IChO 2024 · Theory problem 9

Porphyrin complexes

Part A. Metals in petroleum

Saudi Arabian petroleum, primarily hydrocarbons, contains elements like vanadium in porphyrin complexes, hinting at its biological origins.

Vanadium complex B is presumably formed in petroleum from chlorophyll according to the following scheme:

Figure 1
Step Name
1) Demetallization
2) Hydrolysis
3) Decarbomethoxylation
4) Reduction
5) Aromatization
6) Decarboxylation
7) Metal chelation

9.1 Assign the molecular formulae a–e (given in the answer sheet) to the intermediates I–V. 4.0 pt

Figure 2

9.2 Draw the structures of intermediate A and vanadium complex B using the pre-drawn framework of porphyrin. (If you need a new template to redraw your structure(s), please ask the supervisor.) 7.0 pt

Figure 3

9.3 Indicate the oxidation state of vanadium in B. 1.0 pt

Part B. Porphyrin non-metal complexes

Porphyrins are known to form chelate-type complexes not only with metals but also with some non-metals, for example, with boron and phosphorus. Octaethylporphyrin (\(\ce{H2OEP}\)) is often used to model natural porphyrins and to study porphyrin complexes. It has the following structure and can be represented as:

Figure 4

To produce dicationic planar boron porphyrin complex \(\ce{[X3]^2+}\), either \(\ce{BCl3}\) or another binary compound X1 can be used as a precursor:

Figure 5

9.4 Draw the structures of X1, intermediate complex X2 and the product \(\ce{[X3]^2+}\). 3.0 pt

Figure 6

Phosphorus forms porphyrin cationic complexes \(\ce{[Y1]^{n+}}\), \(\ce{[Y2]^{n+}}\), and \(\ce{[Y3]^{n+}}\). The complex \(\ce{[Y3]^{n+}}\) has one less plane of symmetry compared to \(\ce{[Y1]^{n+}}\) and \(\ce{[Y2]^{n+}}\) (ignore ethyl substituents in OEP ligand when considering symmetry). Pyridine (Py) is used here as a basic solvent, “Hal” is one of the halogens: F, Cl, or Br:

Figure 7

9.5 Indicate: a) the charge “\(n+\)” of phosphorus porphyrin complexes; b) the number of planes of symmetry \(N(\sigma)\) that complex \(\ce{[Y1]^{n+}}\) has. 2.0 pt

Figure 8

9.6 Draw the structures of \(\ce{[Y1]^{n+}}\), \(\ce{[Y2]^{n+}}\), and \(\ce{[Y3]^{n+}}\). 4.0 pt

Figure 9

Part C. Porphyrin sandwiches

Porphyrin sandwich-type complexes form when there are several porphyrin rings aligned relative to each other. Examples are \(\ce{Zr(OEP)2}\), \(\ce{Eu2(OEP)3}\), \(\ce{Bi2(OEP)2Cl2}\) that have at least 3 planes of symmetry each. Note that the hole size in the OEP ligand is ca. \(4.0\ \text{Å}\), and the average metal–N distances in these complexes are \(2.4,\ 2.5,\ 2.3\ \text{Å}\), respectively.

9.7 Suggest the structures of \(\ce{Zr(OEP)2}\), \(\ce{Eu2(OEP)3}\), and \(\ce{Bi2(OEP)2Cl2}\). You may use a simplified representation of the OEP ligand. 3.0 pt

Figure 10

Another type of porphyrin sandwich-type complex is metal–porphyrin dimers with single or multiple metal–metal bonds. For example, \(\ce{[Ru(OEP)]2}\) has a Ru–Ru double bond (considering only d-electrons). Below is the molecular orbital diagram for an eclipsed dimer of the type \(\ce{[M(porphyrin)]2}\) (interaction between \(\mathrm d_{x^2-y^2}\) is not considered as these orbitals lie high in energy):

Figure 11

9.8 Draw the \(\sigma(\mathrm d_{z^2}+\mathrm d_{z^2})\), \(\pi(\mathrm d_{xz}+\mathrm d_{xz})\), \(\pi(\mathrm d_{yz}+\mathrm d_{yz})\), and \(\delta(\mathrm d_{xy}+\mathrm d_{xy})\) bonding orbitals according to the specified coordinate system. 4.0 pt

Figure 12

9.9 Calculate the metal–metal bond order in the following complexes in eclipsed conformation: \(\ce{[Mo(OEP)]2}\), \(\ce{[Ir(OEP)]2}\), \(\ce{[Re(OEP)]2^{+}}\). 3.0 pt

Figure 13

The eclipsed conformation is generally preferred for metal–porphyrin dimers with 7, 8, and 9 d-electrons. Otherwise, these complexes are more stable in a staggered conformation, where the porphyrin rings are rotated relative to each other by \(45^\circ\).

