Physical Chemistry Third Edition

848 20 The Electronic States of Diatomic Molecules

Term symbols can be written for the ground levels of homonuclear diatomic

molecules from inspection of their electron configurations and use of Hund’s first rule.

Consider the ground level of the B 2 molecule, with six electrons. Using the Aufbau

principle, we choose theσg 1 sandσ∗u 1 sspace orbitals for the first four electrons. The

electron spins cancel in these orbitals, so they make no contribution toS. These orbitals

all correspond tom0, so they make no contribution toML. The two degenerate pi

bonding space LCAOMOs are the next available orbitals. By Hund’s first rule, the

remaining two electrons occupy different space orbitals with unpaired spins so that

S1. To determine the value ofΛ, we choose the complex pi orbitals containing the

Φ 1 andΦ− 1 functions instead of theΦ 1 xandΦ 1 yfunctions, so thatm1 for one pi

orbital andm−1 for the other. We find thatML0 andΛ0, corresponding to a

(^3) Σterm.
Both of the real bonding pi orbitals have eigenvalue−1 for the inversion operator,
so their product has eigenvalue+1. The right subscript is g. The same vertical reflection
plane is used for all orbitals to determine whether the term is+or−. A vertical mirror
plane in thexzplane will give eigenvalue+1 for theπu 2 pxorbital and eigenvalue− 1
for theπu 2 pyorbital. The eigenvalues would be reversed for a mirror plane in theyz
plane, and the product gives an eigenvalue of−1 in either case. The term symbol for
the ground state of B 2 is^3 Σ−g. The same result could be obtained in a more compli-
cated way by separately considering the real and imaginary parts of the complex pi
orbitals.
Exercise 20.11
Argue that the same term symbol occurs for a homonuclear diatomic molecule if theπu2,− 1
andπu2,+1 orbitals are considered instead of theπu 2 pxandπu 2 pyorbitals.
An Alternative Set of Configurations
The attractive effect of an electron in a bonding orbital and the repulsive effect of
an electron in an antibonding orbital approximately cancel, so the variational energy
will be almost unchanged if we replace one bonding LCAOMO and one antibonding
LCAOMO by the two atomic orbitals from which these LCAOMOs were constructed.
If this is done, electrons occupying atomic orbitals are counted as nonbonding and are
omitted from bond order calculations.

EXAMPLE20.6

Give the electron configuration for the ground state of C 2 using nonbonding orbitals as much

as possible.

Solution

There are 12 electrons. The electron configuration is

(1sA)^2 (1sB)^2 (2sA)^2 (2sB)^2 (πu 2 px)^2 (πu 2 py)^2

The bond order is equal to 2, with four bonding electrons, no antibonding electrons, and eight

nonbonding electrons.