Molecular Symmetry and Group Theory

Workshop 5 — Chemical bonding


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  • Exercise 2

    What is the shape of the PF5 molecule (use VSEPR theory if you do not know)
    What is the point group?
    Phosphorus has 3s, 3p and empty 3d orbitals available for bonding, and textbooks suggest that P sp3d hybrids may be used in PF5.
    Use the 5 P—F bonds to obtain a representation (Γ) and verify that a set of sp3d hybrids has the appropriate geometry.

    D3h E 2C3 3C2 σh 2S3 v
    Γ


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    D3h E 2C3 3C2 σh 2S3 v
    Order =
     
    Γ    
    A11 1 1 1 1 1
    A21 1 -1 1 1 -1
    E′ 2 -1 0 2 -1 0
    A11 1 1 -1 -1 -1
    A21 1 -1 -1 -1 1
    E″ 2 -1 0 -2 1 0
    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(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 =

    Therefore, the composition of Γ is:   Γ = A1′ + A2′ + E′ + A1″ + A2″ + E″

    D3h E 2C3 3C2 σh 2S3 v    
    A1 1 1 1 1 1 1   x2+y2,z2
    A2 1 1 -1 1 1 -1 Rz  
    E′ 2 -1 0 2 -1 0 (x,y) (x2-y2,xy)
    A1 1 1 1 -1 -1 -1    
    A2 1 1 -1 -1 -1 1 z  
    E″ 2 -1 0 -2 1 0 (Rx,Ry) (xz,yz)

      Which d orbital would be used?  
    Comments:
       

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