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296 9MOLECULAR STRUCTURE
0.56 and N− = 1.10. Both of the bonding and antibonding wavefunctions are
real, therefore ψ± = ψ∗±. �e square of the wavefunction gives the probability
density of �nding the electron at a particular position.�e volume element δV
given in this problem (1.00 pm3) is small enough that it is assumed that the
wavefunction has a constant value over the entire volume element. �erefore
the probability is simply ψ2δV .
(a) At nucleus A x = y = z = 0 the probability is computed as
ψ2+δV = N2+
(πa30)
[e0 + e−2.00 a0/a0]
2
δV
= 0.562
[π × (52.91 pm)3]
[1 + e−2.00]2 × (1.00 pm3) = 8.7 × 10−7
(b) �e probability of �nding the electron at nucleus B must be the same as
the probability of �nding the electron at nucleus A due to the inherent
symmetry of the problem.�erefore P = 8.7 × 10−7 .
(c) �e Cartesian coordinates of the position halfway between A and B are
x = y = 0 and z = 1.00 a0. At this position the probability is computed as
ψ2+δV = N2+
(πa30)
[e−1.00 a0/a0 + e−[((1.00 a0)−(2.00 a0))
2]1/2/a0]
2
δV
= 0.562
[π × (52.91 pm)3]
[e−1.00 + e−1.00]2 × (1.00 pm3) = 3.6 × 10−7
(d) At 20 pm along the bond from nucleus A and 10 pm perpendicularly
the coordinates are z = 20 pm = 0.378... a0 and x2 + y2 = (10 pm)2 =
(0.189... a0)2. At this position the probability is computed as
ψ2+δV = N2+
(πa30)
[e−[(0.189.. . a0)
2+(0.378.. . a0)2]1/2/a0
+e−[(0.189.. . a0)
2+((0.378.. . a0)−(2.00 a0))2]1/2/a0]
2
δV
= 0.562
[π × (52.91 pm)3]
[e−0.422.. . + e−1.63.. .]2 × (1.00 pm3)
= 4.9 × 10−7
�e calculation of the probabilities when the electron occupies the antibonding
orbital follows exactly the same procedure as above, and gives the following re-
sults: (a) P = 1.9 × 10−6 ; (b) P = 1.9 × 10−6 ; (c) P = 0 , which is the expected
result, as the antibonding orbital has a node going through the point halfway
between the two nuclei; (d) P = 5.5 × 10−7
P9B.4 �e bonding and antibondingMOwavefunctions areψ± = N±(ψA±ψB), where
N± is the normalizing factor. �is factor depends on the distance between
the nuclei (Example 9B.1 on page 352), but this just scales the orbital without