Ionic radius — cation - parent atom, anion - parent atom; isoelectronic series: Worked examples

Before any symbol appears, the words and letters we lean on the whole page:

Definition The letters that decide size

$Z$ = number of protons in the nucleus. This is fixed for an element and pulls electrons inward.
$n$ = the principal quantum number: a whole number ( $1, 2, 3, \dots$ ) that labels which shell (energy level / "floor") an electron lives on. $n = 1$ is the innermost floor, closest to the nucleus; larger $n$ means a floor physically farther out. The atom's edge is set by the largest occupied $n$ — the outermost floor that still has electrons.
$S$ = shielding: how much the inner electrons "block" the pull for an outer electron. More electrons crowding around ⇒ more blocking ⇒ larger $S$ .

The electron that sets the edge of the atom feels a leftover pull: $Z_{eff} = Z - S$ Read it as "true pull minus the part that got blocked." Big $Z_{eff}$ ⇒ tight grip ⇒ small radius $r$ . Small $Z_{eff}$ ⇒ loose grip ⇒ big radius.

Definition What "

$r$ " actually means — three different rulers The letter $r$ is a size, but chemists measure "size" in three different experimental ways, and mixing them silently causes errors:

$r_{ion}$ = ionic radius: half the distance between two touching ions in a crystal (e.g. Na⁺ and Cl⁻ in rock salt). This is the ruler for ions.
$r_{cov}$ = covalent radius: half the bond length between two identical atoms sharing a bond (e.g. the Cl–Cl distance in $Cl_{2}$ ). This is the usual ruler for neutral atoms that form bonds.
$r_{vdW}$ = van der Waals radius: half the closest approach of two non-bonded atoms just touching (e.g. two neon atoms bumping). This is the ruler for noble gases and non-bonded contacts, and it is always the largest of the three because nothing is pulling the atoms together.

On this page $r$ means the ionic radius $r_{ion}$ unless a species literally cannot form an ion (a noble gas), where we must switch rulers — see Ex 5 for the reconciliation. Whenever we compare, we make sure the two numbers come from the same ruler, or we flag it.

See Effective Nuclear Charge (Z_eff) and Shielding and Penetration for where $Z_{eff}$ and $S$ come from. This page uses them as tools only.

The scenario matrix

Every ionic-radius question falls into one of these case classes. The examples below are labelled with the cell they hit, and together they cover all of them.

#	Case class	The distinguishing feature	Example that hits it
C1	Cation vs its own parent atom	electrons ↓, same $Z$	Ex 1
C2	Anion vs its own parent atom	electrons ↑, same $Z$	Ex 2
C3	Isoelectronic series, order by size	electrons fixed, $Z$ varies	Ex 3
C4	Two ions, different electron counts (NOT isoelectronic)	must compare shells first	Ex 4
C5	Degenerate / edge input (a species that equals its neighbour, or a noble-gas "zero-charge" member)	charge $= 0$ inside a series	Ex 5
C6	Limiting behaviour (extreme charge, how far can the trend push)	biggest/smallest possible pull	Ex 6
C7	Real-world word problem	translate a physical situation into the rule	Ex 7
C8	Exam-style trap / twist	mixes two rules, tempts a wrong shortcut	Ex 8

The single decision that unlocks every cell:

The figure below draws that fork as a decision tree. Look at the two branches leaving the "Same # of electrons?" question. The left branch is drawn in chalk blue and labelled YES (isoelectronic) → it sends you to a blue box "Count PROTONS (Z)". The right branch is drawn in chalk pink and labelled NO → it sends you to a pink box "Count SHELLS (n)". So whenever the text below says "blue branch" it means same electrons, count protons, and "pink branch" means different electrons, count shells first. Trace your finger down whichever colour matches your problem.

Figure — Ionic radius — cation - parent atom, anion - parent atom; isoelectronic series

Worked examples

Ex 1 — Cation vs parent (cell C1)

Worked example Statement

Magnesium is $Mg = [Ne] 3 s^{2}$ with $r \approx 160$ pm. Predict whether $Mg^{2 +}$ is bigger or smaller, and estimate roughly by how much (data: $r_{Mg^{2 +}} \approx 72$ pm).

Forecast: Guess now — does losing 2 electrons shrink it a little or a lot?

