Download 1 ADVANCED INORGANIC CHEMISTRY (CHE 510) EXAMINATION

ADVANCED INORGANIC CHEMISTRY (CHE 510)
EXAMINATION II
NOVEMBER 20, 2006
Name:_________Key____________
SHOW ALL OF YOUR WORK so that you may get some partial credit. The last
few pages of this exam booklet contain tables that you may need to complete
this exam. Please answer questions IN THE AREA PROVIDED, THE BACK OF
AN EXAM PAGE OR ON THE CLEARLY LABELED SPARE SHEET. No credit will
be given for work written on the tables pages.
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1a. (8 pts) Using a group theoretical approach, construct a MO energy level
diagram for tetrahedral NH4+. Include appropriate symmetry labels and
discuss the logic used as well as any assumptions made in the construction of
the MO diagram. What is the bond order predicted for tetrahedral NH4+ on
the basis of your MO energy level diagram?
First, identify valence orbitals of the atoms involved in the bonding: the
valence orbitals of the nitrogen atom are 2s, 2px, 2py, and 2pz orbitals while
each hydrogen atom has a 1s orbital.
Next, figure out the symmetry of the orbitals involved. Thus, we make
SALC’s of the hydrogen 1s orbitals
H
H
H
H
Td
E
8 C3
3 C2
6 S4
6 σd
Γ1s
4
1
0
0
2
Which reduces to Γ1s = A1 + T2
Looking in the character table, we see that for nitrogen:
2s orbital has a1 symmetry
2px, 2py, and 2pz orbitals have t2 symmetry
Recognizing that nitrogen is more electronegative that hydrogen, we can
suspect that the hydrogen SALC’s will have the higher energy and that the a1
SALC is slightly lower than the t2 SALC’s because of the nodal surfaces that
exist although the atoms are not formally bonded. Thus, we get the diagram
below:
2
N+
NH4+
H4
2t2
t2
2a1
t2
a1
1t2
a1
1a1
B.O. = ½ (8) = 4
1b. (8 pts) Using a group theoretical approach, construct a MO energy level
diagram for square planar NH4+. Include appropriate symmetry labels and
discuss the logic used as well as any assumptions made in the construction of
the MO diagram. What is the bond order predicted for square planar NH4+ on
the basis of your MO energy level diagram?
First, we find the symmetry point group for square planar NH4+:
z
y
H
H
+
x
N
H
H
D4h
3
Next, we generate SALC’s of the hydrogen atom 1s orbitals;
z
y
•
x
D4h
Γ1s
E
4
2 C4
0
C2
0
2 C2' 2 C2''
2
0
i
0
2 S4
0
σh
4
2 σv
2
2 σd
0
We find that Γ1s = A1g + B1g + Eu after applying our reduction formula.
Looking in the D4h character table, we see that for nitrogen:
2s orbital has a1g symmetry
2px and 2py have Eu symmetry
2pz orbital has a2u symmetry
Recognizing that nitrogen is more electronegative that hydrogen, we get the
diagram below:
H
N+
N+
H
H
H
H4
2eu
2a1g
b1g + eu
b1g
a2u + eu
a2u
a1g
1eu
a1g
1a1g
B.O = ½ (# of bonding electrons -# of antibonding electrons) = ½ (6-0) = 3.
Note that electrons in a2u orbital are nonbonding.
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1c. (6 pts) Construct (draw) a Walsh diagram correlating the bonding
molecular orbitals of tetrahedral NH4+ with those of square planar NH4+.
Explain the reasoning behind your diagram.
Our Walsh diagram is shown below. The nodeless a1 - a1g orbital does not
influence shape of the molecule since it is cylindrically symmetrical. As we
can see the t2 set of molecular orbitals in tetrahedral geometry splits into the
a2u and eu set in the square planar structure. The a2u MO is primarily N 3pz in
character and the overlap with the hydrogen SALC’s decreases to essentially
zero, producing a large rise in its energy. Though overlap between the eu (px
and py ) orbitals and the hydrogen orbitals increase slighty, it’s is not enough
to compensate this increase in energy of the a2u orbital because the PX and py
orbitals were already in good overlap in tetrahedral geometry. Thus, when a
molecule possesses a total of eight electrons, tetrahedral structure will be
preferred.
D4h
Td
orbital energy
a2u
t2
eu
a1g
a1
109˚
90˚
Angle (˚)
See Yoshizawa et al. Chem. Phys. 2001, 271, 41-54 for recent discussion of
Walsh diagrams of AH4 molecules.
