1. Electron Counting understanding structure and reactivity
Cr = 6 2*Bz = 12
+ ¾¾
e-count 18
tot. charge 0 ¾¾
2*Bz -
- ¾¾
ox state 0
Rh = 9 3* P® = 6
Cl¾ = 1
+ ¾¾
e-count 16
tot. charge 0 ¾¾
3*P® -
Cl¾ -1
- ¾¾
ox state +1
Peter H.M. Budzelaar

2. Why count electrons ?
Basic tool for understanding structure and reactivity.
Simple extension of Lewis structure rules.
Counting should be “automatic”.
Not always unambiguous Þ don’t just follow the rules, understand them!
Organometallic Chemistry

3. Predicting reactivity
Organometallic Chemistry

4. Predicting reactivity
Most likely associative:
16-e PdII
16-e PdII
18-e PdII
Organometallic Chemistry

5. Predicting reactivity
Organometallic Chemistry

6. Predicting reactivity
Almost certainly dissociative:
16-e Cr(0)
18-e Cr(0)
18-e Cr(0)
Organometallic Chemistry

7. The basis of electron counting
Every element has a certain number of valence orbitals:
1 { 1s } for H
4 { ns, 3´np } for main group elements
9 { ns, 3´np, 5´(n-1)d } for transition metals s px py pz
dxy
dxz
dyz
dx2-y2
dz2
Organometallic Chemistry

8. The basis of electron counting
Every orbital wants to be “used", i.e. contribute to binding an electron pair.
Therefore, every element wants to be surrounded by 1/4/9 electron pairs, or 2/8/18 electrons.
For main-group metals (8-e), this leads to the standard Lewis structure rules.
For transition metals, we get the 18-electron rule.
Structures which have this preferred count are called electron-precise.
Organometallic Chemistry

9. Compounds are not always electron-precise !
The strength of the preference for electron-precise structures depends on the position of the element in the periodic table.
For very electropositive main-group elements, electron count often determined by steric factors.
How many ligands "fit" around the metal?
"Orbitals don't matter" for ionic compounds
Main-group elements of intermediate electronegativity (C, B) have a strong preference for 8-e structures.
For the heavier, electronegative main-group elements, there is the usual ambiguity in writing Lewis structures (SO42-: 8-e or 12-e?). Stable, truly hypervalent molecules (for which every Lewis structure has > 8-e) are not that common (SF6, PF5). Structures with < 8-e are very rare.
Organometallic Chemistry

10. Compounds are not always electron-precise !
The strength of the preference for electron-precise structures depends on the position of the element in the periodic table.
For early transition metals, 18-e is often unattainable for steric reasons
The required number of ligands would not fit.
For later transition metals, 16-e is often quite stable
In particular for square-planar d8 complexes.
For open-shell complexes, every valence orbital wants to be used for at least one electron
More diverse possibilities, harder to predict.
Organometallic Chemistry

11. Prediction of stable complexes
Cp2Fe, ferrocene: 18-e
Very stable.
Behaves as an aromatic organic compound in e.g. Friedel-Crafts acylation.
Cp2Co, cobaltocene: 19-e
Strong reductant, reacts with air.
Cation (Cp2Co+) is very stable.
Cp2Ni, nickelocene: 20-e
Chemically reactive, easily loses a Cp ring, reacts with air.
Organometallic Chemistry

12. If there are not enough electrons...
Structures with a lower than ideal electron count are called electron-deficient or coordinatively unsaturated.
They have unused (empty) valence orbitals.
This makes them electrophilic, i.e. susceptible to attack by nucleophiles.
Some unsaturated compounds are so reactive they will attack hydrocarbons, or bind noble gases.
Organometallic Chemistry

13. Reactivity of electron-deficient compounds
18-e Fe(0) unreactive
16-e Fe(0) very reactive
18-e Fe(0)
Organometallic Chemistry

14. If there are too many electrons...
"Too many electrons" means there are fewer net covalent bonds than one would think.
Since not enough valence orbitals available for these electrons.
An ionic model is required to explain part of the bonding.
The "extra" bonds are relatively weak.
Excess-electron compounds are relatively rare, especially for transition metals.
Often generated by reduction (= addition of electrons).
Organometallic Chemistry

15. Where are the electrons ?
Electrons around a metal can be in metal-ligand bonding orbitals or in metal-centered lone pairs.
Metal-centered orbitals are fairly high in energy.
A metal atom with a lone pair is a s-donor (nucleophile).
susceptible to electrophilic attack.
Organometallic Chemistry

16. Metal-centered lone pairs
Basicity of Cp2WH2 comparable to that of ammonia!
18-e WIV
18-e WVI
Organometallic Chemistry

