Ionic Radius & Isoelectronic Species

What Is 'Ionic Radius'?

In Section 6 we defined the atomic radius of a neutral atom — the distance from the nucleus to the effective edge of its electron cloud. But most of the chemistry you study deals not with neutral atoms but with ions: Na$^{+}$ in salt water, Cl$^{-}$ at your dinner table, Ca$^{2+}$ in your bones, O$^{2-}$ locked inside every oxide you know.

So we need a parallel idea: the ionic radius — the effective size of an ion.

How is it measured?

Unlike neutral atoms, you cannot look at "two ions of the same kind bonded together" — cations and cations repel, anions and anions repel. Instead, we measure ionic radii in ionic crystals, where cations and anions sit right next to each other at a well-defined distance.

For example, in the crystal of sodium chloride (NaCl):

$$d(\text{Na}^{+}\!-\!\text{Cl}^{-}) \;=\; r_{\text{Na}^{+}} + r_{\text{Cl}^{-}} \;=\; 283\ \text{pm}$$

X-ray diffraction tells us the cation-to-anion distance to within a picometre, so we know the sum $r_{+} + r_{-}$ very precisely. The tricky part is: how do we split that single number into two individual radii?

The Landé method — anchor with one ion

In 1920, the German chemist Alfred Landé solved this by studying crystals where the anion is so much larger than the cation that the anions actually touch each other, squeezing the tiny cation into the gap between them. The classic example is LiI: Li$^{+}$ is small (76 pm) and I$^{-}$ is huge (220 pm). In LiI, the iodide ions pack almost like tennis balls in a crate, and the iodide-iodide contact distance is just $2\,r(\text{I}^{-}) = 440$ pm.

From that one anchoring measurement, we get $r(\text{I}^{-}) = 220$ pm. Then, using the I$^{-}$-to-cation distance in other iodides (KI, RbI, CsI), we back out $r(\text{K}^{+})$, $r(\text{Rb}^{+})$, and $r(\text{Cs}^{+})$. Those in turn let us back out chloride, bromide, and fluoride radii from other crystals — and so on, building up a consistent table of ionic radii for the whole periodic table.

A baseline rule of thumb

Ionic radii are always compared at the same coordination number (typically 6, as in NaCl or rock-salt structure). If you see "$r(\text{Na}^{+}) = 102$ pm" quoted, this is the CN-6 value. Higher coordination → slightly larger ionic radius (the ion has more room); lower coordination → smaller. Most Class 11 tables just quote CN-6 values, and that's what we'll use throughout this section.

[JEE Tip] Ionic radii are measured in solids and depend on coordination number. Don't mix values from different sources (different scales: Pauling, Shannon, Ahrens) in one calculation. NCERT uses Pauling's scale, so stick to that.

Cations Are Always Smaller Than the Parent Atom

Here is the first, and most important, observation about ionic radii:

A cation is always smaller than its parent atom.

Consider Group 1:

Atom	$r_{\text{atom}}$ / pm	Ion	$r_{\text{ion}}$ / pm	Ratio
Li	152	Li$^{+}$	76	0.50
Na	186	Na$^{+}$	102	0.55
K	227	K$^{+}$	138	0.61
Rb	248	Rb$^{+}$	152	0.61
Cs	265	Cs$^{+}$	167	0.63

The cation is about half the size of the parent atom — a massive shrinkage.

Two reasons the cation shrinks

The explanation runs on two tracks that reinforce each other.

Reason 1: the valence shell is completely removed.

When a Group-1 atom loses its one valence electron, it loses the only electron in its outermost shell. The new outermost shell is the one below — an older, smaller shell. For sodium:

$$\text{Na}\;\underbrace{1s^{2}\,2s^{2}\,2p^{6}\,3s^{1}}_{\text{three shells}} \;\;\longrightarrow\;\; \text{Na}^{+}\;\underbrace{1s^{2}\,2s^{2}\,2p^{6}}_{\text{two shells; [Ne]}}$$

Going from three shells (Na) to two shells (Na$^{+}$) is a much bigger change than just "losing one electron." The outer shell radius of an atom roughly scales with $n^{2}$, so dropping from $n = 3$ to $n = 2$ can easily slash the radius in half.

