Chemistry 312 Physical Chemistry
Homework Assignment # 7
1. Given that the equilibrium internuclear distance of H+
2 is 106 pm, and that of H2 is 74.1 pm, calculate
the internuclear repulsion energy in each case. Given that the bond dissociation energy De is 2.79 eV for
H+
2 and 4.78 eV for H2 , calculate the electronic energy in each case.
Answer:
The internuclear repulsion energy, in atomic units, is given by
Enuc−rep =
ZA ZB
.
rAB
For H+
2 and H2 the nuclear charges are the same. Therefore,
1 × 1 × 0.529177 × 10−10 m/a0
106 × 10−12 m
=0.499 hartree
=13.60 eV.
1 × 1 × 0.529177 × 10−10 m/a0
Enuc−rep (H2 )=
74.1 × 10−12 m
=0.714 hartree
=19.43 eV.
Enuc−rep (H2+ )=
In these calculations, we have used the conversion factors 1a0 = 0.529177 × 10−10 m, and 1 hartree
= 27.2114 eV. The bond dissociation energy De is the energy difference between the molecule at its
equilibrium geometry and the two atoms at infinite separatation. The energy of the molecule is the sum
of the electronic energy and the nuclear repulsion energy. Therefore,
De = Eel + Enuc−rep
Therefore,
Eel (H2+ )=2.79 − 13.60 = −10.81 eV.
Eel (H2 )=4.78 − 19.43 = −14.65 eV.
2. (a) On the basis of the molecular orbital energy level diagrams for homonuclear diatomic molecules,
explain why Helium is not a diatomic molecule like H2 , O2 , etc. (b) Can we expect the He+
2 molecular
ion to be stable on the basis of molecular orbital theory?
Answer
(a) Helium molecule would have an electronic configuration of (1sσg )2 (1sσu∗ )2 . This means that the molecule
has a bond-order of 0. Therefore, it cannot exist according to the molecular orbital energy level diagrams.
2
∗
(b) He+
2 molecular ion will have the electronic configuration (1sσg ) (1sσu ). Since there are 2 electrons in
the bonding orbital and only one electron in the antiboding orbital, the bond order is 1/2. Therefore, this
species can be expected to be stable at least at low temperatures.
1
3. In a linear variational treatment of a diatomic molecule AB, the following trial function is used:
ϕ = c1 (2sA ) + c2 (2pz,B ),
where the Z-axis is the internuclear axis. The following matrix elements are given for the Hamiltonian
and overlap matrices at a particular value of the internuclear separation rAB :
√
√
√
HAA = HBB = 2 3; HAB = HBA = − 5; SAB = 1/ 5.
(a) Find the variational energies for the bonding and antibonding orbitals.
(b) Assuming that the linear variational parameters c1 and c2 are of comparable magnitude, sketch the bonding
and antibonding orbtitals and clearly mark the positions of the nodes. (You may ignore the presence of
the radial node in the 2s orbital although I would recommend that you try to sketch the molecular orbitals
with the radial node present.)
Answer
(a) The variational energies are
HAA ∓ HAB
1 ∓ S√
√
2 3∓ 5
√ .
=
1 ∓ 1/ 5
W± =
This yields
√
√
2 3+ 5
√ = 10.3117.
W+ =
1 − 1/ 5
√
√
2 3− 5
√ = 0.84855.
W− =
1 + 1/ 5
(b) The bonding orbital is the one corresponding to the lower energy or c1 = c2 . The antibonding orbitals
correspond to the higher energy, or c1 = −c2 . The sketches of the atomic orbitals are shown towards the
end of this file.
4. Sketch qualitative contour maps of the following molecular orbitals for a diatomic molecule AB. Identify
each as a bonding or an anti-bonding molecular orbital. The notation is the same as the one we have been
using for H+
2 and other examples. Unless otherwise specified, the Z-axis is the molecular axis. Ignore
the radial node(s) in the atomic orbitals. (a) 2sA + 2pz,B (b) 2sA − 2pz,B (c) √12 [2sA + 2pz,A ] + 1sB
(d) √12 [2sA − 2pz,A ] + 1sB .
Answer
(a)
(b)
(c)
(d)
Same as the case c1 = c2 in Q. 3.
Same as the case c1 = −c2 in Q. 3.
See next page.
See next page.
2
Q. 3 (c1 = c2) and Q. 4 (a):
−
+
+
A
B
2sA + 2pz,B
+
+
−
Q. 3 (c1 = –c2) and Q. 4(b)
−
+
+
A
B
2sA − 2pz,B
+
−
Q. 4 (c)
+
+
−
+
B
A
+
[2sA + 2pz,A] +1sB
+
−
Q. 4 (d)
+
−
+
+
B
A
−
[2sA − 2pz,A] +1sB
+