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Example #8: EX4 type molecules in Td : CH4 and CCl4 • As with EX3 we can derive the bonding/antibonding MOs in a simple pictorial

approach - just considering the py orbitals on the chlorine atoms of the CCl4 example:

Source: Purcell & Kotz, "Inorganic Chemistry", Holt-Saunders, 1977. • Rearranging the square into a tetrahedron leads to the disappearance of the second

nodal plane in the highest energy combination of the square. In the tetrahedron the three high energy orbitals are energetically degenerate and they transform like T2 in the Td pointgroup.

General rule: The more nodal planes (i.e. sign changes) occur in an MO the higher it

will be in energy. Source: Purcell & Kotz, "Inorganic Chemistry", Holt-Saunders, 1977. • The same procedure can also be done with CH4 and is in fact simpler, as we only

have to deal with s-orbitals on the four hydrogen atoms.

• Shown here are the possible bonding interaction between the central carbon p-orbitals and the set of symmetry related s-orbitals on hydrogen - the corresponding anti-bonding orbitals would result by inverting the p-orbitals:

Source: F.A. Cotton, "Chemical Applications of Group Theory", 3rd Ed. Wiley Interscience, 1990

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The MOs of methane: The HOMO and LUMO are both triply degenerate! (Energies in eV, calculated using SPARTAN at the AM1 level)

(A1) E = -28.88284

(T2) E = -13.30864

(T2) E = -13.30864

(T2) E = -13.30864

(T2) E = 4.65954

(T2) E = 4.65954

(T2) E = 4.65954

(A1) E = 5.18151

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The generic MO scheme for EH4 in Td symmetry is:

Source: F.A. Cotton, "Chemical Applications of Group Theory", 3rd Ed., Wiley Interscience, 1990. • Note that for methane E(A1)* > E(T2)*, i.e. different then in the above scheme. • For CCl4 we have to consider a more general MO scheme that incorporates the MOs

generated by the linear combinations of the chlorine p-orbitals: Source: Purcell & Kotz, "Inorganic Chemistry", Holt-Saunders, 1977.

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The MOs of CCl4:

(A1) E = -28.244603

(T2) E = -18.09583

(E nb) E = -13.53845

(T2 nb) E = -12.53064

(T1 nb) E = -12.37889

HOMO

(A1

*) E = -1.11874 LUMO

(T2

*) E = .11987

Omitted here are the MOs originating from the carbon 1s and chlorine 1s and 2s atomic orbitals. E = orbitals are doubly degenrate in energy T = orbitals are triply degenerate in energy

HOMEWORK: Draw an MO diagram for hypothetical EH4 in D4h symmetry and develop a Walsh diagram (D4h to Td). Why is CH4 tetrahedral?