Atomic - ionic size trends; lanthanide contraction: Exercises

Definition The words and symbols used on this page (built here, not assumed)

$n$ — principal quantum number. A whole number ( $1, 2, 3, \dots$ ) labelling which shell an electron lives in. Picture the shells as nested rings around the nucleus: $n = 1$ is the innermost, tightest ring; each higher $n$ is a bigger ring further out. Bigger $n$ = bigger, roomier orbit — this is exactly why $r \propto n^{2} / Z_{eff}$ grows with $n$ . In the figure below the three black rings are $n = 1, 2, 3$ and the outermost (valence) electron sits on the largest ring.
$Z_{eff} = Z - S$ — effective nuclear charge. The leftover nuclear pull an outer electron actually feels after inner electrons ( $S$ ) block part of the full proton charge ( $Z$ ). In the figure it is the red arrow: shortened because the inner electrons soak up some of the nucleus's pull.
atomic radius / metallic radius. For a metal we cannot measure one lonely atom, so metallic radius ::= half the centre-to-centre distance between two touching, bonded atoms of the same metal. If two copper atoms sit 256 pm apart, the metallic radius is 128 pm. It is not the same as ionic radius (that is for a charged ion in a crystal).
ionic radius. The radius assigned to an ion (a charged atom) inside a crystal — it depends on how many neighbours crowd it (see CN) and on charge.
CN — coordination number ::= how many nearest neighbours surround an ion. Radii depend on it, so we always quote it — here almost always CN 6 (six neighbours arranged as an octahedron). More neighbours push out slightly → larger quoted radius.
Shannon radii. A famous standard table of ionic radii (compiled by R. D. Shannon) measured from real crystals. Because a crystal's spacing shifts with CN and, for transition ions, with spin state (how the $d$ -electrons are arranged — "high-spin" spreads them into more orbitals, nudging the size up), Shannon's table lists a separate value for each CN and spin state. That is why we tag every Shannon value with "CN 6, high-spin" — quoting the number alone would be ambiguous.

L1 — Recognition

Here you only need to recall and point. No calculation heavier than a subtraction.

Problem 1.1

Compute the effective nuclear charge $Z_{eff}$ felt by the outermost electron of two atoms if:

Atom A: $Z = 20$ , shielding constant $S = 15.75$ .
Atom B: $Z = 21$ , shielding constant $S = 16.5$ .

Which atom's outer electron is held more tightly, and does that make it smaller or larger?

Recall Solution

WHAT: $Z_{eff} = Z - S$ is the net pull left over after the inner electrons ( $S$ ) block part of the nuclear charge ( $Z$ ) — the red arrow in the definition figure. Atom A: $Z_{eff} = 20 - 15.75 = 4.25$ . Atom B: $Z_{eff} = 21 - 16.5 = 4.50$ . WHY it matters: Bigger $Z_{eff}$ = stronger inward pull. Atom B (4.50 > 4.25) holds its outer electron more tightly. Since $r \propto n^{2} / Z_{eff}$ , a larger $Z_{eff}$ (denominator up) means a smaller radius. So Atom B is smaller. (This is the Sc-after-Ca situation: one extra proton, one extra poorly-shielding d-electron, so $Z_{eff}$ creeps up and size dips.)

Problem 1.2

Rank these subshells by shielding power (best screener first): $4 f$ , $2 s$ , $3 p$ , $3 d$ .

Recall Solution

WHAT: Shielding power follows how close a subshell penetrates to the nucleus — the closer it sits, the better it blocks the outer electrons from the nucleus. The rule is $s > p > d > f$ . Applying it: $2 s > 3 p > 3 d > 4 f .$ WHY: $s$ electrons dive nearest the nucleus (best block); $f$ electrons are diffuse and poor blockers. This single ranking is the engine behind every trend on this page.

L2 — Application

Now plug the trends into concrete comparisons.

Problem 2.1

Which is smaller, $Fe^{2 +}$ or $Fe^{3 +}$ ? Both come from the same atom. Explain using $Z_{eff}$ .

