Answer to Exercise 6.3-1
6.4 Transition metal complexes
being spherically symmetrical, forms a basis for the totally symmetric representation A1g. The systematic way to find the characters for pand dis as follows.
(i) First study the transformation of the basish e1e2e3j under one symmetry operator from each of the five classes of O. (The one actually used in this step is shown in the first row of Table6.4.)
(ii) Write down the MRs of (T ) for these symmetry operators. This is easily done by inspection.
(iii) Write down the matrices (T )1, taking advantage of the fact that the MRs are orthogonal matrices so that (T )1is just the transpose of (T ).
(iv) Write down the Jones symbols for the operators T1, which again can be done by inspection by just multiplying (T )1into the column matrixjx y z i. So far, we have neglected the other five classes of Ohbecause the variables x, y, and z all change sign under inversion so that the Jones symbols for the operators I(T ) may be obtained from those of T simply by changing the sign of x, y, and z, as is done in Table6.4.
(v) Since the three p functions are just x, or y, or z, multiplied by f(r), the characters of p
can be written down from the Jones symbols for T1.
(vi) The angle-dependent factors in the d orbitals can now be written down using the Jones symbols for T1, which tell us how the variables x, y, and z transform and thus how
6
Figure 6.7. The ML6octahedral complex. Unit vectors , p, p0are oriented parallel to the orthonormal axes x, y, z, which have their origin at M and lie along OX, OY, OZ. The three C2axes that are collinear with the C4axes are along x, y, z, and the second set of six C2axes that bisect the angles between x and y, y and z, and z and x are designated C20. The symmetry planes that contain these C20axes are dplanes because they bisect the angles between C2axes that are normal to one of the three C4principal axes.
functions of x, y and z transform under the symmetry operators of Oh. Since the largest degeneracy in Ohis three, we expect the five-fold degeneracy of the d orbitals in spherical symmetry to be split in a cubic field, but we have no knowledge a priori whether the 5 5 MRs based on the five d functions will be block-diagonal. (All we know in advance is that they will form a representation equivalent to a block-diagonal representation.) In fact, as soon as we have written down the functions into which xy, yz, and zx transform, we notice that these three d orbitals only transform between themselves and never into the two remaining d orbitals. This means that the d orbital MRs are block-diagonal and the five-fold degeneracy of the d orbitals is split into at least three-fold and doubly degenerate subsets. Calling the first set d", we now write down the characters of the MRs based on xy, yz, and zx. It should be emphasized that the d orbitals transform in this way because of the cubic (¼ octahedral) symmetry, and that they will behave differently in different symmetries. In D4h symmetry, for example, the maximum degeneracy is two, so the five d orbitals will transform in a different fashion. In general, one simply studies the transformation of the five d orbitals, and, if subsets emerge, then one can take advantage of this to reduce the arithmetic involved.
(vii) The effect of the function operators on the remaining two d orbitals is given in the next two lines of Table6.4. When a function is a member of a basis set, in general it will transform into a linear combination of the set. In practice, this linear combination often consists of only one term (and then the entries in the corresponding column of the MR are all zero, with the exception of one that is unity). Under some operators, the basis functions transform into linear combinations, and an example of this is the class of 8C3, where, under the chosen operator R(2p / 3 [111]), x2 y2 transforms into y2 z2and 3z2 r2transforms into 3x2 r2. These are not d orbitals but they are linear combinations of d orbitals, for
y2 z2¼ 1=2ðx2 y2Þ 1=2ð3z2 r2Þ, (1)
3x2 r2¼3=2ðx2 y2Þ 1=2ð3z2 r2Þ: (2) When the function operator Rˆ(2p / 3 [111]) acts on the basishx2y2 3z2r2j, (1), (2) Rð2p=3 ½111Þhx^ 2 y2 3z2 r2j
¼ hx2 y2 3z2 r2j 1=2 3=2
1=2 1=2
: (3)
The characters of the MRs for the basis d can now be written down using the transformation of the second subset of d orbitals given in Table 6.4 and eq. (3).
Note that the characters for psimply change sign in the second half of the table (for the classes I{T}); this tells us that it is either a u IR, or a direct sum of u IRs. The characters for both de and d simply repeat in the second half of the table, so they are either g IRs, or direct sums of g IRs. This is because the p functions have odd parity and the d functions have even parity.
