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5.04, Principles of Inorganic Chemistry II Prof. Daniel G. Nocera Lecture 12: Octahedral ML 6 Sigma Complexes
An octahedral complex comprises a central metal ion and six terminal ligands. If the
ligands are exclusively σ-donors, then the basis set for the ligands is defined as
follows,
Ligands that move upon the application of an operation, R, cannot contribute to the
diagonal matrix element of the representation. Since the σ bond is along the
internuclear axis that connects the ligand and metal, the transformation properties
of the ligand are correspondent with that of the M–L σ bond. Moreover, a σ bond
has no phase change within the internuclear axis, hence the bond can only transform into itself (+1) or into another ligand (0).
Oh E 8C 3 6C 2 6C 4 3C 2 (=C 42 ) i 6S 4 8S 6 3 σh 6 σd ΓLσ 6 0 0 2 2 0 0 0 4 2
Γ (^) Lσ = a (^) 1g + t1u + eg
Need now to determine the SALCs of the Lσ basis set. Three different methods will
deliver the SALCs.
Method 1
As we have done previously, the SALCs of Lσ may be determined using the
projection operator. Note that the ligand mixing in Oh is retained in the pure rotational subgroup, O. Can thus drop from Oh → O, thereby saving 24 operations.
O E C 3 C 3 C 3 C 3 C 3 C 3 C 3 C 3 C 2 C 2 C 2 C 2 C 2 C 2 C 4 C 4 C 4 C 4 C 4 C 4 C 2 C 2 C (^2) L 1 → L 1 L 3 L 4 L 4 L 5 L 6 L 5 L 6 L 3 L 2 L 2 L 6 L 5 L 4 L 3 L 1 L 1 L 5 L 3 L 4 L 6 L 1 L 2 L (^2)
The A 1 irreducible representation is totally symmetric. Hence the projection is simply the sum of the above ligand transformations.
5.04, Principles of Inorganic Chemistry II Lecture 12
PA^1 (L 1 ) ~ 4(L 1 + L 2 + L 3 + L 4 + L 5 + L 6 )
and normalizing yields,
1 ψ = ( L + L + L + L + L + L ) 6
a1g 1 2 3 4 5 6
The application of the projection operator for the E irreducible representation furnishes the Eg SALCs.
PE(L 1 ) → (2L 1 – L 3 – L 4 – L 4 – L 5 – L 6 – L 5 – L 6 – L 3 + 2L 1 + 2L 2 + 2L 2 )
→ (4L 1 + 4L 2 – 2L 3 – 2L 4 – 2L 5 – 2L 6 )
and normalizing yields,
ψ (1)^ =
( 2L + 2L − L − L − L − L ) 12 eg 1 2 3 4 5 6
But Eg is a doubly degenerate representation, and therefore there is a another SALC. As is obvious from above, the projection operator only yields one of the two SALCs. How do we obtain the other?
Method 2
The Schmidt orthogonalization procedure can extract SALCs from a non- orthogonal linear combination of an appropriate basis. Suppose we have a SALC, v 1 , then there exists a v 2 that meets the following condition,
v 2 = a v 1 + u
where u is the non-orthogonal linear combination. Multiplying the above equation by v 1 gives,
0 1 v 2 v 1 = a v 1 v 1 + uv (^1)
∴ a = – uv 1
What is the nature of u? Consider using the projection operator on L 3 instead of L 1 ,
5.04, Principles of Inorganic Chemistry II Lecture 12
ψ (^) t(1)1u = 1 ( L 1 − L ) 2 2
Applying the Schmidt orthogonalization method,
PT^1 (L 3 ) ~ 3(L 3 – L 5 ) → ψ t1u =
( L 3 − L 5 )
This wavefunction is orthogonal to ψ (^) t(1)1u^ , hence it is likely a SALC. Can prove this by
applying the Schmidt orthogonalization process and setting this to be u. Solving for a,
a = − uv 1 = − 1 (L 1 − L ) 2 1 (L 3 − L 5 ) 2 2
= − 0
( ) = 0 2
and
v 2 = av 1 + u = 0 •
( L 1 − L 2 ) +
( L 3 − L 5 )
ψ (2)^ =
t1u (^ L 3 −^ L^ ) 2 5
so, as suspected, this is a SALC. And the third SALC of T1u symmetry is the (L4,L6) pair.
ψ (3)^ =
t1u (^ L 4 −^ L^ ) 2 6
5.04, Principles of Inorganic Chemistry II Lecture 12
Method 3
For those SALCs with symmetries that are the same as s, p or d orbitals, may adapt the symmetry of the ligand set to the symmetry of the metal orbitals.
Consider the dz^2 orbital, which is more accurately defined as 2z^2 – x^2 – y^2. Thus the coefficient of the z axis is twice that of x and y and out of phase with x and y. The ligands on the z-axis, L 1 and L 2 , should therefore be twice that and of opposite sign to the equatorial ligands, L 3 ,L 4 ,L 5 ,L 6. This leads naturally to,
ψ e^ (1)g^ ≈ 2L 1 + 2L 2 − L 3 − L 4 − L 5 − L 6
ψ (1)^ =
eg (^ 2L^1 +^ 2L^2 −^ L^3 −^ L^4 −^ L^5 −^ L^6 ) 12
The other SALC of this degenerate set is given by dx^2 –y^2 , which has no coefficient on z, and x and y coefficients that are equal but of opposite sign. By symmetry matching to the orbital,
ψ e(2)g^ ≈ L 3 − L 4 + L 5 − L 6
ψ (2)^ =
e (^) g (^ L^3 −^ L^4 +^ L^5 −^ L^6 ) 2
The other SALCs follow suit.
The t2g d-orbital set (i.e. dxy, dxz, dyz) is of incorrect symmetry to interact with the
Lσ ligand set and thus is non-bonding. This can be seen from the orbital picture.
The Lσ orbitals are directed between the lobes of the t2g d-orbitals,
Only metal orbitals and SALCs of the same symmetry can overlap. In the case of
the octahedral ML 6 σ-complex,
metal SALC metal SALC
s ψ a1g pz ψ t^ ( 11 u^ )
px ψ (1)^ dx^2 –y^2 ψ (2)^
and (dxz, dyz, dxy ) t1u eg (^) are non-bonding
p y ψ t^ ( 11 u^ )^ d z^2 ψ e^ (1 g^ )
5.04, Principles of Inorganic Chemistry II Lecture 12
