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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 293
orbitals that have an accumulation of electron density between nuclei due to
overlap and constructive interference of their component atomic orbitals. A
simple and plausible explanation of this correlation is that enhanced electron
probability between nuclei lowers the potential energy by putting electrons in
a position where they can be attracted to two nuclei at the same time; however,
the source of the reduced energy may be more complicated.
Solutions to exercises
E9B.1(b) �e normalization condition is given by [7B.4c–248], ∫ ψ∗ψ dτ = 1.�e wave-
function is normalized by �nding N such thatψ = N(ψA+λψB+λ′ψ′B) satis�es
this condition.�e wavefunctions ψA, ψB and ψ′B are all real as is N , therefore
∫ Ψ∗Ψ dτ = N2 ∫ (ψA + λψB + λ′ψ′B)2 dτ
= N2[
1
³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ
∫ ψ2A dτ +λ2
1
³¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ
∫ ψ2B dτ +λ′2
1
³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ
∫ ψ′B
2 dτ +2λ
S
³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ
∫ ψAψB dτ
+ 2λ′
S′
³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ
∫ ψAψ′B dτ +2λλ′
0
³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ
∫ ψBψ′B dτ ]
= N2(1 + λ2 + λ′2 + 2λS + 2λ′S)
�e integrals ∫ ψ2A dτ, ∫ ψ2B dτ and ∫ ψ′B
2 dτ are 1 because the wavefunctions
ψA, ψB and ψ′B are normalized; ∫ ψBψ′B dτ = 0 because the exercise speci�es
that these are orthogonal.�e value of this integral must equal 1, therefore the
normalization constant is N = 1/(1 + λ2 + λ′2 + 2λS + 2λ′S)1/2 .
E9B.2(b) �e condition of orthogonality is given by [7C.8–254], ∫ ψ∗i ψ j dτ = 0 for i ≠ j.
�e given molecular orbital, ψ i = 0.727A + 0.144B is real, therefore ψ∗i = ψ i .
�e new linear combination of A and B, which is orthogonal to ψ i must have
the form of ψ j = A + βB, where the coe�cient of wavefunction A is chosen
to be 1 for simplicity. Substitution of these wavefunctions in the condition of
orthogonality gives
∫ ψ∗i ψ j dτ = ∫ (0.727A+ 0.144B) × (A+ βB)dτ
= 0.727
1
³¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹µ
∫ A2 dτ +0.144β
1
³¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹µ
∫ B2 dτ +(0.727β + 0.144)
S
³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ
∫ AB dτ
= 0.727 + 0.144β + (0.727β + 0.144)S
Using S = 0.117 the value of the integral becomes 0.744 + 0.229β. �is value
must be zero for the two wavefunctions to be orthogonal, therefore β = −3.25
and so ψ j = A− 3.25B.