6.4 Transition metal complexes 119
Table6.4.JonessymbolsandcharactersystemsforAOsandMOsintheoctahedralML6complexion. {T}¼ER(pz)R(n)R(z)R(pa){I(T)} E3C28C36C46C20I3h8S66S4 (s)111111111 T{e1e2e3}e1e2e3e1e2e3e2e3e1e2e1e3e2e1e3 (T)100 010 001
2 4
3 5100 010 001
2 4
3 5001 100 010
2 4
3 5010 100 001
2 4
3 5010 100 001
2 4
3 5 (T1 )100 010 001
2 43 5100 010 001 2 43 5010 001 100 2 43 5010 100 001 2 43 5010 100 001
2 43 5 T1 {xyz}xyzxyzyzxyxzyxzxyzxyzyzxyxz (p)310113101 xyxyxyyzxyxyxyxyyzxy yzyzyzzxzxzxyzyzzxzx zxzxzxxyyzyzzxzxxyyz (d")310113101
x2 y2 x2 y2 x2 y2 y2 z2 (x2 y2 )(x2 y2 )x2 y2 x2 y2 y2 z2 (x2 y2 )(x2 y2 ) 3z2 r2 3z2 r2 3z2 r2 3x2 r2 3z2 r2 3z2 r2 3z2 r2 3z2 r2 3x2 r2 3z2 r2 3z2 r2 (d)2210022100 )6202004002 (p)12400000000 ¼2p=3,n¼3½ [111],¼p=2,a¼2½ [110].
(viii) The last step in the construction of Table6.4is to write down the characters of the representations based on the ligand p orbitals labeled , which point towards M, and the ligand p orbitals labeled p or p0, which are normal to unit vectors along the lines joining the ligands to M (Figure6.7). This can be done by the ‘‘quick’’ method of noting how the contours of the basis functions transform under the symmetry operators: those which are invariant, or simply change sign, contribute1, respect-ively, to the character, and the others contribute zero.
Reduction of the representations
From the characters in Table6.4we observe that
s¼ A1g; (4)
p¼ T1u; (5)
d¼ d" d¼ T2g Eg: (6)
The classes for the non-zero characters of , its character system, and reduction, are E 3C2 C4 3h 6d
¼ f6 2 2 4 2g;
cðA1gÞ ¼1=48½1ð1Þð6Þ þ 3ð1Þð2Þ þ 6ð1Þð2Þ þ 3ð1Þð4Þ þ 6ð1Þð2Þ ¼ 1, cðA2gÞ ¼1=48½1ð1Þð6Þ þ 3ð1Þð2Þ þ 6ð1Þð2Þ þ 3ð1Þð4Þ þ 6ð1Þð2Þ ¼ 0,
cðEgÞ ¼1=48½1ð2Þð6Þ þ 3ð2Þð2Þ þ 6ð0Þð2Þ þ 3ð2Þð4Þ þ 6ð0Þð2Þ ¼ 1:
Now
¼ 6 2 0 2 0 0 4 0 0 2f g and
A1g Eg ¼ 3 3 0 1 1 3 3 0 1 1f g;
ðA1g EgÞ ¼ 3 1 0 1 1 3 1 0 1 1f g ¼ T1u: Therefore
¼ A1g Eg T1u: (7)
The non-zero characters for pare
E 3C2
p¼ f12 4g: (8)
The characters for E and 3C2have opposite signs, and so to reach a sum of 48 in the reduction test will be unlikely except for IRs with a negative character for the class of 3C2. Therefore we try first those IRs for which (3C2) is negative. T1g, T2g, T1u, and T2uall have
(3C2)¼ 1, and
cðT1gÞ ¼1=48½1ð3Þð12Þ þ 3ð1Þð4Þ ¼ 1:
Since T2g, T1u, and T2uhave the same characters as T1gfor these classes, they must also occur once in the direct sum, which therefore is
p¼ Tg T2g T1u T2u: (9)
s bonding
We need to find the linear combinations of ligand orbitals of symmetry A1g, Eg, and T1u. Omitting normalization factors, these are
ðA1gÞ ¼ 1ð1Þ þ 1ð4þ 4þ 1Þ
þ 1ð2þ 3þ 5þ 6þ 3þ 5þ 2þ 6Þ þ 1ð2þ 5þ 6þ 3þ 1þ 1Þ
þ 1ð2þ 5þ 4þ 4þ 3þ 6Þ þ 1ð4Þ þ 1ð1þ 1þ 4Þ
þ 1ð5þ 6þ 2þ 3þ 6þ 2þ 3þ 5Þ þ 1ð2þ 5þ 6þ 3þ 4þ 4Þ
þ 1ð5þ 2þ 1þ 1þ 6þ 3Þ,
ðA1gÞ ¼ 1þ 2þ 3þ 4þ 5þ 6; (10)
1ðEgÞ ¼ 21þ 2ð1þ 24Þ 2ð2þ 3þ 5þ 6Þ þ 24 þ 2ð21þ 4Þ
2ð2þ 3þ 5þ 6Þ,
1ðEgÞ ¼ 21 2 3þ 24 5 6: (11) Starting with 2, and then 3, as our arbitrary functions in the subspace with basis vectors (functions) {1 2 3 4 5 6} and projecting as before will simply give the cyclic permutations
