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

WHAT we count: the elements whose differentiating electron (the newest electron that makes each element different — defined above) drops into the 4f subshell. The count: the f subshell has azimuthal quantum number $\ell =3$ (the number that labels its shape, defined above). The number of "compartments" (orbitals) is $2\ell +1=2(3)+1=7$. Each compartment holds 2 electrons (opposite spins), so the 4f subshell holds $7\times 2=14$ electrons. Filling 14 electron-slots = 14 elements, running from ==Ce (Z = 58) to Lu (Z = 71)==. Answer: 14 lanthanides, Ce→Lu, filling the 4f subshell.

Start from the noble-gas core [Xe] (Z = 54, since the lanthanides sit in Period 6, just after Xe). Then the outer electrons distribute over 4f, 5d, 6s: $[\text{Xe}]4f^{1\text{–}14}5d^{0\text{–}1}6s^2$ The $6s^2$ is always full. The $4f$ count runs from 1 to 14. Which members actually have $5d^1$? The range shorthand $5d^{0\text{–}1}$ hides a sharp fact: only four lanthanides carry a $5d^1$ electron in the ground state — $\text{Ce }(4f^{1}5d^{1}6s^2),\text{Gd }(4f^{7}5d^{1}6s^2),\text{Lu }(4f^{14}5d^{1}6s^2)$ plus La ($5d^{1}6s^2$, the gatekeeper with empty 4f). These occur exactly where an empty (4f⁰), half-filled (4f⁷), or full (4f¹⁴) shell is being protected. Every other lanthanide has $5d^0$ — the newest electron sinks straight into 4f.

1. Core = [Xe] = 54 electrons. Remaining $=60-54=6$ electrons. Why start here? The core is chemically inert; only the outer 6 electrons matter.
2. Fill $6s^2$ first, because among 4f/5d/6s the sum $(n+\ell )$ is smallest for 6s ($6+0=6$ vs $7$ for both 4f and 5d — see the boxed rule above). Remaining $=6-2=4$.
3. Next, 4f versus 5d both have $(n+\ell )=7$; the tie-break "smaller $n$ first" puts electrons into 4f ($n=4$) before 5d ($n=5$). Place the 4 into 4f (no empty/half/full stability trick applies at 4). $\Rightarrow 4f^4$, and $5d^0$. Answer: $\text{Nd}=[\text{Xe}]4f^{4}6s^2$.

To ionise, remove the three outermost, weakest electrons: both $6s$ electrons first, then one 4f electron. $[\text{Xe}]4f^{4}6s^2\stackrel{-2e^{-}(6s)}{\to }[\text{Xe}]4f^4\stackrel{-1e^{-}(4f)}{\to }[\text{Xe}]4f^3$. Answer: ${\text{Nd}}^{3+}=[\text{Xe}]4f^3$.

Remaining after core $=64-54=10$ electrons. Naïvely: $6s^2$ then $4f^8$. But $4f^7$ (exactly half-filled — all 7 orbitals singly occupied) is extra stable. So the atom keeps $4f^7$ and puts the 8th outer electron into 5d instead: $\text{Gd}=[\text{Xe}]4f^{7}5d^{1}6s^2$ Answer: $[\text{Xe}]4f^{7}5d^{1}6s^2$; the half-filled $4f^7$ is preserved (see Half-filled and Fully-filled Stability).

Recall $Z_{\text{eff}}$ = (true nuclear charge) − (shielding from inner electrons), the net pull an outer electron feels. Now look at the figure — it plots the ${\text{Ln}}^{3+}$ radius (recall Ln = "any lanthanide," so this is the +3 ion's radius, defined in the box above) against increasing $Z$ (left → right = La → Lu). Two things to observe on it:

1. The curve slopes steadily downward — the marked points fall from La³⁺ ≈ 103 pm at the top-left to Lu³⁺ ≈ 86 pm at the bottom-right. There is no bump: every single step shrinks.
2. The dashed orange arrow traces that smooth, monotone descent — this is the visual signature of the lanthanide contraction.

Why does the graph fall? Each step to the right does two things at once:

• +1 proton in the nucleus → stronger inward pull.
• +1 electron into 4f → should shield the outer electrons and cancel that pull.

The key fact: 4f orbitals are diffuse and poorly-shaped for shielding — they leave "gaps" so the outer (5s, 5p, 6s) electrons still feel almost the full extra proton. In the tug-of-war, the nucleus wins, so $Z_{\text{eff}}$ on the outer electrons keeps climbing and the cloud is pulled inward. Conclusion the figure forces on you: rising $Z_{\text{eff}}$ ⇒ shrinking radius, 14 times over (see Effective Nuclear Charge & Slater's Rules).

• ${\text{Ce}}^{4+}$: Ce = $[\text{Xe}]4f^{1}5d^{1}6s^2$. Follow the correct removal order — 6s first, then 5d, then 4f: strip the two $6s$ electrons, then the single $5d^1$ electron, then the single $4f^1$ electron. That is $2+1+1=4$ electrons removed, leaving ${\text{Ce}}^{4+}=[\text{Xe}]4f^0$ — the ultra-stable empty configuration (like La³⁺). But +4 is not natural for lanthanides, so ${\text{Ce}}^{4+}$ readily grabs an electron to return to the cosy ${\text{Ce}}^{3+}$ ($4f^1$) — i.e. it oxidises other things.
• ${\text{Eu}}^{2+}$: Eu = $[\text{Xe}]4f^{7}6s^2$. Losing only the two $6s$ electrons gives ${\text{Eu}}^{2+}=[\text{Xe}]4f^7$ — the stable half-filled config. So Eu is content to stop at +2. But since +3 is the series norm, ${\text{Eu}}^{2+}$ tends to donate an electron and become ${\text{Eu}}^{3+}$ ($4f^6$) — i.e. it reduces other things.