Reason 2: the proton-to-electron ratio rises.

Before losing the electron, Na has 11 protons and 11 electrons — the $Z/e^{-}$ ratio is exactly 1. After losing one electron, Na$^{+}$ has 11 protons and only 10 electrons. The ratio jumps to $11/10 = 1.10$. Each of the remaining 10 electrons now feels, on average, 10% more "pull per electron" from the nucleus. The electron cloud contracts.

For higher-charge cations, the shrinkage is even more dramatic

The effect grows with charge. Compare magnesium:

• Mg atom: 160 pm
• Mg$^{2+}$: 72 pm ← less than half the atomic radius

Mg$^{2+}$ has lost both $3s$ electrons; it is now [Ne], like Na$^{+}$, but with 12 protons pulling in 10 electrons ($Z/e^{-} = 1.20$). The cloud is pulled even tighter than Na$^{+}$'s.

The general rule

For any atom X and its cation X$^{n+}$:

• Larger positive charge → smaller ion. (Fe$^{3+}$ 65 pm < Fe$^{2+}$ 78 pm.)
• If the cation has lost its entire valence shell, the shrinkage is especially dramatic.

[Board Level Important] A common Board exam question asks "Why is the ionic radius of a cation smaller than that of the parent atom?" — always mention both reasons: (i) loss of the outermost shell, and (ii) increase in the $Z/e^{-}$ ratio. Single-reason answers lose half the marks.

Anions Are Always Larger Than the Parent Atom

The mirror rule is equally clean:

An anion is always larger than its parent atom.

Compare Group 17:

Atom	$r_{\text{atom}}$ / pm	Ion	$r_{\text{ion}}$ / pm	Ratio
F	64	F$^{-}$	133	2.08
Cl	99	Cl$^{-}$	181	1.83
Br	114	Br$^{-}$	196	1.72
I	133	I$^{-}$	220	1.65

Halide anions are 1.6 to 2 times larger than the parent atom — a big expansion.

Two reasons the anion expands

Reason 1: no new shell, but more electrons in the same shell.

When Cl gains an electron, it does NOT start a new outer shell (that would take a huge energy jump). The new electron joins the existing $3p$ sub-shell:

$$\text{Cl}\;1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{5} \;\longrightarrow\; \text{Cl}^{-}\;1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\;=\;[\text{Ar}]$$

So the ion's electron cloud occupies the same $n=3$ shell as the parent atom — but now with one more electron in that shell.

Reason 2: the proton-to-electron ratio falls.

Chlorine has 17 protons and 17 electrons ($Z/e^{-} = 1.00$). Cl$^{-}$ has 17 protons pulling in 18 electrons ($Z/e^{-} = 17/18 = 0.94$). Each electron now feels 6% less pull per electron, on average. Combined with the fact that the extra electron contributes to electron-electron repulsion inside the valence shell, the cloud puffs outward.

The oxide ion — a special warning

The oxide ion O$^{2-}$ has gained two electrons relative to the neutral O atom:

$$\text{O}\;1s^{2}\,2s^{2}\,2p^{4}\,\;(66\ \text{pm}) \;\longrightarrow\; \text{O}^{2-}\;1s^{2}\,2s^{2}\,2p^{6}\;=\;[\text{Ne}]\,\;(140\ \text{pm})$$

The oxide ion is more than twice as large as the neutral O atom. Its $Z/e^{-}$ ratio has crashed from 1.00 to $8/10 = 0.80$, and the electron-electron repulsion in the now-doubly-populated outer shell inflates the cloud still further.

[JEE Tip] The extra-large size of O$^{2-}$ is a recurring theme in inorganic chemistry — it's why many metal oxides are surprisingly stable ionic solids (lots of room for the O$^{2-}$ to sit), and why lattice energies for oxide salts behave differently from those of fluoride salts.

