Problem Set 9 Solutions

Chemistry 460
Dr. Jean M. Standard
Problem Set 9 Solutions
1. Determine the explicit form of the total Hamiltonian operator for the BeH molecule.
The total Hamiltonian operator for BeH (in atomic units) is
H ˆ = − 1 ˆ
2
∇ 2 + ∇ ˆ 2 + ∇ ˆ 2 + ∇ ˆ 2 + ˆ 2
( ∇ 1 2 3 4 5 )
4 4 4 4
− − − − −
+
r Be1
r Be2
4
r BeH
+
1
r 12
+
r Be3
1
r 13
+
r Be4
1
r 14
+
Note that there are five electrons in BeH, so there are five kinetic energy terms for the electrons (the first line).
€ There are no kinetic energy terms for the nuclei (since the Born-Oppenheimer approximation considers the nuclei
to be fixed). In the second line, the electron-nuclear attractions are listed, ten terms in all (five for the attraction
of the electrons to the Be nucleus and five for the attraction of the electrons to the H nucleus). In the third line,
the first term is the nuclear-nuclear repulsion term, and the other ten terms are the electron-electron repulsion
terms.
4
r Be5
1
r 15
+
−
1
r H1
1
r 23
+
−
1
r H 2
1
r 24
+
−
1
r H 3
1
r 25
+
−
1
r 34
+
1
r H 4
−
1
r 35
+
1
r H 5
1
r 45
.
2. The following table gives results from Hartree-Fock (HF) calculations on methane.
Basis Set HF Energy (a.u.)
STO-3G –39.727
3-21G –40.140
6-31G* –40.195
a.) Describe a minimal basis set for methane. That is, explain the types of basis
functions that are employed on each atom, along with the total number of basis
functions.
A minimal basis set for methane consists of one 1s orbital on each H atom and a 1s, 2s, and set of three 2p
orbitals on C. Summarizing, the table below gives the minimal basis set for methane.
Atom/Basis Functions
No. Basis No. Atoms TOTAL
Functions
C -- 1s 2s 2p(3) 5 1 5
H -- 1s 1 4 4
TOTAL NO. BASIS FUNCTIONS = 9

2
2. Continued
b.) Describe the 3-21G basis set for methane.
The 3-21G basis set is a split valence double zeta basis set. For hydrogen atom, a split valence double zeta
basis consists of two 1s orbitals, denoted 1s and 1sʹ′. Note that for the H atom, because the 1s electron is
considered the valence shell a double zeta basis set is used. For the carbon atom, a split valence double zeta
basis set consists of a single 1s orbital, along with two 2s and two each of 2p x , 2p y , and 2p z orbitals, for a
total of 9 basis functions. The basis functions in a split valence double zeta basis set are denoted 1s, 2s,
2sʹ′, 2p(3), and 2pʹ′(3).
For the methane molecule, CH 4 , the 3-21G basis set consists of the basis functions shown in the table
below.
Atom/Basis Functions
No. Basis No. Atoms TOTAL
Functions
C -- 1s 2s 2s' 2p(3) 2p'(3) 9 1 9
H -- 1s 1s' 2 4 8
TOTAL NO. BASIS FUNCTIONS = 17
c.) Describe the 6-31G* basis set for methane. Explain why the HF energies for the 3-
21G and 6-31G* basis sets are lower than the HF energy calculated from the STO-3G
basis set.
The 6-31G* basis set is a split valence double zeta basis set with added polarization functions on nonhydrogen
atoms. For hydrogen atom, the split valence double zeta basis consists of two 1s orbitals,
denoted 1s and 1sʹ′. For the carbon atom, a split valence double zeta basis set consists of a single 1s
orbital, along with two 2s and a double set of 2p orbitals, 2p(3) and 2pʹ′(3). The extra polarization
functions indicated by the * refer to carbon only and add a d(5) set, for a total of 14 basis functions on
carbon.
For the methane molecule, CH 4 , the 6-31G* basis set consists of the basis functions shown in the table
below.
Atom/Basis Functions
No. Basis No. Atoms TOTAL
Functions
C -- 1s 2s 2s' 2p(3) 2p'(3) d(5) 14 1 14
H -- 1s 1s' 2 4 8
TOTAL NO. BASIS FUNCTIONS = 22
As the number of basis functions increases, the description of the wavefunction improves; therefore, the
energy of the system drops (we know that energy of the approximate wavefunction always is greater than or
equal to the exact result by the Variation Principle). Since the number of basis functions increases from
STO-3G to 3-21G to 6-31G*, the energy decreases correspondingly.

