Coordination Compounds

${\left[\text{Ni}{\left(\text{CN}\right)}_4\right]}^{2-}$ is square planar and diamagnetic. ${\left[{\text{NiCl}}_4\right]}^{2-}$ is tetrahedral and paramagnetic. Same metal, same oxidation state, same coordination number. What makes the difference?

The ligand — and specifically how strongly it splits the $d$ orbitals.

Nickel(II) is $3d^8$. Cyanide is a very strong field ligand: it splits the $d$ set so widely that pairing all eight electrons into four orbitals costs less than promoting one into the highest orbital. That frees a $3d$ orbital for $ds{p}^2$ hybridisation, giving a square plane with no unpaired electrons.

Chloride is a weak field ligand. The splitting is small, the electrons stay unpaired according to Hund's rule, no $3d$ orbital is vacated, and the metal uses $s{p}^3$ hybridisation instead — a tetrahedron with two unpaired electrons.

The ligand, not the metal, decides geometry, magnetism and colour together. Once that is grasped, the chapter reduces to one comparison repeated: is the splitting bigger or smaller than the pairing energy?

1. Crystal field splitting and CFSE

Ligands approaching along the axes repel the two axis-directed $d$ orbitals more than the three that lie between axes, so the set splits:

$\text{octahedral}:t_{2g}\text{down by}0.4{\Delta }_o,e_g\text{up by}0.6{\Delta }_o$

The tetrahedral case is inverted, with only four ligands and none pointing directly at any orbital, so

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

which is far too small ever to force pairing. Tetrahedral complexes are therefore always high spin.

For an octahedral complex the electrons fill singly first if ${\Delta }_o<P$ (high spin) and pair up first if ${\Delta }_o>P$ (low spin), where $P$ is the pairing energy. The stabilisation gained is

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

The spectrochemical series ranks ligands by field strength:

${\text{I}}^{-}<{\text{Br}}^{-}<{\text{Cl}}^{-}<{\text{F}}^{-}<{\text{OH}}^{-}<{\text{H}}_2\text{O}<{\text{NH}}_3<\text{en}<{\text{NO}}_2^{-}<{\text{CN}}^{-}<\text{CO}$

Illustration 1

Determine the number of unpaired electrons and the CFSE for ${\left[\text{Fe}{\left({\text{H}}_2\text{O}\right)}_6\right]}^{3+}$ and ${\left[\text{Fe}{\left(\text{CN}\right)}_6\right]}^{3-}$.

Both are iron(III), $3d^5$.

Water is a weak field ligand, so high spin: $t_{2g}^3{e}_g^2$, five unpaired.

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

Cyanide is strong field, so low spin: $t_{2g}^5$, one unpaired.

$\text{CFSE}=-0.4\times 5{\Delta }_o=-2.0{\Delta }_o$

The half-filled high-spin case has zero stabilisation, which is a useful check: any $d^5$ high-spin or $d^0$ or $d^{10}$ configuration gives exactly zero CFSE.

Illustration 2

Explain why ${\left[{\text{CoF}}_6\right]}^{3-}$ is paramagnetic while ${\left[\text{Co}{\left({\text{NH}}_3\right)}_6\right]}^{3+}$ is diamagnetic.

Both are cobalt(III), $3d^6$.

Fluoride is weak field: ${\Delta }_o<P$, so the arrangement is $t_{2g}^4{e}_g^2$ with four unpaired electrons. The metal uses outer $4d$ orbitals, giving an $s{p}^3{d}^2$ outer orbital complex.

Ammonia is strong field: ${\Delta }_o>P$, giving $t_{2g}^6$ with none unpaired, and the metal uses inner $3d$ orbitals for $d^{2}s{p}^3$ — an inner orbital complex.

Magnetic measurement distinguishes them instantly, which is how the two hybridisation schemes were established experimentally in the first place.

2. Colour from the splitting

A $d$-$d$ transition promotes an electron across ${\Delta }_o$, so the complex absorbs light of that energy and appears in the complementary colour. A larger splitting therefore means absorption further into the blue and a colour further towards the red.

${\Delta }_o=h\nu =\frac{hc}{\lambda }$

Strong field ligands give large splittings and so shift the colour: hexaaquacobalt(II) is pink while hexaamminecobalt(II) is much deeper in shade, and the aqua complex of copper turns royal blue on adding ammonia for exactly this reason.

Illustration 3

${\left[\text{Ti}{\left({\text{H}}_2\text{O}\right)}_6\right]}^{3+}$ absorbs at $498$ nm. Find ${\Delta }_o$ in kJ mol${}^{-1}$ and state the observed colour.

