Molecular Symmetry and Group Theory

Workshop 5 — Chemical bonding

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    • Exercise 6

    Use group theory to obtain a qualitative MO energy level diagram for the octahedral complex [CrCl6]3-.

    First you will need to consider which Cr3+ orbitals may be used in bonding.

    i) What is the point group?
    ii) Using the character table (Click here), determine the symmetry of the orbitals on the central atom to be used in bonding (If a group of orbitals have more than one symmetry, put in alphabetical order separated by ",")
    Orbital Symmetry
    3d
    4s
    4p

    iii) Select an appropriate basis to obtain a reducible representation. Reducing this will give irreducible representations (i.e. symmetries) for the orbitals in the basis.
    It is not strictly necessary to know which orbitals on the Cl- ligands are involved: we can instead use six arrows, one pointing from each Cl- towards the central metal atom, to represent σ-bonding orbitals on the chloride ligands and act as a basis for our reducible representation.


    Click for larger image
    Click for larger image

    Oh E 8C3 6C2 6C4 3C2′ i 6S4 8S6 3σh 6σd
    Γ 6

    Oh E 8C3 6C2 6C4 3C2′ i 6S4 8S6 3σh 6σd
    Order =
    		 
    Γ 6    
    A1g 1 1 1 1 11 1 1 1 1
    A2g 1 1 -1 -1 1 1 -1 1 1 -1
    Eg 2 -1 0 0 2 2 0 -1 2 0
    T1g 3 0 -1 1 -1 3 1 0 -1 -1
    T2g 3 0 1 -1 -1 3 -1 0 -1 1
    A1u 1 1 1 1 1 -1 -1 -1 -1 -1
    A2u 1 1 -1 -1 1 -1 1 -1 -1 1
    Eu 2 -1 0 0 2 -2 0 1 -2 0
    T1u 3 0 -1 1 -1 -3 -1 0 1 1
    T2u 3 0 1 -1 -1 -3 1 0 1 -1
    N×χR×χI(A1g)
    Σ(N×χR×χI(A1g)) =
    Σ(N×χR×χI(A1g))/Order =
    N×χR×χI(A2g)
    Σ(N×χR×χI(A2g)) =
    Σ(N×χR×χI(A2g))/Order =
    N×χR×χI(Eg)
    Σ(N×χR×χI(Eg)) =
    Σ(N×χR×χI(Eg))/Order =
    N×χR×χI(T1g)
    Σ(N×χR×χI(T1g)) =
    Σ(N×χR×χI(T1g))/Order =
    N×χR×χI(T2g)
    Σ(N×χR×χI(T2g)) =
    Σ(N×χR×χI(T2g))/Order =
    N×χR×χI(A1u)
    Σ(N×χR×χI(A1u)) =
    Σ(N×χR×χI(A1u))/Order =
    N×χR×χI(A2u)
    Σ(N×χR×χI(A2u)) =
    Σ(N×χR×χI(A2u))/Order =
    N×χR×χI(Eu)
    Σ(N×χR×χI(Eu)) =
    Σ(N×χR×χI(Eu))/Order =
    N×χR×χI(T1u)
    Σ(N×χR×χI(T1u)) =
    Σ(N×χR×χI(T1u))/Order =
    N×χR×χI(T2u)
    Σ(N×χR×χI(T2u)) =
    Σ(N×χR×χI(T2u))/Order =

    Therefore, the composition of Γ is:
    Γ = A1g + A2g + Eg + T1g + T2g + A1u + A2u + Eu + T1u + T2u

    Constructing the molecular orbital diagram - All the required information has been determined above (spectroscopic data can be used to assist in determining the relative energies of the bonding molecular orbitals in [CrCl6]3-).

    i) The central Cr3+ and outer Cl- atomic orbitals in accordance with their relative energies
    ii) Cr3+ atomic orbital symmetries
    iii) Symmetry of Cl- atomic orbitals in Oh point group
    iv) Cr3+ and Cl- valence electrons
    v) Orbitals with the same symmetry, similar energy and size overlap to give bonding and anti-bonding molecular orbitals with the same symmetry as the original atomic orbitals
    vi) Total number of valence electrons inserted into molecular orbitals

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