The universal ordering

For any atom X and its anion X$^{n-}$:

• Higher negative charge → larger ion. (S$^{2-}$ 184 pm > S neutral 104 pm.)
• The added electrons never open a new shell; they crowd the existing outer shell, amplifying mutual repulsion.

Putting cations and anions together, we can summarise the whole picture in one line:

$$r(\text{X}^{n+}) \;<\; r(\text{X atom}) \;<\; r(\text{X}^{n-})$$

The parent atom always sits somewhere in the middle, usually closer to the cation than to the anion on a picometre scale.

Isoelectronic Species — When Electron Counts Are Equal

Here is the most elegant setup in Class 11 chemistry: take several ions (or atoms) that all have exactly the same number of electrons, and watch what happens to their sizes. Such species are called isoelectronic species.

Definition

Two species are isoelectronic if they have the same total number of electrons (and, usually, the same electronic configuration as a noble gas).

A classic isoelectronic set, all of which have 10 electrons ($1s^{2}\,2s^{2}\,2p^{6}$ = [Ne]):

Species	$Z$	$e^{-}$	$Z/e^{-}$	Ionic radius / pm
N$^{3-}$	7	10	0.70	171
O$^{2-}$	8	10	0.80	140
F$^{-}$	9	10	0.90	133
Ne	10	10	1.00	(≈160, vdW)
Na$^{+}$	11	10	1.10	102
Mg$^{2+}$	12	10	1.20	72
Al$^{3+}$	13	10	1.30	54

The master insight

All seven species have exactly the same 10 electrons (and, for Ne and all the ions, exactly the same orbital occupation $1s^{2}\,2s^{2}\,2p^{6}$). What differs is only the nuclear charge $Z$.

• N$^{3-}$ ($Z = 7$): 10 electrons pulled by just 7 protons — the electrons are loosely held, cloud is huge (171 pm).
• Al$^{3+}$ ($Z = 13$): 10 electrons pulled by 13 protons — the electrons are clamped in tight, cloud is tiny (54 pm).

Between the two extremes, the ratio $Z/e^{-}$ rises from 0.70 to 1.30 — a factor of 1.86 — and the radius shrinks by a factor of $171/54 = 3.17$. The same 10 electrons, pulled by successively more protons, live in successively smaller clouds.

The general rule for isoelectronic series

Within an isoelectronic series, radius decreases as nuclear charge increases.

Equivalently: more negative charge $\to$ larger ion; more positive charge $\to$ smaller ion. This is different from the atomic-radius rule (where we had to worry about shell-number as well), because within an isoelectronic series the shell structure is fixed — only $Z$ varies.

Another common isoelectronic set: 18 electrons

The 18-electron (Ar-like) isoelectronic series works the same way:

$$\underbrace{\text{S}^{2-}}_{184\text{ pm}} > \underbrace{\text{Cl}^{-}}_{181\text{ pm}} > \underbrace{\text{Ar}}_{\sim 188\text{ pm, vdW}} > \underbrace{\text{K}^{+}}_{138\text{ pm}} > \underbrace{\text{Ca}^{2+}}_{100\text{ pm}} > \underbrace{\text{Sc}^{3+}}_{75\text{ pm}}$$

And the 2-electron (He-like) series:

$$\text{H}^{-} > \text{He} > \text{Li}^{+} > \text{Be}^{2+}$$

The rule holds every time.

[NEET Important] Isoelectronic ordering is one of the most common MCQ types in NEET (and JEE Main). The trick is to NOT look at the symbol — just compute $Z$ for each species and rank in decreasing order for size. Students who try to reason through "well, O$^{2-}$ has more electrons than F$^{-}$" lose marks. The electron counts are equal by construction.

Ionic Radii Down a Group — The Familiar Pattern Returns

Once we move outside an isoelectronic set, the periodic-table patterns we learned for atomic radii come back with almost the same logic.

