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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

HH

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:

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N+

H4NH4+

1a1

1t2

2a1

2t2

t2

a1

a1

t2

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

+:

HH

H H

N

z

y

x

D4h

+

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Next, we generate SALC’s of the hydrogen atom 1s orbitals;

•

z

y

x

D4h E 2 C4 C2 2 C2' 2 C2'' i 2 S4 σh 2 σv 2 σd Γ1s 4 0 0 2 0 0 0 4 2 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:

N+

1eu

H4

N+

H

H H

H

a1g

b1g + eu

a1g

1a1g

2a1g

a2u + eu

b1g

a2u

2eu

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.

t2 eu

a2u

TdD4h

a1 a1g

orb

ital

en

erg

y

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 6 C4 3C42 i 6 S4 8 S6 3 σh 6 σd

ΓCo 6 0 0 2 2 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 3dx2-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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CO !* orbitals(not involved in " bonding)

E

6CO[Ti(CO)6]2-Ti2-

a1g

eg

t1u

t2geg

a1g

t1u

2eg

t2g

1a1g

1eg

1t1u

2t1u

2a1g

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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.

CO !* orbitals(not involved in " bonding)

E

6CO[Cr(CO)6]Cr

a1g

eg

t1u

t2geg

a1g

t1u

2eg

t2g

1a1g

1eg

1t1u

2t1u

2a1g

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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].

Co

N

N NH3

NH3

Cl

Cl

Co

N

N Cl

NH3

Cl

NH3

Co

N

N Cl

Cl

NH3

NH3

Co

N

NCl

H3N

Cl

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

!t

!O

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 t2g

3eg1 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 t2g

6 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