9.10 Select how the splitting (i.e., the energy between antibonding and bonding orbitals) changes for each type of orbital interaction during the conversion of eclipsed conformation into staggered:

a) increases;
b) remains unchanged;
c) decreases (not to zero);
d) decreases to zero.

4.0 pt

Figure 14

Official English text of the problem.

Answers

Here \(\ce{L^{2-}}\) denotes the octaethylporphyrinato ligand, \(\ce{OEP^{2-}}\). Full structural descriptions and SMILES follow in the Solution.

  • 9.1: \(\boxed{\mathrm{I=b,\ II=e,\ III=c,\ IV=d,\ V=a}}\).
  • 9.2: A is \(13^1\)-deoxopyropheophorbide a: the phytyl ester is hydrolysed, the methoxycarbonyl substituent is removed, and the ring ketone becomes a methylene group. B is its aromatized, decarboxylated oxovanadium(IV) complex; the vinyl group remains unchanged.
  • 9.3: \(\boxed{+4}\).
  • 9.4: X1 is diboron tetrachloride, \(\ce{Cl2B-BCl2}\). X2 is \(\ce{L(BCl2)2}\), with two adjacent N donors per B and no B–B bond. \(\ce{[X3]^2+}\) is planar \(\ce{[LB2]^2+}\), retaining a B–B single bond.
  • 9.5: a) \(\boxed{n+=1+}\); b) \(\boxed{N(\sigma)=5}\).
  • 9.6: \(\boxed{\ce{[Y1]+ = [P(L)Cl2]+}}\), \(\boxed{\ce{[Y2]+ = [P(L)(OH)2]+}}\), and \(\boxed{\ce{[Y3]+ = [P(L)(OH)F]+}}\). Each has two mutually trans axial ligands.
  • 9.7: \(\ce{ZrL2}\): an eight-coordinate double-decker sandwich. \(\ce{Eu2L3}\): a triple-decker sandwich with one eight-coordinate Eu between each pair of adjacent decks. \(\ce{Bi2L2Cl2}\): \(\ce{LBi(\mu-Cl)2BiL}\), with two bridging chlorides.
  • 9.8: Constructive end-on \(\mathrm d_{z^2}\) overlap gives \(\sigma\); constructive side-on overlap in the \(xz\) and \(yz\) planes gives the two \(\pi\) orbitals; four-lobe face-to-face \(\mathrm d_{xy}\) overlap gives \(\delta\).
  • 9.9: \(\boxed{4,\ 1,\ 3.5}\), respectively.
  • 9.10: \(\boxed{\sigma/\sigma^*:b;\quad \pi/\pi^*:b;\quad \delta/\delta^*:d}\).

Solution

SMILES below specify connectivity. Dative bonds are written using ->; coordination geometries are described separately. Formal charges appearing in a SMILES representation are not necessarily oxidation states. Stereochemistry not specified by the supplied drawings is left unspecified.

9.1 — Assignment of molecular formulae

The starting chlorophyll has formula \(\ce{C55H72MgN4O5}\).

Intermediate Formula Choice Identification
I \(\ce{C55H74N4O5}\) b Pheophytin a
II \(\ce{C35H36N4O5}\) e Pheophorbide a
III \(\ce{C33H34N4O3}\) c Pyropheophorbide a
IV \(\ce{C33H34N4O2}\) d Aromatized A
V \(\ce{C32H34N4}\) a Decarboxylated IV, the metal-free ligand of B

The atom balances explain every assignment:

  1. Demetallization: replace the central \(\ce{Mg^2+}\) by two N-bound hydrogens. \[ \ce{C55H72MgN4O5 + 2H+ -> C55H74N4O5 + Mg^2+} \]