Recall Solution

WHAT: Both ions have the same nucleus, $Z = 26$ protons. $Fe^{2 +}$ has 24 electrons; $Fe^{3 +}$ has 23. WHY: Same protons, fewer electrons in $Fe^{3 +}$ ⇒ less mutual shielding ⇒ each remaining electron feels a larger $Z_{eff}$ ⇒ pulled in tighter. Answer: $Fe^{3 +}$ is smaller. The measured Shannon radii (the standard crystal-measured table — recall each value is tagged with its CN and spin state because both shift the number) at CN 6, high-spin are $Fe^{2 +} \approx 0.78$ Å and $Fe^{3 +} \approx 0.645$ Å, confirming the smaller ion is $Fe^{3 +}$ . Watch the mistake it kills: ionic radius is NOT "atomic radius minus a fixed constant"; higher charge means an extra shrink from rising $Z_{eff}$ .

Problem 2.2

The four Ln³⁺ radii below (Å, CN 6) are given out of order — this is the complete set of values you are allowed to use: $0.861, 1.032, 0.947, 0.977.$ They belong to $La^{3 +} (Z = 57)$ , $Pr^{3 +} (Z = 59)$ , $Nd^{3 +} (Z = 60)$ , $Lu^{3 +} (Z = 71)$ . Assign each radius to the correct ion using only the four numbers above.

Recall Solution

WHAT: Across the lanthanides the 4f subshell fills. 4f shields terribly, so $Z_{eff}$ on the outer electrons climbs steadily and the radius shrinks monotonically as $Z$ rises. WHY monotonic: every step adds a proton (full pull) and a 4f electron (almost no shielding) → the pull always wins, so the radius only ever goes down as $Z$ increases. The whole method — no outside numbers needed:

Order the four ions by increasing $Z$ : $La^{3 +} (57) < Pr^{3 +} (59) < Nd^{3 +} (60) < Lu^{3 +} (71)$ .
Sort the four given radii in descending order: $1.032 > 0.977 > 0.947 > 0.861$ .
Because radius decreases as $Z$ increases, map largest radius ↔ smallest $Z$ , straight down the line:

Ion	$Z$	Radius (Å)
$La^{3 +}$	57	1.032
$Pr^{3 +}$	59	0.977
$Nd^{3 +}$	60	0.947
$Lu^{3 +}$	71	0.861

$La^{3 +} = 1.032 > Pr^{3 +} = 0.977 > Nd^{3 +} = 0.947 > Lu^{3 +} = 0.861.$ The whole trick is: radius order is just $Z$ order, reversed — sort the supplied numbers and line them up.

L3 — Analysis

Now reason about why a trend bends or breaks.

Problem 3.1

Metallic radii across the first transition series (pm): Sc 164, Ti 147, V 135, Cr 129, Mn 137, Fe 126, Co 125, Ni 125, Cu 128, Zn 137.

Recall metallic radius = half the centre-to-centre distance between two touching atoms of the metal (built in the definition box above). (a) Describe the overall shape. (b) Explain the rise at Cu and Zn.

Recall Solution

(a) Shape: a fairly steep drop early (Sc→Cr), then a long flat middle (Fe, Co, Ni all ≈125), then a rise at the end (Cu, Zn).

How to read the figure below (alt description): the horizontal axis lists the ten 3d elements left to right (Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn); the vertical axis is metallic radius in pm, from about 115 to 172. The black dotted-marker curve runs Sc(164)→Ti(147)→V(135)→Cr(129), dips up slightly to Mn(137), falls to the flat trough Fe(126)–Co(125)–Ni(125): this is the "drop then flatten" story. From Ni onward a separate red segment climbs Ni(125)→Cu(128)→Zn(137) — the single upward kink the question is about. A red arrow labels that kink "filled d-shell: repulsion up, weak bonding."

