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294 9MOLECULAR STRUCTURE
Normalization of ψ i follows the same logic as in Exercise E9B.1(a). First the
wavefunction is written as ψ i = N(0.727A+ 0.144B) and then the normaliza-
tion constant N is found such that ∫ ψ∗ψ dτ = 1.
∫ ψ∗i ψ i dτ = ∫ [N(0.727A+ 0.144B)]2 dτ
= N2
⎛
⎜⎜⎜
⎝
0.7272
1
³¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹µ
∫ A2 dτ +0.1442β
1
³¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹µ
∫ B2 dτ +(2 × 0.727 × 0.144)
S
³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ
∫ AB dτ
⎞
⎟⎟⎟
⎠
= N2(0.549 + 0.209S)
Using S = 0.117 gives a value of 0.573N2 for the integral, thereforeN = 1/
√
0.573
= 1.32.�erefore the normalized wavefunction is
ψ i = 1.32 × (0.727A+ 0.144B) = 0.960A+ 0.190B
Normalization of ψ j follows a similar procedure as for ψ i , giving N = 0.304
and therefore ψ j = 0.304A− 0.989B .
E9B.3(b) �e energy of the σ∗ antibonding orbital in H2+ is given by [9B.7–355], Eσ∗ =
EH1s + j0/R − ( j − k)/(1 − S). Molecular potential energy curves are usually
plotted with respect to the energy of the separated atoms, therefore the energies
to be plotted are Eσ∗ − EH1s = j0/R − ( j − k)/(1 − S). Using [9B.5d–353],
j0/a0 = 27.21 eV = 1 Eh, the energy for R/a0 = 1 is computed as
Eσ∗ − EH1s =
(1 Eh)
1
− (0.729 Eh) − (0.736 Eh)
(1 − 0.858)
= +1.05 Eh
Similar calculations give the following energies
R/a0 1 2 3 4
(E∗σ − EH1s)/Eh +1.05 +0.340 +0.132 +5.52 × 10−2
�ese data are plotted in Fig. 9.2.�e data are �tted well by the following cubic
(Eσ − EH1s)/Eh = −0.0616(R/a0)3 + 0.6203(R/a0)2 − 2.1385(R/a0) + 2.6288
Note that this cubic equation has no physical meaning, it is only used to draw
the line on the plot above.
E9B.4(b) �e sketch below shows the bonding and the antibonding face-to-face overlap
of two d orbitals, resulting in δ molecular orbitals.�e bonding molecular or-
bital is symmetric with respect to inversion, therefore it is labelled δg, whereas
the antibonding molecular orbital is antisymmetric with respect to inversion,
and is labelled δu.