17. How do you count ? "Covalent" count:
Number of valence electrons of central atom.
from periodic table
Correct for charge, if any.
but only if the charge belongs to that atom!
Count 1 e for every covalent bond to another atom.
Count 2 e for every dative bond from another atom.
no electrons for dative bonds to another atom!
Delocalized carbon fragments: usually 1 e per C
Three- and four-center bonds need special treatment.
Add everything.
There are alternative counting methods (e.g. "ionic count"). Apart from three- and four-center bonding cases, they should always arrive at the same count. We will use the "covalent" count in this course.
Organometallic Chemistry

18. Starting simple... N = 5 3* H¾ = 3 H = 1 H¾ = 1 + ¾¾ e-count 8 + ¾¾
N has a lone pair. Nucleophilic!
C = 4 4* H¾ = 4
+ ¾¾
e-count 8
C = 4 2* H¾ = 2
2* C¾ = 2
+ ¾¾
e-count 8
A double bond counts as two covalent bonds.
Organometallic Chemistry

19. Predicting reactivity
C = 4 + chg = -1
3* H¾ = 3
+ ¾¾
e-count 6
C = 4 2* H¾ = 2
+ ¾¾
e-count 6
"Singlet carbene". Unstable. Sensitive to nucleophiles
(empty orbital)
and electrophiles (lone pair).
Highly reactive, electrophilic.
C = 4 - chg = +1
3* H¾ = 3
+ ¾¾
e-count 8
"Triplet carbene". Extremely reactive as radical, not especially for nucleophiles or electrophiles.
Saturated, but nucleophilic.
Organometallic Chemistry

20. When is a line not a line ? is C = 4 3* H¾ = 3 C¾ = 1 + ¾¾ e-count 8or
B = 3 - chg = +1
3* H¾ = 3
N¾ = 1
+ ¾¾
e-count 8
N = 5 + chg = -1
3* H¾ = 3
B¾ = 1
+ ¾¾
e-count 8
B = 3 3* H¾ = 3
N® = 2
+ ¾¾
e-count 8
N = 5 3* H¾ = 3
+ ¾¾
e-count 8
Organometallic Chemistry

21. "bond" to the allyl fragment
Covalent or dative ?
How do you know a fragment forms a covalent or a dative bond?
Chemists are "sloppy" in writing structures. A "line" can mean a covalent bond, a dative bond, or even a part of a three-center two-electron bond.
Use analogies ("PPh3 is similar to NH3").
Rewrite the structure properly before you start counting.
1 e
2 e
3 e
covalent bond
dative bond
"bond" to the allyl fragment
Pd = 10 Cl¾ = 1
P® = 2
allyl = 3
+ ¾¾
e-count 16
Organometallic Chemistry

22. Handling 3c-2e and 4c-2e bonds
A 3c-2e bond can be regarded as a covalent bond "donating" its electron pair to a third atom.
Rewriting it this way makes counting easy.
B2H6 is often written as
But it cannot have 8 covalent bonds: there are only 12 valence electrons in the whole molecule!
The central B2H2 core is held together by two 3c-2e bonds:
Organometallic Chemistry

23. Handling 3c-2e and 4c-2e bonds
Rewrite bonding in terms of two BH3 monomers:
This is one of the few cases where Crabtree does things differently (for transition metals).
The method shown here is closer to the actual VB description of the bonding.
B = 3 3* H¾ = 3
BH® = 2
+ ¾¾
e-count 8
1 e
2 e
C2H6
BH3NH3
B2H6
Organometallic Chemistry

24. What kind of bridge bond do I have ?
A 3c-2e bond will only form when the central (bridging) atom does not have any lone pairs.
When lone pairs are available, they are preferred as donors.
A methyl group can form one more single bond. After that, it has no lone pairs, so the best it can do is share the Al-C
bonding electrons with a second Al:
Al = 3 3* Me¾ = 3
MeAl® = 2
+ ¾¾
e-count 8
After chlorine forms a single bond, it still has three lone pairs left. One is used to donate to the second Al:
Al = 3 3* Cl¾ = 3
Cl® = 2
+ ¾¾
e-count 8
Organometallic Chemistry

25. 3c-2e vs normal bridge bonds
The orbitals of a 3c-2e bond are bonding between all three of the atoms involved.
Therefore, Al2Me6 has a net Al-Al bonding interaction.
The orbitals involved in "normal" bridges are regular terminal-bridge bonding orbitals.
Thus, Al2Cl6 has strong Al-Cl bonds but no net Al-Al bonding.
Organometallic Chemistry