3
3. Suppose you want to carry out ab initio molecular orbital calculations on the
benzaldehyde molecule, C 6 H 5 CHO.
a). Describe a minimal basis set for benzaldehyde, including the total number of AOs.
A minimal basis set for benzaldehyde consists of a single 1s orbital on each H atom and a 1s, 2s, and set of
three 2p orbitals on C and O. The table below gives the minimal basis set for benzaldehyde.
Atom/Basis Functions
No. Basis No. Atoms TOTAL
Functions
C -- 1s 2s 2p(3) 5 7 35
O -- 1s 2s 2p(3) 5 1 5
H -- 1s 1 6 6
TOTAL NO. BASIS FUNCTIONS = 46
b). Describe the 6-31G(d) basis set for benzaldehyde, including the total number of AOs.
The 6-31G(d) basis set is a split valence double zeta basis set plus polarization functions added to nonhydrogen
atoms. The 6-31G(d) basis set for benzaldehyde consists of the basis functions shown in the table
below.
Atom/Basis Functions
No. Basis No. Atoms TOTAL
Functions
C -- 1s 2s 2s' 2p(3) 2p'(3) d(5) 14 7 98
O -- 1s 2s 2s' 2p(3) 2p'(3) d(5) 14 1 14
H -- 1s 1s' 2 6 12
TOTAL NO. BASIS FUNCTIONS = 124
c). Describe the 6-311G(d,p) basis set for benzaldehyde, including the total number of
AOs.
The 6-311G(d,p) basis set is a split valence triple zeta basis set plus polarization functions added to all
atoms. The 6-311G(d,p) basis set for benzaldehyde consists of the basis functions shown in the table
below.
Atom/Basis Functions
No. Basis No. Atoms TOTAL
Functions
C -- 1s 2s 2s' 2sʺ″ 2p(3) 2p'(3) 2pʺ″(3) d(5) 18 7 126
O -- 1s 2s 2s' 2sʺ″ 2p(3) 2p'(3) 2pʺ″(3) d(5) 18 1 18
H -- 1s 1s' 1sʺ″ p(3) 6 6 36
TOTAL NO. BASIS FUNCTIONS = 180

4
3. Continued
d). Estimate how the CPU time would scale relative to the minimal basis set calculation
for Hartree-Fock single point energy calculations on benzaldehyde using basis sets (b)
and (c).
The CPU time for a Hartree-Fock single point energy calculation scales approximately as the number of
basis functions to the power 3.5, K 3.5 .
Basis Set
No. Basis
Functions
Relative
CPU Time
minimal 46 1
6-31G(d) 124 (124/46) 3.5 = 32
6-311G(d,p) 180 (180/46) 3.5 = 119
4. Consider a linear combination of three gaussian type orbitals, called a contracted gaussian
function,
f (r) =
3
∑ c i G i (r) ,
i=1
where f (r) is an atomic orbital, and the c i s are linear coefficient.
gaussian function, given by € the expression
The function
G i (r) is a
€
€
G i (r) =
⎛⎛
⎜⎜
⎝⎝
2α i
π
3/ 4
⎞⎞
⎟⎟ e −α i r 2 .
⎠⎠
€
The reason that an atomic orbital like f (r) is expressed in terms of a linear combination
of gaussian functions is so € that it can better represent the cusp behavior of a true atomic
orbital in the vicinity of the nucleus.
€
An s-type atomic orbital is represented using the following parameters (in atomic units):
d 1 = 4.44635×10 –1 α 1 = 1.09818×10 –1
d 2 = 5.35328×10 –1 α 2 = 4.05771×10 –1
d 3 = 1.54329×10 –1 α 3 = 2.22766
a.) Plot the three orbitals, G 1 (r) , G 2 (r) , and G 3 (r), as functions of x.
relation that r 2 = x 2 + y 2 + z 2 , and plot a cut for x with y = z = 0.
To do this, use the
The three orbitals are € shown € plotted individually € on the next page. Notice that the functions range from
quite broad € to fairly narrow in extent.
€

5
4. a.) Continued
0.6
0.5
G1(x)
G2(x)
G3(x)
0.4
G(x)
0.3
0.2
0.1
0
0 1 2 3 4
x (angstroms)
b.) Plot the atomic orbital f (r) as a function of x.
orbital s(r) with the form
Compare this to a Slater type atomic
€
€
s(r) = ζ 3 1/ 2
⎛⎛ ⎞⎞
⎜⎜
⎜⎜
⎟⎟
⎝⎝ π ⎟⎟ e −ζ r ,
⎠⎠
with the orbital exponent ζ set to 1.
€
The functions are plotted below. Note that the only significant difference is near the origin, where the
Slater type orbital has € a cusp.
0.6
0.5
s(x)
f(x)
0.4
s(x), f(x)
0.3
0.2
0.1
0
0 1 2 3 4
x (angstroms)

