anomalies. The line zigzags: boron dips below beryllium, oxygen dips below nitrogen, and the same pattern repeats in period 3 (Al below Mg, S below P). You are not expected to explain these in Grade 11 — just know the overall trend still rises, and don't panic when a graph wobbles.

Reactivity of metals

A metal's reactivity (i.e. metallic character) is inversely related to ionization energy. Metals react by losing electrons — so the easier an electron is to remove, the more reactive the metal.

decreases across a period

Less metallic character. The increasing attraction between the valence electrons and the protons in the nucleus makes an atom less likely to lose an electron.

increases down a group

More metallic character. The atomic size is increasing, so the outer energy levels are farther away. The increased shielding effect lets electrons be lost more readily — the atom is more likely to lose an electron.

Which is more reactive: Na or Al?

1Both are in period 3 — across a period, so use Z_eff.

2Al has a larger Z_eff (3 vs 1), so it holds its valence electrons more tightly and has the higher IE (578 vs 496 kJ/mol).

3Reactivity of metals is inversely related to IE.

✓Na is more reactive — it loses its single valence electron far more readily. Al also has to lose three electrons, not one.

Which is more reactive: Li or Cs?

1Both are in group 1 — down a group, so use the number of energy levels / shielding.

2Cs uses 6 energy levels, so its valence electron is far from the nucleus and heavily shielded — its IE is much lower (376 vs 520 kJ/mol).

✓Cs is more reactive. This is exactly why caesium explodes in water and lithium merely fizzes.

Successive Ionization Energies

The first ionization energy removes the first electron; the second removes the second, and so on. Each successive electron takes more energy than the last, because once the first electron is gone the particle is positively charged — and the remaining negative electrons are held more tightly by that positive ion.

And there is a quantum leap. Once all the valence electrons have been removed, the next electron must come out of a full, stable inner shell, and the particle is now much smaller — greater attraction between the protons and the remaining electrons. The energy needed jumps enormously.

Element:

Class example — which has the larger ionization energy: Mg or Mg²⁺?

1Draw the Bohr models. Mg is 2, 8, 2 — 12 protons, 12 electrons, 3 energy levels.

2Mg²⁺ is 2, 8 — still 12 protons, but only 10 electrons and 2 energy levels. It is much smaller.

3Removing an electron from Mg²⁺ means pulling it out of a full, stable octet, from an ion that is already positively charged and smaller.

✓Mg²⁺ has by far the larger ionization energy — 7733 kJ/mol for its next electron, versus 738 kJ/mol for the first electron of neutral Mg.

Two more: Na vs Al, and Be vs Ba

1Na vs Al — same period, so use Z_eff. Al has more protons in the same 3 shells, holds tighter, so Al has the larger IE.

2Be vs Ba — same group, so use the number of energy levels. Be uses 2 shells, Ba uses 6, so Ba's valence electrons are far away and shielded: Be has the larger IE.

Trend: Electron Affinity

Ionic bonding in non-metals — the gain of electrons

The definition

X_(g) + e⁻ → X⁻_(g) + heat

Electron affinity (EA) is the energy released when a gaseous negative ion forms from a gaseous atom. Energy comes out, so the process is exothermic — hence "+ heat" on the product side. It measures the attraction an atom has for electrons other than its own.

EA is specific to non-metals gaining electrons in ionic bonding. It is the mirror image of ionization energy: IE is about metals giving, EA is about non-metals taking.

Trend & explanation

more negative across a period

More exothermic — the value 'increases' in magnitude (ignore the sign). Why? Z_eff increases. As the number of protons increases across a period — while the number of energy levels stays constant — they more strongly attract electrons from other atoms. And non-metals want to complete their octet.

more positive down a group

Less exothermic — the value 'decreases' in magnitude (ignore the sign). Why? There is an increase in the number of energy levels. Because of the large distance from the protons in the nucleus (increased shielding effect), there is less attractive force to add/gain electrons.

The sign trap. Textbooks usually write EA as a negative number because energy is released (F = −328 kJ/mol). "Largest electron affinity" almost always means largest magnitude / most energy released. On this page the bars show the magnitude — the energy released — so taller = stronger attraction for an extra electron.

Show:

Reactivity of the halogens

Reactivity of halogens decreases down a group — the opposite direction to metals, because halogens react by gaining electrons rather than losing them. The higher the electron affinity, the more reactive the atom.

Why? F is very small and has a strong nuclear pulling force acting on electrons nearby, so it releases more energy when an electron is gained — making it the most reactive. Going down the group, the increase in the number of energy levels leads to a decrease in electron affinity: the electron is gained farther from the nucleus, and less energy is released.

Which is more reactive: Cl or Br?

1Both are halogens in group 17 — down a group, so use the number of energy levels.

2Cl uses 3 energy levels; Br uses 4. Br's incoming electron lands further from the nucleus, with more shielding.

3So Cl releases more energy on gaining an electron (349 vs 325 kJ/mol) — a higher EA.

✓Cl is more reactive. Halogen reactivity falls F > Cl > Br > I.

Class examples: N vs F, and F vs I

1N vs F — same period, so use Z_eff. F has more protons in the same 2 shells and needs just one electron to complete its octet: F has the larger EA (328 vs ~0 kJ/mol — nitrogen's half-filled shell makes it unusually reluctant).

2F vs I — same group, so use the number of energy levels. F is tiny (2 shells) and I is large (5 shells), so F releases far more energy: F has the larger EA (328 vs 295 kJ/mol).

Trend: Electronegativity

Covalent bonding — the tug-of-war over shared electrons

The definition

Electronegativity is the attraction an atom has for the shared electrons in a covalent bond — its tendency to pull electron density towards itself. Unlike IE and EA it has no units: it is a relative scale, with fluorine set at 4.0 as the most electronegative element.

