general properties of oerganologametal.ppt

general properties of oerganologametalllic complexes

1. Outline
1. Review on coordination chemistry
2.The 18-electron rule
3.Limitations of 18-electron rule
4.Oxidation number
5.Coordination number and geometry
6.Effect of complexation
7. Differences between metals
Chapter 1. General properties of organometallic
complexes
1.The Organometallic Chemistry of the Transition Metals, Robert H. Crabtree,
3rd Edition, 2001, Chapter 1-2.
2.Organotransition Metal Chemistry, Akio Yamamoto, 1986. Chapter 1-4.
3.Organometallic Chemistry, G. O. Spessard, G. L. Miessler, Prentice-Hall:
New Jersey, 1997, Chapter 1 –3.
References

2. 1. Review on coordination chemistry
Complexesor coordination compoundsare compounds composed
of a metal and ligandswhich donate electrons to the metal, e.g.H
3N: Co:NH
3
H
3N:
:NH
3
H
3NCoNH
3
H
3N
NH
3
NH
3
NH
3
..
..
3+
or
3+NH
3
NH
3
Ligands: a molecule or ion that has at least one electron pair that can be
donated.
The electron pair can be: lone pair, p-bonding electron pair, or
s-bonding electron pair.

3. M
M
M
HH
M
HSiR
3
M
lone pairs, e.g.
p bonding e- pairs, e.g.
s bonding e- pairs, e.g.
H
3N:H
3N:MH
2O:H
2O:M
.. .. Classification of ligands
A). Based on nature of the donating electron pairs, ligands may be
classified as Lone pair donor, p-bonding electron pair donors, s-bonding
electron pair donors.

4. 1) Lone pair donorsML
NH
2-CH
2-CH
2-NH
2 (en)
.. ..
R-O
-
..
..
: L : :-
In terms of bonding:
..
L
..
L
filledempty
Lor
H
3N: :PR
3
:CO
M
Cl
M M
Note:
•Ligands that donate
electron pairs to form M-L
sbond are also called s
donors.
MO description of Metal-Ligand interactionsL
ligand
M
M L
ligand
M L
ligand
Bonding
Antibonding

5. e.g. OR
-R-O
..
..
:
-
M+
(a)
R-O
..
..
M
R-O
-
:
M
+
(b)
MR-O
M
M M
filled
M
empty
or
OR
-
OR
-
OR
-
OR
-
(a)
Note: although ligands
such as OR
-
can form
pbond with a metal,
we usually don't
indicate such
interaction in writing
the structures.
Ligands that donate electrons to metal to form pbond are called pdonor
How many pbonds can an OR form with a transition metal ion?M OR
M OR
M OR
Some lone pair donor ligands may have orbitals to form
M-L pbonds. ===> pdonor, pacceptor.

6. :CO: CO:
CO:
CO: CO:
+
(a)
+
(b)
empty
M M
M M accept e
-
to form pbond (dpppback bonding)
CO is not only a sdonor, but also a pacceptor.
How many pbonds can a CO form with a transition metal ion?
Ligands that accepting electrons from metal to form p bond are called
pacceptor.

7. :CO: CO:
CO:
CO: CO:
+
(a)
+
empty
M M
M M
CO:+
empty
M CO:
M COM COM COM Can CO function as a pdonor?
TWO

8. Types of lone pair donor ligandsstrong
p acceptor
weak
p bonding
strong
p donor
lone pair
donor
CO, PF
3
NH
3, H
-
CH
3
-
Cl
-
, OR
-
s
p

9. Mo
OC
OC
CO
CH
3
2.38 Å
1.99 Åa) b)IR, (CO)
:CO
2149 cm
-1
COH
3B
2178 cm
-1
OCCr
CO
CO
CO
CO
OC
2000 cm
-1 Evidences for dp-ppinteractions.
* Take M-CO complexes as an example for paccepter
Explanation?:CO :COM COM
COM COM
COM COM
-+ 1s
1p
3s
2p
4s
2s
C O
2p
3s
1p
CC 2s
CO
CO
(anti-bonding)

10. *Take M-OR complexes as an example for pdonor.MoMo
OO
O
O
OO
O
O
t-Bu
t-Bu
t-Bu
t-Bu
120
o
O
Nb
O
Tp
Tp
Tp = trypticyl
180
o
In H
2O, O atom has sp
3
hybridization, leading to the A-O-B angle
being about 104.5. However, in the above two complexes the angles
are 120and 180, respectively. Why ?

