Coordination Compounds

Two compounds. Same formula, $\mathrm{PtC}{\mathrm{l}}_2(\mathrm{N}{\mathrm{H}}_3{)}_2$. Same atoms in the same numbers, the same two Pt-Cl bonds and the same two Pt-N bonds, the same molar mass, the same square planar geometry around platinum.

One of them is a front-line anticancer drug that has been used on millions of patients since 1978.

The other does essentially nothing therapeutically, and is simply toxic.

The difference is 90 degrees.

In cis-platin the two chlorides sit on adjacent corners of the square, close enough that once they hydrolyse away, platinum can grip two neighbouring guanine bases on the same strand of DNA. The strand kinks, replication stalls, and the cell dies.

In trans-platin the chlorides sit opposite each other, 180 degrees apart. Platinum can still bind DNA, but never two adjacent bases at once, so no cross-link forms.

A formula cannot tell those two apart. Nothing about $\mathrm{PtC}{\mathrm{l}}_2(\mathrm{N}{\mathrm{H}}_3{)}_2$ distinguishes a drug from a poison, and that is exactly why this chapter's rules exist.

A complex is a Lewis acid-base adduct that survives in solution. The nomenclature exists to name the entity precisely enough that the questions have definite answers, and the two bonding theories exist to answer them.

Valence bond theory answers through hybridisation. Crystal field theory answers by putting a number on the splitting, which is why it can predict colour and explain why the same metal ion is high spin with one ligand and low spin with another.

1. Werner's Theory

Werner faced a puzzle. Cobalt(III) chloride combines with ammonia in several fixed ratios, and the products differ in colour and, decisively, in how much chloride they surrender to silver nitrate.

If all three chlorides were equivalent, every row would read 3. They do not.

Primary valence is ionisable, satisfied by anions, and equal to the oxidation state. Secondary valence is non-ionisable, directed in space, and equal to what we now call the coordination number.

Chloride inside the coordination sphere is bonded to cobalt and cannot be precipitated. Chloride outside is a free ion and can. That one distinction explains the whole table, and Werner proposed it decades before anyone understood bonding.

Illustration 1

Werner also claimed that coordination number 6 means octahedral. He had no X-ray crystallography, no spectroscopy and no bonding theory. How could he possibly know?

He counted isomers. Three geometries can hold six ligands around a centre, and each predicts a different number of isomers for the composition $\mathrm{M}{\mathrm{A}}_4{\mathrm{B}}_2$.

Every $\mathrm{M}{\mathrm{A}}_4{\mathrm{B}}_2$ cobalt(III) complex anyone prepared gave exactly two isomers. Never three.

That is a negative result doing positive work: two of the three candidate geometries predict an isomer nobody has ever isolated, so both are dead. The octahedron predicts precisely what is seen.

The reasoning is worth pausing on, because it is how structure was determined before instruments existed. Count what a geometry requires, then count what nature supplies, and eliminate. Werner received the 1913 Nobel Prize for this argument.

2. The Vocabulary

Ligand. A species that donates a lone pair to the metal. It must have at least one available lone pair, which is why ammonia is a ligand and methane is not.

Coordination number. The number of donor atoms bonded to the metal, not the number of ligands. In $[\mathrm{Co}(\mathrm{en}{)}_3{]}^{3+}$ there are three ligands and a coordination number of six.

Denticity. The number of donor atoms one ligand offers.

Ambidentate ligand. One with two different donor atoms that can bind through either. $\mathrm{N}{\mathrm{O}}_2^{-}$ binds through N or O; $\mathrm{SC}{\mathrm{N}}^{-}$ through S or N.

Chelation. Ring formation when a polydentate ligand binds through more than one donor atom to the same metal.

The chelate effect

Chelate complexes are far more stable than comparable complexes of monodentate ligands, even when the donor atoms are chemically almost identical.

