d- and f-Block Elements

Copper forms two ions. $\mathrm{C}{\mathrm{u}}^{+}$ is $3d^{10}$, a completely filled subshell. $\mathrm{C}{\mathrm{u}}^{2+}$ is $3d^9$, one electron short.

Which one survives in water?

Every rule about filled-subshell stability says $\mathrm{C}{\mathrm{u}}^{+}$. It is the one with the tidy configuration, and it costs less to make.

Drop a copper(I) salt into water and it vanishes on the spot.

$2C{\mathrm{u}}^{+}(\mathrm{aq})\to \mathrm{C}{\mathrm{u}}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$

Half of it becomes the "unstable" $3d^9$ ion and the other half falls out as copper metal. The full subshell loses.

Not by intuition, by arithmetic. Build the cycle and every term is a measured number.

$1186+1958-745-339-2121=-61\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$

Negative, so it happens. The whole verdict turns on the hydration enthalpies: $-593$ kJ mol${}^{-1}$ for $\mathrm{C}{\mathrm{u}}^{+}$ against $-2121$ for $\mathrm{C}{\mathrm{u}}^{2+}$. Water pulls far harder on a doubly charged ion, and that difference pays for the second ionisation with 61 kJ to spare.

Sixty-one is a small margin. Take the water away and the verdict flips: solid $\mathrm{CuCl}$, $\mathrm{CuI}$ and $\mathrm{C}{\mathrm{u}}_2\mathrm{O}$ are all perfectly stable, and copper(I) chemistry is routine in non-aqueous solvents.

Nothing in this chapter is decided by configuration alone. Everything is decided by an energy balance in which the environment gets a vote, and a filled subshell is one term in that balance rather than a trump card.

With that established, the chapter has one organising fact. Transition elements have a partly filled d subshell, and those d electrons lie close in energy to the s electrons above them.

Five properties, one cause. Zinc has none of them, and the reason is that its d subshell is full in every state it adopts.

1. What Counts as a Transition Element

Transition element. One with a partly filled d subshell either in the free atom or in one of its common oxidation states.

That last clause is not decoration. It does the work of including and excluding elements that the shorter definition gets wrong.

Zinc, cadmium and mercury are $d^{10}$ as atoms and $d^{10}$ as their common ions, so they are d-block elements but not transition elements. The consequence is visible: their compounds are colourless, they show only $+2$, they are not useful catalysts, and their ions are diamagnetic. Every characteristic property is absent because the cause is absent.

Scandium is the mirror case. $\mathrm{S}{\mathrm{c}}^{3+}$ is $d^0$, so scandium is a poor transition element in practice, though the atom itself has one d electron.

Illustration 1

Silver is $[\mathrm{Kr}]4d^{10}5s^1$, and $\mathrm{A}{\mathrm{g}}^{+}$ is $4d^{10}$. Both are full. Is silver a transition element?

Apply the short version of the definition and the answer is no, which would put silver alongside zinc.

Apply the full version and the answer is yes. Silver has a genuine, if uncommon, $+2$ state: $\mathrm{Ag}{\mathrm{F}}_2$ and $\mathrm{AgO}$ exist and contain $\mathrm{A}{\mathrm{g}}^{2+}$, which is $4d^9$. A partly filled d subshell in one of its oxidation states is all the definition asks for.

Zinc has no such escape. There is no $\mathrm{Z}{\mathrm{n}}^{3+}$ in any real compound, because the third ionisation enthalpy would have to break into the $3d^{10}$ core and nothing repays it.

Trap. "The d-block minus zinc, cadmium and mercury" is a slogan, not a definition, and it fails on exactly the cases worth asking about. Test each element against the clause.

Electronic configurations

The first row fills 3d after 4s, giving $[\mathrm{Ar}]3d^{1-10}4s^{1-2}$.

Chromium is $3d^{5}4s^1$ and copper is $3d^{10}4s^1$, both taking the extra stability of a half-filled or filled subshell, helped by 4s and 3d lying close in energy.

