Molecular Symmetry and Group Theory Workshop 5 — Chemical bonding

• Exercise 5

Use group theory to obtain a qualitative MO energy level diagram for BCl3.

Which boron 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
2s
2p

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.
For simplicity, assume that bonding involves the 2s and 2p orbitals on boron and the 3p orbitals on chlorine.

Use one 3p orbital on each chlorine pointing directly at boron (we'll call this pσ, see diagram) as a basis for a representation of the chlorine bonding orbitals. Linear combinations with appropriate symmetry can be combined with the boron atomic orbitals.

D3h E 2C3 3C2 σh 2S3 v
Γ 3

Click for larger image D3h E 2C3 3C2 σh 2S3 v
 Order =

Γ 3
A11 1 1 1 11
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″

Constructing the molecular orbital diagram - All the required information has been determined above.

 i) The central B and outer Cl atomic orbitals in accordance with their relative energies ii) B atomic orbital symmetries iii) Symmetry of Cl pσ atomic orbitals in D3h point group iv) B 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

The boron atom has an empty 2p orbital (symmetry a2″) perpendicular to the plane of the molecule that is available for π-bonding with some of the chlorine 3p orbitals. Use a pπ orbital (see diagram) on each chlorine atom as a basis to generate a new representation, and hence produce a new MO energy level diagram including the effects of π-bonding.

D3h E 2C3 3C2 σh 2S3 v
Γ 3

Click for larger image D3h E 2C3 3C2 σh 2S3 v
 Order =

Γ 3
A11 1 1 1 11
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″

Constructing the molecular orbital diagram - All the required information has been determined above.

 i) Previous MO diagram showing the σ-bonding (the non-bonding MO's have been removed as these may change, and the dotted lines have been removed to enable the new changes to be seen more clearly) ii) Symmetry of Cl pπ atomic orbitals in D3h point group iii) Cl valence electrons iv) 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 v) Total number of valence electrons inserted into molecular orbitals

What is the effect of π-bonding on the strength of the B—Cl bonds?