2 - Electron Counting 3l3gu

Electron Counting understanding structure and reactivity Rh = 9 3* P→ = 6 Cl = 1 +  e-count 16 tot. charge 0  3*P→ Cl -1 -  ox state +1

Cr = 2*Bz =

6 12 +  e-count 18 tot. charge 0  2*Bz -  ox state 0

Peter H.M. Budzelaar

Why count electrons ?

• • • •

2

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!

Electron Counting

Predicting reactivity

- C2H4 (C2H4)2PdCl2 CO

3

dissociative (C2H4)PdCl2

CO

? (C2H4)2(CO)PdCl2 associative

(C2H4)(CO)PdCl2 - C2H4

Electron Counting

Predicting reactivity

Most likely associative:

16-e PdII

16-e PdII

18-e PdII

4

Electron Counting

Predicting reactivity

- CO Cr(CO)6 MeCN

5

dissociative Cr(CO)5

MeCN

?

Cr(CO)6(MeCN) associative

Cr(CO)5(MeCN) - CO

Electron Counting

Predicting reactivity Almost certainly dissociative:

16-e Cr(0)

18-e Cr

6

(0)

18-e Cr(0)

Electron Counting

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

dxy 7

px

dxz

py

dyz

pz

dx2-y2

dz2

Electron Counting

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.

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Electron Counting

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 (SO 42-: 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.

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Electron Counting

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.

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Electron Counting

Prediction of stable complexes

2Fe, ferrocene: 18-e Very stable. Behaves as an aromatic organic compound in e.g. Friedel-Crafts acylation.

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2Co, cobaltocene: 19-e Strong reductant, reacts with air. Cation (2Co+) is very stable.

2Ni, nickelocene: 20-e Chemically reactive, easily loses a ring, reacts with air.

Electron Counting

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.

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Electron Counting

Reactivity of electron-deficient compounds Fe(CO)5

18-e Fe(0) unreactive

13

hν - CO

Fe(CO)4 THF

16-e Fe(0) very reactive

Fe(CO)4(THF)

18-e Fe(0)

Electron Counting

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).

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Electron Counting

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 σ-donor (nucleophile). – susceptible to electrophilic attack.

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Electron Counting

Metal-centered lone pairs 2WH2

H+

2WH3+

Basicity of 2WH2 comparable to that of ammonia!

18-e WIV

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18-e WVI

Electron Counting

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.

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Electron Counting

Starting simple... H

H= H =

1 1 +  e-count 2

H

H

H H

H

N= 5 3* H = 3 +  e-count 8

N H

H N

N has a lone pair. Nucleophilic!

H H

H H H

C= 4 4* H = 4 +  e-count 8

C H

H C

H

H

H

C

H C

H H

H

C= 4 2* H = 2 2* C = 2 +  e-count 8

C

H

H

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H

C H

A double bond counts as two covalent bonds.

Electron Counting

Predicting reactivity H H

C H

H H

C H

H

C= 4 + chg = -1 3* H = 3 +  e-count 6 Highly reactive, electrophilic.

C H H C H

H H

C H

H C H H

19

C= 4 - chg = +1 3* H = 3 +  e-count 8

H C H

C= 4 2* H = 2 +  e-count 6 "Singlet carbene". Unstable. Sensitive to nucleophiles (empty orbital) and electrophiles (lone pair). "Triplet carbene". Extremely reactive as radical, not especially for nucleophiles or electrophiles.

Saturated, but nucleophilic.

Electron Counting

When is a line not a line ? H H H H

H C

H H

C

H B

N H H

H H is

C= 4 3* H = 3 C = 1 +  e-count 8

H H H

H B

N H H

B= 3 N= 5 3* H = 3 3* H = 3 N→ = 2 +  +  e-count 8 e-count 8

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H or

H H

H B

B= 3 - chg = +1 3* H = 3 N = 1 +  e-count 8

N H H N= 5 + chg = -1 3* H = 3 B = 1 +  e-count 8

Electron Counting

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. Cl

covalent bond

PPh3 Pd

dative bond

"bond" to the allyl fragment

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Cl 1e

PPh3 Pd

2e

3e

Pd = Cl = P→ = allyl =

10 1 2 3 +  e-count 16

Electron Counting

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

H H

H B

B

H

.

H

H

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: H

H

H B

B

H

H H

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Electron Counting

Handling 3c-2e and 4c-2e bonds Rewrite bonding in of two BH3 monomers: 2e

H B HH

H H B

H

1e

B= 3 3* H = 3 BH→ = 2 +  e-count 8

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.

C2H6

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BH3NH3

B2H6

Electron Counting

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. Me Me

Me Al

Al Me

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 Me Me Me 3* Me = 3 MeAl→ = 2 Al Al +  Me Me Me e-count 8

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Cl Cl

Cl Al

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: Cl

Cl Cl

Al Cl Cl

Al Cl

Al = 3 3* Cl = 3 Cl→ = 2 +  e-count 8

Electron Counting

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.

Me

Al Me Me

Me Me Al

Me

• 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. Cl Cl Al Cl Cl

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Cl

Al Cl

Electron Counting

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... SO3OC SO3 OC Rh Rh OC CO OC CO 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!

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Rh = 9 CH2 = 1 3* CO→ = 6 +  e-count 16

Electron Counting

Handling charges Even overall neutral molecules could have "hidden" charges! A boron atom with 4 bonds would be -1 (c.f. BH4-).

