Chapter 4 · The d- and f-Block Elements

4.3 Lanthanoids, Actinoids and Applications

Two rows set apart at the foot of the periodic table fill the deep-lying $4f$ and $5f$ orbitals. The lanthanoids are a family of near-twins that shrink steadily (the lanthanoid contraction); the actinoids are radioactive and far more varied in oxidation state. Then a tour of how d- and f-block elements are used.

A modern smartphone, an electric car and a wind turbine all depend on elements most people have never heard of. The tiny, powerful magnets in their motors and speakers are neodymium–iron–boron. The red and green phosphors of older colour screens used europium and terbium. Hospitals inject gadolinium complexes to brighten MRI scans, and the "flint" in a cigarette lighter is an alloy of cerium and lanthanum. At the very end of the periodic table, uranium and plutonium fuel nuclear reactors, and a speck of americium in a smoke detector quietly ionises the air.

These are the f-block elements, placed in two rows below the main table. In them the electrons enter the $4f$ (lanthanoids) or $5f$ (actinoids) orbitals, two shells in from the outside. Because those orbitals are buried, adding $f$ electrons changes the chemistry very little. The lanthanoids behave like one big family of $+3$ ions that get slightly smaller at each step, and that steady shrinking, the lanthanoid contraction, reaches forward to the elements after them. The actinoids are different: their $5f$ orbitals reach further out, so they show many more oxidation states, and all of them are radioactive. This section studies both series, compares them, and ends with the uses of the $d$- and $f$-block elements.

The Lanthanoids: Configurations and Oxidation States

📘 Definition 4.8 — Inner transition elements: lanthanoids and actinoids

The f-block elements, or inner transition elements, are those in which the last electron enters an $(n-2)f$ orbital. There are two series:

  • the lanthanoids, the fourteen elements cerium ($Z=58$) to lutetium ($Z=71$) that follow lanthanum, filling $4f$; lanthanum ($Z=57$) is closely similar and is usually discussed with them. The general symbol is Ln.
  • the actinoids, the fourteen elements thorium ($Z=90$) to lawrencium ($Z=103$) that follow actinium, filling $5f$; actinium is discussed with them. The general symbol is An.

They are called inner transition elements because the subshell being filled lies inside the $(n-1)d$ and $ns$ levels that hold the outer electrons.

📐 Rule 4.10 — Configurations of the lanthanoids and their ions
  1. All lanthanoid atoms have $6s^{2}$, and the general configuration is $\mathrm{[Xe]\,4f^{1-14}\,5d^{0-1}\,6s^{2}}$.
  2. Most atoms have no $5d$ electron: $\mathrm{[Xe]\,4f^{\,n}\,6s^{2}}$ with $n=Z-56$ (for example Pr $4f^{3}6s^{2}$, Eu $4f^{7}6s^{2}$, Yb $4f^{14}6s^{2}$). La, Ce, Gd and Lu have one $5d$ electron: La $5d^{1}6s^{2}$, Ce $4f^{1}5d^{1}6s^{2}$, Gd $4f^{7}5d^{1}6s^{2}$, Lu $4f^{14}5d^{1}6s^{2}$. In Gd and Lu this keeps the stable half-filled $4f^{7}$ and full $4f^{14}$ sets.
  3. The $+3$ ions are completely regular: the $6s$ and $5d$ electrons (and one $4f$ electron where there is no $5d$) are removed, leaving $\mathrm{Ln^{3+}}=\mathrm{[Xe]\,4f^{\,n}}$ with $n=Z-57$, from $\mathrm{La^{3+}}$ ($4f^{0}$) to $\mathrm{Lu^{3+}}$ ($4f^{14}$).
ElementLaCePrNdPmSmEuGd
$Z$5758596061626364
atom, beyond [Xe]$5d^{1}6s^{2}$$4f^{1}5d^{1}6s^{2}$$4f^{3}6s^{2}$$4f^{4}6s^{2}$$4f^{5}6s^{2}$$4f^{6}6s^{2}$$4f^{7}6s^{2}$$4f^{7}5d^{1}6s^{2}$
$\mathrm{Ln^{3+}}$$4f^{0}$$4f^{1}$$4f^{2}$$4f^{3}$$4f^{4}$$4f^{5}$$4f^{6}$$4f^{7}$
ElementTbDyHoErTmYbLu
$Z$65666768697071
atom, beyond [Xe]$4f^{9}6s^{2}$$4f^{10}6s^{2}$$4f^{11}6s^{2}$$4f^{12}6s^{2}$$4f^{13}6s^{2}$$4f^{14}6s^{2}$$4f^{14}5d^{1}6s^{2}$
$\mathrm{Ln^{3+}}$$4f^{8}$$4f^{9}$$4f^{10}$$4f^{11}$$4f^{12}$$4f^{13}$$4f^{14}$
📐 Rule 4.11 — Oxidation states of the lanthanoids
  1. The characteristic oxidation state of every lanthanoid is $+3$. The sum of the first three ionisation enthalpies is comparatively low, and it is repaid by the large lattice or hydration enthalpies of the small, highly charged $\mathrm{Ln^{3+}}$ ion.
  2. Some lanthanoids also show $+2$ or $+4$, especially when that state gives an empty ($f^{0}$), half-filled ($f^{7}$) or completely filled ($f^{14}$) $4f$ subshell:
    • $\mathrm{Ce^{4+}}$ ($4f^{0}$) is the most important $+4$ ion. It is a strong oxidising agent, $E^{\circ}_{\mathrm{Ce^{4+}/Ce^{3+}}}=+1.74\ \mathrm{V}$, used in volumetric analysis; it returns to the common $+3$ state. (In theory it could oxidise water, but the reaction is very slow, so cerium(IV) solutions can be stored.) $\mathrm{Tb^{4+}}$ ($4f^{7}$) is also an oxidant.
    • $\mathrm{Eu^{2+}}$ ($4f^{7}$) and $\mathrm{Yb^{2+}}$ ($4f^{14}$) are formed easily and act as reducing agents, changing to the $+3$ state. $\mathrm{Sm^{2+}}$ is another reducing $+2$ ion.
    • Pr, Nd, Tb and Dy form $+4$ compounds only in oxides such as $\mathrm{MO_{2}}$.
  3. Stability of an empty, half-filled or full subshell is a useful guide but not the whole story: $\mathrm{Sm^{2+}}$ ($4f^{6}$) and $\mathrm{Tm^{2+}}$ ($4f^{13}$) also exist, so lattice energies and other thermodynamic and kinetic factors matter too.
✏️ Example 1 — Writing lanthanoid configurations
Using Rule 4.10, write the configurations of (a) cerium and $\mathrm{Ce^{4+}}$, (b) gadolinium and $\mathrm{Gd^{3+}}$, (c) $\mathrm{Eu^{2+}}$ and $\mathrm{Yb^{2+}}$, (d) $\mathrm{Lu^{3+}}$. How many unpaired $4f$ electrons does $\mathrm{Gd^{3+}}$ have, and why is it chosen for MRI contrast agents?
(a) Ce ($Z=58$): $\mathrm{[Xe]\,4f^{1}\,5d^{1}\,6s^{2}}$. Remove four electrons: $\mathrm{Ce^{4+}}=\mathrm{[Xe]}$ ($4f^{0}$), a noble-gas core.
(b) Gd ($Z=64$): $\mathrm{[Xe]\,4f^{7}\,5d^{1}\,6s^{2}}$; $\mathrm{Gd^{3+}}=\mathrm{[Xe]\,4f^{7}}$ ($64-57=7$ ✓).
(c) Eu ($Z=63$) is $\mathrm{[Xe]\,4f^{7}\,6s^{2}}$, so $\mathrm{Eu^{2+}}=\mathrm{[Xe]\,4f^{7}}$. Yb ($Z=70$) is $\mathrm{[Xe]\,4f^{14}\,6s^{2}}$, so $\mathrm{Yb^{2+}}=\mathrm{[Xe]\,4f^{14}}$. Each $+2$ ion keeps a half-filled or full $4f$ set.
(d) and MRI $\mathrm{Lu^{3+}}=\mathrm{[Xe]\,4f^{14}}$ ($71-57=14$). $\mathrm{Gd^{3+}}$ has seven $4f$ orbitals each holding one electron: 7 unpaired electrons, the largest number possible for any ion. Its strong paramagnetism speeds up the relaxation of the water protons near it, which brightens the MRI image. (It is given as a stable complex, because the free ion is toxic.)