In water, oxygen (3.5) pulls much harder on the shared pairs than hydrogen (2.1) does. The electrons spend more time near the oxygen, giving it a partial negative charge δ− and leaving each hydrogen partially positive δ+. That unequal sharing is what makes water a polar molecule.

Trend & explanation

increases across a period

Why? Z_eff increases. The increasing number of protons generates a greater attractive force on SHARED (covalent) bonding electrons — the closer the valence shell is to full, the stronger the pull of that atom on the electrons in a bonding pair.

decreases down a group

Why? There is an increase in the number of energy levels. The shared electrons sit in increasingly higher energy levels at a greater average distance from the nucleus, so there is less attraction of the protons in the nucleus to the shared bonding electrons.

This gives the classic diagonal arrow from the bottom left of the periodic table to the top right: lowest near Fr and Cs (0.7), highest at F (4.0). Open the Trend Map and select electronegativity to see the diagonal light up.

Bond Polarity Tool

Pick two elements to bond together. The tool works out the electronegativity difference (ΔEN) and shows how the shared pair is distributed.

Atom A

—

Atom B

EN of A

—

EN of B

—

ΔEN

—

non-polar covalent

polar covalent

ionic

Class examples:

The cut-offs are guidelines, not laws. ΔEN below about 0.4 is treated as non-polar covalent, 0.4–1.7 as polar covalent, and above 1.7 as ionic. Real bonds sit on a continuous scale — there is no switch that flips at exactly 1.7.

Class examples

H–H vs H–F

1H–H: both atoms have EN 2.1, so ΔEN = 0. The pair is shared perfectly equally — a non-polar covalent bond with no δ+ or δ−.

2H–F: F is 4.0 and H is 2.1, so ΔEN = 1.9. Fluorine pulls the shared pair strongly towards itself.

✓H–F is by far the more polar bond — in fact ΔEN = 1.9 puts it right at the boundary where chemists start calling a bond ionic.

H–F vs H–Br

1F and Br are both in group 17 — down a group, so use the number of energy levels.

2Br uses 4 energy levels to F's 2, so the shared pair sits further from Br's nucleus and is attracted less: EN falls from 4.0 to 2.8.

3ΔEN: H–F = 4.0 − 2.1 = 1.9; H–Br = 2.8 − 2.1 = 0.7.

✓H–F is the more polar bond — the shared pair is pulled much further towards the halogen.

Trend Comparison Tool

Pick two species, pick a property — then justify it before you get the answer

Compare

Species A

Species B

Property

Every example from the class slides

Load one and work through the reasoning.

Atomic radius

Atomic vs ionic radius

Ionization energy

Metal reactivity

Electron affinity & halogen reactivity

Bond polarity — these open the polarity tool

"Explain the Trend" Practice

The exam question is never "what" — it is always "why"

How to answer a "why" question

1Identify the direction. Are the two elements in the same period (across) or the same group (down)?

2Pick the matching cause. Across → Z_eff. Down → number of energy levels / shielding.

3Say what changes. More protons pulling on the same shell — or valence electrons in a higher shell, further away.

4Link it to the property. "…so the atom holds its electrons more tightly, so the ionization energy is higher."

Never cite something that is constant in that direction. Z_eff does not change down a group; the number of energy levels does not change across a period. Using either one in the wrong direction is the single most common way to lose the mark.

Practice — pick the correct reasoning

Score: 0 / 0 ·

Summary

Quick reference for the whole unit

The trends at a glance

Property	What it is	Across →	Down ↓
Atomic radius	Half the distance between two bonded nuclei	decreases	increases
Ionic radius	Radius of the ion — cations smaller, anions larger than the parent	decreases	increases
Ionization energy	Energy required to remove an e⁻ (endothermic, metals)	increases	decreases
Electron affinity	Energy released forming an anion (exothermic, non-metals)	increases	decreases
Electronegativity	Pull on shared electrons in a covalent bond	increases	decreases
Z_eff	Z − S — net nuclear pull on the valence electrons	increases	constant
Metal reactivity	Inversely related to ionization energy	decreases	increases
Halogen reactivity	Follows electron affinity	—	decreases

The two explanations — that's all there is

Rule Direction decides the reason

ACROSS →

Z_eff increases. More protons, same number of energy levels, no extra shielding — a stronger pull on the valence electrons.

DOWN ↓

More energy levels (shielding). Valence electrons sit further from the nucleus and are screened by more core electrons — a weaker pull.

Constant things explain nothing. Z_eff is constant down a group. The number of energy levels is constant across a period.

Common mistakes to avoid

✗Explaining a down-a-group trend with Z_eff. It doesn't change down a group.

✗Explaining an across-a-period trend with shielding or extra energy levels. They don't change across a period.

✗Counting the valence electrons as core electrons in Z_eff = Z − S. Only inner shells shield.

✗Saying a cation is bigger because it "lost repulsion" — losing electrons always makes it smaller; gaining electrons always makes it bigger.

✗Getting the EA sign backwards. More negative = more energy released = stronger attraction for an electron.

✗Assuming metal and halogen reactivity trend the same way down a group. Metals get more reactive, halogens get less — because one loses electrons and the other gains them.

✗Confusing electron affinity (ionic bonding, a measured energy) with electronegativity (covalent bonding, a unitless tendency).

Where to next

See these trends in the real world on the Real-World Applications page — alkali metals in water, halogen reactivity, why water is polar and why noble gases glow but never react — then test yourself with the Kahoot quiz.

Related units: Atomic Structure for electron configurations and Bohr-Rutherford diagrams, and Chemical Bonding — Bohr Model for what happens once those electrons are actually transferred or shared.