Down a group — radii increase (just like atoms)

Going down Group 1 (alkali metal cations):

Cation	Config	Radius / pm
Li$^{+}$	[He] = $1s^{2}$	76
Na$^{+}$	[Ne] = $1s^{2}\,2s^{2}\,2p^{6}$	102
K$^{+}$	[Ar] = [Ne] $3s^{2}\,3p^{6}$	138
Rb$^{+}$	[Kr] = [Ar] $3d^{10}\,4s^{2}\,4p^{6}$	152
Cs$^{+}$	[Xe] = [Kr] $4d^{10}\,5s^{2}\,5p^{6}$	167

The cation grows from 76 pm to 167 pm — more than doubles — just as the neutral atoms did. Same logic: each step down adds a whole new shell that is further from the nucleus.

Going down Group 17 (halide anions):

Anion	Config	Radius / pm
F$^{-}$	[Ne]	133
Cl$^{-}$	[Ar]	181
Br$^{-}$	[Kr]	196
I$^{-}$	[Xe]	220

Same upward trend — each step adds a new shell.

Across a period — it gets complicated

Unlike neutral atoms, we can't just say "ionic radius decreases across a period," because different elements in the same period form ions of different charges.

For instance, across Period 3 we might compare:

• Na$^{+}$ (loses 1 $e^{-}$): 102 pm
• Mg$^{2+}$ (loses 2 $e^{-}$): 72 pm
• Al$^{3+}$ (loses 3 $e^{-}$): 54 pm

These three are an isoelectronic series (all [Ne]), so their radii decrease purely because $Z$ increases. But then:

• Si$^{4+}$: exists only as a theoretical species in some contexts
• P$^{3-}$: 212 pm (now an anion — jumps back up in size)
• S$^{2-}$: 184 pm
• Cl$^{-}$: 181 pm

The "across a period" plot of ionic radius has a huge discontinuity around the middle of the period — cations on the left are small, anions on the right are large, and the jump at the boundary is often 100 pm or more.

So for Class 11 exams, the rules you actually use are:

1. Down a group (same charge ion): radius increases, just like atoms.
2. Within an isoelectronic series: radius decreases with increasing $Z$.
3. For a fixed parent atom: more positive charge → smaller; more negative charge → larger.

Extra wrinkle — lanthanoid and d-block contractions

The lanthanoid contraction we met in Section 6 also shows up for ions. So:

• $r(\text{Zr}^{4+}) \approx r(\text{Hf}^{4+})$ — nearly identical, both [Xe]-core cations with poorly-shielding $4f^{14}$ inside.
• Some post-transition or coinage-metal cations can also be smaller than naive shell-counting predicts because of d-block contraction. For example, $r(\text{Ag}^{+})$ is smaller than one might expect from simply moving one period below Cu$^{+}$ or K$^{+}$.

[JEE Tip] Whenever a problem compares a pre-transition cation with a post-transition or coinage-metal cation, check whether d-block contraction is making the lower ion smaller than expected.

Subtle Points and Classic Exam Traps

Section 7 feels simple at first, but every Board / JEE / NEET paper hides at least one trap in the ionic-radius topic. Here are the most common ones — pre-learn them and you will save several marks.

Trap 1: Comparing ions of different charges

When comparing species with different charges, most students automatically apply the isoelectronic rule and get the wrong answer. Example:

"Arrange in increasing order of size: F$^{-}$, Na$^{+}$, Mg$^{2+}$"

Students who remember just "more $Z$ → smaller" assume this is isoelectronic and rank by $Z$. That happens to be correct here (all three have 10 electrons), and the answer is Mg$^{2+}$ (72) < Na$^{+}$ (102) < F$^{-}$ (133).

But try:

"Arrange in increasing order: F$^{-}$, Cl$^{-}$, Br$^{-}$"

These are NOT isoelectronic (10, 18, and 36 electrons). Now we use the down-a-group rule: F$^{-}$ (133) < Cl$^{-}$ (181) < Br$^{-}$ (196).

Always check isoelectronicity first by counting electrons. If they're equal, use the $Z$ rule. If not, use the periodic trend rules.