The Lanthanoid Contraction and Its Consequences

From lanthanum to lutetium the nuclear charge rises by fourteen units, and fourteen electrons are added, all to the $4f$ subshell. The $4f$ orbitals are diffuse and shield one another, and the outer $5s$, $5p$ and $6s$ electrons, rather poorly. So with every step the outer electrons feel a slightly larger effective nuclear charge, and the atom or ion shrinks a little.

📘 Definition 4.9 — Lanthanoid contraction
The lanthanoid contraction is the steady decrease in the atomic and ionic radii of the lanthanoids with increasing atomic number, from lanthanum to lutetium. It is caused by the imperfect shielding of one $4f$ electron by another, so that the effective nuclear charge experienced by the outer electrons increases across the series. For the $\mathrm{Ln^{3+}}$ ions the radius falls from $103\ \mathrm{pm}$ ($\mathrm{La^{3+}}$) to $86\ \mathrm{pm}$ ($\mathrm{Lu^{3+}}$), almost smoothly. The atomic radii fall too, but less regularly: europium and ytterbium atoms are unusually large because, with their stable $4f^{7}$ and $4f^{14}$ sets, they give only two electrons to metallic bonding.
📐 Rule 4.12 — Consequences of the lanthanoid contraction
  1. The 4d and 5d transition metals have almost the same size. The normal increase in radius on going down a group from period 5 to period 6 is cancelled by the contraction across the lanthanoids that lie in between. Hence zirconium ($160\ \mathrm{pm}$) and hafnium ($159\ \mathrm{pm}$), niobium and tantalum, and molybdenum and tungsten are pairs of near-identical size. Each pair has very similar chemistry, occurs together in nature and is hard to separate.
  2. The lanthanoids are very similar to one another and occur together in the same minerals. Their small differences in size are enough to separate them, for example by ion-exchange methods, but only with effort.
  3. Basicity of the hydroxides decreases from $\mathrm{La(OH)_{3}}$ to $\mathrm{Lu(OH)_{3}}$. As $\mathrm{Ln^{3+}}$ becomes smaller, it polarises the hydroxide ion more and the $\mathrm{Ln{-}OH}$ bond becomes more covalent. $\mathrm{La(OH)_{3}}$ is the most basic and $\mathrm{Lu(OH)_{3}}$ the least.
  4. The 5d metals are unusually dense and have high ionisation enthalpies: their atoms are no larger than the 4d metals above them, but much heavier, and their outer electrons are held more firmly. Yttrium ($\mathrm{Y^{3+}}$, $90\ \mathrm{pm}$) ends up the same size as $\mathrm{Ho^{3+}}$, which is why yttrium is always found with the heavier lanthanoids.
Interactive 1 — The Lanthanoid Contraction

Plot the $\mathrm{Ln^{3+}}$ ionic radius or the atomic radius from La to Lu. Press Play to add one proton and one $4f$ electron at a time and watch the cumulative shrinkage grow. The panel on the right shows why this matters beyond the lanthanoids: in groups 4, 5 and 6 the 5d metal is no bigger than the 4d metal above it. With the atomic radii, notice the two spikes at Eu and Yb.

Lu
element · $\mathrm{Ln^{3+}}$—
radius (pm)—
shrinkage from La—
average per step—
✏️ Example 2 — How big is the contraction, and what does it do to hafnium?
(a) The $\mathrm{Ln^{3+}}$ radius falls from $103\ \mathrm{pm}$ (La) to $86.1\ \mathrm{pm}$ (Lu). Find the total and percentage decrease, and the average decrease per element. (b) The metallic radii of Ti, Zr and Hf are $147$, $160$ and $159\ \mathrm{pm}$. If the increase from Zr to Hf were the same as that from Ti to Zr, what would the radius of Hf be? How much of an increase has the lanthanoid contraction cancelled? (c) Arrange $\mathrm{La(OH)_{3}}$, $\mathrm{Gd(OH)_{3}}$ and $\mathrm{Lu(OH)_{3}}$ in order of decreasing basicity.
(a) Decrease $=103-86.1=16.9\ \mathrm{pm}$, i.e. $\dfrac{16.9}{103}\times100=16.4\%$. Over $14$ steps: $\dfrac{16.9}{14}=1.21\ \mathrm{pm}$ per element, small at each step but large in total.
(b) Ti → Zr: $+13\ \mathrm{pm}$. "Expected" Hf $\approx160+13=173\ \mathrm{pm}$; actual $159\ \mathrm{pm}$. About $14\ \mathrm{pm}$ of expected growth has been cancelled, close to the total contraction across the lanthanoids found in (a).
(c) $\mathrm{La(OH)_{3}}\gt\mathrm{Gd(OH)_{3}}\gt\mathrm{Lu(OH)_{3}}$: the smaller the cation, the more covalent the M–OH bond and the weaker the base.
Meaning Because Zr and Hf are the same size and have the same outer configuration, they are among the hardest pair of elements in the periodic table to separate; hafnium was discovered only in 1923, hidden in zirconium ores.