Ions lose 4s electrons before 3d. Iron is $3d^{6}4s^2$, but $\mathrm{F}{\mathrm{e}}^{2+}$ is $3d^6$, not $3d^{4}4s^2$. This catches almost everyone once and is worth checking every single time.

2. Physical Properties

Transition metals are hard, dense, high-melting and good conductors, far more so than the s-block metals beside them, because both the 4s and the unpaired 3d electrons join the metallic bonding.

Melting points therefore rise towards the middle of the row, where unpaired d electrons are most numerous, and fall away at each end.

Read the bottom row against the curve. Manganese ties chromium for the most unpaired d electrons and melts 660 K lower. Its half-filled $d^5$ configuration is stable enough that those electrons resist being shared into the metallic bond at all, so having them counts for nothing. Technetium does the same thing directly below.

Illustration 2

Mercury is a liquid at room temperature, melting at 234 K, while cadmium directly above it melts at 594 K and zinc at 693 K. Why does the anomaly deepen down the group?

All three are $d^{10}{s}^2$, so none of them puts d electrons into the metallic bond and all three melt low. That accounts for zinc and cadmium.

Mercury goes further because its $6s^2$ pair is contracted and stabilised by relativistic effects, which are significant only for heavy nuclei where inner electrons approach a substantial fraction of the speed of light. The pair is drawn in close, held tightly, and becomes reluctant to delocalise.

So mercury contributes essentially nothing to metallic bonding: not its filled d shell, and barely even its s pair. What holds the liquid together is closer to dispersion forces between atoms than to a metallic lattice.

The same relativistic contraction is why gold is yellow rather than silver-white, and it is the heavy-element cousin of the inert pair effect met in the p-block.

Atomic radii

Across the first row, radius falls at first, then plateaus through the middle, then rises slightly at the end.

The fall comes from rising nuclear charge. The plateau comes from added d electrons screening the 4s electrons rather well, largely cancelling the extra pull. The final rise comes from electron-electron repulsion in the nearly filled d subshell.

Radii of the second and third series are almost identical, which is not what adding a whole period should do. The cause is the lanthanoid contraction, at the end of this chapter.

3. Variable Oxidation States

The 4s and 3d electrons differ so little in energy that a variable number can be used.

Manganese is the extreme case, showing every state from $+2$ to $+7$, because it has exactly seven electrons available in 4s and 3d combined.

Two patterns fall out. The maximum oxidation state rises to $+7$ at manganese and then falls, because after manganese the d electrons begin pairing and become harder to remove. And high oxidation states are stabilised by oxygen and fluorine, which is why manganese(VII) exists as $\mathrm{Mn}{\mathrm{O}}_4^{-}$ and chromium(VI) as $\mathrm{Cr}{\mathrm{O}}_4^{2-}$, never as simple cations.

Illustration 3

Standard reduction potentials for the ${\mathrm{M}}^{2+}/M$ couple across the first row are Ti $-1.63$, V $-1.18$, Cr $-0.91$, Mn $-1.18$, Fe $-0.44$, Co $-0.28$, Ni $-0.25$, Cu $+0.34$, Zn $-0.76$ volts. Find the two anomalies and explain them.

The general drift is towards less negative values, as ionisation enthalpies rise across the row. Two values refuse to follow it.

Manganese at $-1.18$ sits more negative than chromium at $-0.91$, breaking the drift. The reason is on the product side: $\mathrm{M}{\mathrm{n}}^{2+}$ is $3d^5$, half-filled and unusually stable, so manganese gives up two electrons more readily than its position suggests.

Zinc at $-0.76$ sits more negative than copper, nickel and cobalt. Same reason one step further: $\mathrm{Z}{\mathrm{n}}^{2+}$ is $3d^{10}$.

Both anomalies come from the stability of the ion formed, not from the metal.

And copper is the only positive value in the row. That single sign change is why copper does not displace hydrogen from dilute acids, why copper roofs and pipes survive outdoors for centuries, and why copper was one of the first metals humans could obtain and keep.

