Inorganic Chemistry · Part 9 of 9

Applications & Special Topics

Coordination Chemistry, Part 9 · 12 sections · about 23,848 words · CSIR-NET Chemical Sciences, GATE Chemistry & IIT-JAM

The last part answers the question a student is entitled to ask after eight parts of theory: what is any of this for. Haemoglobin is a coordination compound whose behaviour is a spin-state change. Cisplatin is a substitution reaction with a trans-effect synthesis and a kinetic-inertness rationale. An industrial catalytic cycle is oxidative addition and reductive elimination in a loop. Each application here is chosen because it uses something the earlier parts built, and each is traced back to it. The part closes with the master index to all nine. Two layers on every section: a slow, hand-held beginner path and a research-grade advanced/reference path.

The 12 sections in Part 9

  • 1L.1 Oxygen transport and storage — a spin-state change with consequences Free below
  • 2L.2 Other metalloproteins read as coordination compounds
  • 3L.3 Metallodrugs — cisplatin, contrast agents and chelation therapy
  • 4L.4 The elementary steps, with the electron count tracked throughout
  • 5L.5 Three catalytic cycles, worked vertex by vertex
  • 6L.6 Metal–metal bonds, the δ bond, and clusters
  • 7L.7 Crown ethers, cryptands and size-match selectivity
  • 8L.8 Self-assembly and metal–organic frameworks
  • 9L.9 Molecular recognition — and where the subject goes next
  • 10Section-by-section index
  • 11The argument, from Part 1 to Part 9
  • 12The nine parts

L.1 Oxygen transport and storage — a spin-state change with consequences

Free extract

Section 1 of Part 9, reproduced in full from the book — figures and all. No sign-in, no paywall on this section.

Beginner layer

Haem, myoglobin and haemoglobin, told as a story you could repeat to someone else. The chemistry is Part 3’s high-spin/low-spin distinction with a protein wrapped round it.

Oxygen is not very soluble in water. A litre of blood plasma dissolves only a few millilitres of it — nowhere near enough to run a mammal. Evolution’s answer was to build a reversible O₂ ligand-binding site, and the site it built is an iron(II) complex.

The ligand: protoporphyrin IX

Haem is iron(II) bound to protoporphyrin IX, a tetradentate macrocyclic ligand. Three of the descriptors this book has developed apply at once, and they are worth naming explicitly because a question can ask for any of them:

  • Denticity 4 (Part 1) — four pyrrole nitrogens, all N-donors, all in one plane.
  • Charge 2− — the free base loses two NH protons on metallation, so the ligand carries a 2− charge and Fe(II)–porphyrin is neutral overall. This is why haem is not simply washed out of the cell.
  • Macrocyclic (Part 7) — the four donors are pre-organised in a ring, so the complex enjoys the macrocyclic effect on top of the ordinary chelate effect. A macrocycle does not have to pay the conformational entropy cost of wrapping itself round the metal, because it is already wrapped. That is precisely why nature uses a macrocycle for a site that must never let go of its iron.
The real ligands, drawn from SMILES. Left: protoporphyrin IX, the haem ligand — note the two propanoate side chains, which anchor the cofactor to the protein, and the conjugated 18-π-electron inner ring that makes haem intensely coloured. Centre: porphine, the unsubstituted parent, showing the tetradentate N₄ core clearly. Right: histidine, whose imidazole nitrogen supplies the fifth (proximal) ligand and, in myoglobin and haemoglobin, guards the sixth (distal) position without occupying it.

The five-coordinate trick

A metal ion bound by four planar donors has two axial positions left. In myoglobin and haemoglobin, one of them is filled permanently by an imidazole nitrogen from a histidine residue of the protein — the proximal histidine. The other is left empty. That vacant sixth site is the oxygen-binding site.

Proximal histidine: the histidine residue whose imidazole nitrogen is coordinated to the haem iron, on the opposite face from the O₂ site. It is the mechanical link between the metal and the protein.
Distal histidine: a second histidine on the O₂ side, not coordinated to iron. It hydrogen-bonds bound O₂ and sterically discriminates against CO.

What happens on binding — the whole thing in one sentence

Five-coordinate high-spin iron(II), too big for the porphyrin hole and therefore sitting out of plane, binds O₂ at the vacant sixth site; the extra strong-field ligand converts it to six-coordinate low-spin iron(II), which is smaller and drops into the plane, dragging the proximal histidine — and with it the protein — along for the ride.

