Molecular Symmetry and Group Theory

Workshop 3 — Character tables

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  • Exercise 1

    The multiplication table derived earlier for C2v   The A2 irreducible representation
    C2v E C2 σ(xz) σ(yz)
    E E C2 σ(xz) σ(yz)
    C2 C2 E σ(yz) σ(xz)
    σ(xz) σ(xz) σ(yz) E C2
    σ(yz) σ(yz) σ(xz) C2 E
      E C2 σ(xz) σ(yz)
    A2 1 1 -1 -1

    In the table below, these numbers have been substituted for the operations in row 1 and column 1. Complete the multiplication table using the numbers.

    C2v 1 1 -1 -1
    -1 -1

    In the point group C2v, σ(xz)C2 = σ(yz) which becomes -1 × 1 = -1, so multiplication of the numbers allocated is consistent with the original table.

    Is this true for all the numbers in the table? (If not, seek help!)

    Now complete the multiplication table using the assignment:

    E C2 σ(xz) σ(yz)
    1 -1 -1 -1

    C2v 1 -1 -1 -1
    -1 1

    In this case, substituting numbers for σ(xz) and C2 gives -1 × -1 = 1, but the result of the combined operation σ(xz)C2 is σ(yz), which according to the assignment should have the number -1.

    The numbers 1 -1 -1 -1 cannot represent the effects of the group operations satisfactorily, and are therefore not an irreducible representation of the C2v point group. There are in fact only four sets of numbers for which the multiplication table works: these are the characters of the four irreducible representations A1, A2, B1 and B2.

    The number of irreducible representations is always equal to the number of classes of symmetry operations in the group.

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