CHEM 301: Homework assignment #7
Solutions
1. Calculate the energy (in eV) of Coulombic repulsion between two hydrogen nuclei at their equilibrium separation in H2 (74.1 pm). What is the
potential energy for Coulombic attraction of the nuclei to the electron
density between them in this configuration? Can the total energy of the
H2 molecule be lowered? (30%)
The Coulombic repulsion energy between two hydrogen nuclei is:
U=
e2
,
4πεr
where e is the elementary charge (the charge of an H nucleus) and r is
the distance between the H nuclei given in the problem (74.1 pm). The
repulsion energy is then:
2
1.60 · 10−19 C
m = 3.11 · 10−18 J = 19.4 eV.
U=
4 · 3.14 · 8.85 · 10−12 F/m · 74.1 · 10−12
Assuming that all electron density in an H2 molecule is found between
the two nuclei, the attraction of the nuclei to the electrons needs to compensate the nucleus-nucleus repulsion in order for the nuclei to remain at
their equilibrium distance (since there are no other forces acting on the
nuclei). Thus, the potential energy for Coulombic attraction of the nuclei
to the electron density between them should be the same as the Coulombic
repulsion energy between the two nuclei. In reality, some of the electron
density is outside the internuclear region. Thus, the energy of attraction to electrons that are in the internuclear region must be somewhat
higher than what we estimated, so as to compensate for the attraction to
electrons outside the internuclear region that is pulling the nuclei apart.
The equilibrium bond length between two atoms is determined by the
condition that the total energy of the molecule at this distance is minimized. Since the problem states that the H nuclei are at their equilibrium
separation in H2 , the energy of the molecule cannot be lowered.
2. Write down the valence bond wavefunctions for:
• a C–H group of a molecule; (30%)
In general, the valence bond wavefunction is:
ΨC−H = ψC (1) ψH (2) + ψC (2) ψH (1) ,
where the orbital ψC is some orbital on the C atom and the orbital
ψH is some orbital on the H atom. The indices 1 and 2 indicate
the position of the first and the second electron. (Here we have not
included the normalization constant.)
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In order to write down a wavefunction for a C–H bond, we need to
make an assumption regarding the hybridization state of carbon.
For sp3 -hybridized carbon, the bond is formed by pairing the electron
in any of the sp3 orbitals on carbon, for instance,
ψCsp3 (1) =
1
φC2s + φC2px + φC2py + φC2pz ,
2
with the electron in the 1s orbital of hydrogen, φH1s .
Similarly, for an sp2 C atom, the bond is formed by pairing the
electron in any of the sp2 orbitals on carbon, for instance
√
1 ψCsp2 (1) = √ φ2s + 2φ2px
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with the electron in the 1s orbital of hydrogen, φH1s .
Finally, for an sp C atom, the bond is formed by pairing the electron
in either of the sp orbitals on carbon, for instance
r
1
ψCsp(1) =
(φ2s + φ2pz )
2
with the electron in the 1s orbital of hydrogen, φH1s .
• a CO2 molecule. (30%)
The C atom in a CO2 molecule is sp-hybridized. It forms a σ-bond
and a π-bond with each of the O atoms. In general, each valence
bond wavefunction has the form:
ΨC−O = ψC (1) ψO (2) + ψC (2) ψO (1) .
(Here we have not included the normalization constant.)
The two σ-bonds are formed by the sp-orbitals on C:
r
1
ψCsp(1) =
(φ2s + φ2pz ) ,
2
r
1
ψCsp(2) =
(φ2s − φ2pz ) ,
2
and the 2pz orbitals on each of the O atoms.
The π-bonds are formed by the unhybridized 2px and 2py orbitals
on the C atom and the 2px and 2py orbitals on the first and second
O atoms, respectively (see solution to the next question for further
details).
What is the bond angle for the CO2 molecule predicted by valence bond
theory? (10%)
2
The original version of valence bond theory (before the emergence of the
concepts of promotion and hybridization) could not explain the structure
of the CO2 molecule. It would suggest the following electron configuration
for carbon: [He] 2s2 2p1x 2p1y 2p0z . This meant that carbon would only be
able to form 2 bonds. The electron configuration for oxygen would be
[He] 2s2 2p2x 2p1y 2p1z . Thus, each oxygen would also be able to form 2
bonds. Consequently, if each oxygen formed a single bond with carbon,
then carbon could form no more bonds, but each oxygen would still have
one unpaired electron. Furthermore, since in this picture the orbitals on
carbon that have unpaired electrons are 2px and 2py , this would suggest
that the angle between the bonds that carbon forms must always be 90◦
(since 2px and 2py are directed along the x and y axes that are perpendicular to each other).
The introduction of ideas of promotion and hybridization solved these
problems. Promoting an electron from the 2s to the 2pz orbital in carbon
results in the electronic configuration: [He] 2s1 2p1x 2p1y 2p1z . In manyelectron atoms, promotion is energetically unfavorable, since p-orbitals
are higher in energy than s-orbitals of the same shell (this is in contrast
to the case of H-like atoms, where the orbital energy only depends on the
principal quantum number). However, the promoted atom can compensate the energy required for promotion by forming 4 bonds instead of 2,
since each bond formation lowers the total energy of the molecule.
To ensure correct bond lengths and relative bond orientation, we further
need to introduce the idea of hybridization. The C atom in CO2 is sphybridized. The two sp-orbitals are formed by mixing of the 2s orbital and
the 2pz orbital. Thus, they are both cylindrically symmetric with respect
to the z-axis (extend in the positive and negative directions along that
axis). The angle between these orbitals is 180◦ . The two sp-orbitals on
the C atom form σ-bonds with the 2pz orbitals on each of the O atoms.
Both of these bonds are directed along the z-axis and the angle between
them is 180◦ . This is a prediction that agrees with the experimental value.
The remaining two unhybridized orbitals on C, 2px and 2py , can form πbonds with the 2px and 2py orbitals on O, respectively. Since the energy
levels corresponding to 2px and 2py orbitals are the same, they could be
filled by electrons in arbitrary order. Thus, we assume that for one of the
O atoms the electron configuration is [He] 2s2 2p2x 2p1y 2p1z and for the
other it is [He] 2s2 2p1x 2p2y 2p1z . As discussed above, the electrons in the
2pz orbitals of both O atoms participate in the formation of σ-bonds with
sp-orbitals on the C atom. The remaining unpaired electron in the 2py
orbital of the first oxygen atom forms a π-bond with the 2py orbital on
carbon. Similarly, the remaining unpaired electron in the 2px orbital of
the second oxygen atom forms a π-bond with the 2px orbital on carbon.
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