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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 281
E8C.8(b) �e Clebsch–Gordan series, [8C.5–332], is used to combine two spin angular
momenta s1 and s2 to give S = s1 + s2 , s1 + s2 − 1 ..., ∣s1 − s2∣.
(i) Four electrons are treated by �rst combing the angular momenta of two
electrons, then combining the result with the spin of the third electron,
and then with the spin of the fourth. For two electrons, each with s = 1
2 ,
S′ = 1, 0. Combining S′ = 1 with s3 = 1
2 for the third spin gives S
′′ =
1+ 12 , ∣1−
1
2 ∣ =
3
2 ,
1
2 . Combining S
′ = 0 with s3 = 1
2 just gives S
′′ = 1
2 .�e
process is continued with the fourth electron: S′′ = 3
2 with s4 =
1
2 gives
S = 2, 1; S′′ = 1
2 with s4 =
1
2 gives S = 1, 0. In summary the possible
values for the total spin are S = 2, 1, 0 , with multiplicities 5, 3, 0 .
(ii) For �ve electrons the values of S from (i) are combined with a further
spin s5 = 1
2 . For S = 2 the result is 52 ,
3
2 ; S = 1 gives 32 ,
1
2 ; and S = 0
gives just 12 .�e possible values for the total spin for �ve electrons are are
S = 5
2 ,
3
2 ,
1
2 , with multiplicities 6, 4, 2 .
E8C.9(b) �e valence electron con�guration of the V2+ is [Ar] 3d3. With 3 electrons
occupying the 5 d orbitals the Pauli principle places no restrictions on the num-
ber of parallel spins because the 3 electrons can occupy separate orbitals. �e
procedure used in Exercise E8C.8(b) part (i) can therefore be followed, giving
the possible values of S for 3 electrons as S = 3
2 ,
1
2 .�e quantum numberMS ,
which speci�es the z-component of the spin angular momentum, takes values
S , (S−1) . . . −S.�erefore for S = 3
2 , MS = ±
3
2 , ±
1
2 and for S =
1
2 , MS = ±
1
2 .
E8C.10(b) �ese electrons are not equivalent, as they are in di�erent subshells, hence all
the terms that arise from the vector model and the Clebsch–Gordan series are
allowed.�e orbital angular momentum of the p and d electrons are l1 = 1 and
l2 = 2 respectively, and these are combined using L = l1+ l2 , l1+ l2−1, ... ∣l1− l2∣
which in this case gives L = 3, 2, 1.�e spin angular momenta of each electron
is s1 = s2 = 1
2 , and these combine in the same way to give S = 1, 0; these values
of S have spinmultiplicities of 2S+1 = 3, 1.�e termswhich arise are therefore
3F, 1F, 3D, 1D, 3P, and 1P.
�e possible values of J are given by J = L + S , L + S − 1, ..., ∣L − S∣, and
hence for S = 1, L = 3 the values of J are 4, 3, and 2. For S = 1, L = 2 the
values of J are 3, 2, and 1. For S = 1, L = 1 the values of J are 2, 1, and 0. For
S = 0 and any value of L only J = L is possible.�e term symbols are therefore
3F4, 3F3, 3F2, 3D3, 3D2, 3D1, 3P2, 3P1 and 3P0 , and 1F3, 1D2, and 1P1 .
FromHund’s rules, described in Section 8C.2(d) on page 335, the lowest energy
state is the one with the greatest spin, then the greatest orbital angular momen-
tum and then, because the shell is less than half full, the smallest J.�is is 3F2 .
E8C.11(b) (i) 3D has L = 2, S = 1 and so J = 3, 2, 1 only.�ere are 2J + 1 values of MJ ,
which for these three values of J gives 7, 5, 3 states, respectively. (ii) 4D has
L = 2, S = 3
2 , and so J =
7
2 ,
5
2 ,
3
2 ,
1
2 with 8, 6, 4, 2 states, respectively. (ii)
2G
has L = 4, S = 1
2 , and so J =
9
2 ,
7
2 with 10, 8 states, respectively.