cr =6 2*bz =12 + e-count18 tot. charge0 2*bz- - ox state0 rh =9 3* p =6 cl =1 + e-count16 tot....
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Cr = 62*Bz = 12
+ e-count 18
tot. charge 0
2*Bz --
ox state 0
Rh = 93* P = 6Cl = 1
+ e-count 16
tot. charge 0
3*P -Cl -1
- ox state +1
Electron Counting
understanding structure and reactivity
Peter H.M. Budzelaar
Organometallic Chemistry2
Why count electrons ?
• Basic tool for understanding structure and reactivity.• Simple extension of Lewis structure rules.• Counting should be “automatic”.• Not always unambiguous
don’t just follow the rules, understand them!
Organometallic Chemistry3
Predicting reactivity
(C2H4)2PdCl2 (C2H4)(CO)PdCl2
(C2H4)PdCl2
(C2H4)2(CO)PdCl2
?
CO- C2H4
- C2H4CO
dissociative
associative
Organometallic Chemistry4
Predicting reactivity
Most likely associative:
16-e PdII
18-e PdII
16-e PdII
Organometallic Chemistry5
Predicting reactivity
Cr(CO)6 Cr(CO)5(MeCN)
Cr(CO)5
Cr(CO)6(MeCN)
?
MeCN- CO
- COMeCN
dissociative
associative
Organometallic Chemistry6
Predicting reactivity
Almost certainly dissociative:
18-e Cr(0)
16-e Cr(0)
18-e Cr(0)
Organometallic Chemistry7
The basis of electron counting
• Every element has a certain number of valence orbitals:1 { 1s } for H
4 { ns, 3np } for main group elements
9 { ns, 3np, 5(n-1)d } for transition metals
pxs py pz
dxzdxy dx2-y2dyz dz2
Organometallic Chemistry8
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 surroundedby 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 calledelectron-precise.
Organometallic Chemistry9
Compounds are not alwayselectron-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 (SO4
2-: 8-e or12-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 Chemistry10
Compounds are not alwayselectron-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 stableIn 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 Chemistry11
Prediction of stable complexes
Cp2Fe, ferrocene: 18-eVery stable.Behaves as an aromaticorganic compound in e.g.Friedel-Crafts acylation.
Cp2Co, cobaltocene: 19-eStrong reductant,reacts with air.Cation (Cp2Co+) is very stable.
Cp2Ni, nickelocene: 20-eChemically reactive,easily loses a Cp ring,reacts with air.
Organometallic Chemistry12
If there are not enough electrons...
• Structures with a lower than ideal electron count are calledelectron-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 Chemistry13
Reactivity of electron-deficient compounds
Fe(CO)5h
- COFe(CO)4
THF Fe(CO)4(THF)
18-e Fe(0)
unreactive16-e Fe(0)
very reactive18-e Fe(0)
Organometallic Chemistry14
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 Chemistry15
Where are the electrons ?
• Electrons around a metal can be in metal-ligand bonding orbitalsor in metal-centered lone pairs.
• Metal-centered orbitals are fairly high in energy.• A metal atom with a lone pair is a -donor (nucleophile).
– susceptible to electrophilic attack.
Organometallic Chemistry16
Metal-centered lone pairs
Cp2WH2H+
Cp2WH3+
Basicity of Cp2WH2 comparable to that of ammonia!
18-e WIV18-e WVI
Organometallic Chemistry17
How do you count ?
"Covalent" count:
1. Number of valence electrons of central atom.• from periodic table
2. Correct for charge, if any.• but only if the charge belongs to that atom!
3. Count 1 e for every covalent bond to another atom.
4. Count 2 e for every dative bond from another atom.• no electrons for dative bonds to another atom!
5. Delocalized carbon fragments: usually 1 e per C
6. Three- and four-center bonds need special treatment.
7. 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 Chemistry18
Starting simple...
H H
H H H = 1H = 1
+ e-count 2
CH
H
H
H
C
H
H
H
H
C = 44* H = 4
+ e-count 8
NH
H
H
N
H
H
H
N = 53* H = 3
+ e-count 8
N has a lone pair.Nucleophilic!
C
H
H
C
H
H
C C
H
H H
H
C = 42* H = 22* C = 2
+ e-count 8
A double bond countsas two covalent bonds.
Organometallic Chemistry19
Predicting reactivity
H C
H
H
C = 4+ chg = -13* H = 3
+ e-count 6
C
H
H
HHighly reactive,electrophilic.