Trap 2: Ne in the 10-electron series

Neon is isoelectronic with N$^{3-}$, O$^{2-}$, F$^{-}$, Na$^{+}$, Mg$^{2+}$, Al$^{3+}$. But Ne is a neutral atom, and its "radius" is typically quoted as its van der Waals radius (≈160 pm) — which is much larger than F$^{-}$ (133 pm).

So if you naïvely place Ne on the $Z/e^{-}$ ordering, it should sit between F$^{-}$ (0.90) and Na$^{+}$ (1.10) — implying a radius around 110-120 pm. But the actual tabulated 160 pm for Ne makes no sense in that scheme. Why?

Because Ne has no ionic radius. You can't measure Ne in an ionic crystal — it's a noble gas and doesn't form salts. Its 160-pm value is a van der Waals radius, a different beast entirely.

[JEE Trap] If an exam question asks you to include Ne in an isoelectronic-radius ranking, either (a) the question is setting you up to say Ne's value isn't comparable, or (b) it wants the $Z/e^{-}$ prediction, not the tabulated vdW. Read the question's wording carefully.

Trap 3: Ga$^{3+}$ vs Al$^{3+}$

Gallium sits right below aluminium in Group 13. You'd expect Ga$^{3+}$ > Al$^{3+}$ by the down-a-group rule. The numbers:

• Al$^{3+}$: 54 pm
• Ga$^{3+}$: 62 pm

Yes, Ga$^{3+}$ is a little larger — but not by as much as you'd expect (only ≈ 15%). The reason is the d-block contraction we discussed in Note 5: the $3d^{10}$ filling between Al and Ga compresses Ga and all subsequent Period 4 elements.

Trap 4: Fe$^{2+}$ vs Fe$^{3+}$

Iron shows us how charge alone changes the radius dramatically, holding everything else fixed:

• Fe: 126 pm (neutral)
• Fe$^{2+}$: 78 pm (high-spin, CN 6)
• Fe$^{3+}$: 65 pm (high-spin, CN 6)

Going from Fe to Fe$^{2+}$ is a 38% shrinkage; Fe$^{2+}$ to Fe$^{3+}$ is another 17% shrinkage. Higher charge always means smaller radius.

[Board Level Important] This pattern (greater positive charge → smaller radius) is the textbook explanation for why Fe$^{3+}$ salts are more strongly hydrolysed than Fe$^{2+}$ salts in water — the smaller, higher-charged Fe$^{3+}$ cation polarises surrounding water molecules more effectively, enhancing the O-H bond cleavage.

Trap 5: H$^{-}$ vs H — the hydride paradox

The hydride ion H$^{-}$ is surprisingly large. It has 1 proton pulling in 2 electrons ($Z/e^{-} = 0.5$) — half the proton density of Li$^{+}$ (which has 3 protons pulling 2 electrons, $Z/e^{-} = 1.5$). The consequence:

• H atom: 53 pm (Bohr-radius-scale)
• H$^{-}$: ≈ 208 pm (!!)
• Li$^{+}$: 76 pm

H$^{-}$ is bigger than any alkali-metal cation despite being "lower" on the periodic table. Isoelectronic with He ($Z/e^{-} = 1$) and Li$^{+}$ ($Z/e^{-} = 1.5$), it has the smallest $Z/e^{-}$ of the three, so it's the largest of the three. Remember: nothing intrinsic about hydrogen makes its ion small — $Z/e^{-}$ runs the show.

[JEE Tip] The ordering $r(\text{H}^{-}) > r(\text{He}) > r(\text{Li}^{+}) > r(\text{Be}^{2+})$ is a 100% JEE-favourite isoelectronic question.

Trap 6: Cation and anion of the SAME element

What if a single element can form both a cation and an anion? Hydrogen is the obvious case:

• H$^{+}$: ≈ $10^{-6}$ pm (just a bare proton; effectively point-like)
• H (neutral): 53 pm
• H$^{-}$: 208 pm

So for the same element, $r(\text{H}^{+}) \ll r(\text{H}) \ll r(\text{H}^{-})$ by factors of $10^{-5}$ and 4 respectively — the full range of sizes an element can display. Useful to remember when interpreting hydrogen chemistry.

