2 - Electron Counting

7/30/2019 2 - Electron Counting

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Cr = 6

2*Bz = 12

+

e-count 18

tot. charge 0

2*Bz -

-

ox state 0

Rh = 9

3* P = 6

Cl= 1+

e-count 16

tot. charge 0

3*P -Cl -1

-

ox state +1

Electron Counting

understanding structure and reactivity

Peter H.M. Budzelaar


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Electron Countin2

Why count electrons ?

Basic tool for understanding structure and reactivity.

Simple extension of Lewis structure rules.

Counting should be automatic.

Not always unambiguous

dont just followthe rules, understandthem!


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Electron Countin3

Predicting reactivity

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

(C2H4)PdCl2

(C2H4)2(CO)PdCl2

?

CO- C2H4

- C2H4CO

dissociative

associative


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Electron Countin4


Most likely associative:

16-e PdII

18-e PdII

16-e PdII


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Electron Countin5


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

Cr(CO)5

Cr(CO)6(MeCN)

?

MeCN- CO

- COMeCN

dissociative

associative


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Electron Countin6


Almost certainly dissociative:

18-e Cr(0)

16-e Cr(0)

18-e Cr(0)


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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 electronpairs, 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.



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 or12-e?). Stable, truly hypervalent molecules (for which every

Lewis structure has > 8-e) are not that common (SF6, PF

5).

Structures with < 8-e are very rare.



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-planard8 complexes.

For open-shell complexes, every valence orbital wants to be

used forat least one electronMore diverse possibilities, harder to predict.



Prediction of stable complexes

Cp2Fe, ferrocene: 18-eVery 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.



If there are not enough electrons...

Structures with a lower than ideal electron count are called

electron-deficientorcoordinatively 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)

unreactive

16-e Fe(0)

very reactive18-e Fe(0)


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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 -donor (nucleophile). susceptible to electrophilic attack.



Metal-centered lone pairs

Cp2WH2H+

Cp2WH3+

Basicity of Cp2WH2 comparable to that of ammonia!

18-e WIV 18-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 covalentbond 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 = 1

H= 1

+

e-count 2

CH

H

H

H

C

H

H

H

H

C = 4

4* 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 = 4

2* H= 2

2* C= 2

+ e-count 8

A double bond counts

as two covalent bonds.




H C

H

H

C = 4+ chg = -1

3* H= 3

+

e-count 6

C

H

H

H

Highly reactive,electrophilic.

C

H

H

H

H C

H

H

C = 4

- chg = +1

3* H= 3

+ e-count 8

Saturated, but

nucleophilic.

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". Extremely

reactive as radical, notespecially for nucleophiles

or electrophiles.


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Covalent or dative ?

How do you know a fragment forms a covalentor 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.

Cl

Pd

PPh3

covalent

bond

dativebond

"bond" to the

allyl fragment

Cl

Pd

PPh3

1 e 2 e

3 e

Pd = 10Cl= 1

P = 2

allyl = 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 B2H

2core is held together by two 3c-2e bonds:

B

H

H

B

H

H

H

H

HB

H

B

H

HH

H



Handling 3c-2e and 4c-2e bonds

Rewrite bonding in terms of two BH3monomers:

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= 3

BH = 2

+

e-count 8

B

H

HH

H

B

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.

Me

AlMe

Al

Me

MeMe

Me

Cl

AlCl

Al

Cl

ClCl

Cl

Al

Me

MeMe

Me

Al

Me

Me

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* Me= 3

MeAl = 2

+

e-count 8

Al

Cl

ClCl

Cl

Al

Cl

Cl

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* Cl= 3

Cl = 2

+

e-count 8



3c-2e vs normal bridge bonds

The orbitals of a 3c-2e bond are bonding

between all three of the atoms involved.

Therefore, Al2Me

6has a net

Al-Al bonding interaction.

The orbitals involved in "normal" bridges

are regular terminal-bridge bonding orbitals.

Thus, Al2Cl6 has strong Al-Cl bondsbut no net Al-Al bonding.

Al

Me

MeMe

Me

Al

Me

Me

Al

Cl

ClCl

Cl

Al

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 shouldbelong to the metal...

Rh

OC

OC

CO

SO3

-

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

CH2= 1

3* CO = 6

+

e-count 16

RhOC

OC

CO

SO3-



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

2* CO = 4

+ e-count 18

B

Ph2P PPh2Ph2P

Co

COOC

B

Ph2P PPh2Ph2P

Co

COOC


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Electron Countin28

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

Cl

Cl

P = 5

5* Cl= 5

+

e-count 10

P

Cl

Cl

Cl

Cl

Cl

Cl

P Cl

Cl

Cl

Cl

P = 5+ chg = -1

4* Cl= 4

+

e-count 8

SiF62-, SF6, IO6

5- and

noble-gas halides can

be described in a

similar manner.


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Electron Countin29

A few excess-electron examples

HF2

- H only has a single valence orbital,

so it cannot form two covalent H-F bonds!