General Characteristics and Chemistry of the Lanthanoids

📐 Rule 4.13 — Typical reactions of the lanthanoid metals
  • with dilute acids, hydrogen is liberated: $\mathrm{2Ln(s)+6H^{+}(aq)\rightarrow2Ln^{3+}(aq)+3H_{2}(g)}$;
  • with water, the hydroxide forms and hydrogen is liberated: $\mathrm{2Ln(s)+6H_{2}O(l)\rightarrow2Ln(OH)_{3}(s)+3H_{2}(g)}$;
  • they burn in oxygen to give $\mathrm{Ln_{2}O_{3}}$: $\mathrm{4Ln(s)+3O_{2}(g)\rightarrow2Ln_{2}O_{3}(s)}$;
  • heated with nitrogen they give nitrides, $\mathrm{LnN}$; with carbon at about $2773\ \mathrm{K}$, carbides such as $\mathrm{LnC_{2}}$, $\mathrm{Ln_{3}C}$ and $\mathrm{Ln_{2}C_{3}}$; with sulphur, $\mathrm{Ln_{2}S_{3}}$; with halogens, $\mathrm{LnX_{3}}$.

The oxides $\mathrm{Ln_{2}O_{3}}$ and hydroxides $\mathrm{Ln(OH)_{3}}$ are basic, like those of the alkaline earth metals (with the decrease in basicity across the series of Rule 4.12).

Uses. The best-known use is mischmetall, an alloy of about $95\%$ lanthanoid metals (mainly cerium and lanthanum) with about $5\%$ iron and traces of sulphur, carbon, calcium and aluminium. Alloyed with magnesium it is used in bullets and shells and as lighter flints, because it gives sparks when struck. Mixed lanthanoid oxides are used as catalysts in petroleum cracking, some lanthanoid oxides as phosphors in screens, and neodymium in very strong permanent magnets.

✏️ Example 3 — Lanthanoid chemistry in equations
(a) Write balanced equations for the reaction of cerium with dilute hydrochloric acid and of samarium with oxygen. (b) Gadolinium reacts with water to form its hydroxide. Calculate the volume of hydrogen at STP ($22.7\ \mathrm{L\,mol^{-1}}$) released by $3.15\ \mathrm{g}$ of gadolinium ($157\ \mathrm{g\,mol^{-1}}$). (c) Why do $\mathrm{Ce^{4+}}$ salts act as oxidising agents while $\mathrm{Eu^{2+}}$ salts act as reducing agents?
(a) $\mathrm{2Ce(s)+6HCl(aq)\rightarrow2CeCl_{3}(aq)+3H_{2}(g)}$; $\ \mathrm{4Sm(s)+3O_{2}(g)\rightarrow2Sm_{2}O_{3}(s)}$.
(b) Ratio $\mathrm{2Gd(s)+6H_{2}O(l)\rightarrow2Gd(OH)_{3}(s)+3H_{2}(g)}$: $3\ \mathrm{mol}$ $\mathrm{H_{2}}$ per $2\ \mathrm{mol}$ Gd. $n(\mathrm{Gd})=\dfrac{3.15}{157}=0.02006\ \mathrm{mol}$, so $n(\mathrm{H_{2}})=0.03010\ \mathrm{mol}$.
(b) Volume $V=0.03010\times22.7=0.683\ \mathrm{L}$ (683 mL) at STP.
(c) Both ions tend to return to the characteristic $+3$ state. $\mathrm{Ce^{4+}}$ ($4f^{0}$) does so by gaining an electron ($E^{\circ}=+1.74\ \mathrm{V}$), so it oxidises other species; $\mathrm{Eu^{2+}}$ ($4f^{7}$) does so by losing one, so it reduces them.

The Actinoids, a Comparison, and Uses of d- and f-Block Elements

The actinoids are all radioactive. Only thorium and uranium occur in useful amounts in nature; the elements after uranium are made in nuclear reactors or accelerators, and the later ones only a few atoms at a time, with half-lives of minutes or less. That is why far less is known about their chemistry.

📐 Rule 4.14 — Configurations and oxidation states of the actinoids
  1. All actinoids have $7s^{2}$; the general configuration is $\mathrm{[Rn]\,5f^{1-14}\,6d^{0-1}\,7s^{2}}$. The $5f$ and $6d$ levels are so close in energy that the configurations are less regular than for the lanthanoids: thorium is $\mathrm{[Rn]\,6d^{2}\,7s^{2}}$ with no $5f$ electron; Pa, U, Np, Cm and Lr have one $6d$ electron; the others are $\mathrm{[Rn]\,5f^{\,n}\,7s^{2}}$. Half-filled $5f^{7}$ (Am, Cm) and full $5f^{14}$ (No, Lr) sets appear, as in the lanthanoids.
  2. The actinoids show a much greater range of oxidation states than the lanthanoids, because the $5f$, $6d$ and $7s$ electrons have comparable energies and can all be used. The highest state rises from $+4$ (Th) to $+5$ (Pa), $+6$ (U) and $+7$ (Np, Pu), then falls again; from americium onwards $+3$ becomes dominant, and the late elements behave much like lanthanoids.
  3. High oxidation states often occur as oxocations: $\mathrm{UO_{2}^{2+}}$ (uranyl, U(VI)), $\mathrm{PuO_{2}^{2+}}$, $\mathrm{UO_{2}^{+}}$.
ElementAcThPaUNpPuAmCm
$Z$8990919293949596
beyond [Rn]$6d^{1}7s^{2}$$6d^{2}7s^{2}$$5f^{2}6d^{1}7s^{2}$$5f^{3}6d^{1}7s^{2}$$5f^{4}6d^{1}7s^{2}$$5f^{6}7s^{2}$$5f^{7}7s^{2}$$5f^{7}6d^{1}7s^{2}$
states+3+3, +4+3 to +5+3 to +6+3 to +7+3 to +7+3 to +6+3, +4
ElementBkCfEsFmMdNoLr
$Z$979899100101102103
beyond [Rn]$5f^{9}7s^{2}$$5f^{10}7s^{2}$$5f^{11}7s^{2}$$5f^{12}7s^{2}$$5f^{13}7s^{2}$$5f^{14}7s^{2}$$5f^{14}6d^{1}7s^{2}$
states+3, +4+3+3+3+2, +3+2, +3+3