${\Delta }_o=\frac{hc}{\lambda }=\frac{1240}{498}=2.49$ eV per ion

Per mole: $2.49\times 96.5=240$ kJ mol${}^{-1}$

Absorption at $498$ nm is in the green, so the transmitted light is violet.

Titanium(III) is $3d^1$, the simplest possible case: one electron, one transition, one absorption band. That is why it is the standard example in every textbook.

3. Naming a complex

The rules are mechanical once listed, and Advanced expects both directions — formula to name and name to formula.

Two traps recur. Alphabetical order uses the ligand name, not the multiplying prefix, so triammine comes under "a" and not under "t". And an anionic complex often uses the Latin stem: iron becomes ferrate, copper cuprate, silver argentate, lead plumbate and gold aurate.

Illustration 4

Name ${\text{K}}_3\left[\text{Fe}{\left({\text{C}}_2{\text{O}}_4\right)}_3\right]$ and $\left[\text{Co}{\left(\text{en}\right)}_2{\text{Cl}}_2\right]{\text{NO}}_3$, and give the oxidation state of the metal in each.

The first is an anionic complex, so iron takes the Latin stem: potassium trioxalatoferrate(III).

Oxalate is $2-$ and there are three, giving $-6$; the overall charge is $3-$, so iron is $+3$.

The second: dichloridobis(ethylenediamine)cobalt(III) nitrate.

Ethylenediamine takes "bis" because its own name would make "diethylenediamine" ambiguous. Two chlorides give $-2$ against an overall $+1$, so cobalt is $+3$.

Alphabetise on the ligand stem, which puts chlorido before ethylenediamine even though the prefixes read the other way.

4. Structural isomerism

Four kinds recur, and each is defined by what moves.

Ionisation isomers are distinguished experimentally by precipitation: only the one with free sulphate gives a barium sulphate precipitate, and only the one with free bromide gives silver bromide.

Illustration 5

A complex of formula ${\text{CoCl}}_3\cdot 5{\text{NH}}_3$ gives two moles of silver chloride per mole with excess silver nitrate. Deduce its structure and write its name.

Two chlorides are precipitated, so two are outside the coordination sphere as free ions and one is inside as a ligand.

The formula is $\left[\text{Co}{\left({\text{NH}}_3\right)}_5\text{Cl}\right]{\text{Cl}}_2$.

Name: pentaamminechloridocobalt(III) chloride.

This is exactly the experiment Werner used, counting precipitated chloride to establish that some ions are bound and some are not, long before any structural method existed.

5. Stereoisomerism, and how to count it

A tetrahedron has no cis or trans arrangement at all, because every pair of positions is equivalent. This is why a complex showing geometrical isomerism with four ligands must be square planar, which is a standard route to establishing geometry.

Illustration 6

State the number of geometrical and optical isomers of ${\left[\text{Co}{\left(\text{en}\right)}_2{\text{Cl}}_2\right]}^{+}$.

Ethylenediamine is bidentate, so this is the $\text{M}{\left(\text{AA}\right)}_2{\text{B}}_2$ case.

Two geometrical isomers exist: cis and trans.

The trans isomer has a plane of symmetry and is optically inactive. The cis isomer has none, so it exists as a pair of enantiomers.

Total: three stereoisomers — one trans and two cis forms.

A bidentate ligand cannot span trans positions, which is why $\text{M}{\left(\text{AA}\right)}_3$ has no geometrical isomers at all but is always chiral.

6. Stability and the chelate effect

The formation of a complex is described by a stability constant, and successive constants normally decrease. Two structural features raise stability sharply.

Chelation. A ligand binding through two or more donor atoms forms a ring, and such complexes are far more stable than their unidentate equivalents. The reason is entropic: replacing six water molecules with three bidentate ligands releases six particles while consuming only three, so $\Delta S$ is strongly positive.

Ring size. Five- and six-membered chelate rings are the most stable, being free of angle strain.

EDTA is hexadentate, forming six bonds and five rings at once, which is why its complexes are exceptionally stable and why it is used to treat lead poisoning and to soften water.

Illustration 7

Explain why ${\left[\text{Ni}{\left(\text{en}\right)}_3\right]}^{2+}$ is far more stable than ${\left[\text{Ni}{\left({\text{NH}}_3\right)}_6\right]}^{2+}$, although both bind through nitrogen.

Each ethylenediamine replaces two ammonia molecules, so three of them do the work of six.

The enthalpy change is nearly identical, since the same six nitrogen donors are involved.

But the entropy change differs: displacing six aqua ligands with three chelates increases the number of free particles, while displacing six with six leaves it unchanged.