The master cheat-sheet

Put all the ideas together:

Situation	Rule
Atom → cation ($n+$)	shrinks (big!) because valence shell lost + $Z/e^{-}$ up
Atom → anion ($n-$)	grows because $Z/e^{-}$ down + electron repulsion up
Compare ions of different charges for the same element	$r(\text{X}^{n+}) < r(\text{X}) < r(\text{X}^{n-})$, more charge → more extreme
Compare isoelectronic species	higher $Z$ → smaller ion
Compare same-charge ions down a group	larger ion at the bottom (new shell)
Some post-transition / coinage-metal cations	can be smaller than expected (d-block contraction)
Post-lanthanoid element vs its 5d analogue	nearly identical (lanthanoid contraction)

If you can recite those seven rules and recognise which one a problem needs, you can solve any Class 11 ionic-radius question in under 60 seconds.

Solved Examples

Example 1: Cation smaller than parent atom

Why is the ionic radius of $\text{Na}^{+}$ (102 pm) much smaller than the atomic radius of Na (186 pm)? Explain in two sentences.

Solution:

When Na loses its $3s^{1}$ valence electron, it also loses its entire outermost shell ($n = 3$); the new outermost shell of Na$^{+}$ is $n = 2$, which is intrinsically smaller. In addition, the proton-to-electron ratio rises from $11/11 = 1.00$ (Na) to $11/10 = 1.10$ (Na$^{+}$), so the remaining 10 electrons are pulled in more tightly by the unchanged +11 nuclear charge.

Both effects point the same way, and together they shrink the radius from 186 pm to 102 pm — a 45% reduction.

Example 2: Anion larger than parent atom

Calculate the ratio $r(\text{Cl}^{-}) / r(\text{Cl})$ and explain why the ratio is greater than 1.

Solution:

From the tabulated values:

$$\frac{r(\text{Cl}^{-})}{r(\text{Cl})} = \frac{181\ \text{pm}}{99\ \text{pm}} = 1.83$$

The anion is almost twice as large as the parent atom. Two reasons:

1. The added electron goes into the already-occupied $3p$ sub-shell (configuration becomes $3p^{6}$ = [Ar]); it does NOT open a new outer shell. But the outer shell is now more crowded, and the extra electron-electron repulsion pushes the whole cloud outward.
2. The $Z/e^{-}$ ratio falls from $17/17 = 1.00$ to $17/18 = 0.94$; each electron now feels less nuclear pull per electron, so the cloud expands.

Takeaway: anions always grow; the ratio $r_{\text{ion}}/r_{\text{atom}}$ for halide anions is typically 1.5–2.

Example 3: Ranking an isoelectronic series

Arrange in increasing order of ionic radius: $\text{O}^{2-},\ \text{F}^{-},\ \text{Na}^{+},\ \text{Mg}^{2+}$.

Solution:

Step 1 — check that they're isoelectronic.

Species	$Z$	Charge	Electrons
O$^{2-}$	8	-2	10
F$^{-}$	9	-1	10
Na$^{+}$	11	+1	10
Mg$^{2+}$	12	+2	10

All four have 10 electrons — yes, isoelectronic.

Step 2 — apply the isoelectronic rule: within an isoelectronic series, radius decreases as $Z$ increases. So radius rises as $Z$ falls.

$$r(\text{Mg}^{2+}) < r(\text{Na}^{+}) < r(\text{F}^{-}) < r(\text{O}^{2-})$$

Step 3 — verify with actual values: 72 < 102 < 133 < 140 pm. ✓

Answer: $\text{Mg}^{2+} < \text{Na}^{+} < \text{F}^{-} < \text{O}^{2-}$.