  2. Hydrolysis: cleave the phytyl ester, replacing \(\ce{-OC20H39}\) by \(\ce{-OH}\). \[ \ce{C55H74N4O5 + H2O -> C35H36N4O5 + C20H40O} \]

  3. Decarbomethoxylation: replace the ring’s \(\ce{-CO2CH3}\) substituent by H. An atom-balanced hydrolysis/decarboxylation description is \[ \ce{C35H36N4O5 + H2O -> C33H34N4O3 + CO2 + CH3OH}. \]

  4. Reduction: convert the remaining ring ketone into a methylene group. \[ \ce{C33H34N4O3 + 2H2 -> C33H36N4O2 + H2O} \]

  5. Aromatization: remove two hydrogens from the reduced pyrrole ring. \[ \ce{C33H36N4O2 -> C33H34N4O2 + H2} \]

  6. Decarboxylation: \[ \ce{C33H34N4O2 -> C32H34N4 + CO2} \]

  7. Vanadyl chelation: \[ \ce{C32H34N4 + VO^2+ -> C32H32N4OV + 2H+} \]

The \(\ce{H2}\) terms in steps 4–5 express the net redox balances; the actual reducing and oxidizing reagents are not specified.

The corresponding structural SMILES are:

  • I, pheophytin a, using the conventional phytyl connectivity:
    C=CC1=C(C)c2cc3[nH]c(c4c5nc(cc6[nH]c(cc1n2)c(C)c6CC)C(C)=C5C(=O)C4C(=O)OC)C(CCC(=O)OCC=C(C)CCCC(C)CCCC(C)CCCC(C)C)C3C
  • II, pheophorbide a:
    C=CC1=C(C)c2cc3[nH]c(c4c5nc(cc6[nH]c(cc1n2)c(C)c6CC)C(C)=C5C(=O)C4C(=O)OC)C(CCC(=O)O)C3C
  • III, pyropheophorbide a:
    C=CC1=C(C)c2cc3[nH]c(c4c5nc(cc6[nH]c(cc1n2)c(C)c6CC)C(C)=C5C(=O)C4)C(CCC(=O)O)C3C
  • IV, the aromatized derivative of A:
    C=CC1=C(C)c2cc3[nH]c(c4c5nc(cc6[nH]c(cc1n2)c(C)c6CC)C(C)=C5CC4)c(CCC(=O)O)c3C
  • V, the decarboxylated derivative of IV:
    C=CC1=C(C)c2cc3[nH]c(c4c5nc(cc6[nH]c(cc1n2)c(C)c6CC)C(C)=C5CC4)c(CC)c3C

9.2 — Structures of A and B

A: \(13^1\)-deoxopyropheophorbide a, \(\ce{C33H36N4O2}\).

SMILES:
C=CC1=C(C)c2cc3[nH]c(c4c5nc(cc6[nH]c(cc1n2)c(C)c6CC)C(C)=C5CC4)C(CCC(=O)O)C3C

To draw A, retain the orientation and peripheral substitution of the starting chlorophyll, making these changes:

  • Remove Mg and place H on the upper-right and lower-left nitrogens, giving a free-base chlorin.
  • Change the lower-left propionate ester into \(\ce{-CH2CH2COOH}\).
  • Remove the methoxycarbonyl substituent from the fused five-membered ring, replacing it by H.
  • Replace that ring’s \(\ce{C=O}\) group by \(\ce{CH2}\).
  • Retain the upper-left vinyl group, the upper-right ethyl group, all four methyl groups, and the saturated bond between the two peripheral carbons of the lower-left pyrrole ring.

The fused five-membered ring consequently contains a \(\ce{-CH2-CH2-}\) bridge connecting the same two macrocycle positions as in the starting structure.

The formula fixes the extent of reduction: replacing \(\ce{C=O}\) by \(\ce{CH2}\) accounts exactly for the change from \(\ce{C33H34N4O3}\) to \(\ce{C33H36N4O2}\). Additional hydrogenation of the vinyl group would give too many hydrogens.

B: the oxovanadium(IV) complex of V, \(\ce{C32H32N4OV}\).