4. Colour

Most transition metal compounds are coloured, and most compounds of everything else are not.

The usual cause is a d-d transition. In a complex the five d orbitals are no longer degenerate, so an electron can absorb a visible photon and jump from a lower d orbital to a higher one. The colour seen is the complement of the colour absorbed.

This requires a partly filled d subshell. $\mathrm{S}{\mathrm{c}}^{3+}$ is $d^0$ and $\mathrm{Z}{\mathrm{n}}^{2+}$ is $d^{10}$, and both are colourless: no electron to promote, or no vacancy to promote it into. $\mathrm{C}{\mathrm{u}}^{+}$ is $d^{10}$ and colourless while $\mathrm{C}{\mathrm{u}}^{2+}$ is $d^9$ and blue, which is the cleanest demonstration of the rule inside one element.

A necessary correction

The permanganate ion is intensely purple, and manganese in it is $+7$, which is $d^0$.

There is no d electron to promote. The colour cannot be a d-d transition, and saying that it is will be marked wrong.

The colour comes from charge transfer: a photon promotes an electron from an oxygen lone pair into an empty metal orbital. Charge transfer bands are typically a hundred to a thousand times more intense than d-d bands, which is why permanganate colours a solution deeply at concentrations where a copper salt is barely tinted.

Dichromate's orange has the same origin. Being able to say this distinguishes a good answer from a memorised one.

Illustration 4

Which of these are coloured in aqueous solution, and by what mechanism: $\mathrm{S}{\mathrm{c}}^{3+}$, $\mathrm{T}{\mathrm{i}}^{3+}$, $\mathrm{M}{\mathrm{n}}^{2+}$, $\mathrm{Z}{\mathrm{n}}^{2+}$, $\mathrm{C}{\mathrm{u}}^{+}$, $\mathrm{Mn}{\mathrm{O}}_4^{-}$?

Start by counting d electrons, because a d-d transition needs both an electron to promote and a vacancy to promote it into.

$\mathrm{S}{\mathrm{c}}^{3+}$ is $d^0$ — no electron to promote, so colourless. $\mathrm{Z}{\mathrm{n}}^{2+}$ and $\mathrm{C}{\mathrm{u}}^{+}$ are both $d^{10}$ — no vacancy, so colourless as well. That is why zinc salts are white while copper(II) salts are blue: $\mathrm{C}{\mathrm{u}}^{2+}$ is $d^9$ and has one hole to work with.

$\mathrm{T}{\mathrm{i}}^{3+}$ is $d^1$ and is purple, the simplest possible d-d transition.

$\mathrm{M}{\mathrm{n}}^{2+}$ is $d^5$ high spin and is a very pale pink — visible only in concentrated solution. Every d-d transition here would have to flip a spin as well as move an electron, and being doubly forbidden makes it about a hundred times fainter than $\mathrm{T}{\mathrm{i}}^{3+}$.

$\mathrm{Mn}{\mathrm{O}}_4^{-}$ is the interesting one. Manganese is $+7$ here, which is $d^0$, so by the rule above it should be colourless — yet permanganate is the most intensely coloured reagent on the shelf.

Its colour is charge transfer, an electron jumping from an oxygen lone pair onto the metal, and that transition is fully allowed. Charge transfer bands are roughly a thousand times stronger than d-d bands, which is why permanganate and dichromate stain everything while a d-d coloured salt needs real concentration to show. A $d^0$ ion is colourless by the d-d mechanism only.

5. Magnetic Properties

Unpaired electrons make a substance paramagnetic, and the effect can be quantified.

$\mu =\sqrt{n(n+2)}\text{BM}$

This is the spin-only formula because it ignores any contribution from orbital motion, which the surrounding ligands largely quench in the first transition series.

Illustration 5

An iron compound is measured at 5.9 BM. Is the iron $+2$ or $+3$? A second compound, of samarium(III), is measured at 1.5 BM. Check that against the formula.

For iron, invert the formula.