11. 2). p-bonding electron pair donorsCO
M
H
2CCH
2
M
HCCH
M M M
How can these ligands interact with M ?
Take CH
2=CH
2as an example. M
1. e- from p(C2H4) --->s (M)
M
2. e- from dp(M) --->p (C
2H
4)
HH
HH
C
C
p
C
C
p
C
C
C
C
HH
HH
M
Is CH
2=CH
2a pdonor or a pacceptor?

12. Hapticity of ligands:
A ligand may have more than one way to bond to a metal center, e.g.M M
or
M
CH
CH
2
CH
2
H
H
HH
H
(Cp)
H
H
HH
H
M
H
H
HH
H
M
M
In describing the number of atoms (n) attached to a metal, a short
hand h
n
is used. e.g.M
(h
1
-C
3H
5)M
M
H
H
M
CH
2
CH
2
(h

-H
2)M (h

-C
2H
4)M M
Ag
+ M
(h

-C
6H
6)M (h

-C
6H
6)Ag
+ (h

-C
6H
6)M

13. O
CH
3
Fe(CO)
3 Ru Fe(CO)
3
MeO
+ Exercise. Give the hapticity of ligands in following complex

14. Notes: Notation of bridging ligand: 
nCl
M M

2-Cl
Cl
M M

3-Cl
M
C
M M

3-CO
M
O
O
M M

2-OR
R M
HB
H
H
H

h
3
M
M
M M
HHH
B
H
M M
HH
B
H H
B
H
H
H
H
M M

h
4

h
2
h
3

15. 3). s-bonding electron pair donors
Relatively fewer stable complexes are known.
Typical examples:M
H
H
h

-H
2
OCW
OC
CO
PR
3
PR
3
H
H
h

-H-SiR
3
M
H
SiR
3
Mn
OC
OC
H
R
3Si
agostic C-H
Me
2PTi
Me
2P
CH
2
Cl
Cl
M
H
C
Cl
CH
2
H
orM
H
C

16. How can these ligands interact with M? Take H
2as an example. M
H
H
H
H
M
s-bonding p-bonding
Further Notes:
* Relative basicity of electron pairs:
lone pairs > pbonding electron pairs > sbonding electron pairs
* Therefore usual order of binding ability:
lone pair donor > pbonding electron pair donor > sbonding electron pair donor.
Consequence:M
H
H
M
CH
2
CH
2
MPR
3
CH
2
CH
2
CH
2
CH
2
H
H
:PR
3
Is H
2a pdonor or a pacceptor?

17. * Some ligands have several types of electron pairs.
The types of e
-
pairs to be used to bind metal will depend on
metals. e.g.R
CO
H
O
C
R
H
O
C
R
H
low oxidation state M high oxidation state M
M
M

18. B) based on the nature of bonding interaction, ligands may be
classified as soft or hard.
Hard ligands: have low polarizability, especially those containing
period 2 donor atoms N, O, F, e.g.O
2H, NH
3, F
-
Soft ligands:have high polarizability, they include:
a). Those with period three or subsequent donor atoms,
e.g.Cl, Br, S, P.
b). p-acceptors, e.g.CO, CS (carbon sulfide), H
2, CN
-
(cyanide)
c). Those containing p-electrons, e.g.CC CC

