Transcript Electron Countin..

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(0)
6
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.
8
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 (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.
9
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.
10
Electron Counting
Prediction of stable complexes
Cp2Fe, ferrocene: 18-e
Very stable.
Behaves as an aromatic
organic compound in e.g.
Friedel-Crafts acylation.
11
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.
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.
12
Electron Counting
Reactivity of electron-deficient compounds
Fe(CO)5
18-e Fe(0)
unreactive
13
h
- 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 s-donor (nucleophile).
– susceptible to electrophilic attack.
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Electron Counting
Metal-centered lone pairs
Cp2WH2
H+
Cp2WH3+
Basicity of Cp2WH2 comparable to that of ammonia!
18-e WIV
16
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
H
C
H
C=
4
2* H =
2
2* C =
2
+ 
e-count
8
C
H
H
H
H
C
C
H
H
18
H
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
H
C
H
H
C
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
20
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 terms 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
23
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
Al
Cl Cl
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 Cl
Al
Al
Cl Cl
Cl
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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!
26
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
27
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
Cl
Cl
P
Cl
Cl
?
Cl P
Cl
Cl
Cl
P
Cl
Cl
P=
5
5* Cl = 5
+ 
e-count 10
28
SiF62-, SF6, IO65- and
noble-gas halides can
be described in a
similar manner.
Cl
Cl
Cl P
Cl
Cl
Cl
Cl
P=
5
+ chg =
-1
4* Cl = 4
+ 
e-count
8
Electron Counting
A few excess-electron examples
HF2-
H only has a single valence orbital,
This is just an extreme form of
so it cannot form two covalent H-F bonds!
hydrogen bonding. Most other
Write as FH·F , mainly ion-dipole interaction. 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
29
?
F
H
F
H
H
O
F
F
H=
1
1* F =
1
+ 
e-count
2
O
H
O
O
O
H
Electron Counting
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.
30
Electron Counting
Electron-counting exercises
Me2Mg
ZnCl4
ZrCl4
Co(CO)4V(CO)6PdCl(PMe3)3
Ni(PMe3)Cl4
31
Pd(PMe3)4
Pd(PMe3)3
ZnMe42Mn(CO)5V(CO)6
RhCl2(PMe3)2
Ni(PMe3)Cl3
MeReO3
OsO3(NPh)
OsO4(pyridine)
Cr(CO)6
Zr(CO)64+
Ni(PMe3)2Cl2
Cl Pd
Me3P
-
PMe3
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.
32
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
33
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
34
Electron Counting
Examples - main group elements
CCl4
Cl
Cl
Cl
AlCl4-
Cl
Cl
C
Cl
C
Cl
Cl
Cl
Cl
Cl
no chg =
0
4* Cl-C+ = +4
+ 
ox st
+4
COCl2
O
35
Cl
O
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
MnO4-
Pd
Cl
Cl
Mn
O
O
O
O
O
O
36
Cl
O
O
O
O
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
22+
4* O =Mn = +8
+ 
ox st
+7
Electron Counting
Examples - homonuclear bonds
Cl
C2Cl6
Cl
Cl
C
C
Cl
Cl
Cl
Cl
Cl
Cl
Pt2Cl64-
C
C
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Pt
no chg =
0
+
3* Cl C = +3
1* CC =
0
+ 
ox st
+3
Cl
Pt
Cl
Cl
Cl
Pt
Pt
Cl
Cl
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.
37
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
38
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.
39
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
40
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,
p-acceptor ligands (CO) stabilize lower oxidation states
– oxidation states usually change from m to m-2, m or m+2 in reactions
41
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.
Me2Mg
ZnCl4
ZrCl4
Co(CO)4V(CO)6PdCl(PMe3)3
Ni(PMe3)Cl4
42
Pd(PMe3)4
Pd(PMe3)3
ZnMe42Mn(CO)5V(CO)6
RhCl2(PMe3)2
Ni(PMe3)Cl3
MeReO3
OsO3(NPh)
OsO4(pyridine)
Cr(CO)6
Zr(CO)64+
Ni(PMe3)2Cl2
Cl Pd
Me3P
-
PMe3
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
CH4:
C2H4:
C2H2:
AlCl4-:
Me4Zn2-:
OsO4:
43
4
3
2
4
4
4
B2H6:
4 (B)
1 (terminal H)
2 (bridging H)
Electron Counting
p-system ligands
For complexes with p-system ligands, the whole ligand is usually
counted as 1:
Cl
Zr Cl
Cl
Cl Pd
Cl
CN 4
Cyclopentadienyl groups are sometimes counted as 3,
because a single Cp group can replace 3 individual ligands:
OC
Co
CO
H CO
CO
Co
OC
CO
CN 3 or 5
44
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
45
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
46
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.
47
Electron Counting
Example: protonation of WH6(PMe3)3
+
H
H
Protonation gives WH7(PMe3)3+.
H
Me3P
Could that still be
Me3P W HPMe3
a true polyhydride ?
H H
H
48
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.
Electron Counting