Download Assignment 5 (key)

Chem 222
Problem Set 5 Answer key
Intro to Inorganic Chemistry
Summer 2011
Wednesday July 20, 2011
1. (H&S 20.10) (a) In the solid state, Fe(CO)5 possesses a trigonal bipyramidal structure. How
many carbon environments are there?
The trigonal bipyramidal structure places five chemically equivalent CO groups in two different spatial
environments – two COs are in axial positions, and three CO’s occupy equatorial positions.
CO axial
equatorial OC
Fe
CO equatorial
CO equatorial
CO axial
(b) Explain why only one signal is seen in the 13C NMR spectrum of solutions of Fe(CO)5, even at
low temperature.
This observation indicates that, in solution, this five-coordinate complex is undergoing Berry pseudorotation: rapid conversion between a trigonal bipyramidal geometry and a square-based pyramidal
geometry. This has the effect of exchanging all five carbonyl groups between different environments
around the iron, which gives an averaged 13C NMR shift.
2. (from H&S 20.11) Comment on the coordination number and the bond angle data for the iron
complex shown below (what bond angle data would you normally expect for this geometry?).
(Me3Si)3Si
137¡
111¡
Fe
Cl
112¡
(Me3Si)3Si
This iron complex is three-coordinate, which is unusually low. It should have
trigonal planar geometry: the sum of the bond angles is 360°, which is consistent with planarity at Fe,
but the angles are distorted from the usual 120° we’d expect. This is because the two bulky “silyl”
ligands, SiR3– (where R = Si(CH3)3) take up a lot of space in the coordination sphere. Thus the Si-FeSi angle is widened to 137° and the two Si-Fe-Cl angles are compressed. It is probably the size of
these silyl ligands that stabilizes the low coordinate structure.
3. (from H&S 20.14) (a) The salt [Co(bpy)3]3+[Fe(CN)6]3– has an isomer of the formula
[Co(CN)4(bpy)]–[Fe(CN)2(bpy)2]+ What type of isomerism do these two salts demonstrate?
Coordination isomerism (arises from exchange of ligands between two complex ions in a salt).
(b) What other types of isomerism exist for each of the complex ions in (a)? (Draw the structures
and label the isomers appropriately.)
[Co(bpy)3]3+ is chiral, and exists as the two stereoisomers shown below – they are enantiomers.
N
3+
N
N
N
Co
N

N
Co
N
N
3+
N
N
N
N

[Co(CN)4(bpy)]– is limited to the cis diastereomer (a stereoisomer) because the bite angle of the bpy
ligand is too small to allow the nitrogens to be trans to one another.
Chem 222
Intro to Inorganic Chemistry
CN
N
N
Co
NC
CN
trans
CN
N
CN


CN
+
N
Fe
CN
N
CN
N
N
Fe
N
CN
N
+
N
N
Fe
Ğ
N
N
+
N
Summer 2011
CN
cis
cis
+
[Fe(CN)2(bpy)2] exhibits both types of stereoisomerism. It has cis and trans isomers (diastereomers),
and the cis isomer is chiral: it exists as a pair of enantiomers.
4. (H&S 20.20a) The conjugate base (i.e. deprotonated form) of the fluoro-substituted
acetylacetone (aka pentanedione) shown below forms the complex [CoL3], which has mer and
fac isomers. Draw the structures of these isomers, and explain why the labels mer and fac are
used.
H3 C
CF3
The labels mer and fac indicate the relative arrangements of “like” ends of
this unsymmetric chelating ligand. In the fac isomer, the three oxygens close to CH3 (and those close
to CF3) occupy one face of the octahedron, while in the mer isomer, these “like” oxygens each trace
out one edge of the octahedron.
O
O
H3C
F3C
H3C
F3C
O
O
O
O
CF3
O
Co
H3C
O
O
O
F3C
fac
O
CF3
Co
CH3
H3C
O
O
O
CF3
H3C
mer
5. (a) Draw the structure for an ionization isomer of [CoCl(NH3)5]SO4, indicating clearly features
of the inner and outer coordination spheres.
