Question

Consider a molecule with a formula of AB6, that has a trigonal prism geometry. Using the D3n character table, determine the orbitals of the central atom A used sigma bonding. (25 pts)

Answer 1

The orbitals used in sigma bonding while forming a compound of type AB₆ with trigonal prism geometry are,

(i) 2 orbitals of A₁^' symmetry

(ii) 2 orbitals of E^' symmetry

(iii) 2 orbitals of E^'' symmetry.

Explanation:

Symmetry properties and degeneracy of orbitals and bonds can be learned from corresponding character tables by their inspection holding in mind the following transformational properties: Atomic orbital Transforms as S x2+2+z2 Z PZ dza 22, 222-x2-y2 x2-y2 d.242 Oxy dz2 de

See the hybridization for a AB₆ structure for trigonal prism geometry which is sd⁵ hybridization.

You have to now see how the other symmetry orbitals in the D_3h character table are setup.

D3n E 202 3C, on 25, 30, A,' 1 1 1 1 1 1 Az 1 1 1 1 1 1 E 2 - 10 2 - 10 A," 1 1 1 -1 -1 -1 Az" 1 1 -1 -1 -1 1 E” 2 -1 0 -2 1 0 D3n E Ai' 1 Az 1 203 3C2 on 1 1 1 1 1 1 25, 30 1 1 1 - 1 1 1 1 -1 -1 -1 A2" 1 1 -1 -1 -1 1 E” 2 - 10 -2 10 2px 2p, doubly degenerate

So, we know that the p type orbitals will not be involved now as per the table.

I have shown first, we identify the s and d type and we know we need 6 orbitals to sigma bond to 6 incoming ligands.

Dsh E 203 3C2 매 253 3oy x2+y2.22 1 -1 AA A E A1 1 1 2 1 1 2 1 1 -1 1 1 - 1 1 -1 0 1 -1 1 1 2 -1 -1 -2 1 1 - 1 -1 -1 1 0 exy 0-1 RE x.y | 2 R.Ry -1 1 0 | E. 0 2,2