$n(n+2)=5.9^2=34.8\Rightarrow n=5$

Five unpaired electrons means $d^5$. Iron is $[\mathrm{Ar}]3d^{6}4s^2$, so removing three electrons gives $\mathrm{F}{\mathrm{e}}^{3+}$ as $3d^5$. The compound is iron(III), and no chemical test was needed. Running the formula backwards is the standard exam use.

Now samarium. $\mathrm{S}{\mathrm{m}}^{3+}$ is $4f^5$, so five unpaired electrons predict 5.92 BM. The measurement is 1.5.

The formula has failed by a factor of four, and the failure is instructive rather than embarrassing. In the first transition series, 3d orbitals point outwards into the ligands, which lock the electron's orbital motion and leave only spin to be measured. The 4f orbitals of a lanthanoid are buried beneath filled 5s and 5p shells, so ligands never reach them and orbital angular momentum survives, contributing its own term with its own sign.

Trap. The word "spin-only" is a stated limitation, not a decoration. Use the formula freely for the first transition series and never for lanthanoids.

6. Catalysis, Complexes and Alloys

Catalytic behaviour has two origins. Variable oxidation states let a metal accept and release electrons during a reaction, providing a low-energy path. And transition metal surfaces adsorb reactants, holding them close and correctly oriented while weakening their bonds.

Illustration 6

Vanadium(V) oxide catalyses the Contact process. Write what the vanadium actually does.

$\mathrm{S}{\mathrm{O}}_2+{\mathrm{V}}_2{\mathrm{O}}_5\to \mathrm{S}{\mathrm{O}}_3+{\mathrm{V}}_2{\mathrm{O}}_4$

$2{\mathrm{V}}_2{\mathrm{O}}_4+{\mathrm{O}}_2\to 2{\mathrm{V}}_2{\mathrm{O}}_5$

Vanadium falls from $+5$ to $+4$ handing an oxygen to sulphur dioxide, then climbs back to $+5$ by taking one from the air. Add the two equations and the vanadium cancels, leaving $2S{\mathrm{O}}_2+{\mathrm{O}}_2\to 2S{\mathrm{O}}_3$.

That is the whole mechanism, and it is a claim about oxidation states rather than a vague statement about surfaces. A catalyst that could hold only one oxidation state could not run this cycle at all, which is precisely why the useful ones sit in the d-block.

Iron in the Haber process and nickel in hydrogenation are the other standard examples.

Complex formation follows from small size, high charge and empty d orbitals of suitable energy to accept lone pairs. This chapter hands over to Coordination Compounds at exactly this point.

Interstitial compounds form when small atoms such as hydrogen, carbon, nitrogen or boron occupy the gaps in a metal lattice. They are typically non-stoichiometric, harder than the parent metal, and still metallic conductors. Steel is the case everyone has met.

Alloy formation is easy across the d-block because the metals have very similar atomic radii, so one substitutes for another without straining the lattice. This is why brass, bronze and stainless steel exist and why sodium and potassium form no comparable range.

7. Potassium Dichromate

Preparation starts from chromite ore, $\mathrm{FeC}{\mathrm{r}}_2{\mathrm{O}}_4$. Fusing with sodium carbonate in air oxidises chromium to sodium chromate; acidifying converts chromate to dichromate; adding potassium chloride precipitates the less soluble potassium dichromate, purified by crystallisation.

Structure. Two $\mathrm{Cr}{\mathrm{O}}_4$ tetrahedra sharing one corner oxygen, giving a bridging $\mathrm{Cr}-\mathrm{O}-\mathrm{Cr}$ linkage.

The chromate-dichromate equilibrium

$2Cr{\mathrm{O}}_4^{2-}+2{\mathrm{H}}^{+}\rightleftharpoons \mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}+{\mathrm{H}}_2\mathrm{O}$

Yellow chromate dominates in alkali and orange dichromate in acid. This is Le Chatelier applied directly and is a favourite one-mark question.