The trigger: iron moves into the porphyrin plane on binding O₂plane of the four N donorsNNNproximalhistidineFe≈0.4–0.6 ÅDEOXY5-coordinate · high-spin Fe(II), d⁶, S = 2large ion, sits out of plane, ring domedNNNproximalhistidineFeOObent, ≈120°OXY6-coordinate · low-spin Fe(II), d⁶, S = 0smaller ion, drops into the plane, ring flattens+ O₂spin change S = 2 → S = 0
This is Part 3’s high-spin/low-spin distinction doing real biological work. Deoxy haem is high-spin Fe(II) — two electrons occupy the antibonding eg set, the ion is large, and it cannot fit the porphyrin hole, so it sits about 0.4–0.6 Å out of the N₄ plane and domes the ring towards itself. Binding O₂ makes the field strong enough to pair all six d electrons: the complex becomes low-spin, the eg* orbitals empty, the effective radius contracts, and the iron slips into the plane. The proximal histidine is dragged with it — and because that histidine is attached to helix F of the protein, an electronic change at one metal ion becomes a mechanical change in a whole protein. Distances are schematic; the panel is a side-on view.

Read that figure against Part 3 and it is entirely familiar. High-spin d⁶ puts two electrons into the eg set, which points straight at the ligands and is antibonding: metal–ligand distances are long and the effective ionic radius is large. Low-spin d⁶ empties the eg set completely, giving the t2g⁶ closed shell — the most stabilised configuration in the whole octahedral CFSE table, with CFSE = −2.4Δo + 2P. The ion contracts, and now it fits.

DEOXY — high-spin Fe(II) d⁶, S = 2Energy →barycentre (spherical field)egt2gΔo small5-coordinate, weak axial field4 unpaired electrons; e_g occupied and antibonding; large ion, out of plane
OXY — low-spin Fe(II) d⁶, S = 0Energy →barycentre (spherical field)egt2gΔo large6-coordinate, O₂ is a strong-field π-acceptor0 unpaired electrons; e_g empty; small ion, drops into the plane

The magnetic consequence is measurable and is the classic experimental proof: deoxyhaemoglobin is paramagnetic and oxyhaemoglobin is diamagnetic. That single observation, which is Part 6 chemistry, is what pinned down the spin-state change in the first place, and it is why functional MRI can see oxygenated versus deoxygenated blood at all — the BOLD contrast mechanism is literally a paramagnetism measurement.

One haem is a store; four haems are a transport system

Myoglobin has one polypeptide chain and one haem. It sits in muscle and holds oxygen until the muscle needs it. Binding is simple 1:1 equilibrium, so its saturation curve is hyperbolic and it is nearly full even at the low oxygen pressures found in tissue — exactly the behaviour of a store.

Haemoglobin has four chains (α₂β₂) and four haems. Its job is different: pick up oxygen in the lungs at high pO₂, release it in tissue at low pO₂, and do so efficiently — i.e. change its saturation a lot for a modest change in pressure. A hyperbolic curve cannot do that. A sigmoid one can.

Why storage and transport need different curvesfractional saturation YpO₂ / torr →00.250.500.751.0020406080100myoglobin — hyperbolic, n = 1haemoglobin pH 7.4 — sigmoid, n ≈ 2.8haemoglobin, ↓pH / ↑CO₂Bohr effect: curve shifts righttissuelung
Myoglobin has one haem and one binding site: ordinary Langmuir binding, a hyperbolic curve, Hill coefficient n = 1. It is nearly saturated even at tissue oxygen pressures, which is exactly what a store should do. Haemoglobin has four subunits that talk to each other: the curve is sigmoid, n ≈ 2.8, and the steep region is deliberately placed between lung and tissue pressures so that a small drop in pO₂ unloads a large fraction of the cargo. Falling pH and rising CO₂ in working tissue shift the curve further right (Bohr effect), unloading still more where it is needed. Axis values are illustrative and flagged for confirmation.
Cooperativity: binding of a ligand at one site increases the affinity of the remaining sites on the same molecule. It is positive cooperativity that makes the haemoglobin curve sigmoid.

