Lanthanides — electronic configuration, lanthanide contraction, oxidation states (mostly +3): Worked examples

Everything here rests on three prerequisite ideas — click them if any feel shaky:

Aufbau Principle and (n+l) Rule — the order electrons fill.
Half-filled and Fully-filled Stability — why $4 f^{7}$ and $4 f^{14}$ are special.
Effective Nuclear Charge & Slater's Rules — why the atom shrinks.

The scenario matrix

Every lanthanide problem is really one of these case classes. We will hit each cell at least once.

#	Case class	What makes it tricky	Covered by
A	Ordinary atom config (mid-series, no anomaly)	just count into $4 f$	Ex 1 (Nd)
B	Anomaly atom config (empty/half/full- $f$ borrows a $5 d$ )	electron jumps to $5 d$	Ex 2 (Gd)
C	Boundary element (La at start, Lu at end)	is it even a "true" $4 f$ filler?	Ex 3 (La & Lu)
D	The default ion $Ln^{3 +}$	remove the right 3 electrons	Ex 4 (Pr³⁺)
E	Off-default ion (+2 or +4) driven by $f^{0} / f^{7} / f^{14}$	which state is stable & why	Ex 5 (Ce⁴⁺, Eu²⁺, Tb⁴⁺)
F	Degenerate / zero case ( $4 f^{0}$ and $4 f^{14}$ endpoints)	empty and full shells	Ex 6
G	Limiting/quantitative — the contraction as a number	per-step vs total shrink	Ex 7 (radii)
H	Real-world word problem	why Hf is dense / Zr–Hf inseparable	Ex 8
I	Exam-style twist — ranking / odd-one-out	compare many species at once	Ex 9
J	Trap check — the classic misconception	spot the false statement	Ex 10

Cell A — the ordinary atom

Worked example Ex 1. Ground-state configuration of Neodymium (Nd, Z = 60)

Forecast: guess how many electrons sit in $4 f$ before reading on.

Steps:

Start from the noble-gas core: $[Xe]$ is $Z = 54$ . Electrons left $= 60 - 54 = 6$ . Why this step? The Xe core is inert; only the outer 6 electrons do chemistry, so we only place those.
Fill $6 s$ first: it has the lowest $(n + ℓ) = 6$ (see Aufbau Principle and (n+l) Rule). Place $6 s^{2}$ . Remaining $= 6 - 2 = 4$ . Why this step? By the $(n + ℓ)$ rule $6 s$ (sum 6) fills before both $4 f$ and $5 d$ (both sum 7).
Nd is mid-series with no empty/half/full driver, so all 4 remaining electrons go into $4 f$ : $4 f^{4}$ . Why this step? Only La, Ce, Gd, Lu borrow a $5 d^{1}$ . Nd is not on that list → no $5 d$ electron.

Answer: $Nd = [Xe] 4 f^{4} 6 s^{2}$

Verify: electron count $= 54 + 4 + 2 = 60 = Z$ . ✓ And $4$ is between $1$ and $14$ , a legal $4 f$ occupancy.

Cell B — the anomaly

Worked example Ex 2. Ground-state configuration of Gadolinium (Gd, Z = 64)

Forecast: does Gd keep its electron in $4 f$ , or does one jump to $5 d$ ?

Steps:

Core $[Xe] = 54$ ; electrons left $= 64 - 54 = 10$ . Why this step? Same core-subtraction as before — it never changes.
Place $6 s^{2}$ ; remaining $= 8$ . Why this step? $6 s$ always fills first.
Now the choice: put all 8 into $4 f$ ( $4 f^{8}$ ), or put 7 into $4 f$ (half-filled, extra-stable) and let the $8^{th}$ go to $5 d^{1}$ . Why this step? $4 f^{7}$ is a half-filled shell — a genuine stability well (see Half-filled and Fully-filled Stability). Nature pays a small energy to reach it because $4 f$ and $5 d$ energies are almost equal here.
The half-filled route wins: $4 f^{7} 5 d^{1} 6 s^{2}$ .