As an oxidising agent

$\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}\to 2C{\mathrm{r}}^{3+}+7{\mathrm{H}}_2\mathrm{O},E^{\circ }=1.33\text{V}$

The orange solution turns green, since $\mathrm{C}{\mathrm{r}}^{3+}$ is green, and that change is the basis of its use in titrations. It oxidises iodide to iodine, iron(II) to iron(III) and hydrogen sulphide to sulphur.

Illustration 7

Dichromate turns yellow in alkali and orange again on acidification, which sounds like a reversible colour trick. Why can it not be used as an oxidising agent in alkaline solution?

Because the pH change does not merely recolour the ion. It changes what the ion is, and therefore what it can do.

A potential of $+1.33$ V makes dichromate a powerful oxidiser. A potential of $-0.13$ V makes chromate essentially useless as one; chromium(III) in alkali is the species that wants to be oxidised.

So the colour change is the visible half of a much larger change. Acidification is not a cosmetic step in a dichromate titration; it is what creates the oxidising agent. The same logic explains why permanganate titrations specify the acid too.

8. Potassium Permanganate

Preparation starts from pyrolusite, $\mathrm{Mn}{\mathrm{O}}_2$. Fusing with potassium hydroxide in air, or with an oxidising agent such as potassium nitrate, gives green potassium manganate; oxidising that electrolytically, or letting it disproportionate in acid, gives purple potassium permanganate.

Structure. Tetrahedral, with manganese in $+7$ and therefore $d^0$.

As an oxidising agent

The acidic route has $E^{\circ }=1.51$ V, making permanganate one of the strongest common oxidising agents. It is self-indicating, since the first excess drop colours the solution pink.

Titrations must use dilute sulphuric acid, never hydrochloric, because permanganate would oxidise chloride to chlorine and consume itself.

Illustration 8

25.0 mL of 0.0200 M $\mathrm{KMn}{\mathrm{O}}_4$ is used to titrate iron(II). How much iron(II) does it oxidise in acidic solution, and how much would the same volume oxidise at neutral pH?

The permanganate is fixed either way.

$n(\mathrm{Mn}{\mathrm{O}}_4^{-})=0.0250\times 0.0200=5.00\times 10^{-4}\text{mol}$

What changes is how many electrons each ion accepts.

The same burette reading means two different answers, differing by two thirds.

This is why an examiner specifying "in the presence of dilute sulphuric acid" has already told you the answer is $n=5$, and why a titration performed at the wrong pH does not merely give a poor result; it measures a different quantity.

9. Lanthanoids

The fourteen elements after lanthanum fill 4f, giving $[\mathrm{Xe}]4f^{1-14}5d^{0-1}6s^2$.

The $+3$ state dominates throughout, which is why the lanthanoids are chemically so alike and so hard to separate.

Illustration 9

The exceptions to $+3$ are $\mathrm{C}{\mathrm{e}}^{4+}$, $\mathrm{T}{\mathrm{b}}^{4+}$, $\mathrm{E}{\mathrm{u}}^{2+}$ and $\mathrm{Y}{\mathrm{b}}^{2+}$. Four elements out of fourteen. Is there a pattern?

Write each exception as an f configuration.

Every exception lands on $f^0$, $f^7$ or $f^{14}$. Not one is anywhere else.

An element deviates from $+3$ only when doing so buys an empty, half-filled or filled f subshell, and cerium and terbium reach theirs by going up while europium and ytterbium reach theirs by going down.

Once you have the pattern you do not need the list. Look at the element's position, ask which of the three targets is one electron away, and the exception writes itself. $\mathrm{C}{\mathrm{e}}^{4+}$ is also a useful oxidising agent for exactly this reason: it is one electron from $f^0$ and takes it eagerly.

Lanthanoid contraction

Across the series, atomic and ionic radii decrease steadily and substantially, because 4f electrons are diffuse and oddly shaped and shield the nucleus very poorly. Each added proton pulls the outer shells in more than each added f electron pushes them out.

