d- and f-Block Elements

Potassium permanganate colours water an intense purple at concentrations of a few parts per million. Yet manganese in the $+7$ state is $3d^0$ — it has no $d$ electrons whatever. Where does the colour come from?

Not from a $d$-$d$ transition. There is nothing to promote.

The colour comes from a charge transfer transition, in which an electron jumps from a filled oxygen orbital into an empty manganese orbital. Because that transition is fully allowed, it is $100$ to $1000$ times more intense than any $d$-$d$ absorption, which is why permanganate is visible at concentrations where an ordinary transition metal salt looks like water.

The same mechanism explains why dichromate is orange, why chromate is yellow, and why $\text{CuI}$ is white while ${\text{CuI}}_2$ does not exist at all. "Transition metals are coloured because of $d$-$d$ transitions" is only half the story, and Advanced tests the other half.

1. Colour: two mechanisms, not one

Ions with $d^0$ or $d^{10}$ configurations are colourless by the first mechanism: ${\text{Sc}}^{3+}$, ${\text{Ti}}^{4+}$, ${\text{Cu}}^{+}$ and ${\text{Zn}}^{2+}$ all have nothing to promote. If such an ion is nevertheless deeply coloured, charge transfer is the only possible explanation.

Illustration 1

Explain why ${\text{Cu}}^{2+}$ salts are blue but ${\text{Cu}}^{+}$ salts are white, and why ${\text{Zn}}^{2+}$ compounds are colourless.

${\text{Cu}}^{2+}$ is $3d^9$, with one vacancy in the $d$ set, so a $d$-$d$ transition is possible and the ion absorbs in the red, appearing blue.

${\text{Cu}}^{+}$ is $3d^{10}$: the set is full and no promotion within it is possible.

${\text{Zn}}^{2+}$ is likewise $3d^{10}$ and colourless for the same reason.

Zinc is not really a transition metal at all on the strict definition, since neither it nor its common ion has a partly filled $d$ subshell — which is exactly why it shows no variable oxidation state, no colour and no catalytic activity.

2. Variable oxidation states, and which ones survive

The $\left(n-1\right)d$ and $ns$ orbitals lie close in energy, so electrons from both can be lost, and a range of oxidation states results. The range is widest in the middle of the series, where the most electrons are available before the $d$ orbitals begin to contract.

Two rules govern which states are actually stable. Higher oxidation states are favoured with the most electronegative ligands — manganese reaches $+7$ only with oxygen and fluorine, never with chloride. And half-filled and filled configurations are unusually stable, which is why ${\text{Mn}}^{2+}$ ($d^5$) and ${\text{Fe}}^{3+}$ ($d^5$) are so much more common than their neighbours.

Illustration 2

Explain why ${\text{Fe}}^{3+}$ is more stable than ${\text{Fe}}^{2+}$ in aqueous solution, but ${\text{Mn}}^{3+}$ is far less stable than ${\text{Mn}}^{2+}$.

${\text{Fe}}^{3+}$ is $3d^5$, a half-filled set with maximum exchange energy, so removing the third electron is comparatively easy and the product is stabilised.

${\text{Mn}}^{2+}$ is already $3d^5$. Removing a further electron would destroy that half-filled arrangement, so the third ionisation enthalpy of manganese is exceptionally high.

${\text{Mn}}^{3+}$ is therefore a strong oxidant that reverts readily to ${\text{Mn}}^{2+}$.

The same configuration explains both facts, once from the product side and once from the reactant side.

3. Magnetic behaviour

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

with $n$ the number of unpaired electrons. The formula is spin only, ignoring any orbital contribution, which is a good approximation for the first transition series because the ligand field largely quenches the orbital motion. It works less well for the heavier series and for the lanthanoids, where the orbital contribution must be included.

A measured moment identifies the number of unpaired electrons but never a unique ion, since $d^3$ and $d^7$ both give three.

Illustration 3

A complex of a first-row metal has a magnetic moment of $1.73$ BM. Identify the number of unpaired electrons and suggest two ions.

$n\left(n+2\right)=1.73^2=3\Rightarrow n=1$

One unpaired electron arises from $d^1$, $d^9$, or a low-spin $d^5$ or $d^7$.

Candidates: ${\text{Ti}}^{3+}$ ($d^1$) or ${\text{Cu}}^{2+}$ ($d^9$).

The colour usually settles it. Titanium(III) solutions are violet and copper(II) blue, so one further observation identifies the ion uniquely.

4. Trends across the series

Manganese is anomalously low melting for its position, because its half-filled $3d^{5}4s^2$ configuration is so stable that its electrons participate reluctantly in metallic bonding. Zinc is lower still, having no unpaired $d$ electrons at all.

Illustration 4

Explain why chromium has the highest melting point in the first transition series while manganese, immediately next to it, has one of the lowest.

Metallic bonding strength depends on how many electrons are available to the delocalised sea, and unpaired $d$ electrons contribute most.

Chromium is $3d^{5}4s^1$ with six unpaired electrons, the maximum in the series, giving the strongest bonding and a melting point above $2100$ K.