Illustration 2

$\mathrm{N}{\mathrm{H}}_3$ and ethane-1,2-diamine both donate through nitrogen, and both give nickel(II) a coordination number of six. Yet $\log \beta$ is 8.6 for $[\mathrm{Ni}(\mathrm{N}{\mathrm{H}}_3{)}_6{]}^{2+}$ and 18.3 for $[\mathrm{Ni}(\mathrm{en}{)}_3{]}^{2+}$. That is a factor of $10^{10}$ in stability constant. Where does it come from?

Not from the bonds. Six Ni-N bonds are formed in both cases and they are essentially the same bonds, so the enthalpy change is nearly identical.

It comes from counting particles. Write both substitutions honestly, with the displaced water included.

$[\mathrm{Ni}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{2+}+6\mathrm{N}{\mathrm{H}}_3\to [\mathrm{Ni}(\mathrm{N}{\mathrm{H}}_3{)}_6{]}^{2+}+6{\mathrm{H}}_2\mathrm{O}$

Seven particles in, seven out. Entropy barely moves.

$[\mathrm{Ni}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{2+}+3\mathrm{en}\to [\mathrm{Ni}(\mathrm{en}{)}_3{]}^{2+}+6{\mathrm{H}}_2\mathrm{O}$

Four particles in, seven out. Three extra free particles is a substantial entropy gain, and at 298 K a $T\Delta S$ term of that size is worth tens of kilojoules.

The chelate effect is therefore an entropy effect almost in its entirety, which is why it survives even when the chelating ligand forms slightly weaker bonds. Five- and six-membered rings are preferred because they are the least strained, and EDTA forms five of them at once, which is why it is such a formidable complexing agent.

3. IUPAC Nomenclature

The rules are mechanical once applied in the right order.

The Latin stems that matter: ferrate for iron, cuprate for copper, argentate for silver, plumbate for lead, stannate for tin, aurate for gold.

So $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_4\mathrm{C}{\mathrm{l}}_2]\mathrm{Cl}$ is tetraamminedichloridocobalt(III) chloride, and ${\mathrm{K}}_4[\mathrm{Fe}(\mathrm{CN}{)}_6]$ is potassium hexacyanidoferrate(II).

Illustration 3

Work the rules in reverse. Write formulas for pentaamminenitrito-N-cobalt(III) chloride and for tris(ethane-1,2-diamine)cobalt(III) sulphate.

First. Five ammines, one nitrito bound through nitrogen, cobalt(III), and chloride outside as the counter ion.

Charge inside: $+3$ from cobalt, 0 from five ammines, $-1$ from nitrite, giving $+2$. So two chlorides balance it.

$[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_5(\mathrm{N}{\mathrm{O}}_2)]\mathrm{C}{\mathrm{l}}_2$

Second. Three "en" ligands, which is why the prefix is tris and not tri, cobalt(III), sulphate outside.

Charge inside: $+3$, with all three ligands neutral. Sulphate is $2-$. Balancing $+3$ against $-2$ needs the lowest common multiple: two complex ions to three sulphates.

$[\mathrm{Co}(\mathrm{en}{)}_3{]}_2(\mathrm{S}{\mathrm{O}}_4{)}_3$

Trap. The bis, tris and tetrakis prefixes are not stylistic. They exist because "diethylenediamine" would be ambiguous, and using the wrong set is one of the few nomenclature errors that changes what a name means rather than merely how it reads.

4. Isomerism

Structural isomerism

Illustration 4

Two solids share the formula $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_5(\mathrm{N}{\mathrm{O}}_2)]\mathrm{C}{\mathrm{l}}_2$. One is yellow and one is red, and the red one slowly turns yellow on standing. Identify them.

$\mathrm{N}{\mathrm{O}}_2^{-}$ is ambidentate, so these are linkage isomers.

The red form is bound through oxygen, the nitrito-O isomer. The yellow form is bound through nitrogen, the nitrito-N isomer.

The conversion runs one way only, red to yellow, because the nitrogen-bound form is the more stable of the two. Nitrogen is the softer, more polarisable donor and cobalt(III) binds it better.

Note what identified them here: colour and a one-way conversion, not a chemical test. Ionisation isomers are distinguished by precipitating the free ion, hydrate isomers by measuring water content, and linkage isomers by spectra. Each type has its own diagnostic, and questions usually supply exactly one.