Its consequences reach beyond the lanthanoids. Zirconium and hafnium have almost identical radii of 160 and 159 pm despite being a whole period apart, which makes them chemically near-indistinguishable and notoriously difficult to separate. The second and third transition series therefore resemble each other closely throughout, unlike the first and second. And basicity falls across the series, so $\mathrm{La}(\mathrm{OH}{)}_3$ is distinctly more basic than $\mathrm{Lu}(\mathrm{OH}{)}_3$.

10. Actinoids

The actinoids fill 5f, giving $[\mathrm{Rn}]5f^{1-14}6d^{0-1}7s^2$. They differ from the lanthanoids in three ways worth stating.

The syllabus restricts actinoids to electronic configuration and oxidation states, so this is the appropriate depth.

Illustration 10

Lanthanoids are almost always $+3$, while actinoids run from $+3$ through $+7$. Account for the difference.

The answer is how deeply the f orbitals are buried.

In a lanthanoid the 4f orbitals lie inside the filled 5s and 5p shells. They are shielded from the outside world, take essentially no part in bonding, and their electrons are not available for removal. What can be removed is the 6s pair and one further electron, which fixes the oxidation state at $+3$ almost universally.

In an actinoid the 5f orbitals are more spatially extended and sit close in energy to 6d and 7s. Those three sets are near enough to be used together, so 5f electrons are accessible to bonding. Uranium reaches $+6$ and neptunium $+7$.

The same buried-or-not argument settles two other facts in this section. The actinoid contraction is sharper than the lanthanoid contraction, because 5f shields the nuclear charge even more poorly than 4f does. And lanthanoid compounds have colours and magnetic moments barely affected by their ligands, since the 4f electrons responsible are screened from them — which is exactly why the spin-only formula fails for lanthanoids while working well across the 3d series.

Summary

Every characteristic property of the transition elements follows from one fact: a partly filled d subshell whose electrons lie close in energy to the s electrons above. Close in energy gives variable oxidation states; partly filled gives colour and magnetism; both together give catalysis and complex formation.

Nothing here is decided by configuration alone. Copper(I) is $3d^{10}$ and disproportionates in water anyway, by 61 kJ mol${}^{-1}$, because hydration of a doubled charge outweighs a filled subshell, and it stops disproportionating the moment the water is removed.

The definition needs its full wording. A partly filled d subshell in the atom or in a common oxidation state includes silver, through $\mathrm{A}{\mathrm{g}}^{2+}$ in $\mathrm{Ag}{\mathrm{F}}_2$, and excludes zinc, cadmium and mercury, which are $d^{10}$ throughout.

Ions lose 4s before 3d, so $\mathrm{F}{\mathrm{e}}^{2+}$ is $3d^6$. Melting points peak in the middle but dip sharply at manganese, whose stable $d^5$ refuses to join the metallic bond, and mercury is liquid because relativity has stabilised even its $6s^2$ pair.

Reduction potentials drift upward across the row with two anomalies, at manganese and zinc, both caused by the stability of $d^5$ and $d^{10}$ in the ion formed. Copper alone is positive, which is why it survives in air and does not displace hydrogen from acid.

Colour is a d-d transition, except where there are no d electrons: permanganate and dichromate are charge transfer, which is far more intense and is why they colour solutions so deeply.

The spin-only formula $\mu =\sqrt{n(n+2)}$ works for the first transition series and is run backwards to identify an oxidation state. It fails badly for lanthanoids, where buried 4f orbitals keep their orbital angular momentum.

Dichromate oxidises at $+1.33$ V in acid and chromate at $-0.13$ V in alkali, so acidification creates the oxidising agent rather than merely changing the colour. Permanganate accepts 5, 3 or 1 electrons according to pH, so the same volume of the same solution does three different amounts of work.

Lanthanoids are dominated by $+3$, and every exception lands on $f^0$, $f^7$ or $f^{14}$. The lanthanoid contraction makes zirconium and hafnium nearly identical, and the actinoids repeat the pattern with more oxidation states, universal radioactivity and a still larger contraction.