Answer: $Gd = [Xe] 4 f^{7} 5 d^{1} 6 s^{2}$

Verify: $54 + 7 + 1 + 2 = 64 = Z$ . ✓ The $4 f^{7}$ half-shell is preserved — exactly the reason Gd appears in the parent's "anomaly" table.

Cell C — the boundaries

Worked example Ex 3. Configurations of the two end-posts: La (Z = 57) and Lu (Z = 71)

Forecast: which of these two actually has electrons in $4 f$ ?

Steps (La):

Core $[Xe] = 54$ ; left $= 3$ . Place $6 s^{2}$ ; remaining $= 1$ . Why this step? Standard core + $6 s$ first.
With $4 f$ still empty ( $f^{0}$ ), the lone electron prefers $5 d$ over starting a fresh $4 f$ . Why this step? Right at the start, $5 d$ dips below the empty $4 f$ ; nature takes the lower slot → $5 d^{1}$ . La $= [Xe] 5 d^{1} 6 s^{2}$ (technically the header of the group, $4 f^{0}$ ).

Steps (Lu): 3. Core $[Xe] = 54$ ; left $= 17$ . Place $6 s^{2}$ ; remaining $= 15$ . 4. $4 f$ holds a maximum of 14. Fill it fully: $4 f^{14}$ . The last electron ( $15 - 14 = 1$ ) must go to $5 d$ . Why this step? $4 f$ is capacity-limited to 14 (7 orbitals × 2). Once full, the extra electron has nowhere in $4 f$ to go → $5 d^{1}$ . Lu $= [Xe] 4 f^{14} 5 d^{1} 6 s^{2}$ .

Verify: La $54 + 1 + 2 = 57$ ✓; Lu $54 + 14 + 1 + 2 = 71$ ✓. Both carry the tell-tale $5 d^{1}$ — matching the parent's "La, Ce, Gd, Lu" list of $5 d^{1}$ elements.

Cell D — the default +3 ion

Cell E — the off-default ions

Worked example Ex 5. Which ion of Ce, Eu, and Tb is stable

besides +3, and why? Forecast: guess for each element whether the extra-stable ion is +2 or +4.

Ce (Z = 58, atom $= [Xe] 4 f^{1} 5 d^{1} 6 s^{2}$ ):

$Ce^{3 +}$ removes $6 s^{2} + 5 d^{1} = 3$ electrons → $[Xe] 4 f^{1}$ .
Remove one more $4 f$ electron → $Ce^{4 +} = [Xe] 4 f^{0}$ . Why this step? $4 f^{0}$ (empty shell = same as La³⁺, xenon config) is a stability well → +4 becomes reachable. Ce⁴⁺ is a strong oxidiser (it wants that electron back to sit comfortably at +3).

Eu (Z = 63, atom $= [Xe] 4 f^{7} 6 s^{2}$ ): 3. $Eu^{3 +}$ removes $6 s^{2} +$ one $4 f$ → $[Xe] 4 f^{6}$ . 4. But stop one electron short of +3: $Eu^{2 +} = [Xe] 4 f^{7}$ . Why this step? $4 f^{7}$ is half-filled and extra stable, so Eu is content at +2. Eu²⁺ is a reducer (gives away an electron to climb to the still-more-common +3).

Tb (Z = 65, atom $= [Xe] 4 f^{9} 6 s^{2}$ ): 5. $Tb^{3 +} = [Xe] 4 f^{8}$ . Remove one more → $Tb^{4 +} = [Xe] 4 f^{7}$ . Why this step? Going to +4 lets Tb reach the half-filled $4 f^{7}$ well → +4 is stable here; Tb⁴⁺ is an oxidiser.

Verify: Ce⁴⁺: $54 + 0 = 54$ ✓ empty $f$ . Eu²⁺: $54 + 7 = 61 = 63 - 2$ ✓ half-filled. Tb⁴⁺: $54 + 7 = 61 = 65 - 4$ ✓ half-filled. Every "off-default" state lands on $f^{0}$ or $f^{7}$ — exactly the parent's stability table.