C
H
H
H
H C
H
H
C = 4- chg = +13* H = 3
+ e-count 8
Saturated, butnucleophilic.
C
H
H
C = 42* H = 2
+ e-count 6
C
H
H
"Singlet carbene". Unstable.Sensitive to nucleophiles(empty orbital)and electrophiles (lone pair).
C
H
H
"Triplet carbene". Extremelyreactive as radical, notespecially for nucleophilesor electrophiles.
Organometallic Chemistry20
When is a line not a line ?
C C
HH
H H
HH
C = 43* H = 3C = 1
+ e-count 8
B N
HH
H H
HH
is
B N
HH
H H
HH
or B N
HH
H H
HH
B = 33* H = 3N = 2
+ e-count 8
N = 53* H = 3
+ e-count 8
B = 3- chg = +13* H = 3N = 1
+ e-count 8
N = 5+ chg = -13* H = 3B = 1
+ e-count 8
Organometallic Chemistry21
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 properlybefore you start counting.
Cl
Pd
PPh3
covalentbond
dativebond
"bond" to theallyl fragment
Cl
Pd
PPh3
1 e 2 e
3 e
Pd = 10Cl = 1P = 2allyl = 3
+ e-count 16
Organometallic Chemistry22
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:
B
H
H
B
H
H
H
H
HB
HB
H
HH
H
Organometallic Chemistry23
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.
1 e2 e B = 3
3* H = 3BH = 2
+ e-count 8
B
H
HH
HB
H
H
C2H6 BH3NH3 B2H6
Organometallic Chemistry24
What kind of bridge bond do I have ?
A 3c-2e bond will only form when the central (bridging) atomdoes not have any lone pairs.
When lone pairs are available, they are preferred as donors.
MeAl
MeAl
Me
MeMe
Me
ClAl
ClAl
Cl
ClCl
Cl
Al
Me
MeMe
MeAl
Me
Me
A methyl group can formone more single bond. Afterthat, it has no lone pairs, so thebest it can do is share the Al-Cbonding electrons with asecond Al: Al = 3
3* Me = 3MeAl = 2
+ e-count 8
Al
Cl
ClCl
ClAl
Cl
Cl
After chlorine forms a single bond,it still has three lone pairs left.One is used to donateto the second Al:
Al = 33* Cl = 3Cl = 2
+ e-count 8
Organometallic Chemistry25
3c-2e vs normal bridge bonds
• The orbitals of a 3c-2e bond are bondingbetween all three of the atoms involved.
– Therefore, Al2Me6 has a netAl-Al bonding interaction.
• The orbitals involved in "normal" bridgesare regular terminal-bridge bonding orbitals.
– Thus, Al2Cl6 has strong Al-Cl bondsbut no net Al-Al bonding.
Al
Me
MeMe
MeAl
Me
Me
Al
Cl
ClCl
ClAl
Cl
Cl
Organometallic Chemistry26
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...
RhOC
OC
CO
SO3
-
Any alkyl-SO3 groupwould normally be anionic(c.f. CH3SO3
-, the anion of CH3SO3H).
So the negative chargedoes not belong to the metal!
Rh = 9CH2 = 13* CO = 6
+ e-count 16
RhOC
OC
CO
SO3-
Organometallic Chemistry27
Handling charges
Even overall neutral molecules could have "hidden" charges!
A boron atom with 4 bondswould be -1 (c.f. BH4
-).
No other obvious centers ofcharge, so the Co must be +1.
Co = 9+ chg = -13* P = 6CO = 2
+ e-count 16
B
Ph2P PPh2Ph2P
Co
COOC
B
Ph2P PPh2Ph2P
Co
COOC
Organometallic Chemistry28
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- !
?
P
Cl
Cl
Cl
Cl
Cl
Cl
P Cl
Cl
ClCl
P = 55* Cl = 5
+ e-count 10
P
Cl
Cl
Cl
Cl
Cl
Cl
P Cl
Cl
ClCl
P = 5+ chg = -14* Cl = 4
+ e-count 8
SiF62-, SF6, IO6
5- andnoble-gas halides canbe described in asimilar manner.
Organometallic Chemistry29
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.
H = 1- chg = +12* F = 2
+ e-count 4
H = 11* F = 1
+ e-count 2
This is just an extreme form ofhydrogen bonding. Most otherH-bonded molecules haveless symmetric hydrogen bridges.