19. Importance of the concepts of soft and hard ligands:
* Hard ligands tend to form stable complexes with hard metal ions,
(usually those at high oxidation state, e.g. Al
3+
, Fe
3+
, Cu
2+
, Ti
4+
, Pt(IV)).
* Soft ligands tend to form stable complexes with soft metal ions,
(usually those at low oxidation state, e.g. Mn(I), Co(I), Fe(II), Pt(II),
Pt(0) .......)
Examples:
AlF
6
3-
, Ti(OR)
4are very stable complexes, but Pt(0)--F, Pt(0)--OR,
W(0)--F are very rare. ClPt
Cl
Cl
-
stable
Pt(IV)
very rare
(Olefins are soft
base; Pt(II) is soft
but Pt(IV) is hard)

20. Exercise. explain the following facts based on the concept ofSoft-Hard
Acid-Base.
1) W(CO)
6is air stable, but W(NH
3)
6has never been observed.
2).
3). Low oxidation state complexes (most organometallic compounds) are
often air-sensitive, but are rarely water sensitive.NH
2
(NH
3)
5Os
II
2+
- e
- NH
2
(NH
3)
5Os
III
3+
NH
2(NH
3)
5Os
III
3+
slow

21. Further notes on ligands
General presentation of Ligands:
L: form coordination bond with a metal (e.g., CO, PR
3)
X: form covalent bond with a metal (i.e., H, Cl, CH
3)
Types of ligand coordination:
Terminal: Ligand is bound to only one metal center (L-M or X-M)
Bridging (μ): Ligand is attached to different metal centers
Hapticity (η): Ligand attached to a metal center through more than one
atoms
MXL
5
MX
2(L
2) MXL(LX)M
HB
H
H
H

h
3
M
B
H
H
H
H
M M

h
4

22. Further notes on ligands
Chelation: Ligand attached through more than one atom usually separated
by one or moreatoms. Chelating ligands are sometimes classified as being
bidentate(2 points of attachment),tridentate(three points of attachment), or
tetradentate(4 points of attachment).
Kappa convention (κ): The kappa convention is sometimes used to
indicate the coordinatingatoms of a polydentate ligand.

23. 2. The 18 electron rule
1)Thermodynamically stable transition metal organometallic
compounds are formed when the sum of the metal d electrons plus
the electrons supplied by the ligands equals 18.
In this way, the metal formally attains the electronic
configuration of the next noble gas.Ni
CO
CO
OC
OC
Fe CO
OC
OC
CO
CO
Cr COOC
OC
CO
CO
CO
The 18 electron rule = 18 VE rule = inert gas rule = effective
atomic number rule (EAN rule). e.g.
Saturated complexes: 18 VE complexes
Unsaturated complexes: <18 VE complexes
10 + 4x2 = 18
8 + 5x2 = 18
6 + 6x2 = 18
Why 18?

24. 2) Ways to count valence electrons (VE)
# VE = valence e
-
of M (or M
n+
) + e
-
from ligands
Two Models: covalent modeland ionic model
Covalent model: Both M and L are considered as neutral
# VE = valence e
-
of M + e
-
from ligands + charge
e.g.Fe
Cr
H
CO
CO
CO
OC
OC
2 2 x 5e
Fe 8e
18e
5 CO5 x2e
H 1e
Cr 6e
1
-
1e
18e
C
5H
5
For TM, valence e
-
of M = group number

25. For TM, valence e
-
of M = group number.
e.g. Fe, 8; Pt, 10

26. Ionic model: Metal complexes are formed from M
n+
+ L + X
-
# VE = valence e-of M
n+
+ e-from ligands Fe
Cr
H
CO
CO
CO
OC
OC
2 2 x 6e
Fe
2+ 6e
18e
5 CO 5 x 2e
H
-
2e
Cr 6e
18e
C
5H
5
-

27. Exercise:
Determine the valence electron count of the following complexes. CO
OC
OC
CO
CO
Mn Fe Fe
CO
CO
OC
O
C
OC
C
O
O
C
COOC Mn Mn
CO
CO
CO
CO
OC
OC
CO
CO
OC
CO