The chelate effect is essentially entropic, which is why it survives even when the individual bonds are no stronger at all.

7. Metal carbonyls and the EAN rule

Carbon monoxide is a very weak Lewis base yet binds metals strongly, because the bonding is synergic:

Carbon monoxide donates its carbon lone pair into an empty metal orbital, and the metal donates electron density back from a filled $d$ orbital into carbon monoxide's empty antibonding orbital. Each donation reinforces the other, which is why the bond is far stronger than a simple lone-pair donation would give. The consequence is measurable: the $\text{C}-\text{O}$ stretching frequency falls on coordination, because the back-donated electrons occupy an antibonding orbital.

The effective atomic number rule states that stable carbonyls achieve the electron count of the next noble gas:

$\text{EAN}=Z-\left(\text{oxidation state}\right)+2\times \left(\text{number of donor pairs}\right)$

Illustration 8

Verify the EAN rule for $\text{Ni}{\left(\text{CO}\right)}_4$, $\text{Fe}{\left(\text{CO}\right)}_5$ and $\text{Cr}{\left(\text{CO}\right)}_6$, and predict the formula of the cobalt carbonyl.

$\text{Ni}{\left(\text{CO}\right)}_4$: $28-0+8=36$ (krypton)

$\text{Fe}{\left(\text{CO}\right)}_5$: $26-0+10=36$

$\text{Cr}{\left(\text{CO}\right)}_6$: $24-0+12=36$

Cobalt has $Z=27$, an odd number, so $27+2n$ can never equal $36$. It forms a dimer instead, ${\text{Co}}_2{\left(\text{CO}\right)}_8$, with a metal-metal bond supplying the missing electron to each.

Odd atomic number always means dimerisation or an odd-electron radical. Manganese behaves the same way, giving ${\text{Mn}}_2{\left(\text{CO}\right)}_{10}$.

Illustration 9

Predict the hybridisation, geometry and magnetic behaviour of ${\left[\text{Fe}{\left(\text{CN}\right)}_6\right]}^{4-}$ and ${\left[{\text{FeF}}_6\right]}^{3-}$.

${\left[\text{Fe}{\left(\text{CN}\right)}_6\right]}^{4-}$: iron(II), $3d^6$. Cyanide is strong field, so $t_{2g}^6$, two inner $3d$ orbitals vacated, $d^{2}s{p}^3$, octahedral and diamagnetic.

${\left[{\text{FeF}}_6\right]}^{3-}$: iron(III), $3d^5$. Fluoride is weak field, so $t_{2g}^3{e}_g^2$, no $3d$ orbital free, $s{p}^3{d}^2$ using outer orbitals, octahedral and paramagnetic with five unpaired electrons.

Both are octahedral but only one is an inner orbital complex. Geometry and hybridisation are separate questions, and the magnetic moment answers the second.

Illustration 10

Explain why ${\left[\text{Cu}{\left({\text{NH}}_3\right)}_4\right]}^{2+}$ is deep blue while ${\left[\text{Cu}{\left({\text{H}}_2\text{O}\right)}_6\right]}^{2+}$ is pale blue.

Both are copper(II), $3d^9$, and both are coloured by a $d$-$d$ transition.

Ammonia lies higher in the spectrochemical series than water, so it produces a larger ${\Delta }_o$.

The larger gap means absorption at shorter wavelength and, in this case, much stronger absorption, giving the intense royal blue seen when ammonia is added to copper sulphate solution.

This colour change is the standard test for copper(II), and it is a splitting change rather than any change of oxidation state.

Illustration 11

Write the IUPAC name of $\left[\text{Pt}{\left({\text{NH}}_3\right)}_2{\text{Cl}}_2\right]$ and state why both isomers are important.

Name: diamminedichloridoplatinum(II).

It is square planar, so it exists as cis and trans isomers.

The cis isomer is cisplatin, a widely used anticancer drug that binds to adjacent sites on DNA. The trans isomer is therapeutically inactive, because its chlorides are too far apart to bridge the same two bases.

Geometry alone separates a medicine from an inert compound. This is the most cited example of why stereochemistry matters in coordination chemistry.

8. Where coordination chemistry is actually used

Extraction. The Mond process purifies nickel by forming a volatile carbonyl at $330$ K and decomposing it at $450$ K, leaving metal of exceptional purity. Gold and silver are leached from crushed ore as their dicyanido complexes and then displaced by zinc.