Stereoisomerism

Geometrical isomerism requires that different spatial arrangements be genuinely distinct.

Tetrahedral complexes show no geometrical isomerism because all four positions are mutually adjacent, so no cis or trans distinction can be drawn.

Optical isomerism requires a non-superimposable mirror image, and is common in octahedral complexes with bidentate ligands: $[\mathrm{Co}(\mathrm{en}{)}_3{]}^{3+}$, and the cis isomer of $[\mathrm{Co}(\mathrm{en}{)}_2\mathrm{C}{\mathrm{l}}_2{]}^{+}$. The trans isomer of that same complex has a plane of symmetry and is optically inactive, which is the standard discriminating question.

Illustration 5

How many stereoisomers does $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_3\mathrm{C}{\mathrm{l}}_3]$ have, and is either optically active?

This is $\mathrm{M}{\mathrm{A}}_3{\mathrm{B}}_3$, so the choice is how the three chlorides are arranged on the octahedron.

Facial. All three chlorides occupy one triangular face, mutually cis, each 90 degrees from the other two.

Meridional. The three chlorides lie on a meridian, one great circle through the octahedron, so two are trans to each other and the third is cis to both.

There is no third arrangement. Any placement of three ligands on an octahedron either has them all on a face or all on a meridian.

Both are optically inactive. The fac isomer has a three-fold axis and three mirror planes; the mer isomer has a mirror plane containing all three chlorides. Either plane makes the mirror image superimposable.

That last point matters. Geometrical isomerism does not imply optical isomerism, and $\mathrm{M}{\mathrm{A}}_3{\mathrm{B}}_3$ is the cleanest case where two geometrical isomers exist and neither is chiral.

Illustration 6

Two solids share the formula $\mathrm{CoBrS}{\mathrm{O}}_4(\mathrm{N}{\mathrm{H}}_3{)}_5$. One is $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_5\mathrm{Br}]\mathrm{S}{\mathrm{O}}_4$ and the other is $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_5\mathrm{S}{\mathrm{O}}_4]\mathrm{Br}$. Give tests that tell them apart.

Only ions outside the coordination sphere dissociate in water. Everything inside the square brackets is bonded to the cobalt and stays there.

Add barium chloride solution. The first compound has free sulphate, so it gives a white precipitate of $\mathrm{BaS}{\mathrm{O}}_4$. The second has its sulphate bonded to cobalt, so nothing happens.

Add silver nitrate solution. Now it is the other way round: the second compound gives a pale yellow precipitate of AgBr, and the first gives nothing.

The square brackets are therefore not typography. They are a claim about which ions are free in solution, and the claim is directly testable.

This is precisely the reasoning Werner used. Treating $\mathrm{CoC}{\mathrm{l}}_3\cdot \mathrm{nN}{\mathrm{H}}_3$ with silver nitrate and counting how many chlorides precipitated told him how many were ionisable, and by difference how many were bonded to the metal — which is how the idea of a coordination sphere was established before anyone could see one.

5. Valence Bond Theory

The metal provides empty orbitals which hybridise, and each ligand donates a lone pair into one of them.

Inner orbital complexes use $(n-1)d$ orbitals, which requires the d electrons to pair up first and leave two orbitals empty. Outer orbital complexes use $nd$ orbitals, so no pairing is forced.

Illustration 7

$\mathrm{N}{\mathrm{i}}^{2+}$ is $3d^8$. Both $[\mathrm{Ni}(\mathrm{CN}{)}_4{]}^{2-}$ and $[\mathrm{NiC}{\mathrm{l}}_4{]}^{2-}$ have coordination number 4. Predict the geometry of each.

The instinct from organic chemistry is that four groups means tetrahedral. It is right once and wrong once.

The chloride is too weak to disturb the $3d^8$ arrangement, which keeps two unpaired electrons in two different d orbitals. Every 3d orbital is then occupied, none is available to hybridise, and nickel must use $4s$ and $4p$: four orbitals, $s{p}^3$, tetrahedral.