Example 4: The 18-electron isoelectronic series

Which of the following has the largest radius: $\text{S}^{2-},\ \text{Cl}^{-},\ \text{K}^{+},\ \text{Ca}^{2+}$?

Solution:

Each species has 18 electrons (configuration [Ar]), so all four are isoelectronic.

Species	$Z$	Electrons	$Z/e^{-}$	Radius / pm
S$^{2-}$	16	18	0.89	184
Cl$^{-}$	17	18	0.94	181
K$^{+}$	19	18	1.06	138
Ca$^{2+}$	20	18	1.11	100

Within an isoelectronic series, the lowest $Z$ gives the largest radius. Here S$^{2-}$ has the smallest $Z$ (16), so:

Largest = S$^{2-}$ (184 pm).

The ordering is $\text{Ca}^{2+} < \text{K}^{+} < \text{Cl}^{-} < \text{S}^{2-}$ for increasing size.

Example 5: Mixed series — NOT fully isoelectronic

Arrange in increasing order of radius: $\text{N}^{3-},\ \text{P}^{3-},\ \text{As}^{3-}$.

Solution:

Step 1 — count electrons.

• N$^{3-}$: $Z = 7$, electrons = 10 → [Ne]
• P$^{3-}$: $Z = 15$, electrons = 18 → [Ar]
• As$^{3-}$: $Z = 33$, electrons = 36 → [Kr]

These are not isoelectronic. They are, however, the $3-$ ions of three consecutive Group-15 elements — so we use the down-a-group rule, which gives increasing size (new shell at each step).

$$r(\text{N}^{3-}) < r(\text{P}^{3-}) < r(\text{As}^{3-})$$

Step 2 — verify with numbers: 171 < 212 < 222 pm. ✓

Takeaway: when comparing same-charge ions of elements in the same group, use the down-a-group rule, not the isoelectronic rule.

Example 6: Cation of the same element with different charges

Arrange in increasing order: $\text{Fe},\ \text{Fe}^{2+},\ \text{Fe}^{3+}$.

Solution:

For a single element, the rule is simple: more positive charge → smaller ion. The higher the charge, the greater the $Z/e^{-}$ ratio, and the tighter the remaining electrons are held.

Numbers (all at CN 6, high-spin):

• Fe: 126 pm
• Fe$^{2+}$: 78 pm
• Fe$^{3+}$: 65 pm

So:

$$r(\text{Fe}^{3+}) < r(\text{Fe}^{2+}) < r(\text{Fe})$$

Takeaway: the parent atom is the largest; each additional unit of positive charge strips further electrons and shrinks the ion further.

Example 7: Lanthanoid contraction for ions

Compare the ionic radii of $\text{Zr}^{4+}$ and $\text{Hf}^{4+}$. Explain why they are almost identical.

Solution:

• Zr$^{4+}$ (Period 5): 72 pm
• Hf$^{4+}$ (Period 6): 71 pm

Naively, going from Period 5 to Period 6 should make Hf$^{4+}$ larger because a full new shell is added. But the 14 lanthanoid elements between them contribute 14 protons and 14 $4f$-electrons. The $4f$ electrons are very poor shielders, so $Z_{\text{eff}}$ rises substantially through the lanthanoid series.

The extra $Z_{\text{eff}}$ pulls Hf$^{4+}$'s electrons back in, cancelling the expected expansion. The net result is that $r(\text{Zr}^{4+}) \approx r(\text{Hf}^{4+})$.

Takeaway: the lanthanoid contraction is one of the main reasons for the famously similar chemistry of the Zr/Hf, Nb/Ta, and Mo/W pairs.

Example 8: Sum of ionic radii in a crystal

The internuclear distance between $\text{K}^{+}$ and $\text{Cl}^{-}$ in KCl is 314 pm. Given that $r(\text{Cl}^{-}) = 181$ pm, what is $r(\text{K}^{+})$?