28. Rh
Cl
Rh
Cl
Rh
Cl
Rh
Cl O
CH
3
Fe(CO)
3 + O
CH
3
Fe(CO)
3

29. Fe(CO
3
MeO
+ Ru Cr

30. Fe
OC
OC Mn
COOC
OC CO
CO
W

31. Rh Rh
OC
CO (CO)
3(PPh
3)Fe Ir(CO)
2(PPh
3)
Ph
2
P CoCo
N
N
O
O

32. Do the following complexes follow the 18e rule?
ReH
7(PPh
3)
2-
Cr
CO
CO
CO
H
OC
OC
Cr
CO
CO
CO
CO
OC Rh
H
H
H
+

33. R
3PRu
H
H
H
R
3P
PR
3
BH
H
R
3PRu
H
H
H
R
3P
PR
3
BNMe
3
H

34. Additional exerciseIr
N
COOC Ir
OC CO
BPh
3 Ir
OC CO
PPh
3

35. 3. Limitations of the 18 electron rule
Most OMC follow 18e rule. But there are many stable compounds that
do not have 18 valence electrons. e.g.
WMe
6(12e) Pt(PCy
3)
2(14e) [Cu(NH
3)
6]
2+
(21e)
RhCl(PPh
3)
3(16e)
Some general observations:
1).Many main group complexes do not follow 18e rule:
e.g.ZnMe
2(14e), Cp
2Be (12e),
IF
7 (24e), SbF
6
-
(22e)
Question: When the 18e rule works?

36. U
U: 7s
2
5f
3
6d
1
2x8+6 = 22e
LuMe
Lu: 6s
2
4f
14
6d
1
2x5+1+17 = 28e 2).F-block metals do not follow 18e rule. e.g.
Why? because electrons can go to (n-2)f orbitals)

37. 3).Transition metal complexes ===> three classes
Classnumber of valence electrons 18e rule
I ....16, 17, 18, 19, 20 …… not obey
II ....16, 17, 18 not obey
III 18 obey

38. Class I: 3d metalswith weak-field ligands
Class I n(VE) = 18 ~ 22
n(d) n(VE)
TiF
6
2-
0 12
VCl
6
2-
1 13
V(C
2O
4)
3
3-
2 14
Cr(NCS)
6
3-
3 15
Mn(CN)
6
3-
4 16
Fe(C
2O
4)
3
3-
5 17
Co(NH
3)
6
3+
6 18
Co(H
2O)
6
2+
7 19
Ni(en)
32+ 8 20
Cu(NH
3)
6
2+
9 21
Zn(en)
3
2+
10 22

39. Class II: 4d and 5d metals with weak-field ligands
Class II n(VE) 18
n(d) n(VE)
ZrF
6
2-
0 12
WCl
6 0 12
WCl
6
-
1 13
WCl
6
2-
2 14
TcF
6
2-
3 15
OsCl
6
2-
4 16
PtF
6 4 16
PtF
6
-
5 17
PtF
6
2-
6 18
PtCl
4
2-
8 16
Metals in relatively
High oxidation state

40. Class III: Complexes with pgood acceptors
Class III n(VE) = 18
n(d) n(VE)
V(CO)
6
-
6 18
CpMn(CO)
3 7 18
Fe(CN)
6
4-
6 18
Fe(CO)
4
2-
10 18

41. Explanation:LM L
L
L
L
L
Take Oh as an example
MO levels in the absence of pacceptors.
* If Dis small, eg* can be
occupied.(D< pairing energy)
===>
e
g* (antibonding): 0 -4e
t
2g(nonbonding): 0 –6e
a
1g, t
1u, e
g(bonding): 12e
Total e-:
Minimum # of e-:
Maximum # of e-:
3d metal complexes with weak field ligands e.g.H
2O, NH
3and
Cl
-
belong to this class (class I).metal
orbitals
(n-1)d
ns
np
Ligand
orbitals
(a
1g+e
g+t
1u)
(a
1g+e
g+t
1u)
t
1u*
a
1g*
e
g*
t
2g
D