Write the electron change. $Mg = [Ne] 3 s^{2}$ has electrons in the $n = 3$ shell. Removing both $3 s$ electrons gives $Mg^{2 +} = [Ne]$ , whose outermost electrons sit in $n = 2$ . Why this step? Because size is set by the outermost occupied shell (largest $n$ ); if that shell disappears, the edge of the atom jumps inward by a whole floor.
Direction: the entire $n = 3$ shell vanished, so $Mg^{2 +} < Mg$ — a large drop, not a small one. Why this step? This is Reason A (a shell disappears) from the recap above — losing a whole floor dominates.
Cross-check with $Z_{eff}$ : $Z = 12$ stays fixed, but now only 10 electrons share the pull, so shielding $S$ falls and $Z_{eff}$ per electron rises — pulls the remaining cloud tighter still. Why this step? This is Reason B (tighter grip per electron); it confirms the shell argument — both point the same way.

Verify: $160 \to 72$ pm is a drop of $88$ pm, i.e. the ion is $72/160 = 0.45$ of the atom — less than half. Consistent with "whole shell removed." ✅

Ex 2 — Anion vs parent (cell C2)

Worked example Statement

Oxygen is $O = [He] 2 s^{2} 2 p^{4}$ , $r \approx 66$ pm. Predict $O^{2 -}$ 's size relative to $O$ (data: $r_{O^{2 -}} \approx 140$ pm).

Forecast: Same shell ( $n = 2$ ) stays occupied — so will the swell be huge, or modest?

Electron change: add 2 electrons → $O^{2 -} = [He] 2 s^{2} 2 p^{6} = [Ne]$ -configuration. No new shell opens (still $n = 2$ ), but $2 p$ goes from 4 to 6 electrons. Why this step? Unlike Ex 1, the outermost shell number ( $n$ ) does not change — so Reason A is off; the growth must come from Reason B.
Mechanism: $Z = 8$ now grips 10 electrons instead of 8. Per electron the pull weakens ( $Z_{eff} ↓$ ), and the two extra electrons add mutual repulsion, pushing the cloud out. Why this step? This is Reason B running in the "grow" direction — more electrons, same nucleus, looser grip.
Direction: $O^{2 -} > O$ , and because we crammed extra charge into a small tightly-held shell, the proportional swell is big. Why this step? Small ions have high $Z_{eff}$ ; adding electrons there causes a large relative change.

Verify: $66 \to 140$ pm ⇒ ratio $140/66 \approx 2.12$ — the ion is over twice the atom. A big swell, as forecast. ✅

Ex 3 — Isoelectronic series, full order (cell C3)

Worked example Statement

Order by size: $O^{2 -}, F^{-}, Na^{+}, Mg^{2 +}$ .

Forecast: Are these even comparable by the same rule? Count electrons for each first.

Electron counts. $O^{2 -}$ : $8 + 2 = 10$ . $F^{-}$ : $9 + 1 = 10$ . $Na^{+}$ : $11 - 1 = 10$ . $Mg^{2 +}$ : $12 - 2 = 10$ . All 10 electrons → isoelectronic. Why this step? The master fork: same electrons ⇒ we switch to counting protons.
Proton counts (this is $Z$ ): O = 8, F = 9, Na = 11, Mg = 12. Why this step? With electrons and shielding fixed, $Z_{eff} = Z - S \propto Z$ . Whoever has the most protons squeezes the same 10-electron cloud hardest.
Order largest → smallest = fewest protons → most protons: $O^{2 -} > F^{-} > Na^{+} > Mg^{2 +}$ Why this step? Fewest protons (O, 8) ⇒ loosest grip ⇒ biggest; most protons (Mg, 12) ⇒ tightest ⇒ smallest.

Verify (pm): $140 > 133 > 102 > 72$ . Monotonically decreasing as protons increase $8 < 9 < 11 < 12$ . ✅

The figure below draws exactly this series. Notice three things as your eye moves left to right: (1) the four circles get visibly smaller; (2) the number in the centre — the proton count — grows from $8 p$ to $12 p$ ; (3) the colour shifts from chalk blue (the two anions O²⁻, F⁻) to chalk pink (the two cations Na⁺, Mg²⁺), a reminder that all four still share the same 10-electron cloud. The arrow underneath states the rule the picture proves: protons up ⇒ radius down.

Ex 4 — Different electron counts, NOT isoelectronic (cell C4)

Worked example Statement

Which is bigger: $Li^{+}$ or $Br^{-}$ ?

Forecast: Careful — do NOT count protons blindly here. First check electrons.

Electron counts: $Li^{+} = 3 - 1 = 2$ electrons ( $[He]$ , outermost $n = 1$ ). $Br^{-} = 35 + 1 = 36$ electrons ( $[Kr]$ , outermost $n = 4$ ). Why this step? $2 \neq = 36$ — not isoelectronic. The proton-counting rule does NOT apply. Master fork sends us to the pink branch ("count shells").
Compare outermost shells: $Li^{+}$ tops out at $n = 1$ ; $Br^{-}$ fills out to $n = 4$ . Four floors beat one floor by a mile. Why this step? When electron counts differ, the largest occupied $n$ is the dominant size factor — a bigger $n$ means a physically bigger cloud.
Conclusion: $Br^{-} ≫ Li^{+}$ . Why this step? No $Z_{eff}$ subtlety can overcome three extra shells.