(b) Draw the structure for a hydration isomer of [Fe(OH2)6]Cl2, indicating clearly features of the
inner and outer coordination spheres.
Chem 222
Intro to Inorganic Chemistry
Summer 2011
6. (a) (from H&S 21.1) Outline how you would apply crystal field theory to explain why the five dorbitals in an octahedral complex are not degenerate. (Include a labeled splitting diagram.)
The CFT model sees ligands as negative point charges attracted to the positive charge of the metal ion,
but repelled by the d-electrons. As the ligands are brought close to the metal, the d-orbitals are
destabilized by this repulsion. Those d-orbitals with lobes pointing directly at the ligands in an
octahedral field (i.e. six L approaching along x, y , and z axes) are most destabilized by this interaction:
dx2-y2 and dz2. Those d-orbitals with lobes pointing between the approaching ligands are not
destabilized as much: dxy, dxz, dyz (Relative to a spherical crystal field, these latter three are actually
stabilized.)
dz2, dx2-y2 eg
Oh
E
dxy, dxz, dyz t2g
free metal ion metal ion in
spherical
crystal field
metal ion in
octahedral
crystal field
(b) Based on the analogous splitting diagram, as derived from the MO diagram of a simple
octahedral complex, explain why a strong -donor ligand will give a large Oh value relative to a
weaker -donor.
eg*
-antibonding
Oh
E
non-bonding
t2g
metal ion
d-orbitals
ligand donor
orbitals
-bonding
This diagram shows only the d-orbitals interacting with ligand sigma-donor orbitals in an octahedral
complex. (We’ve left out the s and p orbitals on the metal, and the four other sigma bonding and sigma
antibonding sets above and below this diagram in energy.) Two of the d-orbitals are of appropriate
symmetry to interact with two of the ligand orbitals in a sigma fashion. The interaction stabilizes two
of the ligand orbitals (sigma bonding) and destabilizes two of the d-orbitals (sigma antibonding), and
leaves three of the d-orbitals unchanged in energy (non-bonding). If the ligand is a strong sigma donor,
the sigma bonding orbitals will be lower in energy, because a stronger bonding interaction will occur.
Accordingly, the eg* antibonding orbitals will be raised in energy, while the t2g non-bonding orbitals’
energy will remain the same. This increases the separation between the t2g and the eg* sets of d-orbitals
(Oh).
(c) Why does additional π-donation from a ligand such as a halide ion cause the Oh value to
diminish?
As shown in the diagram below, filled p orbitals on halide ligands of the appropriate symmetry can
overlap with empty non-bonding orbitals on the metal in a π-fashion. This interaction stabilizes the
ligands’ orbitals (π) and destabilizes the metal d-orbitals (π*). This gives the t2g set (of d-orbitals) slight
antibonding character – and raises them in energy relative to the eg* set, which makes Oh smaller.
Chem 222
Intro to Inorganic Chemistry
e g*
Summer 2011
-antibonding
Oh
¹ -antibonding
t2g*
E
filled porbitals
¹ -bonding
metal ion
d-orbitals
ligand donor
orbitals
-bonding
7. (a) (RC&O 19.33) The complex [Co(NH3)6]Cl3 is yellow-orange, while [Co(OH2)3F3] is blue.
Suggest an explanation for the difference in colour.
The yellow-orange complex must be absorbing higher energy, blue-violet light, while the blue complex is
absorbing lower energy orange light. This is consistent with colours arising from electronic transitions
between the t2g and eg sets of d-orbitals in these complexes. The hexaammine complex has a larger Oh
(ligand field or crystal field splitting) than the tris(aqua)trifluoro complex, because NH3 is a reasonably
strong field ligand ( a strong sigma donor), while H2O is not quite as strong a sigma donor and F– is a
π-donor ligand – both are weaker-field ligands than NH3. Because the energy gap is larger, a higher
energy light is required to prompt the electronic transition for the hexaammine complex.