Cyanide is strong enough to force those two electrons to pair. That empties one 3d orbital, which can now join the hybridisation as $ds{p}^2$: four orbitals in a plane, square planar, and no unpaired electrons left.

Same metal, same oxidation state, same coordination number. The ligand chose the geometry, and a magnetic measurement alone distinguishes the two.

Where it fails

Valence bond theory gives geometry and magnetism correctly for most complexes and gives no account of colour at all. It cannot explain why one ligand forces pairing and another does not, beyond labelling them strong and weak after the fact. It offers no way to rank ligands and no quantitative prediction of anything.

Those failures are exactly what crystal field theory repairs.

6. Crystal Field Theory

Crystal field theory treats the metal-ligand interaction as purely electrostatic. The ligands are point negative charges, and their approach raises the energy of the metal's d orbitals.

They do not all rise equally.

In an octahedral field, $d_{z^2}$ and $d_{x^2-y^2}$ point directly at the approaching ligands and are pushed up hardest, while $d_{xy}$, $d_{yz}$ and $d_{zx}$ point between them and rise less. The result is a lower triply degenerate $t_{2g}$ set and an upper doubly degenerate $e_g$ set, separated by the crystal field splitting energy ${\Delta }_o$.

High spin against low spin

Once the split exists, filling it is a competition. A fourth electron entering the lower set costs the pairing energy $P$; entering the upper set costs ${\Delta }_o$.

${\Delta }_o>P\Rightarrow \text{low spin}{\Delta }_o<P\Rightarrow \text{high spin}$

That is why the same $\mathrm{F}{\mathrm{e}}^{3+}$ ion is high spin with fluoride and low spin with cyanide. The metal did not change; the splitting did.

The spectrochemical series

${\mathrm{I}}^{-}<\mathrm{B}{\mathrm{r}}^{-}<\mathrm{C}{\mathrm{l}}^{-}<{\mathrm{F}}^{-}<\mathrm{O}{\mathrm{H}}^{-}<{\mathrm{H}}_2\mathrm{O}<\mathrm{N}{\mathrm{H}}_3<\mathrm{en}<\mathrm{N}{\mathrm{O}}_2^{-}<\mathrm{C}{\mathrm{N}}^{-}<\mathrm{CO}$

Note the embarrassment at the strong end: neutral CO outranks negatively charged halides, which a purely electrostatic model cannot explain. That discrepancy is a genuine limitation, repaired only by ligand field theory, beyond this syllabus.

Tetrahedral fields

A tetrahedral field splits the orbitals the other way up, with the lower set doubly degenerate, and the splitting is much smaller.

${\Delta }_t=\frac{4}{9}{\Delta }_o$

Because ${\Delta }_t$ is so small it is essentially never larger than $P$. Tetrahedral complexes are therefore always high spin, and no low spin tetrahedral complex needs considering.

Crystal field stabilisation energy

Each electron in the lower octahedral set is stabilised by $0.4{\Delta }_o$ and each in the upper set destabilised by $0.6{\Delta }_o$.

$\text{CFSE}=\left(-0.4n_{t_{2g}}+0.6n_{e_g}\right){\Delta }_o$

Illustration 8

Compute the CFSE of a high spin $d^5$ octahedral ion, such as $\mathrm{M}{\mathrm{n}}^{2+}$ or $\mathrm{F}{\mathrm{e}}^{3+}$, and say what follows.

High spin $d^5$ places one electron in each of the five orbitals: three in $t_{2g}$ and two in $e_g$.

$\text{CFSE}=\left[-0.4(3)+0.6(2)\right]{\Delta }_o=\left(-1.2+1.2\right){\Delta }_o=0$

Exactly zero. A high spin $d^5$ ion gains nothing whatsoever from the crystal field, whatever the ligand.

Three consequences follow, and all three are observable.

$\mathrm{M}{\mathrm{n}}^{2+}$ forms the weakest complexes of any first-row divalent ion, which is the dip at manganese in the Irving-Williams stability order.

Its complexes are very pale, almost colourless, because every d-d transition would have to change the spin as well as the orbital, and spin-forbidden transitions are extraordinarily weak.

