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Practice problems
1. For all complexes listed below, determine
a) metal oxidation state b) total number of electrons contributed from metal c) total number of electrons contributed from the ligand set d) total electron count of the complex
Please note: use the ionic model unless asked otherwise and comment on any complexes that
do not obey the 18VE rule or have CN < 6.
i) ( 5 ‐Cp) 2 Fe
ii) [( 5 ‐Cp) 2 Co]
Ionic Model Metal oxidation state: 3+ Metal electron count: 6 Ligand electron count: 6 + 6 Total electron count: 18
Co3+
iii) Co 2 (CO) (^8)
Practice problems
iv) Ru( 2 ‐en) 2 H 2
v) Mn(^4 ‐salen)Cl
Ionic Model Metal oxidation state: 3+ Metal electron count: 4 Ligand electron count (s only): 2 + 2 + 2 + 2 + 2 Total electron count: 14
N N
O O
Mn3+
Cl
A total electron count at the metal of just 14 electrons is predicted using the ionic model but
only considering ‐bonds. Both oxide and chloride ligands are capable of ‐donation to the
empty metal orbitals making up the 4 extra electrons to comply with the 18VE rule. The nature
of this ‐donation can be deciphered using the molecules point group symmetry, the
corresponding character table and the method of systematic reduction of non‐shifted/inverted
‐vectors.
vi) Ti( iso ‐propoxide) (^4)
A total electron count at the metal of just 8 electrons is predicted using the ionic model but only
considering ‐bonds. Each isopropoxide ligand is capable of ‐donation to the empty metal
orbitals making up the 10 extra electrons to comply with the 18VE rule. The nature of this ‐
Practice problems This is a square planar d^8 complex thus the d z^2 orbital is filled preventing axial coordination and
therefore precluding a CN=6 geometry in favor of CN=4. As a result of the reduced coordination number,
a total electron count at the metal of 16 electrons is predicted. This prediction only considers ‐bonds
however when in fact the chloride ligand is capable of ‐donation to the unfilled metal orbitals making
up the 2 extra electrons to comply with the 18VE rule. The nature of this ‐donation can be deciphered
using the molecules point group symmetry, the corresponding character table and the method of
systematic reduction of non‐shifted/inverted ‐vectors.
x) Rh(CO)(PPh 3 ) 2 (C 2 H 2 )
This is a square planar d^8 complex thus the d z^2 orbital is filled preventing axial coordination and
therefore precluding a CN=6 geometry in favor of CN=4. As a result of the reduced coordination number,
a total electron count at the metal of 16 electrons is predicted. This prediction only considers ‐bonds
however when in fact the chloride ligand is capable of ‐donation to the unfilled metal orbitals making
up the 2 extra electrons to comply with the 18VE rule. The nature of this ‐donation can be deciphered
using the molecules point group symmetry, the corresponding character table and the method of
systematic reduction of non‐shifted/inverted ‐vectors.
xi) [(NH 3 ) 5 Ru(‐pyrazine)Ru(NH 3 ) 5 ]5+
This represents an unusual example. With a completely neutral ligand set, the overall 5+ charge must be
shared between both metal centers. As this is a symmetric system the obvious choice would be to share
charge equally as +2.5 on each metal center. Alternatively, the charge may be distributed as +2 and +
as above. Without conducting detailed experimental investigations (UV‐vis‐NIR/X‐
Practice problems ray/computational/electrochemical/ etc.) it is difficult to state the true nature of charge localization in
this complex. Thus, either answer would suffice.
- Relative to a spherical ligand field, in a transition metal complex of O h symmetry
The e (^) g orbitals have lobes that point at the ligands and so will increase in energy. The t2g orbitals have lobes that lie between ligands and so will decrease in energy.