B

Ph2P Ph2P Co OC

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PPh2

CO

No other obvious centers of charge, so the Co must be +1.

B

Ph2P Ph2P Co OC

PPh2

Co = 9 + chg = -1 3* P→ = 6 2* CO→ = 4 +  e-count 18

CO

Electron Counting

A few excess-electron examples 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- !

PCl5

Cl Cl Cl

P Cl Cl

Cl

Cl

?

Cl P Cl Cl

P= 5 5* Cl = 5 +  e-count 10

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Cl

Cl

Cl

P Cl

SiF62-, SF6, IO65- and noble-gas halides can be described in a similar manner.

Cl

Cl P Cl

Cl

Cl Cl

P= 5 + chg = -1 4* Cl = 4 +  e-count 8

Electron Counting

A few excess-electron examples H only has a single valence orbital, so it cannot form two covalent H-F bonds! Write as FH·F-, mainly ion-dipole interaction.

HF2-

This is just an extreme form of hydrogen bonding. Most other H-bonded molecules have less symmetric hydrogen bridges. O

F

H

F

H

F

F

H= 1 - chg = +1 2* F = 2 +  e-count 4

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?

F

H

F

H

H

O

F

F

H= 1 1* F = 1 +  e-count 2

O

O

H

H

O

O

Electron Counting

Does it look reasonable ?

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.

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Electron Counting

Electron-counting exercises

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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)2Cl 2 PMe

Ni(PMe3)Cl4

Ni(PMe3)Cl3

Cl Pd Me3P

3

BMe3

Electron Counting

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.

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Electron Counting

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

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Electron Counting

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 results

4. 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 Counting

Examples - main group elements CCl4 Cl

Cl

Cl Cl

C

Cl

AlCl4-

Cl

Cl

Cl

C

Cl

Cl

no chg = 0 4* Cl-C+ = +4 +  ox st +4

COCl2 O

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Cl O

Cl

Cl Cl

Al Cl

Cl

Al Cl

- chg = -1 4* Cl-Al+ = +4 +  ox st +3

Cl C

Cl

C Cl

no chg = 0 2* Cl-C+ = +2 O2-=C2+ = +2 +  ox st +4

Electron Counting

Examples - transition metals PdCl42- Cl

Cl

Cl

Pd Cl

Pd Cl

Cl

O O

Mn O

O O

O O O

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Cl

O

O

MnO4-

Cl

Mn O

O

Mn O

O O

Mn O

2- chg = -2 4* Cl-Pd+ = +4 +  ox st +2

no chg = 0 1* O-Mn+ = +1 3* O2-=Mn2+ = +6 +  ox st +7 - chg = -1 4* O2-=Mn2+ = +8 +  ox st +7

Electron Counting

Examples - homonuclear bonds Cl

C2Cl6

Cl Cl

C

C

Cl Cl

Cl

Pt2Cl64Cl

Cl

Cl Cl

C

C

Cl Cl Cl

Cl Cl

Pt

Cl

Pt Cl

no chg = 0 3* Cl-C+ = +3 1* CC = 0 +  ox st +3

Cl Cl

Cl

Cl

Pt

Cl

Pt Cl

Cl

2- chg = -2 3* Cl-Pt+ = +3 1* PtPt = 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 Counting

Example - handling 3c-2e bonds

Rewrite first, as discussed under "electron counting". The rest is "automatic" (ignore the dative bonds as usual). H H

H

H B

B H

H H

B HH

H

H H B

H

B HH

H H B

H

no chg = 0 3* H-B+ = +3 +  ox st +3

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Electron Counting

Oxidation state and stability

Sometimes you can easily deduce that an oxidation state is "impossible", so the compound must be unstable

MgMe4

Me

Me

Me

Mg Me

Me Mg

Me

Me

Me

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 Counting

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

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Electron 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, π-acceptor ligands (CO) stabilize lower oxidation states – oxidation states usually change from m to m-2, m or m+2 in reactions

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Electron Counting

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.

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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)2Cl 2 PMe

Ni(PMe3)Cl4

Ni(PMe3)Cl3

Cl Pd Me3P

3

BMe3

Electron Counting

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

4

C2H4:

3

C2H2:

2

AlCl4-:

4

Me4Zn2-:

4

OsO4:

4

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CH4:

B2H6:

4 (B) 1 (terminal H) 2 (bridging H)

Electron Counting

π -system ligands For complexes with π-system ligands, the whole ligand is usually counted as 1: -

Cl Pd Cl

Cl

Zr Cl Cl

CN 4

Cyclopentadienyl groups are sometimes counted as 3, because a single group can replace 3 individual ligands:

OC

Co

CO

H CO CO Co OC CO

CN 3 or 5

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Electron Counting

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

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Electron Counting

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

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Electron Counting

Example: protonation of WH6(PMe3)3 Could WH6(PMe3)3 be a true polyhydride ?

H H H Me3P Me3P W PMe3 H H H Count: 18-e (OK). Oxidation state: 6 (OK). CN: 9 (very high). Possible.

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Electron Counting

Example: protonation of WH6(PMe3)3 + Protonation gives WH7(PMe3)3+. Could that still be a true polyhydride ?

H H H Me3P Me3P W HPMe3 H H 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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Electron Counting