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2a. (10 pts) Using group theory, construct a MO energy level diagram
showing sigma bonding only for [Ti(CO)6]2-. Include appropriate
symmetry labels for all orbitals and discuss the logic used as well as any
assumptions made in constructing your MO diagram. Draw the MO
diagram using the template labeled [Ti(CO)6]2- (below).
First, identify valence orbitals of the atoms involved in the bonding. The
valence orbitals of the titanium atom are the five 3d-, 4s-, and three 4p
orbitals. The CO ligands form sigma bonds though their 3σg HOMO (we
showed this in class).
Next, we figure out the symmetry of the orbitals involved. Thus, we make
SALC’s of the CO donor sigma orbitals:
!!
!
!
!
!
!
Oh
E
8 C3
6 C2
ΓCo
6
0
0
6 C4 3C42
2
2
i
6 S4
8 S6
3 σh
6 σd
0
0
0
4
2
Which reduces to ΓCO = A1g + Eg + T1u
Looking in the character table, we see that for titanium:
4s orbital has a1g symmetry
3dz2 and 3dx 2-y2 have eg symmetry
3dxy, 3dxz, and 3dyz have t2g symmetry
4px ,4py , and 4pz orbitals have tiu symmetry
Recognizing that the CO donor electrons must be at lower energy than the
titanium acceptor orbitals (CO carbon is more electronegative that titanium),
we get the diagram on the next page. We see that the t1u orbitals are lower
in energy than the eg, this is because the eg orbitals point directly at the
ligands and are raised in energy due to repulsive interaction.
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[Ti(CO)6]2-
Ti2-
6CO
2t1u
2a1g
CO !* orbitals
(not involved in "
bonding)
2eg
t1u
a1g
E
eg
t2g
t2g
t1u
eg
a1g
1t1u
1eg
1a1g
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2b. (5 pts) [Ti(CO)6]2- displays an IR-active CO stretch (νCO) at 1748 cm-1
while [Cr(CO)6] displays an IR-active CO stretch (νCO) at 2000 cm-1 . Use
the template below to draw a MO energy level diagram for [Cr(CO)6]
that is consistent with its higher CO stretching frequency.
6CO
[Cr(CO)6]
Cr
2t1u
2a1g
CO !* orbitals
(not involved in "
bonding)
2eg
E
t1u
a1g
eg
t2g
t2g
t1u
eg
a1g
1t1u
1eg
1a1g
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2c. (5pts) Rationalize the trend in νCO for [Ti(CO)6]2- and [Cr(CO)6] on the
basis of your MO diagrams and keeping in mind the electroneutrality
principle.
Both Ti2- and Cr are d6 metal centers. The IR data indicate that greater
metal to CO ligand back-bonding interaction occurs in [Ti(CO)6]2- than
[Cr(CO)6]. This can be explained by the fact that titanium is more
electropositive than Cr and has a -2 charge in [Ti(CO)6]2-, which means
more electron density needs to be delocalized onto the CO ligands to
maintain an essentially neutral metal center. Thus, titanium’s valence
orbitals and hence [Ti(CO)6]2-’s molecular orbitals reside at higher
energy than Cr’s valence orbitals and hence [Cr(CO)6]’s molecular
orbitals. As a result, t2g orbitals of [Ti(CO)6]2- are closer in energy to CO
π* orbitals, leading to better overlap and greater (metal d CO π*)
backbonding interaction.
2d. (6 pts) The photoelectron spectrum of gas phase [Mo(CO)6] is shown
below. Use the spectrum to account for the energies of the molecular orbitals
of the octahedral complex. Hint: the ionization energy of CO itself is around
14 eV.
The HOMOs of [Mo(CO)6] are the three t2g orbitals largely confined to the Mo
atom (see MO diagram for related [Cr(CO)6] on previous page for reference )
and their energy can be ascribed to that of the peak with the lowest ionization
energy (close to 8 eV). The group of ionization energies around 14eV is
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probably due to the Mo-CO σ bonding orbitals, as well as bonding orbitals in
CO since the ionization energy of CO itself is around 14 eV.
3a. (8 pts) Name the following compounds according to IUPAC rules:
(a)
[CoCl2(en)(NH3)2]
Diamminedichloroethylenediaminecobalt(II)
(b)
[Co(N3)(NH3)5]SO4
Pentaammineazidocobalt(III) sulfate
(c)
[Ag(NH3)2]PF6
Diamminesilver(I) hexaflurophosphate
(d)
K3[Fe(CN)6]
Potassium hexacyanoferrate(III)
3b. (10 pts) Draw all of the possible isomers for [CoCl2(en)(NH3)2].