The mechanism is the figure you have already seen. When the first O₂ binds to one subunit, that subunit’s iron drops into its porphyrin plane, pulls its proximal histidine, and shifts the helix the histidine belongs to. Those helix movements are transmitted across the subunit interfaces, and they make it easier for the other three subunits to undergo the same change. The whole tetramer switches between two quaternary states:

T state (tense)R state (relaxed)
Oxygen affinityLowHigh
Dominant whenDeoxygenated — tissueOxygenated — lung
Iron positionOut of plane, high-spin, ring domedIn plane, low-spin, ring flat
Subunit interfacesMore salt bridges, constrainedSalt bridges broken, freer
Stabilised byH⁺, CO₂, Cl⁻, 2,3-BPGBound O₂ itself
The T→R transition. Binding O₂ to one subunit shifts the whole tetramer’s equilibrium towards R, and R binds the next O₂ more readily. That is cooperativity, expressed structurally.

The Hill coefficient — putting a number on cooperativity

Y = pn / (P50n + pn)   ⇔   log[Y/(1−Y)] = n log p − n log P50

Plot log[Y/(1−Y)] against log p and the slope is the Hill coefficient n. Read it as follows:

  • n = 1 — independent, non-cooperative sites. Myoglobin.
  • n > 1 — positive cooperativity. Haemoglobin, n ≈ 2.8 at the midpoint.
  • n < 1 — negative cooperativity: binding one ligand makes the next harder.
Note: n is not the number of sites. Haemoglobin has four binding sites but n ≈ 2.8, not 4. n = 4 would require perfectly all-or-nothing binding — the tetramer going straight from empty to full with no intermediates. Real haemoglobin is strongly but not infinitely cooperative, so n falls short of the site count. n is bounded above by the number of sites; it equals it only in the infinitely cooperative limit. Quoting n = 4 for haemoglobin is a standard and costly error.

The Bohr effect

Bohr effect: the oxygen affinity of haemoglobin falls as pH falls and as pCO₂ rises. Physiologically: haemoglobin releases more oxygen exactly where metabolism has made the environment more acidic and more CO₂-rich — i.e. in hard-working tissue.

Chemically it is a coupled equilibrium. Protons and CO₂ (the latter partly as carbamate on N-terminal amino groups) bind preferentially to the T state, stabilising it. Stabilising T means shifting the T⇋R equilibrium away from the high-affinity state, which lowers oxygen affinity and shifts the saturation curve to the right. The organic phosphate 2,3-bisphosphoglycerate does the same thing more powerfully still, binding in the central cavity of the T-state tetramer.

Note the elegance from a coordination-chemistry point of view: the effector molecules never touch the metal. They act entirely by shifting a conformational equilibrium that in turn shifts a ligand-field equilibrium. Allostery is thermodynamic linkage, not chemistry at the metal.

Why carbon monoxide kills

CO is isoelectronic with N₂ and CN⁻, and Part 4 established what that means: it is a strong σ-donor and a good π-acceptor, high in the spectrochemical series, and it stabilises low oxidation states through back-donation. It binds Fe(II) very much harder than O₂ does. Bound CO occupies the very site oxygen needs, and worse: a partly carbonylated haemoglobin is locked towards the R state, so the remaining subunits bind their oxygen too tightly to release it in tissue. Carbon monoxide poisoning is therefore doubly damaging — it removes capacity and sabotages delivery of what capacity remains.

Why carbon monoxide is lethal — and why it is not more lethal stillNNNproximal HisCOlinear Fe–C–O, 180°FREE HAEM in solutionCO binds unhindered; M = K(CO)/K(O₂) ≈ 10⁴NNNproximal HisCOdistal His E7steric clash tilts COCO INSIDE THE PROTEINCO destabilised, O₂ H-bonded: M falls to ≈ 10²
Carbon monoxide is a strong σ-donor and a strong π-acceptor (Part 4): it is near the top of the spectrochemical series and binds Fe(II) far harder than O₂ does. On a bare haem in solution the partition constant M = K(CO)/K(O₂) is of order 10⁴. The protein fights back. Its distal pocket is built to accept the bent Fe–O–O unit and to hydrogen-bond the terminal oxygen, while the linear Fe–C–O preferred by carbon monoxide runs into the distal histidine and must tilt, paying an energy penalty. Net effect: M drops by roughly two orders of magnitude, to of order 10². That is still enough to make CO deadly — but without the protein’s discrimination, air containing any CO at all would be unbreathable. Values flagged for confirmation.