General characteristics. The actinoid metals are silvery and show a variety of crystal structures. They are highly reactive, especially when finely divided: they react with boiling water to give a mixture of oxide and hydride and combine with most non-metals at moderate temperatures. Hydrochloric acid attacks all of them, but most are only slightly affected by nitric acid, because a protective oxide layer forms; alkalis have no action. The ionisation enthalpies of the early actinoids are lower than those of the early lanthanoids, because the $5f$ electrons are more effectively shielded from the nucleus than the $4f$ electrons and are therefore more easily removed. The actinoid contraction, the gradual decrease in the size of atoms and $\mathrm{M^{3+}}$ ions across the series, is greater from element to element than the lanthanoid contraction, because $5f$ electrons shield one another even more poorly. Their magnetic properties are more complicated than those of the lanthanoids.

FeatureLanthanoidsActinoids
orbitals filled$4f$$5f$
configuration$\mathrm{[Xe]\,4f^{1-14}5d^{0-1}6s^{2}}$, fairly regular$\mathrm{[Rn]\,5f^{1-14}6d^{0-1}7s^{2}}$, less regular
oxidation statesmainly $+3$; a few $+2$, $+4$$+3$ to $+7$ in the early members
contractionsmaller step from element to elementlarger step from element to element
radioactivityonly promethiumall
oxocationsnot formed$\mathrm{UO_{2}^{2+}}$, $\mathrm{PuO_{2}^{2+}}$, $\mathrm{NpO_{2}^{+}}$
complex formationweakerstronger ($5f$ orbitals take part more in bonding)
Interactive 2 — f-Block Explorer: Lanthanoids and Actinoids Side by Side

Each column is one element; the dots are its known oxidation states, with $+3$ outlined. Choose a position in the series: the lab builds the configuration of the lanthanoid and the actinoid in that column (from the rules above), and the $f$-electron count and unpaired electrons of the $\mathrm{M^{3+}}$ ion. Press Play to scan the series. Watch the actinoid row fan out to $+7$ at Np and Pu while the lanthanoid row hugs $+3$.

6th
lanthanoid · configuration—
$\mathrm{Ln^{3+}}$ · unpaired—
actinoid · configuration—
$\mathrm{An^{3+}}$ · unpaired—
✏️ Example 4 — Uranium and its oxidation states
(a) Write the configuration of uranium ($Z=92$) and of $\mathrm{U^{4+}}$. (b) What is the highest oxidation state of uranium, and how many electrons does it use? (c) Find the oxidation state of uranium in $\mathrm{UO_{2}^{2+}}$, $\mathrm{UF_{6}}$ and $\mathrm{U_{3}O_{8}}$. (d) How many unpaired electrons does $\mathrm{U^{4+}}$ have?
(a) U: $\mathrm{[Rn]\,5f^{3}\,6d^{1}\,7s^{2}}$. $\mathrm{U^{4+}}$: remove $7s^{2}$, $6d^{1}$ and one $5f$: $\mathrm{[Rn]\,5f^{2}}$.
(b) $+6$: all six electrons beyond the radon core ($5f^{3}6d^{1}7s^{2}$) are used.
(c) $\mathrm{UO_{2}^{2+}}$: $x-4=+2$, $+6$ (the uranyl ion). $\mathrm{UF_{6}}$: $+6$ (the volatile compound used to enrich uranium). $\mathrm{U_{3}O_{8}}$: $3x=16$, $x=+\tfrac{16}{3}$, only an average: $\mathrm{U_{3}O_{8}}$ is a mixed-valence oxide containing uranium in more than one oxidation state (an oxidation number cannot be a fraction for a single atom).
(d) $5f^{2}$: two electrons in two of the seven $5f$ orbitals, with parallel spins: 2 unpaired electrons.
✏️ Example 5 — Comparing the two series
Explain: (a) the actinoids show a much wider range of oxidation states than the lanthanoids; (b) the actinoid contraction is greater from element to element than the lanthanoid contraction; (c) the chemistry of the actinoids is much less fully known than that of the lanthanoids; (d) both $\mathrm{Gd^{3+}}$ and $\mathrm{Cm^{3+}}$ are especially stable.
(a) The $5f$, $6d$ and $7s$ orbitals have very similar energies, so varying numbers of electrons can be removed or used in bonding. The $4f$ orbitals lie much deeper than $5d$ and $6s$, and $4f$ electrons are hard to involve, so the lanthanoids keep to $+3$.
(b) $5f$ orbitals are more diffuse than $4f$ orbitals and shield one another even more poorly, so the effective nuclear charge rises more steeply at each step.
(c) All actinoids are radioactive; many are made only in tiny amounts and have short half-lives. Studying their chemistry needs special laboratories and very small samples.
(d) $\mathrm{Gd^{3+}}$ is $\mathrm{[Xe]\,4f^{7}}$ and $\mathrm{Cm^{3+}}$ ($Z=96$) is $\mathrm{[Rn]\,5f^{7}}$ ($96-89=7$): each has a half-filled $f$ subshell, which is extra stable.

