Worked examples — Atomic - ionic size trends; lanthanide contraction

This page is the training ground for the size-trends topic. We will not introduce new theory — instead we hunt down every kind of question the topic can ask and solve one of each, out loud.

Before we start, one reminder of the single master idea we keep reusing:

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The scenario matrix

Every question this topic throws is one of these cells. The examples afterward each carry the tag of the cell(s) they cover, so by the end nothing is left uncovered.

Cell	Case class	What changes	Which knob wins?
A	Across a 3d series (Sc→Zn)	$Z\uparrow$, electron into inner 3d	$Z_{\text{eff}}$ ↑ mildly → small shrink
B	The end-of-series bump (Cu, Zn)	filled d-shell, $d$–$d$ repulsion	size ticks up
C	Down a group 3d→4d	$n\uparrow$	$n$ wins → size grows
D	Down a group 4d→5d (degenerate!)	$n\uparrow$ but 14 lanthanide protons first	knobs cancel → size ≈ equal
E	Across lanthanides La→Lu	$Z\uparrow$, electron into buried 4f	$Z_{\text{eff}}$ ↑ → steady shrink (accumulates)
F	Ion vs atom / charge change	remove whole shell, $Z_{\text{eff}}$ jumps	cation ≪ atom; higher charge ⇒ smaller
G	Limiting / zero case	La³⁺ start, Lu³⁺ end, per-step → 0	check the extremes behave
H	Real-world word problem	density, separation	apply consequence
I	Exam twist	"explain why", trap comparisons	reason, don't memorise

Now the worked examples.

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Example 1 — Cell A: across a 3d series

Steps.

1. List what changes going Sc→Ti→V→Cr. Each step: $+1$ proton, and the new electron drops into the inner 3d subshell. Why this step? Because the whole answer is "who wins, proton or electron" — first we identify where the electron lands.
2. Rank the two knobs. New electron enters 3d, and $d$ shields poorly. This is the Slater/penetration hierarchy $s>p>d>f$ sketched in the definition box above: a $d$-electron sits in a wide, far-out cloud and barely blocks the nucleus, so it adds little to $S$. Meanwhile $Z$ rises by a full $+1$. Net: $Z_{\text{eff}}$ increases. Why this step? From the master ratio, rising $Z_{\text{eff}}$ with the same $n$ (still $3d/4s$) means shrinking.
3. Read off the order. Size decreases left→right: $\text{Sc}>\text{Ti}>\text{V}>\text{Cr}$. Why this step? Direct application of $r\propto 1/Z_{\text{eff}}$.

Verify: These are metallic radii (metals in the solid), pm: Sc 162, Ti 147, V 134, Cr 128 — strictly decreasing. ✓ The dominant $Z_{\text{eff}}$ shrink of the electron cloud shows through, and the drop is gentle (162→128, only ~34 pm over 3 steps) compared with a main-group period, exactly because d shields poorly. (Cross-check with Periodic Trends — Atomic & Ionic Radii.)

Figure below (s01): the four metallic radii plotted as points falling left→right; the green arrow traces the gentle downhill slope, and the butter-coloured note reminds you why it is gentle (d shields poorly). Read it as "each proton wins a little, so the cloud tightens a little."

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Example 2 — Cell B: the end-of-series bump

Steps.

1. Check where we are in the series. Ni is mid-late; Zn is the last (3d¹⁰4s², d-shell completely full). Why this step? The simple "shrink" rule only holds in the first half; the tail behaves differently.
2. Turn on the two late effects. (i) With 10 d-electrons crammed in, $d$–$d$ electron–electron repulsion puffs the shell out. (ii) A filled d-shell barely helps metallic bonding, so the metal–metal bonds weaken and atoms sit farther apart. Why this step? Both effects push size up, opposing the mild $Z_{\text{eff}}$ pull. Note this bump lives specifically in the metallic radius — it comes from weaker metal–metal bonding, which is exactly the "correction" the free-atom ratio cannot see.
3. Compare. These effects make the metallic radius tick up near the end: $\text{Zn}>\text{Ni}$. Why this step? The bump is a real feature, not a mistake.

Verify: Metallic radii (pm): Ni 124, Cu 128, Zn 134 — a clear rise at the tail. ✓ The naïve "later = smaller" rule fails here, which is the whole point of this cell.

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Example 3 — Cell C: down a group, 3d→4d

Steps.

1. Identify the change. Going Ti→Zr, the outermost occupied shell's principal quantum number rises: Ti's valence sits in the $n=4$ region (4s/3d), while Zr's sits in the $n=5$ region (5s/4d). So $\boxed{n:4\to 5}$ — $n$ increases by one whole shell. Why this step? Down a group, the dominant change is the principal quantum number $n$ — we name it explicitly so it is clear which number is rising (it is $n$, the shell, not the subshell letter).
2. Apply the ratio. $r\propto n^2/Z_{\text{eff}}$. The $n^2$ growth (from $4^2=16$ to $5^2=25$) outpaces the extra $Z_{\text{eff}}$, so size grows. Why this step? This is the "normal" periodic-table behaviour — larger $n$, bigger atom.
3. Conclude. $\text{Zr}>\text{Ti}$.

