Physical Chemistry Third Edition

778 18 The Electronic States of Atoms. II. The Zero-Order Approximation for Multielectron Atoms

so thatMS0 forΨ 3. A similar calculation leads to the valueMS0 forΨ 4. We wrote

the operations and functions explicitly, but could simply have added themvalues and thems

values to obtain the same answers.

The wave functions in Eq. (18.4-2) are eigenfunctions of thêS^2 operator, although we do

not prove this. The eigenvalues are as follows. See Problem 18.9:

̂S^2 α(1)α(2)h ̄^2 (1)(2)α(1)α(2) (18.5-19)

̂S^2 β(1)β(2)h ̄^2 (1)(2)β(1)β(2) (18.5-20)

̂S^2 [α(1)β(2)+β(1)α(2)]h ̄^2 (1)(2)[α(1)β(2)+β(1)α(2)] (18.5-21)

̂S^2 [α(1)β(2)−β(1)α(2)] 0 (18.5-22)

The first three functions correspond to a triplet term (S1). The fourth corresponds to a sin-

glet term (S0). The symmetric spin factors correspond to the triplet, and the antisymmetric

spin factor corresponds to the singlet.

We can now infer the values ofLandSfrom the pattern thatMLranges fromLto−Land

MSranges fromSto−S. We always begin with the largest values ofMLandMSand assign

the states to the largest values ofLandSfirst. Since the only value ofMLis zero for all four

of these wave functions, the only value ofLis zero. The largest value ofMSis 1. This means

that the largest value ofSis 1. A value ofSequal to 1 requires values ofMSequal to 1, 0,

and−1. We assign three states with these values to a^3 S(tripletS) term. There is only one

state remaining, withMS0. We assign it to a^1 S(singletS) term. The only terms are a^3 S

term and a^1 S. SinceL0, the value ofJfor each term is equal to the value ofS. Including

theJvalues, the terms are^3 S 1 and^1 S 0.

EXAMPLE18.3

Enumerate the states in the (1s)(2p) configuration and assign the Russell–Saunders term

symbols.

Solution

With the 1sorbital and one of the 2porbitals we can make wave functions analogous to those

from the (1s)(2s) configuration. We use the complexΦfunctions so that we have definite values

ofm. With the 1sspace orbital and the 2p1 space orbital we have a symmetric space factor

Ψs

1

√

2

[ψ 1 s(1)ψ 2 p 1 (2)+ψ 2 p 1 (1)ψ 1 s(2)] (18.5-23)

and an antisymmetric space factor

Ψa

1

√

2

[ψ 1 s(1)ψ 2 p 1 (2)−ψ 2 p 1 (1)ψ 1 s(2)] (18.5-24)

Both of these space factors correspond toML1. The symmetric space factor combines

with the antisymmetric spin factor, leading to one state withS0. The three triplet spin

factors are all symmetric, so the antisymmetric space factor can combine with any of these,

leading to three states, withMSequal to 1, 0, and−1, corresponding toS1. The 2p 0

orbital combines with the 1sorbital in exactly the same way as the 2p1 orbital except that

ML0 for these wave functions. The 2p,−1 orbital combines with the 1sorbital in exactly

the same way as the 2p1 orbital except thatML−1. This gives a total of 12 states.