- High coordination numbers are favored by
i) high or low oxidation states ii) small or large atomic radii iii) small or bulky ligands
- High spin O electronic configurations are favored by
i) low or high oxidation states ii) first, second or third row transition metals iii) weak or strong field ligands
- The magnitude of o depends most strongly upon which 3 of the following components
i) the metal ion ii) the attaching ligands iii) the counterion iv) the solvent v) the metal oxidation state
- MO theory is a method for determining molecular structure in which electrons are not
assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule. Ligand field theory (LFT) represents an application of molecular orbital (MO) theory to transition metal complexes. For effective overlap to occur between metal atom orbitals and the SALC’s there are two important requisites. Please select these requisites from the following list:
i) Shape ii) Energy iii) Symmetry iv) Size v) Occupancy
Practice problems
- Explain the below trend in CO bond vibrational frequency v (CO) using the Dewar‐Chatt‐
Duncanson model.
v (CO) cm ‐ 1
[Ti(CO) 6 ] 2 ‐ 1748
[V(CO) 6 ]‐^1859
Cr(CO) 6 2000
[Mn(CO) 6 ] +^2100
[Fe(CO) 6 ] 2+ 2204
‐ all complexes have CN = 6 and are of O h symmetry.
‐ all complexes are metal hexacarbonyls
‐ all complexes have 6 d ‐valence electrons
As we progress from Ti(2‐) to V(‐) to Cr(0) to Mn(1+) to Fe(2+) the electron count on the nuclei
is identical however the proton count at the central nuclei increase resulting in a net increase in
electronegativity as we follow the above trend. The increase in electronegativity reduces the
extent of ‐back‐donation from the central metal d xy , d xz and d yz orbitals such that population
of the t 2g ligand SALC orbitals is reduced. Reduced population of the t 2g ligand SALC
orbitals on CO therefore gives rise to a higher bond order between carbon and oxygen
(somewhere between 2 and 3). This can be observed via an increase in the v (CO) stretching
frequency in the IR spectrum. This is most obvious, for example, if we compare the least
electronegative metal center Ti(2‐) where v (CO) = 1748 cm‐^1 relative to the most
electronegative metal given in Fe(2+) where v (CO) = 2204 cm‐^1.
- Using any combination of s , p or d orbitals please demonstrate overlap resulting in
i) a bond has no nodal planes at the internuclear axis and is thus symmetric with respect to C 2 rotation about the bond axis, e.g.
p (^) p p p
bond
bond
Practice problems
ii) a bond is characterized by a single nodal plane at the internuclear axis and is thus asymmetric with respect to C 2 rotation about this axis, e.g.
iii) a bond has two nodal planes at the internuclear axis (one containing the internuclear axis and a second perpendicular to this axis).
- Using molecular orbitals draw the bonding scheme for the following classes of metal
carbene complex, inclusive of any ‐bonding:
i) Metal hydride
Practice problems
vii) Fischer carbyne
viii) Schrock carbyne
ix) Metal alkene
Z
Y
X
Metal dx^2 -dy 2 doublet carbyne sp
bond
Z
Y
X
Metal dyz doublet carbyne p z
-bond
Y
Z
Metal dz^2 doublet carbyne^ sp
bond
X
C R
C R
C R
Z
Y
X
Metal dxy doublet carbyne^ p x
-back-donation
C R
Z
Y
X
Metal dx^2 -dy 2 quartet carbyne^ sp
bond
Z
Y
X
Metal dyz quartet carbyne p z
-bond
Y
Z
Metal dz^2 quartet carbyne^ sp
bond
X
C R R
Z
Y
X
Metal dxy quartet carbyne^ p x
-back-donation
C R C R
C
Practice problems
x) Metal alkyne
xi) Metal allyl
Z
Y
X
Z
Y M
X
Z
Y
X
Metal dx^2 -dy^2 bond^3 -allyl
Metal dxy^3 -allyl
Metal dz^2 3 -allyl
bond
bond
Z
Y
X
Metal dyz bond^3 -allyl
Z
Metal dx^2 -y 2 alkyne
CR
CR
Z
Y X
Metal dyz alkyne
back-donation
CR
CR
bond bond
Z
X
Y
Metal dxy alkyne
CR
CR Metal dyz alkyne
back-donation
bond bond
CR
CR
Z
X
Y
Practice problems
- State the multiplicity of the free ligand for each of the carbene complexes listed below.
Spin, S = n (1/2) where n is the number of unpaired electrons. Multiplicity, M = 2S + 1 Fischer carbyne n = 1; S = 1/2; M = 2 = doublet Schrock carbene n = 2; S = 2; M = 3 = triplet Schrock carbyne n = 3; S = 3/2; M = 4 = quartet Fischer carbene n = 0; S = 0; M = 1 = singlet
- Using the Tolman map below describe similarities/differences between any 3 sets of
ligands.