42. MO levels in the absence of pacceptors.
* If Dis large, eg* can be occupied.
(D> pairing energy)
===>
e
g* (antibonding): 0 e
t
2g(nonbonding): 0 –6e
a
1g, t
1u, e
g(bonding): 12e
Total e-:
Minimum # of e-:
Maximum # of e-:
Complexes of 4d and 5d metals in high oxidation state belong to
this class (class II).metal
orbitals
(n-1)d
ns
np
Ligand
orbitals
(a
1g+e
g+t
1u)
(a
1g+e
g+t
1u)
t
1u*
a
1g*
e
g*
t
2g
D

43. * In the presence of p-acceptors:
Dis large. t
2gare bonding MOs, and prefer to be occupied.
===> e
g* (strongly antibonding): 0 e
t
2g(bonding): 6 e
a
1g, t
1u, e
g(bonding): 12 e
===>Total e:
Since
organometallic
compounds
usually have p-
acceptors, Dis
large and t
2gare
bonding MOs.
They usually
have 18e valence
electrons (class
III).metal
orbitals
(n-1)d
ns
np
Ligand
orbitals
(a
1g+e
g+t
1u)
(a
1g+e
g+t
1u)
t
1u*
a
1g*
e
g*
t
2g
metal
orbitals
(n-1)d
ns
np
Ligand
orbitals
t
1u*
a
1g*
e
g*
p*
t
2g*
t
2g
a
1g+e
g+t
1u
D
D (a
1g+e
g+t
1u)

44. Note,just like the octet rule, the 18-electron rule is not an
absolute requirement. There are manyexceptions.
Common exceptions to the 18 electron rule:
d
8
metals:
The d
8
metals (groups 8 -11) have a
tendency to form square-planar 16
electroncomplexes.
This tendency is weakest for group 8
(Fe(0), Ru(0), and Os(0)) and is very
strong forgroups 10 and 11 (Pd(II),
Au(III)). d
s
p
LM L
L
L
a
1g
a
2u
e
u
x
y
z
a
1g
e
g
b
1g
b
2g
1a
1g
1a
2u
2a
1g
3a
1g
1e
u
2e
u
1b
1g
2b
1g
+1b
2g
1e
g
Because 2b
1gorbital is very high-lying and is usually empty

45. d
0
metals:The high-valent d
0
complexes often have lower
electron counts than 18.
Complexes with bulky ligands:Sterically demanding
ligands will often result in lower than expected electron
counts.

46. > 18 electron complexes: Complexes with formally 19 or 20
electrons are known, but they areusually unstable, or
adopt alternate configurations.

47. The 18 Electron Rule Is Empirically Justified
• The rule is
particularly useful for
Groups 6-8
16 e
-
Compounds 14 e
-
Compounds

48. The oxidation stateof a metal in a complex is simply the charge that
the metal would have on the ionic model.
4. Oxidation number of metals and d electron count
e.g.What is the oxidation number of metals in the following complexes?Fe
H
PR
3
PR
3
R
3P
H
Os
PPh
3
PPh
3
Cl
Cl
OC
OC
WMe
6
H
H
d electron count: # of d electrons in the valence shell
= group # -oxidation state.

49. Problems with O. S. :
a) Ambiguous oxidation statesFe(CO)
3
Fe(CO)
3
or Fe(CO)
3
Fe(0) Fe(II) Fe(IV)
or (Ph
3P)
2Pt (Ph
3P)
2Pt
Pt(0) Pt(II)
Os
PPh
3
PPh
3
OC
OC
CH
2
O
Os
PPh
3
PPh
3
OC
OC
CH
2
O
Os(0) Os(II)
Convention used in this course.