SMILES:
C=CC1=C(C)c2cc3c(C)c(CC)c4c5c6[n]7->[V+2]8(=[O])(<-[n]2c1cc1c(C)c(CC)c(cc7C(C)=C6CC5)[n-]->81)<-[n-]34

To obtain its drawing from A:

  1. Introduce a double bond between the two saturated peripheral carbons of the lower-left pyrrole ring, producing the porphyrin conjugation pattern.
  2. Replace \(\ce{-CH2CH2COOH}\) by \(\ce{-CH2CH3}\).
  3. Remove both inner N–H hydrogens and coordinate all four nitrogens to V.
  4. Add a terminal axial oxo ligand, drawn \(\ce{V=O}\).

The coordination geometry is approximately square pyramidal: four porphyrin nitrogens form the basal donor set and oxygen occupies the axial position. The peripheral vinyl group remains unchanged.

9.3 — Oxidation state of vanadium

The porphyrinato ligand has charge \(-2\), and the oxo ligand has charge \(-2\). Since B is neutral,

\[ x-2-2=0, \qquad \boxed{x=+4}. \]

Thus the metal is vanadium(IV).

9.4 — Boron complexes

Let \(\ce{L^{2-}}\) denote octaethylporphyrinato, whose SMILES is:

CCC1=C(CC)c2cc3[n-]c(cc4nc(cc5[n-]c(cc1n2)c(CC)c5CC)C(CC)=C4CC)c(CC)c3CC

The two inner N–H groups are first deprotonated:

\[ \ce{H2L + 2LiN(SiMe3)2 -> Li2L + 2HN(SiMe3)2}. \]

X1 is diboron tetrachloride, \(\ce{B2Cl4}\).

SMILES: ClB(Cl)B(Cl)Cl

Draw a B–B single bond, with two terminal chlorides on each B. Its reaction with the dilithium ligand is

\[ \ce{Li2L + B2Cl4 -> L(BCl)2 + 2LiCl}. \]

This directly gives the B–B-bonded dichloro complex shown in the problem.

X2 is the bis(dichloroboron) octaethylporphyrin complex, \(\ce{L(BCl2)2}\).

SMILES:
CCC1=C(CC)C2=[N+]3C1=Cc1c(CC)c(CC)c(n1[B-]3(Cl)Cl)C=C1C(CC)=C(CC)C3=[N+]1[B-](Cl)(Cl)n1c(c(CC)c(CC)c1=C3)=C2

Draw two B centres inside the macrocycle. One binds the two adjacent nitrogens on the left, and the other binds the two adjacent nitrogens on the right. Each B also bears two chlorides, giving approximately tetrahedral coordination. There is no B–B bond in X2.

Formation and reduction are balanced by

\[ \ce{Li2L + 2BCl3 -> L(BCl2)2 + 2LiCl}, \]

\[ \ce{L(BCl2)2 + 2M -> L(BCl)2 + 2MCl}, \qquad \ce{M = Na \text{ or } K}. \]

The latter reaction reduces each B from oxidation state \(+3\) to \(+2\), removes one chloride from each, and forms the B–B bond.

\(\ce{[X3]^2+}\) is the B–B-bonded diboron octaethylporphyrin dication, \(\ce{[LB2]^2+}\).

SMILES:
CCC1=C(CC)C2=[N+]3B4B5n6c(c(CC)c(CC)c6=CC6=[N+]5C(=Cc5c(CC)c(CC)c(n54)C=C13)C(CC)=C6CC)=C2

Draw the same adjacent N-pair chelation, remove both remaining chlorides, and retain the B–B single bond. The two B centres and four N donors are coplanar; each B is three-coordinate.

Writing \(\ce{W-}\) for the weakly coordinating borate anion supplied in the problem,

\[ \ce{L(BCl)2 + 2NaW -> [LB2]W2 + 2NaCl}. \]

Chloride abstraction is not a further reduction, so it does not increase the B–B bond order. Charge balance also gives

\[ 2(+2)+(-2)=+2 \]

for the diboron porphyrin dication.

9.5 — Charge and symmetry of the phosphorus complexes

Phosphorus is in oxidation state \(+5\). With a dianionic porphyrinato ligand and two singly charged axial ligands,

\[ q=+5-2-1-1=+1. \]

Therefore,

\[ \boxed{n+=1+}. \]

For Y1, the two identical axial chlorides and the idealized unsubstituted porphyrin framework give \(D_{4h}\) symmetry. Its mirror planes are:

  • the porphyrin plane, \(\sigma_h\);
  • two vertical planes passing through opposite N donors;
  • two further vertical planes bisecting those N–P–N directions.