26. Handling charges
"Correct for charge, if any, but only if it belongs to that atom!"
How do you know where the charge belongs?
Eliminate all obvious places where a charge could belong, mostly hetero-atoms having unusual numbers of bonds.
What is left should belong to the metal...
Any alkyl-SO3 group would normally be anionic (c.f. CH3SO3-, the anion of CH3SO3H).
So the negative charge does not belong to the metal!
Rh = 9 CH2¾ = 1
3* CO® = 6
+ ¾¾
e-count 16
Organometallic Chemistry

27. Handling charges
Even overall neutral molecules could have "hidden" charges!
A boron atom with 4 bonds would be -1 (c.f. BH4-).
No other obvious centers of charge, so the Co must be +1.
Co = 9
+ chg = -1 3* P® = 6
CO® = 2
+ ¾¾
e-count 16
Organometallic Chemistry

28. A few excess-electron examples
PCl5
P would have 10 e, but only has 4 valence orbitals, so it cannot form more than 4 “net” P-Cl bonds. You can describe the bonding using ionic structures ("negative hyperconjugation").
Easy dissociation in PCl3 en Cl2. "PBr5" actually is PBr4+Br- !
SiF62-, SF6, IO65- and noble-gas halides can be described in a similar manner.
P = 5 + chg = -1
4* Cl¾ = 4
+ ¾¾
e-count 8
P = 5 5* Cl¾ = 5
+ ¾¾
e-count 10
Organometallic Chemistry

29. A few excess-electron examples
HF2-
H only has a single valence orbital, so it cannot form two covalent H-F bonds!
Write as FH·F-, mainly ion-dipole interaction.
This is just an extreme form of hydrogen bonding. Most other H-bonded molecules have less symmetric hydrogen bridges.
H = 1 - chg = +1
2* F¾ = 2
+ ¾¾
e-count 4
H = 1 1* F¾ = 1
+ ¾¾
e-count 2
Organometallic Chemistry

30. Does it look reasonable ?
Remember when counting:
Odd electron counts are rare.
In reactions you nearly always go from even to even (or odd to odd), and from n to n-2, n or n+2.
Electrons don’t just “appear” or “disappear”.
The optimal count is 2/8/18 e. 16-e also occurs frequently, other counts are much more rare.
Organometallic Chemistry

31. Electron-counting exercises
Me2Mg Pd(PMe3)4 MeReO3
ZnCl4 Pd(PMe3)3 OsO3(NPh)
ZrCl4 ZnMe42- OsO4(pyridine)
Co(CO)4- Mn(CO)5- Cr(CO)6
V(CO)6- V(CO)6 Zr(CO)64+
PdCl(PMe3)3 RhCl2(PMe3)2 Ni(PMe3)2Cl2
Ni(PMe3)Cl4 Ni(PMe3)Cl3
Organometallic Chemistry

32. Oxidation states
Most elements have a clear preference for certain oxidation states. These are determined by (a.o.) electronegativity and the number of valence electrons. Examples:
Li: nearly always +1. Has only 1 valence electron, so cannot go higher. Is very electropositive, so doesn’t want to go lower.
Cl: nearly always -1. Already has 7 valence electrons, so cannot go lower. Is very electronegative, so doesn’t want to go higher.
Organometallic Chemistry

33. Calculating oxidation states
Rewrite compound as if all bonds were fully ionic/dative, i.e. the electron pairs of each bond go to one end of the bond.
Which end? Use electronegativity to decide.
Ignore homonuclear covalent bonds
No unambiguous choice available
Usually end up with a unique set of charges
The "Lewis structure ambiguity" for hypervalent compounds does not cause problems here
Organometallic Chemistry

34. How do you calculate oxidation states ?
Start with the formal charge on the metal
See earlier discussion in "electron counting"
Ignore dative bonds
Ignore bonds between atoms of the same element
This one is a bit silly and produces counterintuitive results
Assign every covalent electron pair to the most electronegative element in the bond: this produces + and – charges
Usually + at the metal
Multiple bonds: multiple + and - charges
Add
Organometallic Chemistry

35. Examples - main group elements
no chg = 0
4* Cl-¾C+ = +4
+ ¾¾
ox st +4
- chg = -1
4* Cl-¾Al+ = +4
+ ¾¾
ox st +3
no chg = 0
2* Cl-¾C+ = +2
O2-=C2+ = +2
+ ¾¾
ox st +4
Organometallic Chemistry

36. Examples - transition metals
2- chg = -2
4* Cl-¾Pd+ = +4
+ ¾¾
ox st +2
no chg = 0
1* O-¾Mn+ = +1
3* O2-=Mn2+ = +6
+ ¾¾
ox st +7
- chg = -1
4* O2-=Mn2+ = +8
+ ¾¾
ox st +7
Organometallic Chemistry