Verify (pm): $r_{Br^{-}} \approx 196$ , $r_{Li^{+}} \approx 76$ . Indeed $196 > 76$ , ratio $\approx 2.58$ . ✅

Ex 5 — Degenerate / edge input: the noble gas inside a series (cell C5)

Worked example Statement

Place Ne (charge $0$ ) into the 10-electron series and confirm the rule survives the "zero-charge" edge case: order $F^{-}, Ne, Na^{+}$ .

Forecast: A neutral atom sitting between two ions — does the proton rule still work when one member has zero charge?

Electrons: $F^{-} = 10$ , $Ne = 10$ , $Na^{+} = 10$ . All isoelectronic — charge sign is irrelevant to this count. Why this step? The rule keys on electron count, not charge. A charge of $0$ is a perfectly valid series member.
Protons: F = 9, Ne = 10, Na = 11. Why this step? Same logic: $Z_{eff} \propto Z$ , so order by $Z$ .
Order: $F^{-} (9) > Ne (10) > Na^{+} (11)$ . Why this step? The neutral atom is not special — it slots in exactly where its proton count says.
Ruler warning — reconcile the definitions. F⁻ and Na⁺ are measured with the ionic ruler ( $r_{ion}$ ), but Ne, being a noble gas, cannot form a crystal of ions — its tabulated size is a van der Waals radius ( $r_{vdW}$ ), a different, systematically larger ruler (see the definition box up top). So we may trust the ordering (protons don't lie) but must not read the raw numbers as if they were on one scale. Why this step? Otherwise a naive lookup ( $Ne \approx 154$ pm van der Waals) would falsely place Ne above F⁻ — a ruler-mixing artefact, not real chemistry.

Verify (order, using like-for-like ionic-scale estimates): on a consistent ionic scale, $F^{-} \approx 133$ , $Ne \approx 112$ , $Na^{+} \approx 102$ pm — decreasing with $Z$ : $133 > 112 > 102$ . ✅ The zero-charge case obeys the rule once the rulers are matched.

Ex 6 — Limiting behaviour (cell C6)

Worked example Statement

In the 10-electron series, how far can the shrinking go? Compare the two extremes present in the parent's full set: $N^{3 -}$ (fewest protons, 7) vs $Al^{3 +}$ (most protons, 13).

Forecast: Both differ from Ne by 3 charge units. Guess: is the difference in size roughly symmetric, or lopsided?

Both isoelectronic (10 electrons), so $S$ is nearly the same for both. Why this step? Shielding $S$ depends on how the electrons are arranged, and both species have the same 10-electron $[Ne]$ arrangement — so the "blocking" is essentially identical and only $Z$ changes.
Build $S$ from scratch with Slater's grouping (so the number isn't magic). Slater's rules bin the 10 electrons and give each bin a screening weight toward our target — an outer $n = 2$ electron:
- The other electrons in the same $n = 2$ group (there are $8 - 1 = 7$ of them: the $2 s^{2} 2 p^{6}$ minus our own electron) each screen by 0.35.
- The two $1 s$ electrons (one shell deeper, $n = 1$ ) each screen by 0.85.
Add them up: $S = same shell 7 \times 0.35 + inner shell 2 \times 0.85 = 2.45 + 1.70 = 4.15.$ Why this step? Now $S \approx 4.15$ is derived, not quoted — same-shell neighbours block weakly (0.35), the closer inner shell blocks strongly (0.85). Both ions share this identical cloud, so both use this same $S$ .
Compute the effective pulls and their ratio. $Z_{eff} (N^{3 -}) = 7 - 4.15 = 2.85, Z_{eff} (Al^{3 +}) = 13 - 4.15 = 8.85.$ $\frac{Z _{eff} ( Al ^{3 +} )}{Z _{eff} ( N ^{3 -} )} = \frac{8.85}{2.85} \approx 3.1$ Why this step? This ratio measures how much harder Al³⁺ grips the identical cloud than N³⁻ does.
Turn pull into size — WHY $r \sim 1/ Z_{eff}$ . Picture the outer electron in a hydrogen-like orbit around an effective charge $Z_{eff}$ . The nucleus pulls with a force $\propto Z_{eff} / r^{2}$ (Coulomb's law), and for a stable orbit this pull sets the size. Solving the balance for a one-electron atom gives the Bohr result $r = n^{2} a_{0} / Z_{eff}$ , where $a_{0}$ is a fixed length (the Bohr radius). With $n$ the same for both ions, everything except $Z_{eff}$ is constant, so $r \propto 1/ Z_{eff}$ : triple the effective pull ⇒ about one-third the radius. Why this step? This supplies the missing link between "stronger grip" and an actual size ratio, rather than asserting it.
Predict: since Al³⁺ has $\approx 3.1 \times$ the pull, its radius should be roughly $1/3.1 \approx 0.32$ of N³⁻'s. So $N^{3 -}$ is the largest, $Al^{3 +}$ the smallest, with a dramatic span.