Hence

\[ \boxed{N(\sigma)=1+2+2=5}. \]

9.6 — Phosphorus structures

All three cations contain P coordinated to the four porphyrin nitrogens, with two additional ligands on opposite sides of the porphyrin plane.

Species Identification Axial ligands
\(\ce{[Y1]+}\) Dichlorido(octaethylporphyrinato)phosphorus(V) Cl and Cl
\(\ce{[Y2]+}\) Dihydroxido(octaethylporphyrinato)phosphorus(V) OH and OH
\(\ce{[Y3]+}\) Fluorido(hydroxido)(octaethylporphyrinato)phosphorus(V) F and OH

Their SMILES, using covalent, charge-separated representations of the P–N bonds, are:

  • \(\ce{[Y1]+}\):
    CCC1=C(CC)C2=[N+]3C1=Cc1c(CC)c(CC)c4n1[P-]31(Cl)(Cl)n3c(c(CC)c(CC)c3=CC3=[N+]1C(=C4)C(CC)=C3CC)=C2
  • \(\ce{[Y2]+}\):
    CCC1=C(CC)C2=[N+]3C1=Cc1c(CC)c(CC)c4n1[P-]31(O)(O)n3c(c(CC)c(CC)c3=CC3=[N+]1C(=C4)C(CC)=C3CC)=C2
  • \(\ce{[Y3]+}\):
    CCC1=C(CC)C2=[N+]3C1=Cc1c(CC)c(CC)c4n1[P-]31(O)(F)n3c(c(CC)c(CC)c3=CC3=[N+]1C(=C4)C(CC)=C3CC)=C2

The initial oxidative insertion can be balanced as

\[ \ce{H2L + PCl3 + Cl2 -> [P(L)Cl2]Cl + 2HCl}. \]

Pyridine acts as a base and captures the acid.

Hydrolysis replaces the axial chlorides with hydroxido ligands. A net balance including carbonate is

\[ \ce{[P(L)Cl2]+ + Na2CO3 + H2O -> [P(L)(OH)2]+ + 2NaCl + CO2}. \]

The acid workup establishes the protonated hydroxido form shown as Y2.

The decisive information for Y3 is its reversible conversion, under base, to the neutral oxo–halido complex. Thus Y3 must retain one OH group:

\[ \ce{[P(L)(OH)Hal]+ + OH- -> P(L)(O)Hal + H2O}. \]

Since the last substitution reagent is \(\ce{KF}\), \(\ce{Hal = F}\). An overall atom-balanced description of the oxalyl bromide/fluoride sequence is

\[ \ce{[P(L)(OH)2]+ + (COBr)2 + KF -> [P(L)(OH)F]+ + KBr + HBr + CO + CO2}. \]

This is a net balance, not a claim about every elementary step.

The final acid–base interconversion is therefore

\[ \ce{[P(L)(OH)F]+ + OH- <=> P(L)(O)F + H2O}. \]

The neutral product is fluorido(oxo)(octaethylporphyrinato)phosphorus(V). Its charge-separated SMILES is:

CCC1=C(CC)C2=[N+]3C1=Cc1c(CC)c(CC)c4n1[P-]31([O-])(F)n3c(c(CC)c(CC)c3=CC3=[N+]1C(=C4)C(CC)=C3CC)=C2

In the requested drawing, represent its axial oxygen as \(\ce{P=O}\).

Finally, different axial groups in Y3 remove the horizontal mirror plane that would exchange them. The four vertical mirror planes remain in the idealized geometry, explaining the stated loss of exactly one plane.

9.7 — Sandwich structures

The approximate cavity radius is

\[ r\approx \frac{4.0}{2}=2.0\ \text{Å}. \]

All three supplied metal–N distances exceed this radius, so the metals must lie outside the individual N-donor planes in the proposed structures. For an ideal planar, centred N-donor square,

\[ h=\sqrt{d_{\mathrm{M-N}}^2-r^2}. \]

Metal \(d_{\mathrm{M-N}}\) Approximate height \(h\) above an N-donor plane
Zr \(2.4\ \text{Å}\) \(1.3\ \text{Å}\)
Eu \(2.5\ \text{Å}\) \(1.5\ \text{Å}\)
Bi \(2.3\ \text{Å}\) \(1.1\ \text{Å}\)

These are geometric estimates using the approximate cavity size.