Analysis. EDTA titration determines the hardness of water directly. Qualitative tests are almost all complex formation: the deep blue of ${\left[\text{Cu}{\left({\text{NH}}_3\right)}_4\right]}^{2+}$, the blood red of the iron thiocyanate complex, and Tollens' reagent ${\left[\text{Ag}{\left({\text{NH}}_3\right)}_2\right]}^{+}$.

Biology. Haemoglobin carries oxygen on an iron(II) centre held in a porphyrin ring; chlorophyll uses magnesium(II) in the same kind of ring; vitamin B12 uses cobalt in a corrin ring. In every case the metal is held by a chelating macrocycle that keeps it in the right oxidation state and geometry.

Illustration 12

Explain why carbon monoxide is so much more toxic than its concentration suggests, in coordination terms.

Oxygen binds reversibly to the iron(II) of haemoglobin, which is what allows it to be released in the tissues.

Carbon monoxide binds at the same site but roughly $200$ times more strongly, because it accepts back donation from the iron into its antibonding orbital while oxygen does so far less effectively.

The resulting carboxyhaemoglobin does not release its ligand, so those sites are permanently blocked.

A small partial pressure therefore disables a large fraction of the carrier. The mechanism is exactly the synergic bonding that makes metal carbonyls stable in the first place.

Illustration 13

Explain how the Mond process achieves such high purity, and why it works for nickel but not for iron in the same plant.

Nickel reacts with carbon monoxide at about $330$ K to give volatile $\text{Ni}{\left(\text{CO}\right)}_4$, which distils away from the solid impurities.

Heating the vapour to about $450$ K decomposes it back to pure nickel and carbon monoxide, which is recycled.

Iron does form a carbonyl, but only under far higher pressure, so at the mild conditions used it stays behind with the residue.

Selectivity comes from the difference in formation conditions, not from any difference in stability of the final metals.

Summary

• The ligand decides everything: ${\left[\text{Ni}{\left(\text{CN}\right)}_4\right]}^{2-}$ is square planar and diamagnetic, ${\left[{\text{NiCl}}_4\right]}^{2-}$ tetrahedral and paramagnetic.
• Octahedral splitting puts $t_{2g}$ down $0.4{\Delta }_o$ and $e_g$ up $0.6{\Delta }_o$; ${\Delta }_t=\frac{4}{9}{\Delta }_o$, so tetrahedral complexes are always high spin.
• High spin when ${\Delta }_o<P$, low spin when ${\Delta }_o>P$; $\text{CFSE}=\left(-0.4n_{t_{2g}}+0.6n_{e_g}\right){\Delta }_o$.
• $d^0$, $d^{10}$ and high-spin $d^5$ all give zero CFSE — a fast check on any calculation.
• Spectrochemical series: ${\text{I}}^{-}<{\text{Br}}^{-}<{\text{Cl}}^{-}<{\text{F}}^{-}<{\text{OH}}^{-}<{\text{H}}_2\text{O}<{\text{NH}}_3<\text{en}<{\text{CN}}^{-}<\text{CO}$.
• Colour comes from promotion across ${\Delta }_o$; a stronger ligand gives a larger gap and a different observed colour.
• Naming: ligands alphabetically by stem, anionic complexes take the Latin $-ate$ stem (ferrate, cuprate, aurate).
• Structural isomerism: ionisation, hydrate, linkage and coordination, each distinguished by a precipitation or conductivity test.
• Tetrahedral complexes show no geometrical isomerism, so a four-coordinate complex that does must be square planar.
• $\text{M}{\left(\text{AA}\right)}_2{\text{B}}_2$ has cis and trans, with only the cis optically active; $\text{M}{\left(\text{AA}\right)}_3$ has no geometrical isomers but is always chiral.
• The chelate effect is entropic: three bidentate ligands release six waters while consuming three particles.
• EDTA is hexadentate, forming five rings at once, and is used against lead poisoning and hard water.
• Carbonyl bonding is synergic, and the back donation into an antibonding orbital lowers the $\text{C}-\text{O}$ stretching frequency.
• EAN $=Z-\text{oxidation state}+2\times \text{donor pairs}$; odd $Z$ forces dimers such as ${\text{Co}}_2{\left(\text{CO}\right)}_8$ and ${\text{Mn}}_2{\left(\text{CO}\right)}_{10}$.
• Mond process purifies nickel through a volatile carbonyl; gold is leached as ${\left[\text{Au}{\left(\text{CN}\right)}_2\right]}^{-}$; EDTA measures water hardness.
• Carbon monoxide poisons because it binds haemoglobin about $200$ times more strongly than oxygen, and irreversibly.
• Cisplatin works and its trans isomer does not, because only adjacent chlorides can bridge two DNA bases.