And $\mathrm{F}{\mathrm{e}}^{3+}$, also high spin $d^5$, has no crystal field preference between geometries, so it adopts whatever the ligands' size and charge dictate rather than what the field would prefer.

A CFSE of zero is not an absence of an answer. It is the answer, and it predicts three separate observations.

Colour and magnetism, explained

Colour is now straightforward. An electron absorbs a photon whose energy matches ${\Delta }_o$ and jumps from the lower set to the upper. Since ${\Delta }_o$ depends on the ligand, the same metal gives different colours with different ligands, which valence bond theory could never explain.

A $d^0$ or $d^{10}$ ion has no such transition available, which is why $\mathrm{S}{\mathrm{c}}^{3+}$ and $\mathrm{Z}{\mathrm{n}}^{2+}$ complexes are white.

Illustration 9

$[\mathrm{Ti}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{3+}$ absorbs most strongly at 500 nm and looks violet. Find ${\Delta }_o$ in kJ mol${}^{-1}$, then predict how the colour changes if water is replaced by ammonia.

Convert the absorbed wavelength to an energy per mole.

$E=\frac{hc}{\lambda }=\frac{(6.626\times 10^{-34})(3.00\times 10^8)}{500\times 10^{-9}}=3.98\times 10^{-19}\mathrm{J}$

${\Delta }_o=(3.98\times 10^{-19})(6.022\times 10^{23})=2.39\times 10^5\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}=239\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$

That is comparable to a chemical bond, from an interaction the theory models as nothing more than point charges pushing on orbitals.

The colour follows from what was not absorbed. Removing green from white light leaves red and blue, which the eye reads as violet.

Now swap the ligand. Ammonia sits above water in the spectrochemical series, so ${\Delta }_o$ increases, so the absorbed photon must carry more energy, so the absorption moves to shorter wavelength. The same reasoning explains why $[\mathrm{Cu}({\mathrm{H}}_2\mathrm{O}{)}_4{]}^{2+}$ is a pale blue and $[\mathrm{Cu}(\mathrm{N}{\mathrm{H}}_3{)}_4{]}^{2+}$ is the deep royal blue of the standard test.

Trap. Larger ${\Delta }_o$ means a shorter absorbed wavelength, and the colour you see is the complement of the colour absorbed. Two inversions in a row, and dropping either one gives an answer that is exactly backwards.

Magnetism follows from the unpaired electron count after the high spin or low spin decision, using the spin-only formula from the previous chapter.

Illustration 10

$[\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{4-}$ is diamagnetic, while $[\mathrm{Fe}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{2+}$ has a magnetic moment near 4.9 BM. Both are octahedral iron(II). Account for the difference and give the unpaired electron count in each.

Iron(II) is $d^6$ in both complexes, so the electron count cannot be what separates them. What differs is how those six electrons are arranged, and the ligand decides that.

Cyanide sits high in the spectrochemical series and is a strong-field ligand, so the splitting ${\Delta }_o$ exceeds the pairing energy $P$. Electrons pair up in the lower set rather than climb:

$t_{2g}^6{e}_g^0\Rightarrow n=0\text{unpaired, diamagnetic}$

Water is a weak-field ligand, so ${\Delta }_o<P$ and the electrons spread out instead:

$t_{2g}^4{e}_g^2\Rightarrow n=4\text{unpaired}$

Check that against the measurement using the spin-only formula:

$\mu =\sqrt{n(n+2)}=\sqrt{4(6)}=\sqrt{24}=4.90\mathrm{BM}$

which is the quoted value.

So the metal and its oxidation state fix how many d electrons there are, and the ligand fixes how they sit. Same ion, same geometry, opposite magnetic behaviour — and a magnetic measurement is therefore a direct read-out of where a ligand falls in the spectrochemical series.

7. Importance of Coordination Compounds

In qualitative analysis, complex formation makes many tests work. Silver chloride dissolves in ammonia as the diamminesilver(I) ion, and copper(II) gives the deep blue tetraamminecopper(II) ion.