⚠ Common mistakes & exam traps

  • Do not say the iron is oxidised on binding O₂. The functional cycle is Fe(II)⇋Fe(II). Iron(III) haemoglobin — methaemoglobin — does not bind O₂ and is physiologically useless; the body maintains a reductase specifically to reverse its formation. (The honest subtlety in the electronic description is discussed in the advanced layer below.)
  • The Hill coefficient is not the number of subunits. n ≈ 2.8 for a four-subunit protein.
  • Myoglobin is a store, not a transporter. It cannot be cooperative: cooperativity requires more than one interacting site, and myoglobin has one.
  • The distal histidine is not coordinated to iron. Only the proximal one is. Writing haemoglobin as a bis(imidazole) complex is wrong — it would leave nowhere for O₂ to bind.
  • The Bohr effect shifts the curve right, not down. The maximum saturation is unchanged; what changes is the pressure needed to achieve it.

Deoxyhaemoglobin has a measured magnetic moment of about 5.4 μB per haem, oxyhaemoglobin essentially zero. Deduce the spin states and comment. Medium

Identify the ion. Iron(II) is d⁶ in both forms — oxygenation does not change the oxidation state.
Apply the spin-only formula (Part 6). μs.o. = √[n(n+2)] μB. For n = 4 unpaired electrons this gives √24 = 4.90 μB; for n = 0 it gives 0.
Match. 5.4 exceeds 4.90, which is entirely normal for a high-spin d⁶ ion: the ground term is 5T2g, a T term, so there is an unquenched orbital contribution (Part 6) and the observed moment sits above spin-only. So deoxy is high-spin, S = 2.
Oxy. Zero moment means no unpaired electrons: low-spin d⁶, t2g⁶, S = 0, and the ground term is 1A1g.
Comment. The whole allosteric machine rests on this: the spin change alters the effective radius of the iron, which alters where it sits relative to the porphyrin plane, which moves a helix. Part 3 explains the spin change; Part 6 measures it.
Advanced / reference layer

The electronic description of bound O₂ — where the accepted picture is firm and where it is still argued; and the quantitative shape of cooperativity.

What exactly is bound: Fe(II)–O₂, or Fe(III)–superoxide?

This is the one place in L.1 where the honest answer is more interesting than the textbook one, and where a careless statement in an exam script is genuinely wrong rather than merely incomplete.

Three descriptions have been proposed for the Fe–O₂ unit in oxyhaemoglobin:

ModelFormal descriptionPredicted spinStatus
PaulingFe(II) (low-spin, d⁶) with neutral O₂ bound end-on and bentDiamagnetic — both partners closed-shell in the bound stateThe description almost universally taught, and the one to give unless a question explicitly asks for more
WeissFe(III) (low-spin, d⁵, S = ½) bound to superoxide O₂⁻ (S = ½)Diamagnetic only if the two spins are antiferromagnetically coupled to a singletStrongly supported by spectroscopic and computational evidence for substantial charge transfer from iron to O₂
McClure–Goddard (ozone model)A more covalent picture in which neither integer assignment is right; the Fe–O–O unit is treated as a single delocalised entityDiamagneticA serious third position in the literature
All three predict the observed diamagnetism, which is why magnetism alone cannot settle the question.

What is not in dispute, and is what an examiner is testing:

  1. The complex is diamagnetic, S = 0 overall.
  2. O₂ binds end-on and bent (η¹, Fe–O–O angle near 120°), not side-on and not linear.
  3. There is substantial charge transfer from iron towards O₂: the O–O stretching frequency of bound dioxygen falls into the range characteristic of superoxide, well below that of free O₂. The bound ligand really does carry significant negative charge.
  4. Reversibility requires that the charge transfer not go to completion. If genuine Fe(III) and free superoxide were released, the protein would auto-oxidise to methaemoglobin on every breath.

How to answer. Say: “formally described as low-spin Fe(II) bound to O₂, but with substantial Fe→O₂ charge transfer, so the unit has appreciable Fe(III)–superoxide character; the complex is diamagnetic either way, and the O–O stretching frequency supports the superoxide contribution.” That sentence is correct under any of the three models and shows you know why the question exists.

Note: This is a genuine instance of the non-innocent ligand problem introduced in Part 4. When charge is substantially delocalised between metal and ligand, the oxidation state stops being a physical observable and becomes a bookkeeping convention. Bound O₂, bound NO and dithiolene ligands are the standard examples. The lesson is the same one Part 4 drew: an oxidation state is a useful fiction, and it is worth knowing which fictions are load-bearing.

