Cr =6 2Bz =12 + e-count18 tot. charge0 2Bz- - ox state0 Rh =9 3* P =6 Cl =1 + e-count16 tot. charge0 3*P - Cl -1 - ox state+1

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!


Predicting reactivity

(C2H4)2PdCl2 (C2H4)(CO)PdCl2

(C2H4)PdCl2

(C2H4)2(CO)PdCl2

?

CO- C2H4

- C2H4CO

dissociative

associative



Most likely associative:

16-e PdII

18-e PdII

16-e PdII



Cr(CO)6 Cr(CO)5(MeCN)

Cr(CO)5

Cr(CO)6(MeCN)

?

MeCN- CO

- COMeCN

dissociative

associative



Almost certainly dissociative:

18-e Cr(0)

16-e Cr(0)

18-e Cr(0)


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


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.


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.


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.


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.


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.


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)


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).


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.


Metal-centered lone pairs

Cp2WH2H+

Cp2WH3+

Basicity of Cp2WH2 comparable to that of ammonia!

18-e WIV18-e WVI


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.


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.



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.


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


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


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


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


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


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


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-


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


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.


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


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.


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

-


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.


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


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


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


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


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.


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


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.


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


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


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.


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)


-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


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


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


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


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.

Documents

Cr =6 2*Bz =12 + e-count18 tot. charge0 2*Bz- - ox state0 Rh =9 3* P =6 Cl =1 + e-count16 tot. charge0 3*P - Cl -1 - ox state+1

Cr =6 2Bz =12 + e-count18 tot. charge0 2Bz- - ox state0 Rh =9 3* P =6 Cl =1 + e-count16 tot. charge0 3*P - Cl -1 - ox state+1