Transcript Electron Count Oxidation State Coordination Number

Electron Count
Oxidation State
Coordination Number
• Basic tools for understanding
structure and reactivity.
• Doing them should be “automatic”.
• Not always unambiguous  don’t just
follow the rules, understand them!
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Counting Electrons
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The basis of counting electrons
• 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
• Every orbital wants to be “used", i.e. contribute to binding
an electron pair.
• Therefore, every element wants to be surrounded
by 2/8/18 electrons.
• The strength of the preference for electron-precise
structures depends on the position of the element in the
periodic table.
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The basis of counting electrons
• Too few electrons:
An empty orbital makes the compound very electrophilic,
i.e. susceptible to attack by nucleophiles.
• Too many electrons:
There are fewer covalent bonds than one would think (not
enough orbitals available). An ionic model is required to
explain part of the bonding. The "extra" bonds are
relatively weak.
• Metal-centered (unshared) electron pairs:
Metal orbitals are fairly high in energy. A metal atom with
a lone pair is a strong s-donor (nucleophile) and
susceptible to electrophilic attack.
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Use a localized (valence-bond) model
to count electrons
H2
H
H
Every H has 2 e. OK
H
CH4
H has 2 e, C 8. OK
H
C
H
H
H
NH3
N
N has 8 e. Nucleophile! OK
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H
Counting Electrons
H
4
C2H4
C has 8 e. OK
H
H
C
C
H
H
singlet CH2
H
C has only 6 e, and an empty pz orbital:
extremely reactive ("singlet carbene").
Unstable. Sensitive to nucleophiles and
electrophiles.
H
H
triplet CH2
C
C has only 6 e, is a "biradical" and extremely
reactive ("triplet carbene"), but not especially
for nucleophiles or electrophiles.
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C
Counting Electrons
H
5
CH3+
C has only 6 e, and an empty pz orbital:
extremely reactive. Unstable. Sensitive to
nucleophiles.
H
H
C
H
CH3-
H
C has 8 e, but a lone pair. Sensitive to
electrophiles.
C
H
H
ClCl has 8 e, 4 lone pairs. OK Somewhat sensitive
to electrophiles.
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Counting Electrons
Cl
6
BH3
H
B has only 6 e, not stable as monomer,
forms B2H6:
H
B
H
B2H6
H
B has 8 e, all H's 2 (including the bridging
H!). 2-electron-3-center bonds! OK
H
H
B
B
H
H
H
AlCl3
Al has only 6 e, not stable as monomer,
forms Al2Cl6:
Cl
Cl
Al
Cl
Al2Cl6
Al has 8 e, all Cl's too (including the
bridging Cl!). Regular
2-electron-2-center bonds! OK
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Counting Electrons
Cl
Cl
Al
Cl
Cl
Al
Cl
Cl
7
2 MeAlCl2  Me2Al2Cl4
Me
Me
Cl
Al
Cl
Cl
Cl
Al
Cl
Cl
Al
Me
Al
Cl
Cl
Me
2-electron-3-center bonds are a stopgap!
H3B·NH3
N-B: donor-acceptor bond (nucleophile NH3 has
attacked electrophile BH3).
H
Organometallic chemists are "sloppy" and write
H
H
H
H
H
H
H
H
H
.
Writing
or
would be more
H B N
H B N
H B N
H
H
H
H
H
H
correct (although the latter does not reflect the
“real” charge distribution).
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H
H
Counting Electrons
H
B
H
N
H
8
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
Cl
(hyperconjugation).
Cl
Easy dissociation in PCl3 en Cl2.
Cl
Cl
Cl
P
Cl
P
Cl
Cl
Cl
?
Cl
HF2Write as FH·F-, mainly ion-dipole interaction.
F
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H
F
?
F
H
Counting Electrons
F
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How do you count?
1. Number of valence electrons
(from periodic table)
2. Correct for charge, if any
(only if it 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
5. Add
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Counting Electrons
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Examples: counting electrons
H
BH4-
H
B H
H
B
4H
tot
OK
=
=
=
=
3
1
4
8
C O
H
C
= 4
1=O = 2
2H
tot
OK
-
Cl
Cl Pd NH3
Cl
Pd
=10
-
=1
3Cl = 3
1NH3 = 2
tot
=16
could have additional 2 e
(Pd-Cl p-bond?)