Solution:

In an ionic crystal, the internuclear distance $d$ is the sum of the two ionic radii:

$$d(\text{K}^{+}\!-\!\text{Cl}^{-}) = r(\text{K}^{+}) + r(\text{Cl}^{-})$$

So,

$$r(\text{K}^{+}) = d - r(\text{Cl}^{-}) = 314 - 181 = \mathbf{133\ \text{pm}}$$

(The widely tabulated value for K$^{+}$ is about 138 pm; our result differs by a few picometres because different scales and coordination assumptions can give slightly different radius values. The method is still correct.)

Example 9: Reverse — finding an unknown anion radius

In MgO crystal, the $\text{Mg}^{2+}\!-\!\text{O}^{2-}$ internuclear distance is 212 pm. Given $r(\text{Mg}^{2+}) = 72$ pm, calculate $r(\text{O}^{2-})$.

Solution:

$$r(\text{O}^{2-}) = d - r(\text{Mg}^{2+}) = 212 - 72 = \mathbf{140\ \text{pm}}$$

This matches the tabulated value of 140 pm and shows why O$^{2-}$ is much larger than neutral O (66 pm).

Example 10: The hydride trap

Which of the following has the largest radius? (a) $\text{H}^{-}$ (b) $\text{He}$ (c) $\text{Li}^{+}$ (d) $\text{Be}^{2+}$.

Solution:

All four are isoelectronic with 2 electrons ($1s^{2}$ = [He]).

Species	$Z$	$e^{-}$	$Z/e^{-}$	Radius
H$^{-}$	1	2	0.50	≈208 pm
He	2	2	1.00	≈140 pm (vdW)
Li$^{+}$	3	2	1.50	76 pm
Be$^{2+}$	4	2	2.00	45 pm

Within an isoelectronic series, smaller $Z$ gives larger radius. So:

Answer: H$^{-}$ (≈ 208 pm).

Takeaway: the hydride ion is unusually large because its $Z/e^{-}$ ratio is very small.

Example 11: A multi-step ordering

Arrange in increasing order of radius:

$$\text{Na}^{+}, \text{Mg}^{2+}, \text{Al}^{3+}, \text{Si}^{4+}, \text{P}^{3-}, \text{S}^{2-}, \text{Cl}^{-}$$

Solution:

Count electrons:

Species	$Z$	Charge	Electrons
Na$^{+}$	11	+1	10
Mg$^{2+}$	12	+2	10
Al$^{3+}$	13	+3	10
Si$^{4+}$	14	+4	10
P$^{3-}$	15	-3	18
S$^{2-}$	16	-2	18
Cl$^{-}$	17	-1	18

We have two isoelectronic groups: the first four (10 electrons, [Ne]-like) and the last three (18 electrons, [Ar]-like). Any 18-electron species is larger than any 10-electron species.

• 10-electron group: Si$^{4+}$ (41 pm) < Al$^{3+}$ (54) < Mg$^{2+}$ (72) < Na$^{+}$ (102)
• 18-electron group: Cl$^{-}$ (181) < S$^{2-}$ (184) < P$^{3-}$ (212)

So the final increasing order is:

$$\text{Si}^{4+} < \text{Al}^{3+} < \text{Mg}^{2+} < \text{Na}^{+} < \text{Cl}^{-} < \text{S}^{2-} < \text{P}^{3-}$$

Example 12: Explain the anomaly

Explain why the ionic radius of $\text{Ag}^{+}$ (126 pm) is smaller than that of $\text{K}^{+}$ (138 pm), even though Ag is in Period 5 and K is in Period 4.

Solution:

If only shell count mattered, Ag$^{+}$ should be larger than K$^{+}$ because Ag lies one period below K. However, Ag$^{+}$ has a filled $4d^{10}$ subshell, and these $d$-electrons shield the nuclear charge poorly. As a result, the effective nuclear attraction on the outer electron cloud of Ag$^{+}$ is larger than simple shell counting would suggest.

This inward pull causes a d-block contraction, making Ag$^{+}$ smaller than expected — and in fact smaller than K$^{+}$.

Takeaway: poor shielding by filled d-subshells can make some heavier cations smaller than lighter ones.