PF 3 and P(OCH 3 ) 3 have similar cone angles (100‐ 105 ). PF 3 is more electron withdrawing due to
the greater electronegativity of the F atom which lowers the PF ‐accepting orbital. This
reduces ‐back donation to the trans CO ligand resulting in a larger CO bond order and a higher
frequency v (CO) stretch (2110 vs 2080 cm‐^1 ). Likewise, P(Tol) 3 and P(4‐ClPh) 3 have a similar cone angle
(145) but the Cl substituent is electron withdrawing resulting in a larger CO bond order and a
higher frequency v (CO) stretch (2065 vs 2075 cm‐^1 ). Similarly, P( t Bu) 3 and P(C 6 F 5 ) 3 have a similar cone
angle (180‐ 185 ) but the perfluorobenzene substituent is both ‐delocalized and electron withdrawing
resulting in a lower energy ‐accepting ^ orbital and subsequently a larger CO bond order and
a higher frequency v (CO) stretch (2055 vs 2095 cm‐^1 ).
Practice problems
- Draw resonance structures for the following two complexes.
(OC) 5 Cr Ph
OMe
Me(Cp) 2 Ta Ph
(OC) 5 Cr Ph
OMe (OC) 5 Cr Ph
OMe
Me(Cp) 2 Ta Ph
- Draw the X 2 and L bonding modes of a metal‐alkene complex.
- Complete the following diagram by drawing lobes and nodes of each orbital and filling electrons.
Energy
allyl anion (ionic model)
allyl radical (covalent model)
Practice problems
The CuCl 2 reagent is used as a co‐catalyst to regenerate Pd(II) from Pd(0) to re‐initiate the
catalytic cycle
2Cu(II) + Pd(0) → 2Cu(I) + Pd(II)
Pd Cl
H 2 O Cl
2
Metal oxidation state: + Total electron count: 16 electron
Pd H 2 O
H 2 O H
Cl 6
Metal oxidation state: 2+ Total electron count: 16 electron
Practice problems
- Using MO models for metal carbonyl and metal phosphine complexes explain the
difference observed in v (CO) by FTIR spectroscopy for the trans ‐CO ligand of the following complexes.
The M‐P bond can be represented with a particular geometry in a generic representation as is shown below. A bond is formed from donation of the lone pair on the P atom ( sp orbital) to one of the empty e g metal orbitals ( dz^2 or dx^2 ‐ dy^2 ) and back‐bonding occurs from one of the filled t 2g set of metal orbitals ( dxy , dxz or dyz ) to the PR anti‐bonding ‐orbitals.
The PF 3 ligand has a lower energy * ligand orbital than the P(t^ Bu) 3 ligand due to the greater electronegativity of F vs. C. Therefore the PF 3 ligand is a better ‐acceptor than the P(tBu) 3 ligand. As a result of this energy difference (shown in MO diagram below) the PF 3 ligand has a greater back‐donation from the Ni metal than the P(tBu) 3 ligand and thus decreases the back‐donation to the trans CO ligand relative to the analogous P(tBu) 3 complex. This decreased back‐donation to the trans CO ligand in the PF 3 complex results in a stronger CO bond and a higher energy stretching frequency in the IR spectrum.
Practice problems
- For the following bonding modes of the alkyne ligand complete the table below
indicating the charge of the ligand and the number of electrons donated (in both the ionic and covalent models). Also describe the bonding using the L (^) a X (^) b formalism.
- Using a molecular orbital bonding scheme and the ligand atomic orbitals describe how
transition metal back‐bonding to the butadiene ligand can favor the X 2 bonding mode over the L 2 bonding mode.
Energy
M
M
X 2
L 2
The HOMO ( 2 ) is fully occupied in the butadiene ligand and forms a ‐bond with an empty metal d
orbital of appropriate symmetry and energy. The LUMO ( 3 ) orbital is empty and available for ‐back
donation from a filled metal orbital of appropriate symmetry and energy. As can be seen in the
schematic above, occupation of the LUMO ( 3 ) via ‐back donation will increase the bond order
between C 2 C 3 and decrease the bond order between C 1 C 2 and C 3 C 4 of the butadiene ligand. Thus ‐
back donation favors the MX 2 bonding motif.