Cell F — the degenerate endpoints

Worked example Ex 6. The two "empty" and "full"

$f$ ions: what do La³⁺ and Lu³⁺ have in common? Forecast: what unusual property do ions with $4 f^{0}$ and $4 f^{14}$ share?

Steps:

$La^{3 +}$ : La $= [Xe] 5 d^{1} 6 s^{2}$ ; remove all three outer electrons → $[Xe] 4 f^{0}$ . Why this step? La has no $4 f$ electrons to start with, so its +3 ion is a bare xenon core.
$Lu^{3 +}$ : Lu $= [Xe] 4 f^{14} 5 d^{1} 6 s^{2}$ ; remove $6 s^{2} + 5 d^{1}$ → $[Xe] 4 f^{14}$ . Why this step? The $4 f$ shell is already full and buried; the three removable electrons are $6 s$ and $5 d$ .
Both have no unpaired $4 f$ electrons ( $f^{0}$ has zero, $f^{14}$ is fully paired). Why this step? An empty shell and a full shell both have every possible electron paired or absent → zero unpaired spins.

Consequence: $La^{3 +}$ and $Lu^{3 +}$ are colourless and diamagnetic, while the ions in between (with partly-filled $f$ ) are coloured and paramagnetic.

Verify: La³⁺ $= 54$ electrons (Xe), Lu³⁺ $= 54 + 14 = 68$ . Unpaired-electron count: $f^{0} \to 0$ , $f^{14} \to 0$ . ✓ These are the true degenerate/limiting cases of the series.

Cell G — the contraction as a number

Worked example Ex 7. Average shrink per element from La³⁺ (103 pm) to Lu³⁺ (86 pm)

Forecast: guess the per-element drop in picometres — is it big or tiny?

Steps:

Total contraction $= 103 - 86 = 17$ pm. Why this step? Radius decreases, so it's (start − end); this is the accumulated shrink.
Number of steps from La to Lu. La, Ce, ..., Lu is 15 elements → $15 - 1 = 14$ gaps. Why this step? Shrink happens between neighbours; $n$ elements have $n - 1$ intervals.
Average per step $= 17/14 \approx 1.21$ pm. Why this step? Total shrink ÷ number of gaps = mean shrink per neighbour.

Answer: about 1.2 pm per element, tiny individually but ~17 pm overall — enough to cancel a whole period's expected growth.

Verify: $103 - 86 = 17$ ; $17/14 = 1.214 \dots \approx 1.2$ pm ✓. This matches the parent's "small per step (~1 pm), substantial in total."

See how the near-linear slope of the plot below turns "1 pm each" into a big total drop:

Figure — Lanthanides — electronic configuration, lanthanide contraction, oxidation states (mostly +3)

Cell H — the real-world word problem

Worked example Ex 8. Why is hafnium (Hf) almost the same size as zirconium (Zr), and so dense?

Forecast: Zr is right above Hf; going down a group atoms usually get bigger. Guess whether Hf is bigger, smaller, or equal.

Given (from parent): $r (Zr) \approx 160$ pm, $r (Hf) \approx 159$ pm.

Steps:

Normally Hf (period 6) should be noticeably larger than Zr (period 5) — extra shell. Why this step? Down a group, a new principal shell is added → bigger radius, the default expectation.
But Hf sits right after the 14 lanthanides. It has inherited the entire lanthanide contraction (~17 pm of shrink) that happened just before it. Why this step? The contraction raised $Z_{eff}$ (poor $4 f$ shielding, see Effective Nuclear Charge & Slater's Rules), pulling everything after it inward.
The expected growth and the inherited shrink nearly cancel: $r (Hf) \approx r (Zr)$ . Difference $= 160 - 159 = 1$ pm — negligible. Why this step? Two opposite effects of similar size → near-zero net change.
Density $= \frac{mass}{volume}$ . Hf's atomic mass ( $\approx 178.5$ ) is far larger than Zr's ( $\approx 91.2$ ), but their volumes are nearly equal. Why this step? Same size but ~2× heavier atoms packs far more mass into the same space → much higher density.