Figure — Atomic - ionic size trends; lanthanide contraction

Why the early drop: each step adds a proton but the new electron enters inner $3 d$ (poor shield) → $Z_{eff}$ rises → radius falls. Why the flat middle: growing $d$ – $d$ electron–electron repulsion starts to push back against the rising nuclear pull; the two nearly balance. (b) Why Cu, Zn rise — two effects:

Maximal $d$ – $d$ repulsion: the $3 d$ shell is now nearly/completely full ( $3 d^{10}$ at Zn), so electron–electron repulsion is largest, swelling the atom.
Weak metallic bonding: a filled $3 d^{10}$ shell contributes almost nothing to the metallic bond, so bonds weaken and atoms sit farther apart → larger metallic radius. Mn note: the small bump at Mn (137) comes from the extra stability/repulsion of the half-filled $3 d^{5}$ configuration — a preview of the same electron-repulsion logic.

Problem 3.2

Atomic radii (pm): Ti 147, Zr 160, Hf 159. Explain why Zr > Ti but Hf ≈ Zr rather than Hf > Zr.

Recall Solution

Ti → Zr (3d → 4d): the principal quantum number $n$ (which shell — bigger $n$ = bigger ring, from the definition box) jumps up; since $r \propto n^{2} / Z_{eff}$ , a larger $n$ makes the atom bigger. So Zr > Ti — the expected down-a-group increase. Zr → Hf (4d → 5d): you'd again expect an increase from larger $n$ . But between Zr's row and Hf's row sit the 14 lanthanides (La→Lu). Filling 4f added 14 protons whose pull is barely shielded (4f is the worst shielder). This lanthanide contraction shrinks the atom by almost exactly the amount the larger $n$ would have grown it. Result: the two effects cancel ⇒ Hf ≈ Zr (159 ≈ 160). Same story gives Nb ≈ Ta and Mo ≈ W.

L4 — Synthesis

Combine several ideas to reach a quantitative or multi-step conclusion.

Problem 4.1

The lanthanide ionic radii (CN 6) go from La³⁺ = 1.032 Å to Lu³⁺ = 0.861 Å across 14 steps (La→Lu, i.e. from $Z = 57$ to $Z = 71$ ). (a) Total contraction? (b) Average shrink per element step? (c) Use these numbers to argue why the per-step shrink is called "small" yet the effect is "large".

Recall Solution

(a) Total contraction = $1.032 - 0.861 = 0.171$ Å $= 17.1$ pm. (b) There are $71 - 57 = 14$ increments in $Z$ from La to Lu. $per step = \frac{0.171 A ˚}{14} \approx 0.01221 \overset{˚}{A} \approx 1.22 pm .$ (c) Each step shrinks the ion by only ~1.2 pm — utterly tiny, easy to dismiss. But it happens the same direction 14 times in a row (monotonic, because 4f never shields well), so it accumulates to a substantial 17 pm. That accumulated 17 pm is exactly the size increase that 4d→5d would have produced — hence it cancels it, giving Zr≈Hf. Small cause, big consequence, because it never reverses.

Problem 4.2

Os has atomic mass ≈ 190.2 u and packs into a very small volume because of the contraction. Given a hypothetical metal X of the same molar volume as Os but atomic mass 96 u (like Mo), and Os density = 22.59 g/cm³, estimate X's density. What single idea does this illustrate?

Recall Solution

WHAT: density $ρ = \frac{mass}{volume}$ . If molar volume is the same for both, density scales directly with molar mass: $\frac{ρ _{X}}{ρ _{Os}} = \frac{M _{X}}{M _{Os}} = \frac{96}{190.2} .$ $ρ_{X} = 22.59 \times \frac{96}{190.2} \approx 22.59 \times 0.5047 \approx 11.40 g/cm^{3} .$ Idea illustrated: the lanthanide contraction keeps the volume of 5d metals small (Os is nearly as compact as a 4d metal), while the mass keeps climbing across the periods. Same small volume + much bigger mass ⇒ Os's density is roughly double. That is why Os and Ir are the densest elements.