Write as FHF-, mainly ion-dipole interaction.

H = 1

- chg = +1

2* F= 2+

e-count 4

H = 1

1* F= 1

+

e-count 2

This is just an extreme form of

hydrogen bonding. Most other

H-bonded molecules have

less symmetric hydrogen bridges.

O

O

H

O

O

H

OH

O

?F H F

F H F

F H F

F H F


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Electron Countin30

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 orn+2. Electrons dont just appear or disappear.

The optimal count is 2/8/18 e. 16-e also occurs frequently, other

counts are much more rare.


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Electron Countin31

Electron-counting exercises

Me2Mg Pd(PMe

3)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(PMe

3)2

Ni(PMe3)2Cl

2

Ni(PMe3)Cl4 Ni(PMe3)Cl3 Cl PdMe3P

PMe3

BMe3

-


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Electron Countin32

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 doesnt want to go lower.

Cl: nearly always -1.

Already has 7 valence electrons, so cannot go lower.Is very electronegative, so doesnt want to go higher.


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Electron Countin33

Calculating oxidation states

Rewrite compound as if all bonds were fully ionic/dative, i.e. the

electron pairs of each bond go to one endof the bond.

Which end? Use electronegativity to decide.

Ignore homonuclear covalent bondsNo unambiguous choice available

Usually end up with a unique set of charges

The "Lewis structure ambiguity" for hypervalent compounds does not cause

problems here


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Electron Countin34

How do you calculate oxidation states ?

1. Start with the formal charge on the metal

See earlier discussion in "electron counting"

2. Ignore dative bonds

3. Ignore bonds between atoms of the same element

This one is a bit silly and produces counterintuitive results4. 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

5. Add


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Electron Countin35

Examples - main group elements

CCl4

COCl2

AlCl4-

C

Cl

Cl

Cl

Cl

Cl

Cl

Cl

C

Cl

no chg = 0

4* Cl-C+ = +4+

ox st +4

Al

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Al

Cl

- chg = -1

4* Cl-Al+ = +4+

ox st +3

O C

Cl

Cl

O C

Cl

Cl

no chg = 02* Cl-C+ = +2

O2-=C2+ = +2

+

ox st +4


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Electron Countin36

Examples - transition metals

MnO4-

PdCl42-

2- chg = -2

4* Cl-Pd+ = +4

+

ox st +2

no chg = 0

1* O-Mn+ = +13* O2-=Mn2+ = +6

+

ox st +7

Cl

Pd

ClCl

Cl Cl

Cl Cl

Pd

Cl

O

Mn

OO

O

O

Mn

OO

O

O

Mn

OO

O

O

Mn

OO

O

- chg = -1

4* O2-=Mn2+ = +8+

ox st +7


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Electron Countin37

Examples - homonuclear bonds

Cl

CCl

Cl

C

Cl

Cl

Cl Cl

CCl

Cl

C

Cl

Cl

ClC2Cl6

Pt2Cl64-

no chg = 03* Cl-C+ = +3

1* CC = 0

+

ox st +3

Cl Pt Pt

Cl

Cl

ClCl

ClCl Pt

Cl

Cl

Cl

Cl

Cl

Pt

2- chg = -2

3* Cl-Pt+ = +3

1* 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.


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Electron Countin38

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

HH

BH H

H

no chg = 0

3* H-B+ = +3

+ ox st +3


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Electron Countin39

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 = 0

4* Me-Mg+ = +4

+

ox st +4

But Mg only has 2 valence electrons!

Any compound containing Mg4+ will not be stable.


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Electron Countin40

Significance of oxidation states

Oxidation states are formal.

They do notindicate 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


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Electron Countin41

Normal oxidation states

For group n orn+10: never >+n or


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Electron Countin42

Oxidation-state exercises

Me2Mg Pd(PMe

3)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(PMe

3)2

Ni(PMe3)2Cl

2

Ni(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.


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Electron Countin43

Coordination number and geometry

The coordination numberis the number of atoms directly bonded to

the atom you are interested in

regardless of bond orders etc

often abbreviated as CN

CH4: 4

C2H

4: 3

C2H

2: 2

AlCl4

-: 4

Me4Zn2-: 4

OsO4: 4

B2H

6: 4 (B)

1 (terminal H)

2 (bridging H)


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Electron Countin44

-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


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Electron Countin45

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 lowcoordination numbers (


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Electron Countin46

Coordination number and geometry

C.N. "Normal" geometry

2 linear or bent

3 planar trigonal, pyramidal; "T-shaped" often ford8 14-e

4 tetrahedral; square planar often ford8 16-e5 square pyramidal, trigonal bipyramidal

6 octahedral

Exceptions can be expected for abnormal electron counts or forligands with unusual geometric requirements


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E l t ti f WH (PM )


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Example: protonation of WH6(PMe3)3

Protonation gives WH7(PMe

3)3

+.

Could that still be

a 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 electrons

but must form 7 W-H bonds !

This is almost certainly

a dihydrogen complex.

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2 - Electron Counting