3.3.3 · D5d-Block (Transition Metals) & f-Block
Question bank — Atomic - ionic size trends; lanthanide contraction
Recall the two-knob picture before you start:
Recall The one ratio everything hangs on
Atomic size scales as , where is the net pull an outer electron actually feels after the inner electrons ( = shielding) block part of the nuclear charge . Bigger shell number → bigger atom; bigger effective pull → smaller atom. Shielding power by subshell: ====. An electron sits in a diffuse cloud and blocks the nucleus worst.
True or false — justify
Every step across a 3d series adds an electron, so the atom must grow.
False. The added electron enters the inner 3d shell (poor shielding), while the added proton pulls hard, so rises and the atom gently shrinks.
The lanthanide contraction is a large shrink at each element.
False. Each step is only ~1 pm; the effect is dramatic only because it accumulates over all 14 lanthanides (La→Lu).
5d atoms are larger than the 4d atoms above them because their principal quantum number is larger.
False. The 14 intervening lanthanides contract the atom by exactly enough to cancel the -increase, so 5d ≈ 4d (Zr≈Hf, Nb≈Ta, Mo≈W).
Down a group, 3d→4d radius increases.
True. No lanthanides sit between them, so the larger wins uncontested and 4d > 3d.
Fe³⁺ is smaller than Fe²⁺.
True. Both have 26 protons, but Fe³⁺ has one fewer electron, so shielding drops, per electron rises, and the remaining cloud is pulled tighter.
Os and Ir are the densest metals mainly because their atoms are unusually heavy for their volume.
True. Lanthanide contraction shrinks their volume while atomic mass keeps climbing, so mass/volume (density) becomes enormous.
La(OH)₃ is a stronger base than Lu(OH)₃.
True. La³⁺ is the largest lanthanide ion (lowest charge density), so it holds OⁿH loosely and releases OH⁻ easily → more basic; Lu³⁺ is smallest → least basic.
Across a period in the main-group elements, radius drops faster than across a 3d series.
True. In main-group the added electron enters the same outer shell (poor mutual shielding at similar ), so climbs steeply; in a d-series it hides in an inner shell, softening the drop.
The metallic radius decreases monotonically all the way from Sc to Zn.
False. It decreases then levels, and ticks up near Cu/Zn because growing d–d repulsion plus a filled d-shell that barely bonds weakens the metallic bonding.
f electrons are far from the nucleus on average, so they should shield very well.
False. "Far on average" is not the same as "in the way." Their diffuse, oddly-shaped clouds leave the nucleus largely exposed to outer electrons, so they shield worst.
Ionic radius equals atomic radius minus a fixed constant.
False. Making a cation removes a whole outer shell and raises on what remains, so the shrink is large and depends on charge, not a constant offset.
Spot the error
"Zr and Hf are hard to separate because they have nearly the same atomic mass."
The reason is nearly-equal size (from lanthanide contraction), not mass. Equal size + equal charge → equal charge density → nearly identical chemistry.
"Because 4f fills across the lanthanides, the 4f electrons themselves grow larger and push the atom outward."
Backwards. The 4f electrons shield poorly, so the added protons win and the whole atom contracts; nothing pushes outward.
", since shielding adds to the nuclear pull."
Sign error. Shielding opposes the nucleus: . Inner electrons cancel part of the charge an outer electron feels.
"Since d-electrons shield poorly, adding them makes the atom grow across the series."
Poor shielding means the proton's pull is not cancelled, so rises and the atom shrinks, not grows.
"Cu³⁺ would be larger than Cu²⁺ because higher charge means more electrons stripped means a looser cloud."
Reversed. Higher positive charge means fewer electrons and higher per electron, so higher-charge cations are smaller: Cu³⁺ < Cu²⁺.
"The lanthanide contraction affects only the lanthanides themselves."
Its most important effect is on the elements after them — the entire 5d row (Hf onward) inherits the shrink, which is why 5d ≈ 4d.
"Down group 4, radius increases smoothly: Ti < Zr < Hf."
The last inequality fails. Ti < Zr, but Zr ≈ Hf because the lanthanide contraction cancels the expected 4d→5d increase.
Why questions
Why does the radius drop across a 3d series but only gently?
Each step's added electron lands in the inner (n−1)d shell, which shields poorly, so creeps up slowly rather than jumping — a soft contraction.
Why do the 5d transition metals have such high densities?
Lanthanide contraction squeezes 14 extra protons' worth of pull into the atom, shrinking its volume while mass keeps rising, so mass/volume is very large.
Why does hydroxide basicity fall from La(OH)₃ to Lu(OH)₃?
The Ln³⁺ ion shrinks steadily, so its charge density rises, the M–O bond becomes more covalent, and OH⁻ is held more tightly (released less easily) → weaker base.
Why is a cation always smaller than its parent atom?
Removing outer electrons often empties the outermost shell entirely and lowers total shielding, so the remaining electrons feel a higher and are pulled inward.
Why does the metallic radius rise slightly at Cu and Zn?
The nearly-filled d-shell contributes little to metallic bonding, and d–d electron–electron repulsion grows, so bonds weaken and the atoms sit farther apart.
Why do we always quote a coordination number with ionic radii?
An ion's effective radius depends on how many neighbours pack around it; the same ion measures differently at CN 6 versus CN 8, so a bare number is ambiguous.
Why does the simple picture explain both "shrink across" and "grow down"?
Moving across raises (denominator up → smaller); moving down raises (numerator up → larger). Each trend is one knob turning.
Edge cases
Zn has a completely filled 3d¹⁰4s²; does it still count as "contracted" by the d-electrons?
The filled d-shell still shields poorly, so stays high, but its non-participation in metallic bonding is why Zn's metallic radius ticks up despite that high .
Sc³⁺ has no d electrons at all — does the "d-shielding" story apply to it?
Not to Sc³⁺ itself; it's an [Ar] core ion. The d-shielding story governs the neutral atoms and lower cations across the series, where d electrons are actually present.
Is the atomic radius of a lone gas-phase atom the same as the metallic radius used in these trends?
No. Metallic radius is half the internuclear distance in the solid metal; a free-atom radius (van der Waals or covalent) differs, so trends must compare like with like.
At the very end of the lanthanides (Yb, Lu), does the contraction ever reverse?
No net reversal in ionic radius — it decreases monotonically La³⁺→Lu³⁺. Small irregularities appear in metallic radii (e.g. Eu, Yb) from half/fully-filled 4f stability affecting bonding, not from any regrowth of the ion.
If 4f shielded as well as an s subshell, what would happen to Zr vs Hf?
The 14 added protons would be well-screened, so the atom would not contract; Hf would then be clearly larger than Zr (as alone predicts) and the two would be easy to separate.
Does the actinide series show an analogous contraction?
Yes — the "actinide contraction" from filling 5f, which is even more diffuse; it is real but complicated by relativistic effects and variable oxidation states, so it is less textbook-clean than the lanthanide case.
See also: f-Block (Lanthanides & Actinides), Density, Melting Point Trends in Transition Metals, Basic Character of Oxides & Hydroxides, Separation of Lanthanides (Ion-exchange).