Applications of d- and f-block elements. Iron and steels are the most important construction materials of all. Alloy steels and special alloys use Cr, Mn, Ni, V, Mo and W; titanium alloys are used in aircraft and implants; chromium plating protects and decorates. Titanium(IV) oxide is the main white pigment in paints, and $\mathrm{MnO_{2}}$ is used in dry cells. The batteries industry needs Zn and Ni (and Cd). Group 11 metals (Cu, Ag, Au) are the coinage metals. Transition metals and their compounds are vital catalysts: $\mathrm{V_{2}O_{5}}$ for the oxidation of $\mathrm{SO_{2}}$ in making sulphuric acid; $\mathrm{TiCl_{4}}$ with $\mathrm{Al(CH_{3})_{3}}$ (the Ziegler catalyst) for making polythene; iron catalysts for ammonia (Haber process); nickel for the hydrogenation of fats; and nickel complexes for polymerising alkynes and other organic compounds such as benzene. Silver bromide is light-sensitive and was the basis of photographic film. Among the $f$-block elements, lanthanoids give mischmetall, catalysts, phosphors, lasers and NdFeB magnets, and gadolinium contrast agents; uranium and plutonium are nuclear fuels, and americium-241 is used in smoke detectors.

✏️ Example 6 — Matching elements to uses
Name the d- or f-block element (or compound) used for each purpose, and say which property of the element makes it suitable: (a) catalyst for converting $\mathrm{SO_{2}}$ to $\mathrm{SO_{3}}$; (b) lighter flints; (c) nuclear fuel in most power reactors; (d) catalyst for making polythene; (e) the white pigment in most paints.
(a) $\mathrm{V_{2}O_{5}}$: vanadium switches easily between $+5$ and $+4$, taking oxygen from $\mathrm{O_{2}}$ and passing it to $\mathrm{SO_{2}}$ (variable oxidation state).
(b) Mischmetall (Ce, La with a little Fe) alloyed with Mg: the very reactive lanthanoids burn in air as sparks when the alloy is scraped.
(c) Uranium (uranium-235, as $\mathrm{UO_{2}}$ fuel enriched via volatile $\mathrm{UF_{6}}$): its nuclei undergo fission and release energy.
(d) The Ziegler catalyst, $\mathrm{TiCl_{4}}$ with $\mathrm{Al(CH_{3})_{3}}$: titanium's vacant $d$ orbitals bind ethene molecules and join them into chains.
(e) $\mathrm{TiO_{2}}$: $\mathrm{Ti^{4+}}$ is $d^{0}$, so there is no $d$–$d$ absorption (white), and the oxide scatters light very strongly and is chemically inert.
⚠️ Common mistakes with the f-block

1. Giving every lanthanoid a $5d^{1}$ electron. Only La, Ce, Gd and Lu have one; the rest are $4f^{n}6s^{2}$. The $\mathrm{Ln^{3+}}$ ions, however, are always $4f^{\,Z-57}$.

2. Using the spin-only formula for lanthanoids. It works for 3d ions, not for $4f$ ions, whose orbital contribution is large.

3. Mixing up the direction of the contraction's effect. The lanthanoid contraction makes the 5d metals as small as the 4d metals (Zr ≈ Hf), not smaller than the 3d metals.

4. Saying lanthanoid hydroxides become more basic across the series. They become less basic, because the cation becomes smaller.

🚀 Beyond the board syllabus (enrichment only) — separating the lanthanoids

Enrichment only: no 2026 syllabus names the separation methods (JEE Main, NEET and JEE Advanced list the lanthanoid contraction behind them); not required by the CBSE syllabus.

Because the $\mathrm{Ln^{3+}}$ ions differ only slightly in size, they are separated industrially by ion exchange or solvent extraction, which magnify tiny differences by repeating a partition many times. On a cation-exchange resin eluted with a complexing agent, the smallest ion, $\mathrm{Lu^{3+}}$, forms the strongest complex and comes off first, and $\mathrm{La^{3+}}$ comes off last; yttrium appears in the eluate near holmium, exactly where its radius places it. Cerium is usually removed first by oxidising it to $\mathrm{Ce^{4+}}$, whose chemistry is quite different.