Picket-fence porphyrins — the control experiment

A simple iron(II) porphyrin in solution does not reversibly bind O₂: two of them meet, form a μ-peroxo bridge Fe–O–O–Fe, and go on to an irreversibly oxidised μ-oxo dimer Fe–O–Fe. The protein prevents this simply by burying each haem in its own hydrophobic pocket so that no two irons can meet.

Collman’s picket-fence porphyrins reproduce this synthetically: bulky amide “pickets” on one face of the porphyrin create a protected cavity, a hindered base occupies the other axial site, and the resulting complex binds O₂ reversibly in ordinary solution. It is the clean demonstration that the protein contributes steric protection and a hydrophobic environment, not any special electronic magic — the coordination chemistry is doing the work.

On oxygen binding and cooperativity.

  • M. F. Perutz, ‘Stereochemistry of cooperative effects in haemoglobin’, Nature, 1970 — the structural origin of the T→R transition. Volume and page numbers to be confirmed.
Med
Explain, in terms of crystal field theory, why the iron in deoxyhaemoglobin cannot fit into the porphyrin hole but the iron in oxyhaemoglobin can.
Show solution
Both are Fe(II), d⁶. In deoxy the field is weak (only five ligands, and the sixth position empty), so the complex is high-spin: t2g⁴ eg². The two eg electrons occupy orbitals (dx²−y², d) that point directly at the ligands and are strongly antibonding, so the metal–nitrogen distances are long and the effective ionic radius is large — larger than the porphyrin N₄ hole. The iron is therefore displaced towards the proximal histidine and the ring domes. Binding O₂ adds a sixth, strong-field π-acceptor ligand; Δo now exceeds the pairing energy P, so the complex becomes low-spin t2g⁶. With eg empty there is no antibonding population, the Fe–N distances contract, the effective radius falls, and the ion now fits the hole and moves into the plane. This is exactly the Δo vs P criterion from Part 3, applied to a protein.
Hard
A haemoglobin variant is engineered in which the proximal histidine is replaced by a residue that cannot coordinate. Predict qualitatively what happens to (a) O₂ binding and (b) cooperativity.
Show solution
(a) O₂ binding. The iron loses its permanent axial ligand. Two things follow: the sixth-site chemistry is no longer confined to one face (the vacant fifth site can be filled by water, by another donor, or by a second haem), and the ligand field on the proximal side is much weaker, which destabilises the low-spin oxygenated form. Reversible binding is likely to be lost or badly degraded, and irreversible oxidation to an Fe(III) form becomes far more probable. (b) Cooperativity. Essentially abolished. The proximal histidine is the mechanical linkage that converts the iron’s in-plane movement into a helix movement and hence into a signal at the subunit interfaces. Break that linkage and the four haems become four independent binding sites: the curve reverts towards hyperbolic and n falls towards 1. Note that the four subunits are still physically joined — cooperativity fails not because the protein falls apart but because the signal-transmission path is severed.
Hard
At the steepest part of its binding curve haemoglobin has a Hill coefficient of about 2.8. Calculate the ratio of pO₂ needed to go from 10% to 90% saturation, and compare with a non-cooperative carrier (n = 1).
Show solution
Rearrange the Hill equation: Y/(1−Y) = (p/P50)n. At Y = 0.9, Y/(1−Y) = 9; at Y = 0.1, it is 1/9. So (p90/p10)n = 9/(1/9) = 81, giving p90/p10 = 811/n. For n = 2.8: 811/2.8 = exp[(ln 81)/2.8] = exp(4.394/2.8) = exp(1.569) ≈ 4.8. For n = 1: 811 = 81. Interpretation. A cooperative carrier goes from nearly empty to nearly full over a fivefold change in oxygen pressure; a non-cooperative one needs an eighty-onefold change. Since lung and tissue pO₂ differ by only about a factor of five, cooperativity is not a refinement — it is the difference between a carrier that works and one that does not. This calculation is a standard CSIR-NET numerical.

Read the rest of Part 9

The remaining 11 sections of this part — L.2 Other metalloproteins read as coordination compounds, L.3 Metallodrugs — cisplatin, contrast agents and chelation therapy, L.4 The elementary steps, with the electron count tracked… — and all nine parts of Coordination Chemistry are part of ChemVidya Full Access, along with the other books, 55 Study Notes and 6,000+ practice questions.

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