3.3.8 · D5d-Block (Transition Metals) & f-Block
Question bank — Lanthanides — electronic configuration, lanthanide contraction, oxidation states (mostly +3)
True or false — justify
Lanthanides all have the general form , i.e. every one carries a electron.
False. Only La, Ce, Gd, Lu keep a (driven by empty/half/full 4f stability); the rest push that electron into 4f, so most are .
Because La has configuration with no 4f electron, La itself is not counted among the 14 lanthanides.
Depends on the definition. Strictly, the 14 elements whose differentiating electron enters 4f run Ce→Lu, so La is excluded. But by IUPAC "lanthanoid" convention La is included as the group namesake (making 15 elements) even though its 4f is empty — both usages are common, so always check which count a question expects.
The lanthanide contraction means the atoms are losing electrons as we cross the series.
False. Electrons are being added (one 4f per step); the atom shrinks because the added proton's pull outweighs the poor shielding of the added 4f electron.
Across the series the effective nuclear charge felt by the outer electrons stays roughly constant.
False. It rises steadily — each new proton adds full pull ( up by 1), but the buried 4f electron shields it badly ( barely rises), so on the outer shell climbs and the cloud contracts.
and have the same electron configuration.
True. Both are — Gd () loses its 5d + two 6s; Eu () loses only its two 6s. The shared half-filled explains why both are stable species.
+3 is the common oxidation state because lanthanides most readily gain three electrons.
False. They lose three electrons (two 6s and one 5d/4f). Lanthanides are metals — they form cations, not anions.
The 4f electrons are the ones responsible for lanthanide chemistry varying wildly across the row.
False. 4f electrons are buried and poorly shielding, so they barely touch bonding — that is exactly why all 14 look nearly identical chemically.
Spot the error
"Ce⁴⁺ is stable, which proves the +4 state is common and normal for lanthanides."
Error: +4 is the exception. Ce⁴⁺ is stable only because it reaches the empty configuration; it remains a strong oxidiser, meaning it wants to fall back to the +3 state.
"Since Hf sits one period below Zr, Hf should be noticeably larger — going down a group always increases size."
Error: the 14-element contraction between them eats up the expected increase. So (≈160 pm), making them nearly inseparable.
"Basicity of the hydroxides increases La(OH)₃ → Lu(OH)₃ because the metals get more metallic."
Error: basicity decreases. The ion shrinks across the series, so the M–OH bond strengthens and OH⁻ is released less easily — less basic, not more.
"Eu is anomalous because it borrows a 5d electron to reach a stable ."
Two errors: Eu keeps no 5d electron (), and its stable configuration is the ==half-filled ==, not .
"The contraction is huge per step — about 15 pm each element."
Error: it is ~1 pm per step (recall ). It only looks large because it accumulates over 14 steps ().
"To make Ln³⁺ we remove three 4f electrons because 4f is the outermost subshell being filled."
Error: 4f is buried inside 5s5p. We remove the two 6s electrons first (outermost) and then one 5d-or-4f electron — leaving .
Why questions
Why are the lanthanides placed below the main body of the periodic table rather than inside it?
Because inserting all 14 f-block elements into Period 6 would make the table impractically wide; they are pulled out to keep it compact.
Why do 4f electrons shield so poorly compared to, say, 3p electrons?
The 4f orbitals are contracted and buried beneath the 5s and 5p shells, so their density sits inside the outer electrons rather than between them and the nucleus — they cannot screen the added nuclear charge from the outer electrons.
Why does the lanthanide contraction matter for elements that are not lanthanides (Hf, Ta, W)?
These 5d metals come immediately after the whole 4f series, so they inherit all the accumulated shrinkage — making them anomalously small and dense, and nearly twins of the 4d metals above them.
Why is +3 so uniquely favoured rather than +2 or +4 for most lanthanides?
The two 6s and one 5d/4f electrons are loosely held and cheap to remove; going further to +4 would tear into the buried, well-shielded 4f electrons, which costs too much energy (except when +4 hits a stable /).
Why does Ce show +4 but the neighbouring Nd does not readily?
Ce⁴⁺ reaches the specially stable empty ; Nd⁴⁺ would leave , an unremarkable configuration offering no such stabilisation, so it stays +3.
Why are Zr and Hf so hard to separate but easy to confuse?
Nearly identical radii (from the contraction) give them almost identical charge/size ratios, so their chemistry — bonding, solubility, complexation — is virtually the same.
Edge cases
At the very start of the series, why is La () an exception rather than following ?
With an empty 4f, the 4f and 5d orbitals are so close in energy that the electron finds the 5d slightly favourable — 4f only sinks below 5d as the nuclear charge climbs at higher Z.
At the very end, why does Lu carry a ()?
The 4f subshell is already completely full (), so the next differentiating electron has nowhere to go but 5d — some argue Lu is really a d-block element for this reason.
What happens to and radius right at the half-filled point (Gd)?
The trend is smooth and continuous — no visible kink in radius. The half-filled affects configuration choice (5d¹ retained), not the steady contraction of radius.
Is the +3 state ever not the ground preference — e.g. does any lanthanide favour +2 in its normal chemistry?
The default is always +3; +2 (Eu²⁺, Yb²⁺) appears only as a reducing species stabilised by reaching or , and it readily oxidises back to +3.
For the degenerate limiting cases , , — why do these three configurations keep reappearing as "special"?
They carry extra exchange energy stability. Exchange energy is an extra lowering of energy that occurs whenever two electrons of the same spin can swap places among equal-energy orbitals — the more such same-spin pairs, the more stable the atom. Empty (, nothing to disturb), exactly half-filled (, all seven spins parallel = maximum exchange pairs), and completely full (, a closed symmetric shell) each maximise this stabilisation, so ions that can reach them (Ce⁴⁺; Eu²⁺/Gd³⁺/Tb⁴⁺; Yb²⁺/Lu³⁺) gain an unusual foothold. See Half-filled and Fully-filled Stability.
Recall One-line self-test before you close this page
Cover everything: (a) which four elements have ? (b) during the contraction, does the atom gain or lose electrons? (c) why is Ce⁴⁺ only an "exception"? If any answer is a bare word without a reason, reopen the relevant group above. La, Ce, Gd, Lu ::: It gains electrons — one 4f electron is added at every step; the atom still shrinks because the added proton's pull outweighs that poorly-shielding 4f electron ::: because it merely reaches the empty configuration and still acts as a strong oxidiser, wanting to fall back to +3.