2ðEgÞ ¼ 22 3 4þ 25 6 1 (12) and
3ðEgÞ ¼ 23 4 5þ 26 1 2: (13)
6.4 Transition metal complexes 123
There cannot be three LI basis functions of Eg symmetry, so we must choose, from eqs. (11), (12), and (13), two LI combinations that are orthogonal in the ZOA and which will overlap satisfactorily with the M atom orbitals of Egsymmetry. A suitable choice that meets these three requirements is
1ðEgÞ 2ðEgÞ ¼ 1 2þ 4 5, (14)
3ðEgÞ ¼ 23 4 5þ 26 1 2: (15)
Exercise 6.4-1 Verify that eqs. (11), (12), and (13) are linearly dependent and that eqs. (14) and (15) are orthogonal in the ZOA.
Continuing with the T1urepresentation,
1ðT1uÞ ¼ 31 1ð24þ 1Þ þ 1ð2þ 5þ 6þ 3þ 21Þ
1ð2þ 5þ 3þ 6þ 24Þ
34þ 1ð21þ 4Þ 1ð2þ 5þ 6þ 3þ 24Þ þ 1ð5þ 2þ 21þ 6þ 3Þ,
1ðT1uÞ ¼ 1 4: (16)
Since Pˆ(T1u)1¼ 1 4, the other two LI linear combinations of ligand orbitals that form bases for T1uare, by cyclic permutation,
2ðT1uÞ ¼ 2 5 (17)
and
3ðT1uÞ ¼ 3 6: (18)
We now have the bonded MOs
(10) a1g¼ a1½ðn þ 1Þs þ b1½1þ 2þ 3þ 4þ 5þ 6, (19) (14) eg¼ a2½ndx2y2 þ b2½1 2þ 4 5, (20) (15) eg0¼ a3½nd3z2r2 þ b3½23þ 26 1 2 4 5, (21)
(16) t1u¼ a4½ðn þ 1Þpx þ b4½1 4, (22)
(17) t1u0¼ a5½ðn þ 1Þpy þ b5½2 5, (23)
(18) t1u00¼ a6½ðn þ 1Þpz þ b6½3 6: (24) The orbitals occur in bonding and antibonding pairs, according to whether ai, bihave the same sign or opposite sign. Rough sketches of contours ofj j2in the bonding AOs are shown in Figure6.8.
Figure 6.8. Approximate charge density prior to bonding in overlapping atomic orbitals that form
-type molecular orbitals in ML6: (a) t1u,x; (b) eg; (c) e0g. The actual charge density in the molecule would require a quantum-chemical calculation. Only the relevant halves of the ligand p orbitals are shown in some figures. Atom centers may be marked by small filled circles for greater clarity. As usual, positive or negative signs show the sign of , like signs leading to an accumulation of charge density and therefore chemical bonding. The ring depicting the region in which the d3z2r2orbital has a negative sign has been shaded for greater clarity, but this has no other chemical significance apart from the sign.
6.4 Transition metal complexes 125
p bonding
We now seek linear combinations of ligand orbitals p and p0that form bases for the IRs T1u, T2g, T2u, and T1g(eq. (9)). The characters for T1uin O are {3 1 0 1 1}, so with p1as our arbitrary function in the p, p0subspace,
1ðT1uÞ ¼ 3ðp1Þ 1ðp4 p1 p4Þ þ 1ðp2þ p5þ p10 p10þ p6 p3Þ These ligand p orbitals are symmetrically disposed to point along the OZ axis. Since the OX and OY axes are equivalent to OZ in Ohsymmetry, we may write down by inspection
2ðT1uÞ ¼ p20þ p3þ p50þ p6¼ px (26) and
3ðT1uÞ ¼ p10þ p60 p40 p30¼ py: (27) The MOs that form bases for T1uare therefore
t1u, x¼ a7½ðn þ 1Þpx þ b7 px, (28)
t1u, y¼ a8½ðn þ 1Þpy þ b8 py, (29)
t1u, z¼ a9½ðn þ 1Þpz þ b9 pz: (30)
As an example, the MO t1u,zis shown in Figure6.9(a). The character system for (T2g) is {3 1 0 1 1}, and so
Figure 6.9. Atomic orbitals that form the p molecular orbitals: (a) t1u,z; (b) t2g,zx.