50. Consider complexes W(CO)
6and WH
6(PMe
3)
3.
Which one would you expect to have a higher positive charge on W?
b) charge density and formal oxidation states
There is no strict correlation between charge density and
formal oxidation states!Oxidation states in organometallic complexes
are merely formalisms that may bear little resemblance to the actual positive
charge on the metal.
Another example

51. Any Uses of formal oxidation states?
Oxidation number usually can not be higher than
group member!
==> predict if a compound/intermediate is possible
==> Help to formulate the structure of a compound.
e.g.
(1) Are the following species possible?
WMe
6TiMe
6VH
7(PMe
3)
2
(2) WH
6(PMe
3)
3+ HBF
4------> [WH
7(PMe
3)
3] BF
4
Which of the following is unlikely the structure for [WH
7(PMe
3)
3]
+
?W
H
H
+
H
H
H
P
P
P
H
H
W
H
H
+
H
H
H
P
P
P
H
H
W
H
H
+
H
H
H
P
P
P
H
H
(a)
(b) (c)

52. Co
Cl
NH
3
NH
3
Cl
H
3N
H
3N ReH
9
2-
six-coordinated 9-coordinated 5. Coordination number (C.N.) and geometry
a)It is easy to define C.N. for complexes with lone pair donors.
* Monodentate ligand L :
C.N. = # of L present = # of atoms bound to metal
= # of electron pairs involved in M-L sbonds.
e.g.

53. •Polydentate ligands:
C.N. = # of atoms bound to metal
= # of electron pairs involved in M-L sbonds.
# of L presentCo
Cl
CH
3
P
Ph
2
Ph
2
P
Ir
Ph
2P
Cl
PPh
3
Ph
3P
H
4-coordinated 6-coordinated

54. b)For Organometallic compounds, it is difficult to define C.N. e.g.Fe Fe
OC
CO
CO
# of L
Fe
Fe
OC
CO
CO
# of M-L
Fe
Fe
OC
CO
CO
# of e pairs in M-L bond
C.N. = 1
C.N. = 5
C.N. = 3
C.N. = 1
C.N. = 4
C.N. = 2
2 + 3
2
2
+ 3
+ 3
Convention used in this course.
We normally adopt the e-pairs in M-L bonds as # of C. N..

55. Fe
OC
OC
M
+
H
+
MH
+ Exercise. What is the coordination number of following complexes?

56. Further notes on coordination numbers:
a).For TM, C.N. ≤ 9 , why?
b).d
n
<==> C.N. and geometry
d
n
C.N. geometrical structure
d
6
6 prefer octahedral
d
8
4 prefer square planar
d
0
, and d
10
4 prefer tetrahedral
c).Each C.N. is associated with one or more geometries.

57. Further notes on coordination numbers:
a).For TM, C.N. ≤ 9 , why?
TM has 9 valence orbitals ((n-1)d, ns, np).
b).d
n
<==> C.N. and geometry
d
n
C.N. geometrical structure
d
6
6 prefer octahedral
d
8
4 prefer square planar
d
0
, and d
10
4 prefer tetrahedral
c).Each C.N. is associated with one or more geometries.

58. Some common coordination numbers and geometries
2 linear
M Mn(CH
2SiMe
3)
2
3 trigonal M Al(mesityl)
3
T-shape M Rh(PPh
3)
3
+
4 square planarM RhCl(CO)(PPh
3)
2
tetrahedron
M
Ni(CO)
4 d
8
Prefered d
n
d
8
d
0
,
, d
5
, d
10
d
0
,
, d
5
, d
10

59. 5
Trigonal
bipyramidal M
Fe(CO)
5
square
pyramidal
M
Co(CNPh)
5
2+
6 Octahedron M Mo(CO)
6
trigonal
prism M
WMe
6 (benzonitrile)
d
8
,
,d
6
d
6
,
,d
7
d
6
,
,d
3
d
0

60. 6. Effect of complexation
M-L:
Complexation of L on M may cause:
* the change of electron density distribution on L
* new reactivity of L: unreactive ==> reactive
Examples:
a.Change the electron density on L.
σ donation will reduce electron-density of L.
π-accepting will increase electron-density of L.H
2CCH
2
Fe
OC
OC
Fe
OC
OC
R
N. R.+ Nu
-
but,
+ + R
-
Complexation reduce electron-density of olefin