Exercises

I
Lanthanoid configurations and oxidation states. Atomic numbers: La 57, Ce 58, Nd 60, Sm 62, Eu 63, Gd 64, Tb 65, Er 68, Yb 70, Lu 71.
  • 1 Write the configurations of $\mathrm{Nd}$, $\mathrm{Sm^{3+}}$, $\mathrm{Tb}$ and $\mathrm{Er^{3+}}$.
  • 2 Which lanthanoid atoms (including lanthanum) have one $5d$ electron?
  • 3 How many unpaired electrons are there in $\mathrm{Ce^{3+}}$, $\mathrm{Eu^{3+}}$, $\mathrm{Gd^{3+}}$, $\mathrm{Yb^{3+}}$ and $\mathrm{Lu^{3+}}$?
  • 4 Which of $\mathrm{La^{3+}}$, $\mathrm{Ce^{4+}}$, $\mathrm{Nd^{3+}}$, $\mathrm{Yb^{2+}}$ and $\mathrm{Lu^{3+}}$ are diamagnetic?
  • 5 Why is $+3$ the characteristic oxidation state of the lanthanoids?
  • 6 Name one lanthanoid that shows $+4$ and one that shows $+2$, and explain each by its configuration.
  • 7 Why are aqueous solutions of $\mathrm{Ce^{4+}}$ used as oxidising agents in analysis?
  • 8 Identify the lanthanoid ion with the configuration $\mathrm{[Xe]\,4f^{7}}$ and charge (a) $+2$, (b) $+3$, (c) $+4$.
1 $\mathrm{Nd}$ $\mathrm{[Xe]\,4f^{4}\,6s^{2}}$; $\mathrm{Sm^{3+}}$ $\mathrm{[Xe]\,4f^{5}}$; $\mathrm{Tb}$ $\mathrm{[Xe]\,4f^{9}\,6s^{2}}$; $\mathrm{Er^{3+}}$ $\mathrm{[Xe]\,4f^{11}}$.
2 La ($5d^{1}6s^{2}$), Ce ($4f^{1}5d^{1}6s^{2}$), Gd ($4f^{7}5d^{1}6s^{2}$) and Lu ($4f^{14}5d^{1}6s^{2}$).
3 $\mathrm{Ce^{3+}}$ $4f^{1}$: 1; $\mathrm{Eu^{3+}}$ $4f^{6}$: 6; $\mathrm{Gd^{3+}}$ $4f^{7}$: 7; $\mathrm{Yb^{3+}}$ $4f^{13}$: 1; $\mathrm{Lu^{3+}}$ $4f^{14}$: 0. (Seven $f$ orbitals: up to $f^{7}$ every electron is unpaired; beyond, $n=14-$ number of $f$ electrons.)
4 $\mathrm{La^{3+}}$ ($f^{0}$), $\mathrm{Ce^{4+}}$ ($f^{0}$), $\mathrm{Yb^{2+}}$ ($f^{14}$) and $\mathrm{Lu^{3+}}$ ($f^{14}$). $\mathrm{Nd^{3+}}$ ($f^{3}$) is paramagnetic.
5 Removing three electrons (the $6s^{2}$ and one $5d$ or $4f$ electron) needs a comparatively low total ionisation enthalpy, which is more than repaid by the high lattice or hydration enthalpy of the small $\mathrm{Ln^{3+}}$ ion. The $4f$ electrons beyond that are held tightly and are hard to remove.
6 Cerium: $\mathrm{Ce^{4+}}$ is $4f^{0}$, a noble-gas core. Europium: $\mathrm{Eu^{2+}}$ is $4f^{7}$, half-filled. (Also Tb⁴⁺ $f^{7}$; Yb²⁺ $f^{14}$.)
7 $\mathrm{Ce^{4+}}$ readily gains an electron to return to the common $+3$ state; $E^{\circ}_{\mathrm{Ce^{4+}/Ce^{3+}}}=+1.74\ \mathrm{V}$ is high, so it oxidises many species (for example $\mathrm{Fe^{2+}}$) completely, and its solutions keep well.
8 (a) $Z=54+7+2=63$: $\mathrm{Eu^{2+}}$. (b) $64$: $\mathrm{Gd^{3+}}$. (c) $65$: $\mathrm{Tb^{4+}}$.
II
The lanthanoid contraction. $\mathrm{Ln^{3+}}$ radii (pm): La 103, Ce 101, Eu 94.7, Gd 93.8, Lu 86.1.
  • 1 What is the lanthanoid contraction, and what causes it?
  • 2 Why are the atomic radii of zirconium and hafnium almost the same? What practical problem does this cause?
  • 3 Find the decrease in $\mathrm{Ln^{3+}}$ radius from Ce to Gd and from Gd to Lu. Is the contraction roughly uniform?
  • 4 Arrange $\mathrm{La^{3+}}$, $\mathrm{Eu^{3+}}$ and $\mathrm{Lu^{3+}}$ in order of increasing ionic radius.
  • 5 Arrange $\mathrm{La(OH)_{3}}$, $\mathrm{Nd(OH)_{3}}$ and $\mathrm{Yb(OH)_{3}}$ in order of increasing basicity, and explain.
  • 6 The metallic radii of europium and ytterbium are much larger than those of their neighbours. Explain.
  • 7 Why do molybdenum and tungsten (and niobium and tantalum) have such similar chemistry?
1 The steady decrease in atomic and ionic radii from La to Lu. Each step adds a proton and a $4f$ electron; $4f$ electrons shield the outer electrons poorly, so the effective nuclear charge rises and the outer electrons are pulled in.
2 The lanthanoids come between Zr (period 5) and Hf (period 6); their contraction cancels the increase in size expected on adding a shell, so Hf ($159\ \mathrm{pm}$) is the same size as Zr ($160\ \mathrm{pm}$). With the same size and outer configuration their chemistry is almost identical, so they occur together and are very difficult to separate.
3 Ce → Gd: $101-93.8=7.2\ \mathrm{pm}$ over 6 steps ($1.2$ per step). Gd → Lu: $93.8-86.1=7.7\ \mathrm{pm}$ over 7 steps ($1.1$ per step). Yes, roughly uniform.
4 $\mathrm{Lu^{3+}}\ (86.1)\lt\mathrm{Eu^{3+}}\ (94.7)\lt\mathrm{La^{3+}}\ (103\ \mathrm{pm})$.
5 $\mathrm{Yb(OH)_{3}}\lt\mathrm{Nd(OH)_{3}}\lt\mathrm{La(OH)_{3}}$. Smaller cations polarise $\mathrm{OH^{-}}$ more, making the M–OH bond more covalent and less ready to release $\mathrm{OH^{-}}$.
6 Eu ($4f^{7}6s^{2}$) and Yb ($4f^{14}6s^{2}$) keep their stable half-filled or full $4f$ sets and contribute only the two $6s$ electrons to metallic bonding (the others give three). Weaker bonding means larger atoms.
7 Because of the lanthanoid contraction, W is the same size as Mo (both $139\ \mathrm{pm}$) and Ta the same as Nb (both $146\ \mathrm{pm}$); with the same size and outer configuration, each pair behaves almost identically.
III
Properties and reactions of the lanthanoids. La $138.9$, $\mathrm{H_{2}}$ $2.016\ \mathrm{g\,mol^{-1}}$; $E^{\circ}$: $\mathrm{Ce^{4+}/Ce^{3+}}$ $+1.74$, $\mathrm{Fe^{3+}/Fe^{2+}}$ $+0.77$, $\mathrm{Eu^{3+}/Eu^{2+}}$ $-0.35$, $\mathrm{Yb^{3+}/Yb^{2+}}$ $-1.05\ \mathrm{V}$.
