CHE 610
Spring 2007
Problem Set 2 Key
1. Determine the electron configuration (t2gx egy or ex t2y), number of unpaired electrons,
approximate µeff, ligand-field stabilization energy (in ΔO or ΔT) for each of these
complexes.
a) [Cr(NH3)6]3+
t2g3 eg0, 3 unpaired electrons
µeff ≈ 3.87 µB or 3.87 BM (spin-only estimate: µeff ≈ 2[S(S+1)]1/2 ≈ [n(n+2)]1/2 where n = number
of unpaired electrons or S = electron spin quantum number = n/2)
CFSE (or LFSE) = -1.2ΔO
b)
[Co(NH3)6]3+
t2g eg , 0 unpaired electrons
µeff = 0
CFSE (or LFSE) = -2.4ΔO + 2P
6
0
c)
[Fe(OH2)6]2+
t2g eg , 4 unpaired electrons
µeff = 4.90 µB
CFSE (or LFSE) = -0.4ΔO (Why no P (pairing energy)? Because there is one set of paired
electrons in both the unsplit d6 and high-spin octahedral d6 configurations.)
4
2
d)
[Mo(CO)6]
t2g eg0 , 0 unpaired electrons
µeff = 0
CFSE (or LFSE) = -2.4ΔO + 2P
6
e)
[Mn(OH2)6]2+
t2g3 eg2, 5 unpaired electrons
µeff ≈ 5.92 µB
CFSE (or LFSE) = 0
f) [MnF6]3t2g eg1, 4 unpaired electrons
µeff ≈ 4.90 µB
CFSE (or LFSE) = -0.6ΔO
3
g) Tetrahedral [FeCl4]2e3 t23, 4 unpaired electrons
µeff ≈ 4.90 µB
CFSE (or LFSE) = -0.6Δt
h)
Tetrahedral [NiBr4]2e t2 , 2 unpaired electrons
µeff ≈ 2.83 µB
CFSE (or LFSE) = -0.8Δt
4
4
i)
Square planar [Ni(CN)4]2-
The standard t2gx egy and ex t2y labels don’t apply to square planar. In D4h, the d orbitals break into
4 levels (lowest to highest energy) with the symmetry labels: = dyz, xz = eg, dz2 = a1g, dxy = b2g and
dx2-y2 = b1g In one convention, the energy difference between the lowest and highest levels is ΔSP
(≈ 1.3 ΔO) and the gaps between levels are (lowest to highest) Δ1, Δ2, and Δ3,
The electron configuration is eg4 a1g,2 b2g2 b1g0, 0 unpaired electrons
µeff = 0
If Δ1 = Δ2 = Δ3,CFSE (or LFSE) could be expressed as ≈ -4(0.5ΔSP) –2(0.17ΔSP) + 2(0.17ΔSP) =
-2ΔSP ≈ -2.6 ΔO, but that’s a rough approximation.
j) [Fe(CN)6]4t2g eg0, 0 unpaired electrons
µeff = 0
CFSE (or LFSE) = -2.4ΔO + 2P
6
k) [Fe(CN)6]3t2g5 eg0, 1 unpaired electron
µeff = ≈ 1.73 µB
CFSE (or LFSE) = -2.0ΔO + 2P
l)
[Ru(bipy)3]2+
t2g6 eg0, 0 unpaired electrons
µeff = 0
CFSE (or LFSE) = -2.4ΔO + 2P
m) [Pt(MeCN)6]4+
t2g eg0, 0 unpaired electrons
µeff = 0
CFSE (or LFSE) = -2.4ΔO + 2P
6
2. Rationalize the following values of oxide lattice enthalpies (kJ/mol). All compounds have
the solid-state rock-salt structure.
CaO (3460), TiO (3878), VO (3913), MnO (3810), FeO (3921), CoO (3988), NiO (4071).
The lattice enthalpy increases with increasing atomic number because of the general (V2+ is an
exception) decrease in ionic radius of M2+ -- the smaller the ion, the stronger the ionic bonding to
the oxide lattice. Superimposed on this trend, the lattice enthalpy parallels the CFSE of the metal
ion in an octahedral oxide (high-spin) environment. A larger CFSE leads to a larger lattice
enthalpy, i.e., a stronger metal oxide lattice.
Oxide
dn
CFSE
CaO
TiO
VO
MnO
FeO
CoO
NiO
0
2
3
5
6
7
8
0
-0.8ΔO
-1.2ΔO
0ΔO
-0.4ΔO
-0.8ΔO
-1.2ΔO
Ionic radius M2+
(pm)
114
100
93
97.0
92.0
88.5
83.0
Lattice enthalpy
(kJ/mol)
3460
3878
3913
3810
3921
3988
4071
Lattice enthalpy of Octahedral Metal Oxides
4200
Lattice enthalpy (kJ/mol)
4100
4000
3900
3800
3700
3600
3500
3400
0
1
2
3
4
5
d electron count
6
7
8
9
3. Predict the structure of [Cr(OH2)6]2+. (Consider probable distortion.)
Cr(II) is d4,. t2g3 eg1, and H2O is a weak-field ligand. The complex will undergo a Jahn-Teller
distortion to reduce the degeneracy of the ground state. An octahedral complex typically
undergoes a tetragonal (D4h) distortion that results in splitting of the t2g3 and eg1 levels. Often an
elongation along the z axis accompanied by a slight compression in the xy plane takes place.
There is no energetic advantage to splitting t2g3 because the dxy orbital is destabilized twice as
much as the dxz and dxy orbitals are stabilized. On the other hand, the eg1 splitting leads to a
filled, stabilized dz2 orbital and an empty, destabilized dx2-y2 orbital, a net stabilization.
Simple view of tetragonal distortion
b1g
eg
a1g
E
b2g
t2g
eg
D4h
Oh
Term-splitting view of tetragonal distortion
5E
5T
2g
g
eg
5B
2g
E
5
5A
1g
Eg
5B
1g
t2g
Oh
D4h
4. Use Tanabe-Sugano diagrams to estimate ΔO and B
Ni(II) is d8,. H2O is a weak-field ligand; NH3 is slightly stronger-field
[Ni(OH2)6]2+(absorptions at 8500, 15400, 26000 cm-1)
[Ni(NH3)6]2+ (absorptions at 10750, 17500, 28200 cm-1)
See scanned pages.