Manganese is $3d^{5}4s^2$. Its half-filled $d$ set is so stable that those electrons participate only reluctantly, and its melting point falls to about $1519$ K.

The very stability that makes ${\text{Mn}}^{2+}$ so favourable weakens the metal itself, which is one configuration explaining two quite different observations.

The electrode potentials are similarly irregular. ${\text{Mn}}^{2+}$ and ${\text{Zn}}^{2+}$ have anomalously negative values because forming those particular ions is unusually favourable — $d^5$ and $d^{10}$ respectively.

Illustration 5

Explain why copper is the only first-row transition metal with a positive $E^{\circ }\left({\text{M}}^{2+}/\text{M}\right)$, and what follows for its reaction with acids.

Copper has an exceptionally high sum of sublimation and ionisation enthalpies, and its hydration enthalpy is not large enough to compensate.

Converting copper metal to the aqueous ion is therefore unfavourable, giving $E^{\circ }=+0.34$ V.

Consequently copper does not liberate hydrogen from dilute acids, since that would require a negative potential.

It dissolves only in oxidising acids, where nitrate or concentrated sulphate provides the driving force rather than the proton.

5. Catalysis, alloys and interstitial compounds

Transition metals catalyse for two distinct reasons. Variable oxidation states allow the metal to accept and release electrons in a cycle, as vanadium does in the contact process. Surface adsorption on a partially filled $d$ band weakens the bonds in adsorbed molecules, as nickel does in hydrogenation.

Interstitial compounds form when small atoms — hydrogen, carbon, nitrogen, boron — occupy the gaps in a metal lattice. They are non-stoichiometric, harder than the parent metal, retain metallic conductivity, and have much higher melting points. Steel is the everyday example.

Alloys form readily because the metals have similar radii, so one can replace another in the lattice without strain.

Illustration 6

Explain why interstitial carbides such as those in steel are harder and higher melting than the pure metal, yet still conduct electricity.

The small carbon atoms occupy the interstices without displacing the metal atoms, so the metallic lattice and its delocalised electrons remain intact — hence the conductivity.

But they pin the metal layers, preventing them from sliding over one another, which is what makes a pure metal soft and malleable.

Blocking that slip raises hardness and, because more energy is needed to disrupt the lattice, the melting point too.

Composition is variable rather than fixed, which is why such compounds are written with non-integral formulas and why steel comes in a continuum of grades.

Illustration 7

Give two distinct reasons why transition metals are good catalysts, with an example of each.

Variable oxidation states. The metal can accept electrons from one reactant and pass them to another, cycling between two states. Vanadium in the contact process alternates between $+5$ and $+4$ as it transfers oxygen from air to sulphur dioxide.

Surface adsorption. A partly filled $d$ band forms weak bonds to adsorbed molecules, concentrating them on the surface and weakening their internal bonds. Finely divided nickel adsorbs hydrogen and alkene together in catalytic hydrogenation.

Zinc does neither, having a filled $d$ shell in both the metal and its only ion, which is why it is catalytically inert while its neighbours are not.

6. Permanganate and dichromate

Potassium permanganate is made from pyrolusite. The ore is fused with potassium hydroxide and an oxidising agent to give the green manganate, which is then oxidised further:

$2{\text{MnO}}_2+4\text{KOH}+{\text{O}}_2\to 2{\text{K}}_2{\text{MnO}}_4+2{\text{H}}_2\text{O}$

$3{\text{MnO}}_4^{2-}+4{\text{H}}^{+}\to 2{\text{MnO}}_4^{-}+{\text{MnO}}_2+2{\text{H}}_2\text{O}$

the second step being a disproportionation, though industrially the oxidation is done electrolytically to avoid losing a third of the manganese.

Potassium dichromate comes from chromite ore by fusion, acidification and crystallisation. In solution the chromate and dichromate ions interconvert with pH:

$2{\text{CrO}}_4^{2-}+2{\text{H}}^{+}\rightleftharpoons {\text{Cr}}_2{\text{O}}_7^{2-}+{\text{H}}_2\text{O}$

so the solution is yellow in base and orange in acid. Dichromate has $n=6$ in every medium, unlike permanganate, which is why it is preferred as a primary standard.

Illustration 8

A dichromate solution turns yellow when sodium hydroxide is added and orange again on acidification. Explain, and state which species is the oxidant.

Adding base removes hydrogen ions, shifting the equilibrium towards chromate, which is yellow.

Acidifying restores them and the orange dichromate returns.

Dichromate is the oxidising species; chromate is not, which is why every dichromate titration is carried out in acidic solution.

The colour change is a shift of equilibrium, not a redox change. Chromium remains at $+6$ throughout, which is the point most often missed.

Illustration 9

Write the reaction of acidified permanganate with oxalic acid and with iodide, and give the n-factor in each case.