\(\ce{Zr(OEP)2}\): bis(octaethylporphyrinato)zirconium(IV).

Draw two parallel, cofacial porphyrin decks with Zr between their centres. Zr binds all four N donors of each deck, giving coordination number eight. A \(45^\circ\) offset between the N-donor squares gives a square-antiprismatic arrangement with four vertical mirror planes.

SMILES:
CCC1=C(CC)c2cc3c(CC)c(CC)c4cc5[n]6->[Zr+4]789%10(<-[n]2c1cc1c(CC)c(CC)c(cc6C(CC)=C5CC)[n-]->71)(<-[n]1c2cc5c(CC)c(CC)c(cc6[n]->8c(cc7c(CC)c(CC)c(cc1C(CC)=C2CC)[n-]->97)C(CC)=C6CC)[n-]->%105)<-[n-]34

\(\ce{Eu2(OEP)3}\): tris(octaethylporphyrinato)dieuropium(III).

Draw three parallel decks in the sequence

\[ \text{porphyrin–Eu–porphyrin–Eu–porphyrin}. \]

Each Eu binds the four nitrogens of each adjacent deck. The middle porphyrin coordinates both metals, one on each face. An arrangement with aligned outer decks and a middle deck rotated by \(45^\circ\) satisfies the required symmetry.

SMILES:
CCC1=C(CC)c2cc3c(CC)c(CC)c4cc5[n]6->[Eu+3]789%10(<-[n]2c1cc1c(CC)c(CC)c(cc6C(CC)=C5CC)[n-]->71)(<-[n-]34)<-[n]12->[Eu+3]34567(<-[n]%11c%12cc%13c(CC)c(CC)c(cc%14[n]->3c(cc3c(CC)c(CC)c(cc%11C(CC)=C%12CC)[n-]->43)C(CC)=C%14CC)[n-]->5%13)<-[n]->83c4cc5c(CC)c(CC)c(cc1C(CC)=C(CC)c2cc1c(CC)c(CC)c(cc3C(CC)=C4CC)[n-]->9->61)[n-]->%10->75

\(\ce{Bi2(OEP)2Cl2}\): di-\(\mu\)-chlorido-bis[(octaethylporphyrinato)bismuth(III)].

Draw two parallel porphyrins, each binding one Bi displaced toward the space between the decks. Join the two Bi atoms through two bridging chlorides, giving a central \(\ce{Bi2Cl2}\) four-membered ring:

\[ \ce{LBi(\mu-Cl)2BiL}. \]

Each Bi has four N donors and two chloride contacts. A symmetric arrangement with eclipsed decks has three mirror planes: the plane of the \(\ce{Bi2Cl2}\) ring, a perpendicular plane through the Bi–Bi axis, and the plane midway between the porphyrin decks. No Bi–Bi bond is required.

SMILES:
CCC1=C(CC)c2cc3c(CC)c(CC)c4cc5[n]6->[Bi+3]78(<-[Cl-]->[Bi+3]9%10%11(<-[Cl-]->7)<-[n]7c%12cc%13c(CC)c(CC)c(cc%14[n]->9c(cc9c(CC)c(CC)c(cc7C(CC)=C%12CC)[n-]->%109)C(CC)=C%14CC)[n-]->%11%13)(<-[n]2c1cc1c(CC)c(CC)c(cc6C(CC)=C5CC)[n-]->81)<-[n-]34

The oxidation-state balances are, respectively,

\[ +4+2(-2)=0,\qquad 2(+3)+3(-2)=0,\qquad 2(+3)+2(-2)+2(-1)=0. \]

The SMILES encode coordination connectivity; they do not encode the deck rotations just described.

9.8 — Bonding orbitals

Take the metal–metal axis as \(z\), pointing right in the printed coordinate system. Use two colours or \(+\)/\(-\) signs to denote orbital phase. In every bonding combination, overlapping lobes in the internuclear region must have the same phase.

