Transcript Slide 1

Lecture 23
Coordination Chemistry
1) Electron counting in transition metal complexes
•
“Neutral atom” method.
(a) Write down the number of electrons in the neutral atom valence shell, which
corresponds to its group number: 3 … 12.
(b) Account for the ionic charge if the species is not neutral. This will reduce (positive
ion) or increase (negative ion) the number of electrons available in the metal
valence shell;
(c) Add the number of electrons contributed by ligands (see table on p. 3).
(A)
Complex (A)
Cl
NH3
Number of e’s
Pd
H3N
Cl
(B)
H3N
NH3
Pd
H3N
NH3
Complex (B)
2+
Number of e’s
Pd
10
Pd
10
charge
0
charge
-(+2)
2 Cl
2·1=2
4N
4·2=8
2N
2·2=4
Total
= 16
Total
= 16
2) “Oxidation state” method
• “Oxidation state” method.
(a) Write down the number of electrons in the neutral atom valence shell, which
corresponds to its group number: 3 … 12.
(b) Determine the formal oxidation state of the metal. This will reduce (positive
oxid. state) or increase (negative oxid. state) the number of electrons
contributed by the metal
(c) Add the number of electrons contributed by ligands.
Number of e’s
Cl
2+
NH3
Pd
NH3
•
Cl
Pd
10
Oxid. state
-(+2)
2 Cl-
+2·2=4
2N
+2·2=4
Total
= 16
If you are given a structure with delocalized bonds, counting electrons may
be problematic:
S
S
Ni(0), Ni(I), Ni(II), Ni(III) or Ni(IV)
Ni
compound?
S
S
3) Counting electrons: ligand contributions
•
Each ligand is characterized by the number of electrons it can contribute to the metal
valence shell. If a ligand is bridging between n metal atoms, the number of electrons
it donates to a single atom is n times less. Here you have some data for the
“oxidation state” / “neutral atom” methods:
L
Charge
# of e’s donated
H
-1/0
2/1
ReH92-
Alkene
0/0
2/2
Ir(C2H4)4Cl
X (F, … I)
-1/0
2/1
PtCl62-
Alkyne
0/0
2,4
W(PhC≡CPh)3CO
m-X (F,.. I)
-1/0
4/3
Rh2Cl2(C2H4)4
=CR2
-2/0
4/2
(tBuCH2)3Ta=CHtBu/..
OR
-1/0
2/1
Fe(OPh)63-
≡CR
-3/0
6/3
(tBuCH2)3W≡CtBu
NR2
-1/0
2/1
Ti(NMe2)4
CO
0/0
2/2
Fe(CO)5
Alk
-1/0
2/1
PtMe62-
C≡NR
0/0
2/2
Fe(CNtBu)5
h1-Ar, Vin
-1/0
2/1
(CO)5MnPh
h1-N2
0/0
2/2
[Ru(NH3)5(N2)]2+
h1-Allyl
-1/0
2/1
[Pd(PMe3)3C3H5]+
N≡CR
0/0
2/2
[Cu(NCMe)4]+
h3-Allyl
-1/0
4/3
Ni(C3H5)2
=NR
-2/0
4/2
ReCl3(PPh3)2NMe
h1-Cp
-1/0
2/1
HgCp2
≡N
-3/0
6/3
OsNCl5
h5-Cp
-1/0
6/5
FeCp2
PR3 (NR3)
0/0
2/2
Ni(PF3)4
L
Chg
# of e’s donated
h6-Arene
0/0
6/6
Cr(C6H6)2
m-Alk
-1/0
2/1
Al2Me6
h2-Arene
0/0
2/2
CpRh(PMe3)L
m-CO
0/0
2/2
Co2(CO)8
4) 18 electron rule
•
For d-elements with 9 orbitals in their valence shell (s, three p and five d orbitals)
normally there is a maximum of 18 electrons in the shell (ns2(n-1)d10np6 configuration):
[V(CO)6]-; Cr(C6H6)2; [Mn(CO)5]-; [Fe(OH2)6]2+; [Co(CN)6]3-; Ni(CO)4; [Cu(NH3)4]+
•
If this number is not reached, the species is coordinatively unsaturated and tend to
add more ligands. It is reflected in higher reactivity and therefore the difficulty with
which coordinatively unsaturated compounds can be isolated.
•
There are multiple examples of transition metal compounds with less (group 3, 4 and
10) or more than 18 electrons in the metal shell:.
Ti
PhH2C
8e
•
CH2Ph
CH2Ph
Me
OAr
CH2Ph
Me
Me
W
ArO
W
OAr
9e
Me
Me
H3N
Pd
Cl
Cl
NH3
Co
Ni
Me
12 e
16 e
19 e
Ru
But these are 18 electron complexes:
W
CO
CO
20 e