With oxalic acid: $2{\text{MnO}}_4^{-}+5{\text{C}}_2{\text{O}}_4^{2-}+16{\text{H}}^{+}\to 2{\text{Mn}}^{2+}+10{\text{CO}}_2+8{\text{H}}_2\text{O}$

With iodide: $2{\text{MnO}}_4^{-}+10{\text{I}}^{-}+16{\text{H}}^{+}\to 2{\text{Mn}}^{2+}+5{\text{I}}_2+8{\text{H}}_2\text{O}$

In both, manganese goes from $+7$ to $+2$, so its n-factor is $5$.

The n-factor depends on the medium, not on the reducing agent. In neutral solution both reactions would stop at manganese dioxide, with an n-factor of $3$.

7. Lanthanoids and actinoids

The lanthanoids are so alike that they were separated only with great difficulty, and their near-identical radii are why zirconium and hafnium behave as twins. The few departures from $+3$ are explained entirely by $f$-subshell stability: cerium reaches $+4$ because that gives $4f^0$, and europium drops to $+2$ because that gives $4f^7$.

Actinoid chemistry is far richer because the $5f$ orbitals are less buried and can participate in bonding, so uranium alone shows $+3$, $+4$, $+5$ and $+6$.

Illustration 10

Explain why ${\text{Ce}}^{4+}$ is a good oxidising agent and ${\text{Eu}}^{2+}$ a good reducing agent.

Cerium in the $+4$ state is $4f^0$. Reverting to ${\text{Ce}}^{3+}$ gives $4f^1$, but the $+4$ state itself is accessible only because the empty $f$ shell is stable — and in aqueous solution the $+3$ state is preferred, so cerium(IV) readily accepts an electron.

Europium in the $+2$ state is $4f^7$, a stable half-filled shell, which is why it exists at all. But $+3$ remains the normal lanthanoid state, so europium(II) readily gives up an electron.

Both anomalies come from $f$-subshell stability, and both revert to $+3$ in solution — which is why the two ions behave as opposite reagents.

Illustration 11

Explain why the atomic radii of the first transition series change so little from vanadium to copper, unlike the steady contraction seen across a typical period.

Across a normal period the added electrons enter the outermost shell and screen poorly, so $Z_{eff}$ rises steadily and the atom contracts.

In a transition series the added electrons enter the inner $\left(n-1\right)d$ shell, which screens the outer $ns$ electrons quite effectively.

The increase in nuclear charge is therefore largely cancelled, and the radius changes only slightly across the middle of the series.

Towards the end the $d$ electrons begin to repel each other appreciably, and the radius even rises slightly at copper and zinc.

Illustration 12

Explain why ${\text{MnO}}_4^{-}$ and ${\text{CrO}}_4^{2-}$ are both intensely coloured despite both metals being $d^0$, and why ${\text{VO}}_4^{3-}$ is much paler.

All three are coloured by ligand-to-metal charge transfer, since neither has any $d$ electrons to promote.

The energy of that transition depends on how easily the metal is reduced. Manganese(VII) is the most strongly oxidising of the three, so its transition is lowest in energy and absorbs in the visible most strongly.

Vanadium(V) is the least oxidising, so its charge transfer band lies further into the ultraviolet and the ion appears only faintly coloured.

Colour intensity tracks oxidising power across the series, which is a useful qualitative check on any charge-transfer explanation.

Summary

• Colour has two causes: weak $d$-$d$ transitions needing a partly filled set, and intense charge transfer needing none.
• ${\text{MnO}}_4^{-}$ and ${\text{Cr}}_2{\text{O}}_7^{2-}$ are $d^0$ yet intensely coloured — charge transfer is the only explanation.
• $d^0$ and $d^{10}$ ions are colourless by the first mechanism: ${\text{Sc}}^{3+}$, ${\text{Ti}}^{4+}$, ${\text{Cu}}^{+}$, ${\text{Zn}}^{2+}$.
• Variable oxidation states arise because $\left(n-1\right)d$ and $ns$ are close in energy; the range is widest in the middle.
• High oxidation states need electronegative ligands: manganese reaches $+7$ with oxygen and fluorine, never with chloride.
• $d^5$ stability explains both why ${\text{Fe}}^{3+}$ is favoured and why ${\text{Mn}}^{3+}$ is not.
• $\mu =\sqrt{n\left(n+2\right)}$ BM is spin only and works well for the first series, less well for heavier ones.
• Radii change little across the series because $d$ electrons screen the outer shell effectively.
• Melting points peak near chromium; manganese is anomalously low because its half-filled set resists metallic bonding.
• Copper alone has a positive $E^{\circ }\left({\text{M}}^{2+}/\text{M}\right)$, so it does not liberate hydrogen from dilute acids.
• Chromium melts highest (six unpaired electrons); manganese's stable $d^5$ set makes it one of the lowest.
• Interstitial compounds are non-stoichiometric, harder and higher melting, yet remain conducting.
• Chromate and dichromate interconvert with pH, not by redox; chromium stays at $+6$ and only dichromate oxidises.
• Permanganate's n-factor is $5$ in acid but $3$ in neutral solution; dichromate's is always $6$.
• Lanthanoids are almost uniformly $+3$; cerium reaches $+4$ for $4f^0$ and europium drops to $+2$ for $4f^7$.