H
H2CO
PdCl3(NH3)-
= 2
= 8
PMe3
Cl Ru Cl
Me3P
CH2
Ru
= 8
2Cl
= 2
2PMe3 = 4
1CH2 = 2
tot
=16
could have additional 2 e
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Counting is not always trivial
Cl
Cl Pd
Cl
Cl
is
NH3
Cl Pd
Cl
NH3
Pd
=10
2=2
3Cl = 3
1CH2 = 1
tot
=16
could have additional 2 e
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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.
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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:
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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Calculating the
formal oxidation state
1. Start with the formal charge on the metal
2. Ignore dative bonds
3. Ignore bonds between atoms of the same element
(this one is a bit silly)
4. Assign every covalent electron pair to the most
electronegative element in the bond: this produces
+ and – charges (usually + at the metal)
5. Add
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Examples: oxidation states
CCl4
charge C= 0
4C-Cl:
C+-Cl- =+4
tot
=+4
COCl2
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AlCl4-
charge Al = -1
4Al-Cl:
Al+-Cl- = +4
tot
= +3
charge C= 0
2C-Cl:
C+-Cl- =+2
1C=O:
C2+-O2- =+2
tot
=+4
MnO4-
PdCl42-
charge Pd = -2
4Pd-Cl:
Pd+-Cl- = +4
tot
= +2
charge Mn = -1
4Mn=O:
Mn2+-O2- = +8
tot
= +7
Counting Electrons
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Examples: oxidation states
C2Cl6
charge C = 0
3C-Cl:
C+-Cl- =+3
tot
=+3
trivalent carbon ?
MgMe4
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Pt2Cl64-
charge Pt = -2
3Pt-Cl:
Pt+-Cl- = +3
tot
= +1
univalent Pt ?
charge Mg = 0
4Mg-Me:
Mg+-Me- = +4
tot
= +4
impossible, Mg has only
2 valence electrons!
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The significance of
an oxidation state ?
Oxidation states are formal.
However, they do give an indication whether a
structure or composition is reasonable (apart from
the M-M complication).
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Acceptable oxidation states
For group n or n+10:
–
–
–
–
–
–
–
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never >+n or <-n (except group 11: frequently +2 of +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, pacceptor 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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Coordination number
Simply the number of atoms directly bonded to the
atom you are interested in, regardless of
bond orders etc.
CH4:
4
B2H6:
4 (B)
C2H4:
3
1 (terminal H)
C2H2:
2
2 (bridging H)
AlCl4 :
4
Me4Zn2-: 4
OsO4:
4
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Coordination Number
For complexes with p-system ligands, the whole
ligand is usually counted as 1:
Cl
Cl Pd
Cl
Zr Cl
Cl
C.N. 4
Cyclopentadienyl groups are sometimes counted as 3,
because a single Cp group can replace 3 individual ligands:
OC
Co
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CO
H CO
CO
Co
OC
CO
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Coordination Number
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. So are
very low coordination numbers (<4) together with a
“too-low” electron count.
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Coordination number and
coordination geometry
C.N.
2
3
4
5
6
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"Normal" geometry
linear or bent
planar trigonal, pyramidal, "T-shaped"
square planar, tetrahedral
square pyramid, trigonal bipyramid
octahedron
Counting Electrons
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Illustration:
protonation of WH6(PMe3)3
Could WH6(PMe3)3 be
H
Me3P H H
Me3P W PMe3
H H
H
?
Count W: 18 VE (OK), oxidation state 6 (OK),
coordination number 9 (very high). Possible.
Protonation gives WH7(PMe3)3+.
+
Could that be
H
Me3P H H
Me3P W HPMe3
H H
H
?
Count W: 18 VE (OK), oxidation state 8 (too high),
coordination number 10 (extremely high). W+ must form 7
covalent bonds using only 5 electrons. That will not work!
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Exercises
Give electron count and oxidation state for the following
compounds. Draw conclusions about their (in)stability.
Me2Mg
ZnCl4
ZrCl4
Co(CO)4V(CO)6PdCl(PMe3)3
Ni(PMe3)Cl4
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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+
Cl Pd
Me3P
PMe3
Ni(PMe3)2Cl2
Counting Electrons
-
BMe3
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