Verify: Atomic (metallic) radii (pm): Ti ≈ 147, Zr ≈ 160. Zr larger by ~13 pm. ✓ Cell C behaves as expected — contrast this with Cell D next.

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Example 4 — Cell D: the degenerate case, 4d→5d

Steps.

1. Spot who sits before Hf. Between Zr's row and Hf's row lie the 14 lanthanides (La→Lu), which added 14 protons whose 4f electrons shield terribly. Why this step? This is the one place where an invisible row changes everything — you must remember the lanthanides are inserted before the 5d block.
2. Weigh the two knobs. Going 4d→5d: $n$ up (wants bigger) but 14 poorly-shielded extra protons drive $Z_{\text{eff}}$ up (wants smaller). These two nearly cancel. Why this step? This is a degenerate cell — the knobs balance, so we can't read the answer off one rule.
3. Estimate why it balances (a back-of-envelope). In the normal step Ti→Zr ($n:4\to 5$), the shell-growth alone bought about $+13$ pm (Example 3). Going 4d→5d we get that same $n$-driven urge to grow by roughly $+13$ pm. But the 14 lanthanide protons, shielded by lazy 4f electrons, pull the outer shell in by almost exactly the same amount. Compare the observed jumps: Ti→Zr grew $+13$ pm, whereas Zr→Hf grew only $160\to 159=-1$ pm. So the contraction "ate" about $13-(-1)=14$ pm of expected growth — roughly $1$ pm per lanthanide proton, which is why 14 of them wipe out one full shell's worth of expansion. Why this step? A number turns "they cancel" into "here is how much each side is worth, and the tally comes out near zero" — no hand-waving.
4. Conclude. $\text{Zr}\approx \text{Hf}$ (a near-tie), so they are chemical twins — hardest pair to separate.

Verify: Atomic radii (pm): Zr ≈ 160, Hf ≈ 159 — difference ≈ 1 pm, essentially equal. ✓ The expected $\sim 13$ pm growth (from Cell C) is cancelled by the $\sim 14$ pm lanthanide pull-in. Same charge $+4$ and same size ⇒ same charge density ⇒ near-identical chemistry. The naïve rule fails, and lanthanide contraction is the reason.

Figure below (s02): two atoms drawn as pastel discs of nearly the same radius — Zr (lavender) and Hf (mint). Above them the two competing arrows: "n up → wants BIGGER (+13 pm)" (butter) and "+14 lanthanide protons → wants SMALLER (−14 pm)" (coral). The double-headed slate arrow between the discs shows the two pulls cancelling into an almost-equal size.

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Example 5 — Cell E + G: across (and at the ends of) the lanthanides

Steps.

1. Total contraction. $\Delta r=1.032-0.861=0.171\text{A˚}=17.1\text{pm}$. Why this step? The "total shrink" is just end minus start — this is the number consequences care about.
2. Per-step average. From La³⁺ to Lu³⁺ is $14$ atomic-number steps ($Z=57\to 71$). Average $=17.1/14\approx 1.22\text{pm}$ per step. Why this step? Divides the total over the steps to expose why it feels small yet matters.
3. Interpret the two extremes (Cell G). At the start (La³⁺, empty 4f) the atom is largest; at the end (Lu³⁺, full 4f¹⁴) it is smallest. The per-step limit ($\sim 1$ pm) is nearly the smallest change we ever discuss — it is the accumulation over 14 that produces the dramatic Zr≈Hf result. Why this step? Checks both limiting inputs and the "tiny-but-accumulating" logic.

Verify: $0.171\times 10=1.71$ → $17.1$ pm total ✓. $17.1/14=1.221\dots$ pm ✓ — tiny per step, exactly matching the "~1 pm" claim in the parent note. (This same steady shrink is what makes Separation of Lanthanides (Ion-exchange) necessary — the ions differ by only a hair.)

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Example 6 — Cell F: ions vs atoms and charge changes

Steps.

1. Fe atom → Fe²⁺. Removing 2 electrons strips the outer 4s shell entirely. Same 26 protons now pull on fewer electrons ⇒ $Z_{\text{eff}}$ per electron jumps ⇒ much smaller. Why this step? A cation is not "atom minus a constant" — it loses a whole shell, so the drop is large.
2. Fe²⁺ → Fe³⁺. Remove one more electron (a 3d one). Still 26 protons, one fewer electron ⇒ $Z_{\text{eff}}$ per remaining electron rises again ⇒ even smaller. Why this step? Higher positive charge ⇒ tighter pull ⇒ smaller ion.
3. Order. $\text{Fe}>{\text{Fe}}^{2+}>{\text{Fe}}^{3+}$.