Orbital Precise drawing description Nodal planes containing the M–M axis
\(\sigma(\mathrm d_{z^2}+\mathrm d_{z^2})\) On each metal, draw two lobes along \(z\) and an opposite-phase torus around the metal in the \(xy\) plane. The two facing axial lobes have the same phase and overlap end-on. None
\(\pi(\mathrm d_{xz}+\mathrm d_{xz})\) Draw four lobes on each metal in the \(xz\) plane, between the \(x\) and \(z\) axes. Match phases across the internuclear region both above and below the \(z\) axis; the upper and lower overlap regions have opposite phases. \(yz\)
\(\pi(\mathrm d_{yz}+\mathrm d_{yz})\) Draw the analogous four-lobed orbitals in the \(yz\) plane. Match phases across the internuclear region on both sides along \(y\). \(xz\)
\(\delta(\mathrm d_{xy}+\mathrm d_{xy})\) Draw four lobes around each metal in its \(xy\) plane, between the \(x\) and \(y\) axes. Corresponding lobes on the two metals overlap face-to-face with matching phases. Adjacent quadrants have opposite phases. \(xz\) and \(yz\)

The overall sign of an atomic orbital is arbitrary. For the \(\pi\) combinations, choose the relative atomic-orbital phases so that the “+” combinations printed in the question are constructive.

9.9 — Metal–metal bond orders

Each porphyrinato ligand contributes charge \(-2\). The supplied MO ordering is

\[ \sigma<\pi<\delta<\delta^*<\pi^*<\sigma^*, \]

where each \(\pi\) level contains two orbitals. Bond order is

\[ \mathrm{BO}=\frac{N_{\mathrm{bonding}}-N_{\mathrm{antibonding}}}{2}. \]

Complex Metal oxidation states Total d-electrons Occupancy Bond order
\(\ce{[Mo(OEP)]2}\) Two Mo(II) \(2(6-2)=8\) \(\sigma^2\pi^4\delta^2\) \((8-0)/2=\boxed{4}\)
\(\ce{[Ir(OEP)]2}\) Two Ir(II) \(2(9-2)=14\) \(\sigma^2\pi^4\delta^2(\delta^*)^2(\pi^*)^4\) \((8-6)/2=\boxed{1}\)
\(\ce{[Re(OEP)]2^{+}}\) Oxidation-state sum \(+5\) \(2(7)-5=9\) \(\sigma^2\pi^4\delta^2(\delta^*)^1\) \((8-1)/2=\boxed{3.5}\)

For the Re cation, the average oxidation state is \(+2.5\); assigning the positive charge to a particular metal is unnecessary.

9.10 — Effect of a \(45^\circ\) rotation

\(\sigma/\sigma^*\): b, remains unchanged.

The \(\mathrm d_{z^2}\) orbital is rotationally symmetric about the metal–metal axis. Rotating a porphyrin around \(z\) therefore leaves the \(\sigma\) interaction unchanged.

\(\pi/\pi^*\): b, remains unchanged.

Both members of the degenerate \(\{\mathrm d_{xz},\mathrm d_{yz}\}\) pair must be considered. Rotation changes their basis according to

\[ \begin{pmatrix} \mathrm d'_{xz}\\ \mathrm d'_{yz} \end{pmatrix} = \begin{pmatrix} \cos\theta&\sin\theta\\ -\sin\theta&\cos\theta \end{pmatrix} \begin{pmatrix} \mathrm d_{xz}\\ \mathrm d_{yz} \end{pmatrix}. \]

This mixes the two orbitals without changing the available \(\pi\) subspace or its bonding–antibonding splitting. Although an individual same-label overlap decreases, cross-overlap with the other \(\pi\) orbital compensates for it.

\(\delta/\delta^*\): d, decreases to zero.

For the retained \(\mathrm d_{xy}\)-based interaction, the overlap varies as

\[ S_\delta(\theta)\propto\cos(2\theta). \]

Thus, at \(45^\circ\),

\[ S_\delta(45^\circ)\propto\cos90^\circ=0. \]

Positive and negative overlap contributions cancel, so the \(\delta\) and \(\delta^*\) levels become nonbonding and degenerate in the stated model.

This also explains why 7, 8, or 9 d-electrons favour eclipsing: those occupancies give a net \(\delta\)-bonding contribution, which is lost on staggering.

The solution was checked against the official solution.