Verify (pm): $r_{N^{3 -}} \approx 146$ , $r_{Al^{3 +}} \approx 54$ . Ratio $146/54 \approx 2.70$ — same order of magnitude as the predicted $\approx 3.1$ from $Z_{eff}$ (the simple Bohr model overshoots slightly, as expected for a rough estimate). ✅

Ex 7 — Real-world word problem (cell C7)

Worked example Statement

A materials engineer wants a small $+ 2$ cation to fit into a tight crystal site of radius $\approx 75$ pm. She's choosing between $Mg^{2 +}$ and $Ca^{2 +}$ . Which fits, and why does the reasoning reduce to our rule? (Data: $r_{Mg^{2 +}} \approx 72$ pm, $r_{Ca^{2 +}} \approx 100$ pm.)

Forecast: Same charge $+ 2$ , same group — so what breaks the tie?

Configurations: $Mg^{2 +} = [Ne]$ (outermost $n = 2$ ). $Ca^{2 +} = [Ar]$ (outermost $n = 3$ ). Why this step? Same charge, but $Ca^{2 +}$ retains a higher outer shell — so this is a different electron count comparison (10 vs 18), the C4 / pink branch.
Shells decide: $Ca^{2 +}$ 's $n = 3$ edge sits farther out than $Mg^{2 +}$ 's $n = 2$ edge ⇒ $Ca^{2 +}$ is bigger. Why this step? More occupied shells (larger $n$ ) ⇒ larger ion, independent of the identical charge.
Fit check: the site is $75$ pm. $Mg^{2 +}$ ( $72$ pm) fits; $Ca^{2 +}$ ( $100$ pm) is too big. Why this step? Translate the size ranking back into the physical requirement.

Verify: $72 < 75 < 100$ — $Mg^{2 +}$ fits, $Ca^{2 +}$ does not. ✅ The choice reduced to "fewer shells ⇒ smaller ion."

Ex 8 — Exam-style trap (cell C8)

Worked example Statement

An exam claims: " $S^{2 -}$ has 18 electrons and $Ca^{2 +}$ has 18 electrons, so they're the same size." Find the flaw and give the correct order.

Forecast: Spot the tempting-but-wrong shortcut before reading on.

Verify the isoelectronic claim: $S^{2 -} = 16 + 2 = 18$ , $Ca^{2 +} = 20 - 2 = 18$ . Yes, both 18 electrons. Why this step? The claim's premise is true — that's what makes the trap seductive.
Expose the flaw: "same electrons ⇒ same size" is false. Same electrons means we must now compare protons, not stop. S has 16 protons, Ca has 20. Why this step? This is exactly the parent note's steel-man mistake: isoelectronic ⇒ compare $Z$ (blue branch), and $Z$ differs a lot here.
Correct order: fewer protons ⇒ looser grip ⇒ bigger. $16 < 20$ , so $S^{2 -} > Ca^{2 +} .$ Why this step? Ca's 20 protons squeeze the same 18 electrons far harder than S's 16.

Verify (pm): $r_{S^{2 -}} \approx 184$ , $r_{Ca^{2 +}} \approx 100$ — very different, $184 > 100$ . ✅ The "same size" claim is badly wrong.

Recall Self-test: name the cell, then solve

Rank $Se^{2 -}$ , $Br^{-}$ , $Rb^{+}$ , $Sr^{2 +}$ . Which cell is this? ::: C3 — isoelectronic series (all 36 electrons, Kr-like). Give the order. ::: $Se^{2 -} > Br^{-} > Rb^{+} > Sr^{2 +}$ (protons 34 < 35 < 37 < 38; fewer protons = bigger).

Connections

Effective Nuclear Charge (Z_eff) — the $Z_{eff} = Z - S$ used in every example.
Shielding and Penetration — where $S$ (and Slater's rules for it) come from.
Atomic Radius Trends — the neutral-atom sizes every comparison starts from.
Ionization Energy — tightly-held small cations resist further ionization.
Lattice Energy — ion sizes from Ex 7 feed directly into crystal packing.
Electron Affinity — energetics of forming the anions in Ex 2 and Ex 3.