L5 — Mastery

Full-chain reasoning, including a case where you must justify direction and magnitude.

Problem 5.1

Predict and justify which hydroxide is more basic: La(OH)₃ or Lu(OH)₃. Then generalise the rule and state one real-world use of the underlying size trend.

Recall Solution

Chain of reasoning:

Across La→Lu the lanthanide contraction shrinks the ion: La³⁺ (1.032 Å) → Lu³⁺ (0.861 Å).
Charge density $= \frac{charge}{size}$ . Same $+ 3$ charge on a smaller ion ⇒ Lu³⁺ has the higher charge density.
Higher charge density ⇒ Lu³⁺ pulls harder on the oxygen of an O–H group ⇒ the M–OH bond becomes more covalent ⇒ the ion holds onto OH⁻ more tightly and releases it less readily.
"Basic" means readily giving up OH⁻. Lu(OH)₃ gives it up less readily ⇒ less basic. Answer: La(OH)₃ is more basic; Lu(OH)₃ is least basic. General rule: basicity of Ln(OH)₃ decreases La→Lu as charge density rises. Real-world use: the same steady, tiny size change makes Ln³⁺ ions differ just enough in charge density that they can be separated by ion-exchange chromatography — see Separation of Lanthanides (Ion-exchange). The smaller (later) ions bind the resin differently and elute in a predictable order.

Problem 5.2 (capstone)

A student claims: "Since Cu sits at the end of the 3d series where $Z_{eff}$ is highest, Cu must be the smallest atom in the 3d row." Refute or support with the data from Problem 3.1, and reconcile it with the $r \propto n^{2} / Z_{eff}$ picture.

Recall Solution

Refute. The data: Ni 125, Cu 128, Zn 137 — Cu is not the smallest; Fe/Co/Ni (≈125) are. Where the student's logic breaks: $Z_{eff}$ is highest late in the row, and the naive $r \propto n^{2} / Z_{eff}$ (a one-electron model) does predict shrinking. But that formula ignores electron–electron repulsion, which is not one-electron physics. By Cu ( $3 d^{10} 4 s^{1}$ ) the $d$ -shell is full: (i) $d$ – $d$ repulsion is maximal, swelling the atom, and (ii) the filled, low-lying $3 d^{10}$ barely helps metallic bonding, so atoms sit farther apart. These two effects overpower the rising $Z_{eff}$ , so the metallic radius rises at Cu and Zn. Reconciliation: $r \propto n^{2} / Z_{eff}$ is the leading term (it explains the early drop), but the full story needs the repulsion + bonding correction the parent note flagged. The smallest 3d atoms are the mid-series ones where nuclear pull and repulsion best balance.

Recall Quick self-check (close everything and answer)

What does $n$ stand for and what does bigger $n$ do to size? ::: The principal quantum number (which shell); bigger $n$ = bigger ring = larger radius. What is metallic radius? ::: Half the centre-to-centre distance between two touching bonded atoms of a metal. Why do Shannon radii come tagged with CN and spin state? ::: Both change the crystal spacing, so the same ion has different tabulated radii for different CN/spin. Radius order of Ln³⁺ vs $Z$ ? ::: Exactly reversed — higher $Z$ , smaller ion (monotonic, because 4f shields worst). Why Hf ≈ Zr, not Hf > Zr? ::: Larger $n$ would grow it, but the intervening 14-lanthanide contraction cancels the growth. Per-step lanthanide shrink from Problem 4.1? ::: ≈ 1.2 pm (17.1 pm total over 14 steps). Why is La(OH)₃ more basic than Lu(OH)₃? ::: La³⁺ is larger ⇒ lower charge density ⇒ releases OH⁻ more easily. Why isn't Cu the smallest 3d atom despite highest $Z_{eff}$ ? ::: Full $3 d^{10}$ maximises $d$ – $d$ repulsion and weakens metallic bonding, swelling the atom. What does CN 6 mean? ::: Coordination number 6 — the ion has six nearest neighbours (octahedral); radii are quoted for a stated CN.