In metallurgy, gold and silver are extracted by cyanide leaching, dissolving as soluble cyanido complexes and later displaced by zinc. Nickel is purified by the Mond process through volatile $\mathrm{Ni}(\mathrm{CO}{)}_4$, which decomposes on heating to leave pure metal.

In biology, haemoglobin carries oxygen on an iron centre, chlorophyll is a magnesium complex and vitamin ${\mathrm{B}}_{12}$ is a cobalt complex. All three are chelates of large ring ligands.

In medicine, EDTA sequesters calcium and magnesium in water softening and treats lead poisoning by chelating the metal for excretion. Cis-platin is the anticancer drug this chapter opened on, and the trans isomer is inactive.

Illustration 11

Lead poisoning is treated by giving the patient EDTA. Why does a chelating agent remove lead without stripping the body of every other metal it needs?

Because stability constants differ enormously, and the treatment exploits the gap.

Lead binds more than seven orders of magnitude more tightly than calcium.

So the drug is administered as the calcium salt, $\mathrm{CaN}{\mathrm{a}}_2\mathrm{EDTA}$, already saturated with the metal it is allowed to lose. On meeting lead, the exchange runs strongly forward, releasing harmless calcium and locking up the lead as a stable, water-soluble complex the kidneys excrete.

Give free EDTA instead and it would strip calcium out of the blood indiscriminately, which is fatal. The chelate effect is doing the therapeutic work, and the choice of counter-ion is what keeps it selective.

Summary

A complex is a Lewis acid-base adduct that survives in solution, and the chapter exists to answer three questions about it: what shape, what colour, what magnetism. Cis-platin and trans-platin share a formula and differ by 90 degrees, which is the difference between a drug and a poison.

Werner separated primary from secondary valence to explain why $\mathrm{CoC}{\mathrm{l}}_3\cdot 4N{\mathrm{H}}_3$ gives only one mole of silver chloride, and established the octahedron by isomer counting: a planar hexagon and a trigonal prism each predict three $\mathrm{M}{\mathrm{A}}_4{\mathrm{B}}_2$ isomers, and only two are ever found.

Coordination number counts donor atoms, not ligands. The chelate effect is almost entirely entropic, since three bidentate ligands displacing six waters turns four particles into seven, and it is worth about $10^{10}$ in stability constant for nickel.

Nomenclature runs cation first, ligands alphabetically, anionic ligands in -o, bis and tris when the ligand name already carries a multiplier, and -ate on a Latin stem for anionic complexes.

Structural isomerism comes in four kinds, each with its own diagnostic. Geometrical isomerism gives cis and trans for $\mathrm{M}{\mathrm{A}}_4{\mathrm{B}}_2$ and fac and mer for $\mathrm{M}{\mathrm{A}}_3{\mathrm{B}}_3$, never anything for tetrahedral, and it does not imply optical activity: both $\mathrm{M}{\mathrm{A}}_3{\mathrm{B}}_3$ isomers are achiral.

Valence bond theory assigns hybridisation, and the ligand chooses it: $[\mathrm{NiC}{\mathrm{l}}_4{]}^{2-}$ is $s{p}^3$ tetrahedral and paramagnetic while $[\mathrm{Ni}(\mathrm{CN}{)}_4{]}^{2-}$ is $ds{p}^2$ square planar and diamagnetic. It says nothing about colour and cannot rank ligands, which is exactly what crystal field theory supplies.

Crystal field theory splits the d orbitals by ${\Delta }_o$ and lets ${\Delta }_o$ against $P$ decide high spin or low spin. ${\Delta }_t=\frac{4}{9}{\Delta }_o$, so tetrahedral complexes are always high spin. CFSE is $(-0.4n_{t_{2g}}+0.6n_{e_g}){\Delta }_o$, and it is exactly zero for high spin $d^5$, which is why $\mathrm{M}{\mathrm{n}}^{2+}$ complexes are the weakest and palest of the row.

Colour is the ${\Delta }_o$ transition: $[\mathrm{Ti}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{3+}$ absorbs at 500 nm, giving ${\Delta }_o=239$ kJ mol${}^{-1}$, and a stronger ligand shifts the absorption to shorter wavelength while the eye sees the complement of whatever was absorbed.