O
O
H
O
O
H
OH
O
?F H F
F H F
F H F
F H F
Organometallic Chemistry30
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 Chemistry31
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)6
4+
PdCl(PMe3)3 RhCl2(PMe3)2 Ni(PMe3)2Cl2Ni(PMe3)Cl4 Ni(PMe3)Cl3
Cl PdMe3P
PMe3
BMe3
-
Organometallic Chemistry32
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 Chemistry33
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 bondsNo unambiguous choice available
Usually end up with a unique set of chargesThe "Lewis structure ambiguity" for hypervalent compounds does not cause
problems here
Organometallic Chemistry34
How do you calculate oxidation states ?
1. Start with the formal charge on the metalSee earlier discussion in "electron counting"
2. Ignore dative bonds
3. Ignore bonds between atoms of the same elementThis one is a bit silly and produces counterintuitive results
4. Assign every covalent electron pair to the most electronegative element in the bond: this produces + and – chargesUsually + at the metal
Multiple bonds: multiple + and - charges
5. Add
Organometallic Chemistry35
Examples - main group elements
CCl4
COCl2
AlCl4-
CCl
Cl
Cl
Cl
Cl
Cl
Cl
CCl
no chg = 04* Cl-C = +4
+ ox st +4
AlCl
Cl
Cl
Cl
Cl
Cl
Cl
AlCl
- chg = -14* Cl-Al = +4
+ ox st +3
O C
Cl
Cl
O C
Cl
Cl
no chg = 02* Cl-C = +2O2-=C2+ = +2
+ ox st +4
Organometallic Chemistry36
Examples - transition metals
MnO4-
PdCl42-
2- chg = -24* Cl-Pd = +4
+ ox st +2
no chg = 01* O-Mn = +13* O2-=Mn2+ = +6
+ ox st +7
Cl
Pd
ClCl
Cl Cl
Cl Cl
Pd
Cl
O
Mn
OO
O
O
MnOO
O
O
Mn
OO
O
O
MnOO
O
- chg = -14* O2-=Mn2+ = +8
+ ox st +7
Organometallic Chemistry37
Examples - homonuclear bonds
Cl
CCl
Cl
C
Cl
Cl
Cl Cl
CCl
Cl
C
Cl
Cl
ClC2Cl6
Pt2Cl64-
no chg = 03* Cl-C = +31* CC = 0
+ ox st +3
Cl Pt Pt
Cl
Cl
ClCl
ClCl Pt
Cl
Cl
Cl
Cl
Cl
Pt
2- chg = -23* Cl-Pt = +31* PtPt = 0
+ 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 Chemistry38
Example - handling 3c-2e bonds
Rewrite first, as discussed under "electron counting".The rest is "automatic" (ignore the dative bonds as usual).
HB
HB
H
HH
HB
H
HH
HB
H
H
B
H
HHB
H H
H
no chg = 03* H-B = +3
+ ox st +3
Organometallic Chemistry39
Oxidation state and stability
Sometimes you can easily deduce that an oxidation state is "impossible", so the compound must be unstable
MgMe4
Mg
Me Me
MeMe
Mg
Me Me
MeMe
no chg = 04* Me-Mg = +4
+ ox st +4
But Mg only has 2 valence electrons!Any compound containing Mg4+ will not be stable.
Organometallic Chemistry40
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 Chemistry41
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 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,
-acceptor ligands (CO) stabilize lower oxidation states– oxidation states usually change from m to m-2, m or m+2 in reactions
Organometallic Chemistry42
Oxidation-state 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)6
4+
PdCl(PMe3)3 RhCl2(PMe3)2 Ni(PMe3)2Cl2Ni(PMe3)Cl4 Ni(PMe3)Cl3
Cl PdMe3P
PMe3
BMe3
-
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.
Organometallic Chemistry43
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 Chemistry44
-system ligands
For complexes with -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:
Cl PdCl
Cl
-
ZrClCl CN 4
H
CoCOOC
COCO
CoOC CO
CN 3 or 5
Organometallic Chemistry45
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 Chemistry46
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 Chemistry47
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.
W PMe3Me3P
Me3P HH H
HHH
Organometallic Chemistry48
Example: protonation of WH6(PMe3)3
Protonation gives WH7(PMe3)3+.
Could that still bea true polyhydride ?
W PMe3Me3P
Me3P HH H
HHH
H
+
Count: 18-e (OK).
Oxidation state: 8 (too high).
CN: 10 (extremely high).
Virtually impossible.
W+ has only 5 electronsbut must form 7 W-H bonds !
This is almost certainlya dihydrogen complex.