37. Examples - homonuclear bonds
no chg = 0
3* Cl-¾C+ = +3
1* C¾C = 0
+ ¾¾
ox st +3
2- chg = -2
3* Cl-¾Pt+ = +3
1* Pt¾Pt = 0
+ ¾¾
ox st +1
Artificial, abnormal formal oxidation states. If you want to say something about stability, pretend the homonuclear bond is polar (for metals, typically with the + end at the metal you are interested in). For the above examples, that would give C(+4) and Pt(+2), very normal oxidation states.
Organometallic Chemistry

38. Example - handling 3c-2e bonds
Rewrite first, as discussed under "electron counting". The rest is "automatic" (ignore the dative bonds as usual).
no chg = 0
3* H-¾B+ = +3
+ ¾¾
ox st +3
Organometallic Chemistry

39. Oxidation state and stability
Sometimes you can easily deduce that an oxidation state is "impossible", so the compound must be unstable
no chg = 0
4* Me-¾Mg+ = +4
+ ¾¾
ox st +4
But Mg only has 2 valence electrons! Any compound containing Mg4+ will not be stable.
Organometallic Chemistry

40. Significance of oxidation states
Oxidation states are formal.
They do not indicate the "real charge" at the metal centre.
However, they do give an indication whether a structure or composition is reasonable.
apart from the M-M complication
They have more meaning when all bonds are relatively polar.
i.e. close to the fully ionic description used for counting
Organometallic Chemistry

41. Normal oxidation states
For group n or n+10:
never >+n or <-n (except group 11: frequently +2 or +3)
usually even for n even, odd for n odd
usually ³ 0 for metals
usually +n for very electropositive metals
usually 0-3 for 1st-row transition metals of groups 6-11, often higher for 2nd and 3rd row
electronegative ligands (F,O) stabilize higher oxidation states, p-acceptor ligands (CO) stabilize lower oxidation states
oxidation states usually change from m to m-2, m or m+2 in reactions
Organometallic Chemistry

42. Oxidation-state exercises
Calculate oxidation states for the metal in the complexes below. From this and the electron count (done earlier), draw conclusions about expected stability or reactivity.
Me2Mg Pd(PMe3)4 MeReO3
ZnCl4 Pd(PMe3)3 OsO3(NPh)
ZrCl4 ZnMe42- OsO4(pyridine)
Co(CO)4- Mn(CO)5- Cr(CO)6
V(CO)6- V(CO)6 Zr(CO)64+
PdCl(PMe3)3 RhCl2(PMe3)2 Ni(PMe3)2Cl2
Ni(PMe3)Cl4 Ni(PMe3)Cl3
Organometallic Chemistry

43. Coordination number and geometry
The coordination number is the number of atoms directly bonded to the atom you are interested in
regardless of bond orders etc
often abbreviated as CN
CH4: 4
C2H4: 3
C2H2: 2
AlCl4-: 4
Me4Zn2-: 4
OsO4: 4
B2H6: 4 (B)
1 (terminal H)
2 (bridging H)
Organometallic Chemistry

44. p-system ligands
For complexes with p-system ligands, the whole ligand is usually counted as 1:
Cyclopentadienyl groups are sometimes counted as 3, because a single Cp group can replace 3 individual ligands:
CN 4
CN 3 or 5
Organometallic Chemistry

45. Common coordination numbers
The most common coordination numbers for organometallic compounds are:
2-6 for main group metals
4-6 for transition metals
Coordination numbers >6 are relatively rare, as are very low coordination numbers (<4) together with a “too-low” electron count.
Abnormally high coordination numbers are found for "polyhydrides", where there is often ambiguity between "hydride" and "dihydrogen" descriptions
the low steric requirements of H make this possible
example given later on
Organometallic Chemistry

46. Coordination number and geometry
C.N. "Normal" geometry
2 linear or bent
3 planar trigonal, pyramidal; "T-shaped" often for d8 14-e
4 tetrahedral; square planar often for d8 16-e
5 square pyramidal, trigonal bipyramidal
6 octahedral
Exceptions can be expected for abnormal electron counts or for ligands with unusual geometric requirements
Organometallic Chemistry

47. Example: protonation of WH6(PMe3)3
Could WH6(PMe3)3 be a true polyhydride ?
Count: 18-e (OK).
Oxidation state: 6 (OK).
CN: 9 (very high).
Possible.
Organometallic Chemistry

48. Example: protonation of WH6(PMe3)3
Protonation gives WH7(PMe3)3+. Could that still be a true polyhydride ?
Count: 18-e (OK).
Oxidation state: 8 (too high).
CN: 10 (extremely high).
Virtually impossible.
W+ has only 5 electrons but must form 7 W-H bonds !
This is almost certainly a dihydrogen complex.
Organometallic Chemistry