Normal expectation: going down a group (Zr → Hf) adds a whole electron shell, so the radius should jump up. What actually intervenes: Hf sits immediately after the entire lanthanide series (Ce→Lu). All 14 tiny contractions accumulate between Zr's period and Hf. This built-up shrinkage almost exactly cancels the expected "go-down-a-period-get-bigger" increase. Result: $r(\text{Zr})\approx r(\text{Hf})$, so Zr and Hf are chemically nearly twins. What the figure shows: two overlapping circles drawn to scale — the orange Zr disc (≈160 pm) and the magenta Hf disc (≈159 pm). Notice the double-headed arrows marking each radius are essentially the same length: the circles are nearly identical in size even though Hf lies a whole period below Zr. That visual "no size jump" is the whole point. Consequence: they are extremely hard to separate (also Nb/Ta, Mo/W), and Hf ends up unusually dense (same small volume, much larger mass).

1. Total contraction $=103-86=17$ pm. Why subtract? The radius falls, so the drop is (start − end).
2. Steps from La to Lu: La→Ce→…→Lu. La³⁺ to Lu³⁺ spans 14 filling steps. Average per step $=\frac{17}{14}\approx 1.21$ pm.
3. Estimate $r({\text{Gd}}^{3+})$: Gd is 7 steps beyond La. $r\approx 103-7\times 1.21\approx 103-8.5=94.5$ pm. Answer: ≈ 1.21 pm/step; $r({\text{Gd}}^{3+})\approx 94.5$ pm (real value ≈ 94 pm — the linear estimate is excellent because the contraction is remarkably smooth).

(a) Remaining after core $=65-54=11$. Fill $6s^2$ (→ 9 left), then place the 9 into 4f: $4f^9$. Why $[\text{Xe}]4f^{9}6s^2$ and not $[\text{Xe}]4f^{8}5d^{1}6s^2$? A $5d^1$ electron is only "borrowed" when doing so lets 4f sit at a specially stable count — empty (f⁰), half-filled (f⁷) or full (f¹⁴). That is what happens at Gd (protects $4f^7$) and Lu (protects $4f^{14}$). At Tb, promoting one electron to 5d would give $4f^8$ — which is neither half-filled nor full, so there is no stabilising payoff to justify the promotion. The plain $4f^9$ arrangement is lower in energy, so the ground state is $[\text{Xe}]4f^{9}6s^2$. (b) Remove 4 electrons in the order 6s→4f (Tb has no 5d electron to strip): $6s^2$ (2) then two 4f electrons → $[\text{Xe}]4f^7$. ${\text{Tb}}^{4+}=[\text{Xe}]4f^7$ (c) $4f^7$ is the half-filled super-stable configuration — the same reason ${\text{Gd}}^{3+}$ and ${\text{Eu}}^{2+}$ are stable. Reaching $4f^7$ makes the extra (fourth) electron worth removing. (d) Since +3 is the norm and Tb "over-oxidised" to +4, ${\text{Tb}}^{4+}$ wants to grab an electron back to ${\text{Tb}}^{3+}$ ($4f^8$). Therefore ${\text{Tb}}^{4+}$ is an oxidiser (mirror image of ${\text{Ce}}^{4+}$).

1. Ionic size order (from the lanthanide contraction — smaller as $Z$ rises): $r({\text{La}}^{3+})>r({\text{Gd}}^{3+})>r({\text{Lu}}^{3+})$, i.e. 103 pm > 94 pm > 86 pm.
2. Link size to bond strength: a smaller cation has its positive charge packed into a tinier volume → higher charge density → it pulls the oxygen of O–H more strongly → the M–O bond is stronger and more covalent → the ion holds onto OH⁻ tightly and releases it less readily.
3. Less free OH⁻ released = weaker base. So basicity tracks increasing ion size: bigger ion → looser OH⁻ → stronger base.
4. Applying this to our three ions (La³⁺ largest, Lu³⁺ smallest): ${\text{La(OH)}}_3>{\text{Gd(OH)}}_3>{\text{Lu(OH)}}_3(\text{strongest}\to \text{weakest base})$ Answer: La(OH)₃ is the strongest base, Lu(OH)₃ the weakest — a direct fingerprint of lanthanide contraction. This size trend is also what makes ion-exchange separation possible (see Separation of Lanthanides — Ion Exchange).

In the actinides the 5f orbitals extend further out and lie close in energy to 6d/7s, so several electrons are removable, giving variable oxidation states. In the lanthanides the 4f is buried and inert, so only the outer three electrons leave → almost exclusively +3. (More in Actinides — 5f Series.)

The lanthanides are 14 near-twins filling a buried 4f shell → nearly all +3, steadily shrinking (contraction), which secretly shrinks Hf/Ta/W and makes the whole family hard to separate.