NH3
Cl
N
Co
N
NH3
N
NH3
N
Cl
Cl
Cl
Cl
Co
N
Cl
NH3
Cl
N
Cl
Co
N
Co
NH3
H3N
NH3
N
NH3
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4. (10 pts) The important structural role of Zn2+ in biological systems can be
attributed to its electronic preference for octahedral over tetrahedral
geometry. True or False? Explain your reasoning
False. Zn2+ has d10 electron configuration. The LFSE is zero for both
octahderal and tetrahedral geometries (below). Thus, octahedral site
stabilization energy is zero hence there is no electronic preference for
octahedral geometry over tetrahedral geometry. The geometry adopted by
Zn2+ generally depends on the steric requirements of the ligands and
thermochemical considerations.
Td
Oh
!O
!t
d
LSFE = 0
LSFE = 0
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5. (15 pts) Bearing in mind the Jahn-Teller theorem, rank the following
compounds in terms of their degree of deviation from idealized octahedral
structure: [Cr(CN)6]4-, [Cu(OH2)6]2+, and [Cr(OH2)6]3+. Explain your
reasoning and indicate what type of distortion can be expected (use drawings
as required).
[Cr(OH2)6]3+ < [Cr(CN)6]4- < [Cu(OH2)6]2+
Cr2+ = d4 ion, Cr3+ = d3 ion, and Cu2+ = d9 ion. H2O is a weak field ligand
while CN- is a strong field ligand. Hence, [Cr(CN)6]4- is a low spin complex
while only one electron configuration is possible in the case of [Cu(OH2)6]2+
or [Cr(OH2)6]3+. While a d3 ion is not subject to Jahn-Teller distortion since
no stabilization is gained, a Jahn-Teller (J-T) distortion, which lowers the
symmetry, removes orbital degeneracy, and leads to a more stable complex,
is expected for both low spin d4 - (axial compression) and d9 electron
configurations (axial compression or elongation). J-T distortion results in
greater splitting of the eg orbitals than the t2g orbitals, hence deviation from
idealized octahedral is greater for [Cu(OH2)6]2+ than [Cr(CN)6]4- (see your
textbook for splitting diagrams)
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6. (12 pts) Explain the differences in values of the ligand field splittings (∆ =
∆o or ∆T) for the cobalt complexes below:
Complex
∆ (cm-1 )
[Co(NH3)6]3+
22,900
[Co(H2O)6]3+
18,200
[Co(NH3)6]2+
10,200
[Co(NH3)4]2+
5,900
NH3 exerts a stronger ligand field than H2O hence ∆ is greater for
[Co(NH3)6]3+ than [Co(H2O)6]3+ , both of which contain Co3+ ions. While
[Co(NH3)6]3+ and [Co(NH3)6]2+are both hexaammine complexes, an increase
in ionic charge (from Co2+ to Co3+) will draw the ligands more closely in and
thereby increase electrostatic repulsion. Consequently, ∆ for [Co(NH3)6]3+ is
greater than for [Co(NH3)6]2+. Clearly, NH3 is not a sufficiently stronger
ligand field than H2O to overcome the effect of the increased charge in
[Co(H2O)6]3+ versus in [Co(NH3)6]2+. The greater the number of ligands, the
greater the perturbation of the d orbitals. Thus, six coordinate complexes
have greater values of ∆ than the tetrahedral complex, [Co(NH3)4]2+
(remember that ∆t = 4/9∆o ).
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7. (12 pts) For each of the following pair of complexes, identify the one that
has the larger ligand field stabilization energy (LFSE). Explain your
reasoning and where possible, show your work.
(a)
[Mn(OH2)6]2+ or [Fe(OH2)6]3+
Both Mn2+ and Fe3+ are isoelectronic (d5) ions. Since H2O is a weak
field ligand, both [Mn(OH2)6]2+ and [Fe(OH2)6]3+ are high spin
complexes. Hence both have LFSE = 0.
(b)
[Cr(OH2 )6]2+ or [Mn(OH2)6]2+
Both are high-spin complexes.
While Mn2+ is a d5 ion and hence LFSE = 0, Cr2+ is a d4 ion and
hence [Cr(OH2)6]2+ has t2g3eg1 configuration and LFSE = [(-0.4 x 3)
+ (0.6 x 1)]∆o = -0.6 ∆o .
Thus, [Cr(OH2)6]2+ has the larger LFSE.
(c)
[Ru(CN)6 ]3- or [Fe(CN)6]3-
Both Ru3+ and Fe3+ are d6 ions that belong to the same group. Both
complexes are low spin and hence have t2g6 electron configuration.
However, LFSE increases down the group (as d orbital size increases,
allowing for better overlap) and hence the ruthenium complex will
have the higher LFSE
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