(b) One of the above complexes is diamagnetic, while the other is paramagnetic, with a µ eff of
approximately 5.0 µB at room temperature. Which is which, and why?
Both complexes contain Co3+, which has a d6 configuration. If the complex is diamagnetic, it has no
unpaired electrons, which corresponds to a low spin d6 arrangement. The µeff value for the
paramagnetic complex is close to that predicted for four unpaired electrons by the spin-only formula
µeff = (n(n+2))1/2, 4.90µB. This corresponds to a high spin d6 arrangement. The compound with the
larger Oh value is most likely to be low spin – hence the hexaammine complex is diamagnetic, while
the aqua/fluoro complex, with its smaller Oh must be the high spin complex with four unpaired
electrons.
eg
low spin
Oh
Oh
eg
high spin
t2g
t2g
8. (RC&O 19.17) For which member of the following pairs of complex ions would Oh be greater?
Explain your reasoning.
(a) [MnF6]2– and [ReF6]2–
Mn and Re are both in group 7. For metals in the same triad with
analogous ligand sets and oxidation states, the ligand field splitting will be larger for the heavier
element(s). (This is because M-L bonding tends to strengthen going down the triad – see 7(b) above.)
(b) [Fe(CN)6]4– and [Fe(CN)6]3–
The first complex has Fe2+ and the second complex contains Fe3+.
For metal complexes with identical M and ligand set, the complex with the metal in the higher
oxidation state will have larger ligand field splitting. (This is because oxidation of the metal lowers the
d-orbitals in energy, which puts the t2g d-orbital set lower relative to the eg set.)
9. Explain the following: magnetic measurements indicate that [CoCl4]2– has three unpaired
electrons, [NiBr4]2– has two unpaired electrons, and [Ni(CN)4]2– is diamagnetic. Include
coordination geometries and labeled splitting diagrams in your answer. What µeff values would
you expect?
Chem 222
Intro to Inorganic Chemistry
Summer 2011
2–
2+
7
[CoCl4] contains Co , which is d . It is probably tetrahedral. Placing seven electrons into the Td dorbital splitting diagram does give an arrangement with three unpaired electrons. It should have a µeff
of around 3.87 µB (the spin-only estimated value).
recall: dxy, dxz, dyz
t2
E
Td
e
dx2-y2, dz2
Both nickel complexes contain four-coordinate Ni2+, which is d8. We might expect them both to have Sq.
Pl. geometry, since that is favoured for d8 complexes, but the fact that they have different numbers of
unpaired electrons (two, for which the spin-only µeff is calculated to be 2.83µeff, and zero, for which
µeff = 0 µB) indicates that they must have different geometries, and hence different d-orbital splitting.
Two unpaired electrons is consistent with a tetrahedral d8 arrangement for the chloro complex, while
no unpaired electrons is consistent with a square planar arrangement for the cyano complex. (For the
nickel chloride complex, the ligand field splitting is small enough that having four electrons in the t2 set
is not too destabilizing. For the nickel cyano complex, the ligand field splitting becomes large enough
that the square planar geometry is preferred, since it keeps the electrons at lower energy.)
recall: dx2-y2
Oh
t2
E
Td
e
Td d8
Sq Pl d8
Definitions/Concepts: inner/outer coordination sphere, coordination geometry, Kepert model,
electroneutrality principle, isomerism, stereoisomerism, diastereomer, enantiomer, optical isomers,
chirality, structural isomerism, ionization isomers, hydration isomers, linkage isomers, coordination
isomers, MO diagram for simple ML6 complex with only -bonding, d-orbital splitting, Oh, Td, -donor
ligand, π-donor ligand, π-acceptor ligand, weak field ligand, strong field ligand, spectrochemical series,
high spin, low spin, crystal field theory for Oh, Td, Sq Pl geometries, max, CO, effective magnetic
moment, diamagnetic, paramagnetic, pairing energy.