Symmetry Adapted Linear Combinations (SALCs)

Question 1

a. PdCl₄^2- (square planar)

Point Group = D_4h

Reduces to A_1g + B_1g + E_u

b. SiCl₄

Point Group = T_d

Reduces to A₁ + T₂

c. cis-PtF₂Cl₂^2-

Point Group = C_2v

Reduces to A₁ + B₁ for both F and Cl

d. SF₆

Point Group = O_h

Reduces to A_1g + E_g + T_1u

Question 2

a. PdCl₄^2- (square planar)

b. SiCl₄

c. cis-PtF₂Cl₂^2-

d. SF₆

Molecular Orbital Diagrams (σ-bonding)

Question 1

a. trans-SF₂Cl₄

Point Group = D_4h

Γ_σ(F) reduces to A_1g + A_2u

Γ_σ(Cl) reduces to A_1g + B_1g + E_u

b. NH₄⁺

Point Group = T_d

Reduces to A₁ + T₂

c. SF₆

Point Group = O_h

Reduces to A_1g + E_g + T_1u

Molecular Orbital Diagrams (π-bonding)

Question 1

a. SiF₄

Point Group = T_d

All sets of vectors lie along a C₃ rotation axis, therefore the p orbitals cannot be separated into subsets

Reduces to E + T₁ + T₂

Note: Si used s, p_x, p_y and p_z orbitals for σ-bonding

b. GeH₂Cl₂

Point Group = C_2v

Only the Cl atoms can participate in π-bonding. The Cl atoms lie along a C₁ rotation axis, therefore the p orbitals can be separated into perpendicular and parallel to the Cl-Ge-Cl plane.

Γ_π⊥ reduces to A₂ + B₂

Γ_π // reduces to A₁ + B₁

Hint: Draw the SALCs to determine the relative order

Note: Ge used s, p_x, p_y and p_z orbitals for σ-bonding

c. SnCl₆^2-

Point Group = O_h

All sets of vectors lie along a C₄ rotation axis, therefore the p orbitals cannot be separated into subsets

Reduces to T_1g + T_2g + T_1u + T_2u

Question 2

a. π-bonding perpendicular to the plane

Point Group = D_4h

The vectors lie along a C₂ rotation axis, there the p orbitals can be separated into one set of four perpendicular to the plane and one set of four parallel to the plane.

Recall that d block elements use ns, np and (n-1)d orbitals for bonding, and that (n-1)d orbitals are lower in energy than ns and np orbitals.

Reduces to E_g + A_2u + B_2u

b. π-bonding parallel to the plane

Point Group = D_4h

The vectors lie along a C₂ rotation axis, there the p orbitals can be separated into one set of four perpendicular to the plane and one set of four parallel to the plane.

Recall that d block elements use ns, np and (n-1)d orbitals for bonding, and that (n-1)d orbitals are lower in energy than ns and np orbitals.

Reduces to A_2g + B_2g + E_u

Molecular Orbital Diagrams (π-bonding in Planar Molecules)

Question 1

a. C₄H₄

Point Group = D_4h

C₄H₄ is anti-aromatic.

b. B₃N₃H₆ (Borazine)

Point Group = D_3h

B₃N₃H₆ is aromatic.

c. C₈H₈^2-

Point Group = D_8h

C₈H₈^2- is aromatic.