Verify: Radii (pm): Fe atom (metallic) ≈ 126, Fe²⁺ ≈ 78 (CN-6 ionic), Fe³⁺ ≈ 65 (CN-6 ionic). Strictly decreasing, and note atom→ion is a huge drop (~48 pm), not a small constant. ✓ Kills the mistake "ionic = atomic − constant."

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Example 7 — Cell H: real-world word problem (density)

Steps.

1. Write the definition. Density $\rho =\frac{\text{mass}}{\text{volume}}$. To get high $\rho$ we want large mass and small volume. Why this step? Density is a ratio — we must control both top and bottom.
2. Mass side. Os packs ~190 g/mol of protons+neutrons into each atom — far more than Fe's ~56. Numerator is large. Why this step? Establishes the "heavy" half.
3. Volume side (the size-trend punchline). Os is a 5d metal, so it should be much bigger than a 3d metal — but the lanthanide contraction shrank the entire 5d row back down. So Os stays small; volume per atom is not much bigger than a 3d metal's. Why this step? This is the topic's contribution: mass climbed but volume did not climb proportionally.
4. Combine. Large mass ÷ stubbornly small volume ⇒ very high density. Os (~22.6 g/cm³) beats Fe (~7.9 g/cm³) by roughly 3×. Why this step? Puts numerator and denominator together.

Verify: $22.6/7.9\approx 2.86$ — Os is ~2.9× denser than Fe. ✓ (See also Density, Melting Point Trends in Transition Metals.) The size trend is essential: without contraction, Os would be fatter and thus far less dense.

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Example 8 — Cell I: the exam twist (the trap comparison)

Why this is a Cell I (twist), not a Cell H: there is no measured quantity to compute — the whole task is to catch a plausible-sounding but false chain of reasoning and repair it. The examiner rewards the why, not a number.

Steps.

1. Grant the true part. Yes — by lanthanide contraction La³⁺ > Nd³⁺ > Lu³⁺, so Lu³⁺ is the smallest. Why this step? You must separate the correct fact from the wrong conclusion glued to it — that separation is the exam skill being tested.
2. Find the broken link. The trap says "smallest ion ⇒ most basic." But basicity depends on how easily the hydroxide releases OH⁻. A smaller, denser ion grips the O of OH harder, making M–OH more covalent, so it holds on to OH⁻ ⇒ less basic, not more. Why this step? The statement inverted the cause–effect direction; naming charge density fixes it.
3. Give the correct order. Basicity: ${\text{La(OH)}}_3>{\text{Nd(OH)}}_3>{\text{Lu(OH)}}_3$ (largest ion = weakest grip = most basic). The claim is False. Why this step? Delivers the demanded verdict plus the ranking.

Verify: Charge density $\propto 3/r$ with CN-6 radii $1.032,0.983,0.861$ Å gives $2.907,3.052,3.484$ Å⁻¹ — strictly rising La→Lu, so basicity strictly falling ⇒ La(OH)₃ most basic, Lu(OH)₃ least. ✓ Matches the parent note. (Connects to Basic Character of Oxides & Hydroxides.)

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Recall Quick self-test (cover the answers)

Which cell does "Zr vs Hf" belong to, and what's special about it? ::: Cell D — the degenerate case where higher $n$ (worth +13 pm) and the 14-lanthanide-proton pull-in (−14 pm) cancel, giving near-equal size. Total lanthanide contraction (La³⁺→Lu³⁺, CN-6) and per-step average? ::: ~17.1 pm total; ~1.2 pm per step (14 steps). What does "CN-6" mean when quoting an ionic radius? ::: Coordination number 6 — the ion is surrounded by 6 nearest neighbours; radii must be quoted at a fixed CN to be comparable. Why is Fe³⁺ smaller than Fe²⁺? ::: Same 26 protons, one fewer electron ⇒ higher $Z_{\text{eff}}$ per electron ⇒ tighter pull. Why does Os have huge density but Fe doesn't? ::: Lanthanide contraction keeps Os small (volume ↓) while mass ↑ ⇒ mass/volume large. In which cell does the naïve "later = smaller" rule fail across a 3d series? ::: Cell B — Cu, Zn tick up due to d–d repulsion and a filled d-shell that barely bonds. What's the difference between the ratio's radius and a metallic radius? ::: The ratio $r\propto n^2/Z_{\text{eff}}$ describes a free atom/ion; metallic radius also depends on metal–metal bond strength (which causes the Cell B bump). Why do d-electrons shield worse than s or p (Slater/penetration)? ::: They sit in a wide, far-out cloud and barely penetrate to the nucleus, so they block little of the pull ($s>p>d>f$).

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