Download Problem set 1

424/525 Advanced Transition Metal Chemistry 2008
Answers to problem set 2
1. Draw the 3 most common 7-coordinate geometries.
The bulky, rigid tridentate ligand NNN is a good ligand for Ln3+ ions. Its rigidity means that it always
coordinates in such a way as to occupy a triangular face:
M
N
N
N
NNN is sufficiently bulky that only 4 other ligands can bind to the metal. How many possible geometric
isomers for Ln(NNN)L4 (L = monodentate ligand) are there for each coordination geometry?
[Q from 2004 midterm]
Count the number of different types of triangular face on each coordination geometry. The pentagonal
bipyramid has one, the capped trigonal prism three, and the capped octahedron four.
2. Ln(NO3)3(18-crown-6) complexes have structures in which the 12 O atoms take up an icosahedral
geometry around the lanthanide ion (“18-crown-6” is a crown ether, cyclo-{CH2CH2O}6). The 6 O
atoms of the crown take up a boat rather than a chair conformation. By drawing the structure, explain
why.
Ln
If the crown ether takes up a chair conformation, only 3 coordination sites are available “above” and
“below” the crown ether. However, the boat conformation (above) provides 4 sites above (2 chelating
NO3 ligands) and 2 below (1 NO3 ligand), allowing all ligands to bond fully.
3. Explain why tungsten has the highest mp, bp and heat of atomization of any element.
[Q from 2004 midterm]
See graph of atomization enthalpies from notes. Group 6, 5d metal. All most-bonding orbitals of band
filled; not far enough across d-block for atom to contract too significantly; 5d rather than 4d because
d-orbitals more expanded & available for bonding.
424/525 Advanced Transition Metal Chemistry 2008
4. Determine the enthalpies of formation for WCl, WCl3 and WCl6 using a Born-Haber cycle,
calculating lattice energies using the Kapuskinskii equation.
Lattice energies: ΔU(0 K) = (1.07 × 105) ν | z+ | | z– | / r+ + r–
WCl: ΔU(0 K) = 1.07 × 105 × 2 × 1 × 1 / (130 + 167) = 720 kJ mol-1
WCl3: ΔU(0 K) = 1.07 × 105 × 4 × 3 × 1 / (98 + 167) = 4800 kJ mol-1
WCl6: ΔU(0 K) = 1.07 × 105 × 7 × 6 × 1 / (74 + 167) = 18600 kJ mol-1
ΔHf(WCln) = ΔHa(W) + n/2 ΔHa(Cl2) + ΣIE(W) – nEA(Cl) – ΔU
ΔHf(WCl) = 860 + 0.5×244 + 770 – 349 – 720 = 680 kJ mol-1
ΔHf(WCl3) = 860 + 1.5×244 + (770+1700+2300) – 3×349 – 4800 = 150 kJ mol-1
ΔHf(WCl6) = 860 + 3×244 + (770+1700+2300+3400+4600+5900) – 6×349 – 18600 = –400 kJ mol-1
(i)
WCl does not exist. What particularly large factor is mitigating against its formation?
High ΔHa(W); compare Cs, <10% of the value for W.
(ii)
WCl3 and WCl6 do exist. Find out their actual structures, and describe them. Do either of
these fit the description of a “simple ionic solid”?
WCl3 has a cluster structure, formula better written [W 6Cl12]Cl6. Retention of some M-M bonding
means effectively not all of the ΔHa need be spent. WCl6 is a covalent molecule. Neither fit the
description of simple ionic solid.
(iii)
What do the Born-Haber cycle and Kapuskinskii equation fail to take into account?
M-M bonding, covalent bonding, alternative lattice geometries, etc. Question was intended to illustrate
that the approach is primarily applicable to compounds that form simple ionic solids.
IEn(W) 770, 1700, 2300, 3400, 4600, 5900 kJ mol-1; ΔHa(W) 860 kJ mol-1; ΔHa(Cl2) 244 kJ mol-1;
EA(Cl) 349 kJ mol-1; r+ 130 pm, r3+ 98 pm, r6+ 74 pm, r– 167 pm.
5. The Mo-Mo distance in [Mo6Cl8]4+ is shorter than the Nb-Nb distance in [Nb6Cl12]2+. Account for this
by calculating the M-M bond order in each case.
424/525 Advanced Transition Metal Chemistry 2008
[Q from 2004 midterm]
[Mo6Cl8]4+ corresponds to a [Mo6]12+ core. Total d electrons = 6 × 6 – 12 = 24 electrons. 12 M-M bonds
in octahedron so bond order = 1.
[Nb6Cl12]2+ corresponds to a [Nb6]14+ core. Total d electrons = 6 × 5 – 14 = 16 electrons. 12 M-M bonds
in octahedron so bond order = 8/12 = 2/3.
6. Rationalise the following Mo-Mo bond distances:
[Mo2(O2CMe)4]
[Mo2(SO4)4]4–
[Mo2Cl8]4–
[Mo2(SO4)4]3–
[Mo2(HPO4)4]2–
[Mo2(TPP)2]
209 pm
211 pm
214 pm
217 pm
223 pm
224 pm
Ph
N
N
2–
Ph
N
Ph
TPP = meso-tetraphenylporphyrin
N
Ph
–
Bond orders 4, 4, 4, 3.5, 3, 4 respectively. MeCOO “bite” better matched to Mo-Mo bond than SO42–;
some Cl-Cl repulsion in chloride case. TPP ligand bulky, large L-L repulsion lengthens bond.
7. The complex [OsO2(OH)4]2– has a linear OOsO group. Construct a simplified MO diagram for the
π-bonding in this complex assuming that OOsO lies along the z-axis and only the dxy, dxz and dyz
orbitals of the metal are involved. Do you expect the complex to be paramagnetic?
non-bonding dxy
dxz, dyz and
dxy orbitals
px and py orbitals
of 2 oxygen atoms
O hybridized sp. 2 each of bonding, non-bonding and antibonding MOs, as the picture is repeated in
xz and yz planes. Compound is diamagnetic; all electrons are paired.