Molecular Symmetry and Group Theory

Workshop 5 — Chemical bonding

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    • Exercise 3

    From a conventional viewpoint we might expect all eight bonding electrons in CH4 to be indistinguishable with identical energies. UV photoelectron spectroscopy experiments, however, indicate that while six valence electrons have an ionization energy of 14 eV, two are more strongly bound (23 eV).

    Use the following procedure to develop a qualitative MO energy level diagram for CH4:

    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 (i.e. The C 2s and 2p orbitals) (N.B. By convention lower case symbols e.g. a1 are used to label the symmetry of orbitals)
    Orbital Symmetry
    2s
    2p

    As the 2s orbital is spherically symmetric, it is unaffected by any operation. This is true of all s orbitals, which as a result always belong to the totally symmetric irreducible representation (the first one in any character table). The symmetry of the 2s orbital in Td is therefore a1.

    iii) Select appropriate orbitals on the outer atoms (H 1s) and use these as a basis to obtain a reducible representation. Reducing this will give irreducible representations (i.e. symmetries) appropriate to the orbitals in the basis.

    Use the 4 H 1s orbitals in the CH4 molecule as a basis to obtain a representation in the tetrahedral point group Td, and reduce this to its component irreducible representations.

    Td E 8C3 3C2 6S4 6σd
    Γ 4


    Click for larger image

    Td E 8C3 3C2 6S4 6σd
    Order =
    		 
    Γ 4    
    A1 1 1 1 1 1
    A2 1 1 1 -1 -1
    E 2 -1 2 0 0
    T1 3 0 -1 1 -1
    T2 3 0 -1 -1 1
    N×χR×χI(A1)
    Σ(N×χR×χI(A1)) =
    Σ(N×χR×χI(A1))/Order =
    N×χR×χI(A2)
    Σ(N×χR×χI(A2)) =
    Σ(N×χR×χI(A2))/Order =
    N×χR×χI(E)
    Σ(N×χR×χI(E)) =
    Σ(N×χR×χI(E))/Order =
    N×χR×χI(T1)
    Σ(N×χR×χI(T1)) =
    Σ(N×χR×χI(T1))/Order =
    N×χR×χI(T2)
    Σ(N×χR×χI(T2)) =
    Σ(N×χR×χI(T2))/Order =

    Therefore, the composition of Γ is:   Γ = A1 + A2 + E + T1 + T2

    Reducing to find the constituent irreducible representations tells us the symmetry of linear combinations of atomic orbitals on the outer atoms that can be combined with atomic orbitals of the same symmetry on the central atom, and thus enables an energy level diagram to be constructed. Group theory and considerations of symmetry alone, however, cannot help us in determining the energies of atomic orbitals, the degree of overlap and the strength of bonding interactions, for which we must resort to quantum mechanics or experimental data.

    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 CH4.

    i) The central C and outer H atomic orbitals in accordance with their relative energies
    ii) C atomic orbital symmetries
    iii) Symmetry of H atomic orbitals in Td point group
    iv) C and H 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

    How many bonds are indicated?
    Comments:
    Why do we not see different C—H bond lengths corresponding to the different bonding MO's?

    [Note that the experimental data given above provides strong evidence for the validity of MO theory.]

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