21.1 The BeH 2 Molecule and thespHybrid Orbitals 869
we do not discuss the higher-energy orbitals. The LCAOMOs are all sigma orbitals,
since all of the atomic orbitals in the basis set correspond tom0. The molecular
orbitals are numbered from the lowest to the highest energy, and are labeled g or u
depending on the eigenvalue of the inversion operator. Since there are six electrons in
the molecule, all three of these LCAOMOs are occupied in the ground-state.The orbital region of the 3 (^) u orbital
The orbital region of the 2 (^) gorbital
The orbital region of the 1 (^) g orbital.
Figure 21.2 The Orbital Regi-
ons of Two Bonding Orbitals
of the BeH 2 Molecule.Positive
regions are in black and negative
regions are in gray.
The 1σgorbital is nearly the same as the 1sBe atomic orbital. This orbital is an
eigenfunction of all of the symmetry operators that belong to the molecule with all
eigenvalues equal to+1. The 2σgLCAOMO is a completely bonding orbital and is
also an eigenfunction of all of the relevant symmetry operators with eigenvalues+1.
The 3σuorbital has a nodal surface through the Be atom, but also counts as a bonding
orbital, because it has no nodal surfaces between the atoms. It is also an eigenfunction
of the relevant symmetry operators, but some of the eigenvalues are equal to−1. Each
bonding orbital corresponds to a one-half order bond between the beryllium atom and
each of the hydrogen atoms, because the pair of bonding electrons moves over both
bonds. We have two single bonds. Figure 21.2 depicts the orbital regions of the 2σg
and 3σgmolecular orbitals.
Exercise 21.1
Determine the eigenvalues of all of the relevant symmetry operators for the 3σuLCAOMO.
The orbitals that result from a Hartree–Fock–Roothaan calculation are calledcanon-
ical orbitals. There is only one possible set of such orbitals for a given molecule and
a given basis set, except that different linear combinations of degenerate orbitals can
occur. Two of the canonical orbitals of BeH 2 extend over the entire molecule and do not
conform to the elementary chemical notion that bonding electrons are shared between
two nuclei. We can transform these orbitals to another set of orbitals that nearly conform
to this notion.
The wave function of the molecule must be antisymmetrized, which is done by
constructing a 6×6 Slater determinant in which the three space LCAOMOs are used
to form six spin orbitals. The first three rows of this determinant are:
ψ 1 σg(1)α(1)ψ 1 σg(1)β(1)ψ 2 σg(1)α(1)ψ 2 σg(1)β(1)ψ 3 σu(1)α(1)ψ 3 σu(1)β(1)
ψ 1 σg(1)α(1)ψ 1 σg(2)β(2)ψ 2 σg(2)α(2)ψ 2 σg(2)β(2)ψ 3 σu(2)α(2)ψ 3 σu(2)β(2)
ψ 1 σg(3)α(3)ψ 1 σg(3)β(3)ψ 2 σg(3)α(3)ψ 2 σg(3)β(3)ψ 3 σu(3)α(3)ψ 3 σu(3)β(3)
etc.
The other three rows contain the same spin orbitals as the first three rows, but with
electron labels 4, 5, and 6. Each column of the determinant contains one spin orbital.
A useful property of determinants is stated as property (5) of determinants in
Eq. (B-100) of Appendix B: If each element of a column is replaced by that element
plus the same constant times the corresponding element of one of the other columns,
the value of the determinant is unchanged.
Exercise 21.2
Using expansion by minors, show that the equality in Eq. (B-100) is valid.
We replaceψ 2 σgwithc 1 (ψ 2 σg+ψ 3 σu) andψ 3 σuwithc 2 (ψ 2 σg−ψ 3 σu) in every row of
the determinant, wherec 1 andc 2 are constants. According to property (5) and property