1ðT2gÞ ¼ 3ðp1Þ 1ðp1Þ 1ðp2þ p5þ p6 p3Þ þ 1ðp2 p5þ p3 p6Þ þ 3ðp4Þ 1ðp4Þ 1ðp5 p2 p3þ p6Þ þ 1ðp5þ p2 p6þ p3Þ, giving
1ðT2gÞ ¼ p1 p4þ p3 p6¼ pzx: (31) By inspection, the linear combinations of ligand AOs in the yz and xy planes are
2ðT2gÞ ¼ p2 p30 p5 p60¼ pyz (32) and
3ðT2gÞ ¼ p10 p20þ p40 p50¼ pxy: (33) The MOs of T2gsymmetry are therefore
t2g, xy¼ a10½ndxy þ b10 pxy, (34)
t2g, yz¼ a11½ndyz þ b11 pyz, (35)
t2g, zx¼ a12½ndzx þ b12 pzx: (36)
As an example, t2g,zx is shown in Figure 6.9(b). The character system (T1g) is { 3 1 0 1 1} and so
1ðT1gÞ ¼ 3ðp1Þ 1ðp1Þ þ 1ðp2þ p5þ p6 p3Þ 1ðp2 p5þ p3 p6Þ þ 3ðp4Þ 1ðp4Þ þ 1ðp5 p2 p3þ p6Þ 1ðp5þ p2 p6þ p3Þ, giving
1ðT1gÞ ¼ p1 p3 p4þ p6: (37) There is no metal orbital of T1gsymmetry, so eq. (37) represents a non-bonding MO, t1g,y. The three degenerate MOs of T1gsymmetry are therefore
t1g, x¼ b13½p2þ p30 p5þ p60, (38)
t1g, y¼ b14½p1 p3 p4þ p6 , (39)
t1g, z¼ b15½p10þ p20þ p40þ p50: (40) Finally, the character system (T2u) is {3 1 0 1 1} so that
1ðT2uÞ ¼ 3ðp1Þ 1ðp1Þ 1ðp2þ p5þ p6 p3Þ þ 1ðp2 p5þ p3 p6Þ
3ðp4Þ þ 1ðp4Þ þ 1ðp5 p2 p3þ p6Þ 1ðp5þ p2 p6þ p3Þ,
6.4 Transition metal complexes 127
giving
1ðT2uÞ ¼ p1 p2þ p4 p5: (41) There is no metal orbital of T2usymmetry, so eq. (41) represents a non-bonding MO, t2u,z. By inspection, the three degenerate MOs of T2usymmetry are
t2u, x¼ b16½p20þ p3 p50þ p6, (42)
t2u, y ¼ b17½p10 p60 p40 p30, (43)
t2u, z¼ b18½p1 p2þ p4 p5: (44)
A schematic energy-level diagram is shown in Figure6.10. To draw an accurate energy-level diagram for any specific molecule would require an actual quantum chemical calculation. The energy levels are labeled by the appropriate group theoretical symbols for the corresponding IRs. When the same IR occurs more than once, the convention is used that energy levels belonging to the same IR are labeled 1, 2, 3, beginning at the lowest level.
In summary, in ascending order, there are
(i) the orbitals: 1a1g, 1t1u, 1eg, fully occupied by twelve electrons;
(ii) the mainly ligand p orbitals: 1t2g, 2t1u, t2u, t1g, which hold twenty-four electrons;
(iii) the metal d orbitals 2t2g(or d") (with a small mixture of mainly non-bonding ligand p) and 2eg(or d) plus ligand ;
Figure 6.10. Schematic energy-level diagram for ML6complexes.
For example, [MoCl6]3has a total of thirty-nine valence electrons from the molyb-denum 4d5 5s and six chlorine 3p5 configurations, and the 3 charge on the ion.
Its electron configuration is therefore 12 p24 (2t2g)3. Thus the MO theory of these ML6complex ions confirms that the electrons of prime importance are those occupying the t2gand eglevels on the metal, as predicted by crystal-field theory. However, the MO theory points the way to the more accurate calculation of electronic structure and properties.