  • 1 Write balanced equations for the reactions of lanthanum with water, of neodymium with oxygen and of cerium with chlorine.
  • 2 What is mischmetall? Give its approximate composition and two uses.
  • 3 Why are the absorption bands of $\mathrm{Ln^{3+}}$ ions much narrower than those of 3d transition-metal ions?
  • 4 What mass of hydrogen is released when $2.78\ \mathrm{g}$ of lanthanum dissolves completely in dilute acid?
  • 5 Most $\mathrm{Ln^{3+}}$ ions are coloured, but $\mathrm{La^{3+}}$ and $\mathrm{Lu^{3+}}$ are colourless. Explain.
  • 6 Which is the stronger reducing agent, $\mathrm{Eu^{2+}}$ or $\mathrm{Yb^{2+}}$? Use the $E^{\circ}$ values.
  • 7 Show that $\mathrm{Ce^{4+}}$ can oxidise $\mathrm{Fe^{2+}}$, and find the volume of $0.0500\ \mathrm{mol\,L^{-1}}$ $\mathrm{Ce^{4+}}$ needed for $25.0\ \mathrm{mL}$ of $0.100\ \mathrm{mol\,L^{-1}}$ $\mathrm{Fe^{2+}}$.
1 $\mathrm{2La(s)+6H_{2}O(l)\rightarrow2La(OH)_{3}(s)+3H_{2}(g)}$; $\mathrm{4Nd(s)+3O_{2}(g)\rightarrow2Nd_{2}O_{3}(s)}$; $\mathrm{2Ce(s)+3Cl_{2}(g)\rightarrow2CeCl_{3}(s)}$.
2 An alloy of about $95\%$ lanthanoid metals (mostly Ce and La), about $5\%$ iron and traces of S, C, Ca and Al. Used (with magnesium) in lighter flints and in bullets and shells.
3 The $4f$ orbitals are deep inside the ion, shielded by the filled $5s$ and $5p$ subshells, so the surroundings (ligands, vibrations) hardly disturb their energies; the $f$–$f$ transitions occur at sharply defined energies. The $3d$ orbitals are on the outside of a 3d ion and interact strongly with the ligands, broadening the bands.
4 $\mathrm{2La(s)+6H^{+}(aq)\rightarrow2La^{3+}(aq)+3H_{2}(g)}$. $n(\mathrm{La})=\dfrac{2.78}{138.9}=0.02001\ \mathrm{mol}$; $n(\mathrm{H_{2}})=1.5\times0.02001=0.03002\ \mathrm{mol}$; mass $=0.03002\times2.016=0.0605\ \mathrm{g}$.
5 $\mathrm{La^{3+}}$ is $4f^{0}$ and $\mathrm{Lu^{3+}}$ is $4f^{14}$: there is no $f$–$f$ transition possible, so they absorb no visible light.
6 $\mathrm{Yb^{2+}}$: its couple has the more negative $E^{\circ}$ ($-1.05$ against $-0.35\ \mathrm{V}$), so it gives up its electron more readily.
7 $\mathrm{Ce^{4+}(aq)+Fe^{2+}(aq)\rightarrow Ce^{3+}(aq)+Fe^{3+}(aq)}$, $E^{\circ}=1.74-0.77=+0.97\ \mathrm{V}\gt0$. Ratio $1:1$: $n(\mathrm{Fe^{2+}})=2.50\ \mathrm{mmol}$, so $V=\dfrac{2.50}{0.0500}=50.0\ \mathrm{mL}$.
IV
Actinoids, comparison and applications. Atomic numbers: Th 90, Pu 94, Am 95, Cm 96.
  • 1 Write the configurations of Th, Pu, Am and Cm.
  • 2 State the highest oxidation states of Th, Pa, U and Np. What pattern do they show?
  • 3 Why do the actinoids show a wider range of oxidation states than the lanthanoids?
  • 4 Why is the actinoid contraction greater from element to element than the lanthanoid contraction?
  • 5 Give the oxidation state of the actinoid in $\mathrm{UO_{2}^{2+}}$, $\mathrm{PuO_{2}^{2+}}$, $\mathrm{NpO_{2}^{+}}$ and $\mathrm{ThO_{2}}$.
  • 6 State three differences between the lanthanoids and the actinoids.
  • 7 Name the d-block catalyst used in (a) the contact process for sulphuric acid, (b) the Haber process, (c) the hydrogenation of vegetable oils, (d) the manufacture of polythene.
  • 8 Smoke detector. Americium-241 has a half-life of $432$ years and decays by first-order kinetics (Chapter 3). What percentage of it is left in a detector after $10.0$ years? Why are the actinoids handled only in specialised laboratories?
1 Th $\mathrm{[Rn]\,6d^{2}\,7s^{2}}$; Pu $\mathrm{[Rn]\,5f^{6}\,7s^{2}}$; Am $\mathrm{[Rn]\,5f^{7}\,7s^{2}}$; Cm $\mathrm{[Rn]\,5f^{7}\,6d^{1}\,7s^{2}}$.
2 Th $+4$, Pa $+5$, U $+6$, Np $+7$: in each case all the electrons beyond the radon core; the maximum rises to Np (and Pu) and then falls.
3 The $5f$, $6d$ and $7s$ levels are close in energy, so different numbers of electrons can be used; the $4f$ electrons of the lanthanoids are buried and seldom used.
4 The $5f$ electrons shield each other from the nuclear charge even less effectively than $4f$ electrons do, so each added proton pulls the electrons in more.
5 $\mathrm{UO_{2}^{2+}}$ $+6$; $\mathrm{PuO_{2}^{2+}}$ $+6$; $\mathrm{NpO_{2}^{+}}$ $+5$; $\mathrm{ThO_{2}}$ $+4$.
6 Any three: actinoids are all radioactive (only Pm among lanthanoids); actinoids show many more oxidation states (up to $+7$); actinoid contraction is larger per step; actinoids form oxocations such as $\mathrm{UO_{2}^{2+}}$; actinoids form complexes more readily; actinoid configurations are less regular.
7 (a) $\mathrm{V_{2}O_{5}}$; (b) finely divided iron; (c) nickel; (d) $\mathrm{TiCl_{4}}$ with $\mathrm{Al(CH_{3})_{3}}$ (Ziegler catalyst).
8 Fraction left $=\left(\tfrac12\right)^{10.0/432}=e^{-0.693\times10.0/432}=0.984$, i.e. $98.4\%$. All actinoids are radioactive (many intensely) and toxic, so they need shielded, contained facilities.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks (a) What is meant by the lanthanoid contraction? (b) Give two of its consequences. (c) Why do zirconium and hafnium have almost identical properties?