Answer: near-identical radii → Zr and Hf are chemically twins (hard to separate); Hf packs a big mass into a small volume → very dense (~13.3 g cm⁻³).

Verify: $∣160 - 159∣ = 1$ pm (essentially equal) ✓; mass ratio $178.5/91.2 \approx 1.96 \approx 2$ ✓ — a roughly doubled mass in the same volume explains the density jump.

Cell I — the exam ranking twist

Worked example Ex 9. Rank these Ln³⁺ ions by ionic radius (largest → smallest): Ce³⁺, Nd³⁺, Gd³⁺, Lu³⁺.

Forecast: guess the order before using any rule.

Steps:

All are the same +3 charge, so charge does not decide the order — size does. Why this step? Comparing radii fairly needs the same charge; here it's already fixed at +3.
Radius falls monotonically with increasing $Z$ across the series (lanthanide contraction). List their $Z$ : Ce (58) < Nd (60) < Gd (64) < Lu (71). Why this step? Every extra proton with poor $4 f$ shielding raises $Z_{eff}$ → smaller ion. So higher $Z$ ⇒ smaller radius.
Reverse the $Z$ order to get largest-first radius.

Answer: $r (Ce^{3 +}) > r (Nd^{3 +}) > r (Gd^{3 +}) > r (Lu^{3 +})$ .

Verify: $Z$ increases $58 < 60 < 64 < 71$ , and radius is a decreasing function of $Z$ here, so the radius order is strictly the reverse ✓. (Consistent with $La^{3 +} = 103$ pm shrinking to $Lu^{3 +} = 86$ pm.)

Cell J — the trap check

Worked example Ex 10. Odd-one-out: which statement is FALSE?

(i) Most lanthanides show +3 as the main state. (ii) Ce⁴⁺ is stable because it reaches an empty $4 f^{0}$ shell. (iii) In the lanthanide contraction the atom shrinks because it is losing electrons. (iv) $La^{3 +}$ and $Lu^{3 +}$ are both colourless.

Forecast: pick the false one before reading the analysis.

Steps:

(i) True — the buried $4 f$ electrons resist removal, so losing $6 s^{2} +$ one more (giving +3) is the natural default.
(ii) True — $4 f^{0}$ is a stability well; that is why +4 is reachable for Ce and why Ce⁴⁺ oxidises.
(iii) FALSE — during the contraction electrons are being added to $4 f$ , not removed. The atom shrinks because the added proton's pull (poorly shielded by $4 f$ ) wins the tug-of-war, not because electrons leave. Why this step? This is the classic trap from the parent's [!mistake] callout: "contraction ≠ removing electrons."
(iv) True — $f^{0}$ and $f^{14}$ have no partly-filled $f$ shell → no $f$ – $f$ transitions → colourless.

Answer: statement (iii) is false.

Verify: across La→Lu the $4 f$ count rises from $0$ to $14$ (electrons added), directly contradicting "losing electrons." ✓

Recall Rapid self-test (cover the answers)

Config of Nd atom? ::: [Xe] 4f⁴ 6s² Config of Gd atom (and why anomalous)? ::: [Xe] 4f⁷ 5d¹ 6s² — keeps the half-filled 4f⁷ Config of Pr³⁺? ::: [Xe] 4f² Which off-default ion does Tb favour and why? ::: Tb⁴⁺ = [Xe]4f⁷ (half-filled), an oxidiser Average radius shrink per element La³⁺→Lu³⁺? ::: about 1.2 pm (17 pm over 14 gaps) Why is Hf ≈ Zr in size? ::: Inherited lanthanide contraction cancels the expected down-group growth Rank radii Ce³⁺, Gd³⁺, Lu³⁺ ::: Ce³⁺ > Gd³⁺ > Lu³⁺ (radius falls as Z rises) The classic false claim about contraction? ::: That the atom shrinks by losing electrons — it actually gains them