61. Another example,Mn
OC
CO
CO Mn
OC
CO
CO
H
H
H
H
HH
H
H
H
H
HH
H
H
E
E
H
HH
H
H
+
+
E
+
- H
+
Nu
-
N. R.
E
+
N. R.
but,
NaBH
4
(H
-
)
Complexation reduces the e-density on C
6H
6.
Reactivity towards Nu
-
: increase
Reactivity towards E
+
: decrease

62. Change the electron density distribution on LM:NN:
s-donation
:NN:
:CO:
 = 0  = 0
unreactive
MCOMNN
Upon complexation
NN:
M:CO:
CO:
p- backdonation
CONN
M CO
M NN
overall
MCOMNN
+ +
Nu
- E
+
Nu
- E
+
M
M
MM
(e
-
to on both atom)
(e
-
to mainly C atom)

+

+

-

-

-

-

63. 7. Differences between metals
1) Electronegativity differences:
Sc Ti V Cr Mn Fe Co Ni Cu
1.3 1.5 1.6 1.6 1.6 1.8 1.9 1.9 1.9
Y Zr Nb Mo Tc Ru Rh Pd Ag
1.2 1.3 1.6 2.1 1.9 2.2 2.3 2.2 1.9
La Hf Ta W Re Os Ir Pt Au
1.1 1.3 1.5 2.3 1.9 2.2 2.2 2.3 2.5
Moving from left to right, the electronegativity of the elements increases
substantially.

64. 2). Trend in the stability of high oxidation states.
From left to right decrease
From top to bottom increase
e.g.Ti(II) unstable Fe(II) stable
Ti(IV) stable Fe(VIII) not exist
Hf(IV) very stableOs(VIII) stable
Stability of high oxidation states
Early transition metals are electropositive, so they readily lose all their
electrons to give d
0
centers (e.g. Zr(IV), Ta(V)). Low-valent early transition
metals, such as Ti(II) and Ta(III), are easily oxidized.
Late transition metals are more electronegative, thus they prefer lower
oxidation states (i.e., Rh(I) compared to Rh(III)).

65. 2). Trend in the relative energy of d orbitals and backdonation.
Question 1: Which of the following species has d electrons of highest energy?
(1)(a) Ti(0)(b) Ti(II) (c) Ti(III)
(2) (a) Ti
2+
(b) Fe
2+
(c) Ni
2+
(d) Zn
2+
(3) Fe
Ru
Os

66. Question 2. How would you explain the following trend in the
IR dada of (C-O) (in cm
-1
)?
a). V(CO)6 Fe(CO)
5 CO [Ag(CO)]
+
1976 2023 2057 2204
b) Cr(CO)
6 2000
W(CO)
6 1998

67. Effects of changing net ionic charge, ligands, and metal on the pbasicity of a metal
carbonyl, as measured by (CO) values (cm-1) of the highest frequency band in the IR
spectrum
Changing Metal
V(CO)
6
1976
Cr(CO)
6
2000
Mn
2
(CO)
10
2000
Fe(CO)
5
2023
Co
2
(CO)
8
2044
Ni(CO)
4
2057
Changing Net Ionic Charge in an Isoelectronic Series
[Ti(CO)
6
]
2-
1747
[V(CO)
6
]
-
1860
Cr(CO)
6
2000
[Mn(CO)
6
]
+
2090
[Fe(CO)
6
]
2+
2204
Replacing p-Acceptor CO groups by Non-p-Acceptor Amines
[Mn(CO)
6
]
+
2090
[(MeNH
2
)Mn(CO)
5
]
+
2043
[(en)Mn(CO)
4
]
+
2000
[(tren)Mn(CO)
3
]
+
1960

68. M
H
H
H
H
M
2. e- from dp(M) ---> s (H2) 1. e- from s(H2) --->s (M)
H
H
M
s(H
2)
s*(H
2)
Bonding picture of M(H
2) Question 3. Compounds of the formula MH
4P
3(M = Fe, Ru and Os, P = PR
3)
are known to have the following structure.P FeH
P
P
H
HH
PRuH
P
P
H
HH
POs
H
P
P
H
H
H
why not
POsH
P
P
H
HH
?