(a) The steady decrease in atomic and ionic radii from lanthanum to lutetium, caused by the poor shielding of the outer electrons by $4f$ electrons as the nuclear charge increases.
(b) Any two: 4d and 5d metals of a group have nearly the same radii; the lanthanoids are hard to separate; the basicity of $\mathrm{Ln(OH)_{3}}$ decreases from La to Lu; 5d metals are very dense.
(c) Hf comes just after the lanthanoids; the contraction cancels the expected increase in size, so Zr ($160\ \mathrm{pm}$) and Hf ($159\ \mathrm{pm}$) have the same size and the same outer configuration.

E2. JEE Main Numerical value. The total number of unpaired electrons in the ions $\mathrm{Ce^{3+}}$, $\mathrm{Gd^{3+}}$ and $\mathrm{Yb^{3+}}$ ($Z$ = 58, 64, 70) is ______.

Configurations $\mathrm{Ln^{3+}}=\mathrm{[Xe]\,4f^{\,Z-57}}$: $4f^{1}$, $4f^{7}$, $4f^{13}$.
Unpaired $1$, $7$ and $14-13=1$.
Answer $1+7+1=\mathbf{9}$.

E3. NEET Which of the following lanthanoid ions is a good oxidising agent in aqueous solution? (A) $\mathrm{Eu^{2+}}$  (B) $\mathrm{Ce^{4+}}$  (C) $\mathrm{Yb^{2+}}$  (D) $\mathrm{La^{3+}}$

Idea An ion above $+3$ is reduced back to $+3$, so it oxidises. $\mathrm{Eu^{2+}}$ and $\mathrm{Yb^{2+}}$ go up to $+3$ (reducing agents); $\mathrm{La^{3+}}$ is already $+3$ and stable.
Answer (B) $\mathrm{Ce^{4+}}$ ($E^{\circ}=+1.74\ \mathrm{V}$).

E4. JEE Main Which actinoid shows the highest oxidation state, $+7$? (A) Th  (B) U  (C) Np  (D) Cm

Idea The maximum state equals the electrons beyond [Rn] up to Np: Th $+4$, U $+6$, Np $+7$; after Pu the maximum falls (Cm mostly $+3$).
Answer (C) neptunium (plutonium also reaches $+7$).

E5. JEE Advanced One or more correct. Choose the correct statement(s). (A) The actinoid contraction is greater from element to element than the lanthanoid contraction. (B) The actinoids show a smaller range of oxidation states than the lanthanoids. (C) $\mathrm{La(OH)_{3}}$ is more basic than $\mathrm{Lu(OH)_{3}}$. (D) All the actinoids are radioactive.

(A) True $5f$ electrons shield even more poorly than $4f$.
(B) False The actinoids show a larger range ($+3$ to $+7$); the lanthanoids are mostly $+3$.
(C) True $\mathrm{La^{3+}}$ is the largest $\mathrm{Ln^{3+}}$ ion, so its hydroxide is the most ionic and most basic.
(D) True · Answer Correct: (A), (C) and (D).

E6. Board · Case study A lighter flint is made of mischmetall alloyed with magnesium, and MRI contrast agents contain gadolinium(III) complexes. (a) What is the composition of mischmetall, and why does it spark when scraped? (1 mark) (b) Write the configuration of $\mathrm{Gd^{3+}}$ ($Z=64$) and explain why the $+3$ state is the most stable one for gadolinium. (2 marks) (c) Which of $\mathrm{La^{3+}}$ and $\mathrm{Nd^{3+}}$ is coloured in solution? (1 mark)

(a) About $95\%$ lanthanoid metals (mainly Ce, La), $5\%$ iron and traces of S, C, Ca, Al. Small fragments scraped off are so reactive that they ignite in air.
(b) $\mathrm{Gd^{3+}}=\mathrm{[Xe]\,4f^{7}}$. Removing the $6s^{2}$ and $5d^{1}$ electrons leaves the stable half-filled $4f^{7}$; removing a fourth electron would break it, and lattice/hydration enthalpies favour $+3$ anyway.
(c) $\mathrm{Nd^{3+}}$ ($4f^{3}$, $f$–$f$ transitions possible); $\mathrm{La^{3+}}$ is $4f^{0}$ and colourless.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Clean energy runs on magnets: wind turbines and electric-car motors use neodymium–iron–boron, $\mathrm{Nd_{2}Fe_{14}B}$. How much neodymium does the energy transition need, and why is it so hard to produce and recycle?

Why it matters. Mining and separation of rare earths is concentrated in very few countries, and separating one lanthanoid from its neighbours produces large volumes of acidic and slightly radioactive waste. India has large monazite sands along the Kerala and Odisha coasts and is building its own magnet supply chain.

What to do.

  1. Calculate the molar mass of $\mathrm{Nd_{2}Fe_{14}B}$ (Nd $144.2$, Fe $55.85$, B $10.81\ \mathrm{g\,mol^{-1}}$) and the percentage of neodymium by mass.
  2. A large direct-drive wind turbine may contain about $2.0$ tonnes of magnet, and an electric-car motor about $1.5\ \mathrm{kg}$. Estimate the neodymium in each (treat the magnet as pure $\mathrm{Nd_{2}Fe_{14}B}$).
  3. Use the lanthanoid contraction to explain why separating neodymium from praseodymium (its neighbour, also found in monazite) takes many repeated extraction steps.
  4. Find out one method being developed to recycle rare earths from old magnets or hard-disk drives, and what limits it today.

Tools: Calculator; Interactive 1 (compare the radii of $\mathrm{Pr^{3+}}$ and $\mathrm{Nd^{3+}}$); reports of the International Energy Agency or national geological surveys on critical minerals.

Hand in: Your calculations, a labelled flow diagram "ore → separated $\mathrm{Nd}$ → magnet → recycling", and a short paragraph on one environmental and one supply risk.

Going further. Real magnets replace a few per cent of the neodymium by dysprosium to keep them strong at high temperatures. Dysprosium is a heavier, scarcer lanthanoid. Why might engineers try to design motors that need less of it?

Instructor note: $M=2(144.2)+14(55.85)+10.81=1081.1\ \mathrm{g\,mol^{-1}}$; Nd $=288.4/1081.1=26.7\%$. Turbine: $0.267\times2000\approx534\ \mathrm{kg}$ Nd; car motor: $0.267\times1.5\approx0.40\ \mathrm{kg}$. $\mathrm{Pr^{3+}}$ ($99\ \mathrm{pm}$) and $\mathrm{Nd^{3+}}$ ($98.3\ \mathrm{pm}$) differ by less than $1\%$ in radius and have identical charge, so each extraction stage enriches only slightly and hundreds of stages are used. Recycling routes include hydrogen decrepitation of whole magnets and hydrometallurgical leaching; collection and disassembly costs are the main obstacles.

🧪 Self-check quiz

0/5 correct

Q1. The configuration of $\mathrm{Gd^{3+}}$ ($Z=64$) is

Q2. The lanthanoid contraction is caused by

Q3. Which pair of elements has almost identical atomic radii?

Q4. Which hydroxide is the most basic?

Q5. Actinoids show more oxidation states than lanthanoids mainly because

← 4.2 Important Compounds: Potassium Dichromate and Potassium PermanganateChapter 5 — Coordination Compounds →