6
5. A minimal basis set Hartree-Fock calculation on LiH is performed. The atomic orbital
basis functions used in the calculation are
f 1 = 1s Li
f 2 = 2s Li
f 3 = 1s H .
Thus, the molecular orbitals ( φ i ) of LiH are written as a linear combination of these three
atomic orbitals,
€
€
3
φ i = ∑ c µi f µ .
µ=1
Given below are the coefficients for the MOs obtained from the Hartree-Fock calculation.
€
MO (i)
c 1i c 2i c 3i
1 1.000 0.000 0.000
2 0.000 0.493 0.820
3 € 0.000 € –0.820 € 0.493
a.) How many occupied MOs are there
Since LiH has a total of 4 electrons, there will be 2 occupied MOs.
b.) Based on the MO coefficients, describe the molecular orbitals of LiH.
orbitals make up each of the molecular orbitals
Which atomic
The first occupied MO is defined by the expression
φ 1 = c 11 f 1 + c 21 f 2 + c 31 f 3
= 1.0 f 1sLi + 0.0 f 2sLi + 0.0 f 1sH .
Since two of the coefficients are zero, we see that the first MO consists solely of the 1s core orbital on Li,
φ 1 = f 1sLi .
€
€
The second occupied MO is defined by the expression
φ 2 = c 12 f 1 + c 22 f 2 + c 32 f 3
= 0.0 f 1sLi + 0.493 f 2sLi + 0.820 f 1sH .
In this case, only one of the coefficients is zero. Thus, the second MO consists of a combination of the 2s
valence orbital on Li € and the 1s valence orbital on H, φ 2 = 0.493 f 2sLi + 0.820 f 1sH . This appears to be a
bonding MO; since both coefficients are positive, the two atomic orbitals add to form a bonding orbital.
€

7
5 b.) continued
The third MO (which is unoccupied) is defined by the expression
φ 3 = c 13 f 1 + c 23 f 2 + c 33 f 3
= 0.0 f 1sLi − 0.820 f 2sLi + 0.493 f 1sH .
Here, we see that, like the second MO, the third MO consists of a combination of the 2s valence orbital on
Li and the 1s valence € orbital on H, φ 3 = −0.820 f 2sLi + 0.493 f 1sH . This appears to be an antibonding MO;
since one coefficient is positive and the other is negative, the atomic orbitals subtract to form an
antibonding MO with a node along the internuclear axis.
€
c.) The molecular orbital energies are ε 1 = –2.426 a.u., ε 2 = –0.618 a.u., and
a.u. Do these results agree qualitatively with what you would expect
ε 3 = 0.085
The energies of the AOs in atomic units are
€
€
ε 1sLi = − 2.478
ε 2sLi = − 0.196
ε 1sH = − 0.500.
€
€
What do these atomic orbital energies suggest about the AOs that mix to form MOs
€
We might expect the first MO, which represents something closely resembling the core 1s orbital on Li, to
be much lower in energy than the second MO, which represents the bonding MO. This is why we say that
the core atomic orbitals do not contribute to the bonding – they are generally too low in energy to
effectively combine with the valence atomic orbitals to form the bonds. Since MO1 consists solely of the
core 1s atomic orbital on Li, we would expect that the orbital energy of MO1 would be nearly the same as
the energy of the 1s atomic orbital, and it is. It is also much lower in energy than the other AOs or MOs.
We also would generally expect a bonding MO to have an orbital energy that is less than zero. A negative
orbital energy generally indicates a favorable interaction (since it is dominated by the Coulomb interaction,
which is negative in sign for an attractive interaction). The second MO consists of a linear combination of
the 2s orbital on Li and the 1s orbital on H, forming a bonding orbital. The energy of this MO is similar
to the energies of the two atomic orbitals from which it is formed, but the MO energy is a bit lower due to
the favorable bonding interaction, so this also makes sense.
Finally, we would expect an antibonding MO to have an orbital energy that is greater than zero. A positive
orbital energy generally indicates an unfavorable interaction. The third MO consists of a linear combination
of the 2s orbital on Li and the 1s orbital on H, forming an antibonding orbital. The energy of this MO is
higher than the energies of the two atomic orbitals from which it is formed, so this also makes sense.
€

8
5. Continued
d.) Describe the bonding in LiH. Does LiH have a single bond or a double bond Is the
bond polar or nonpolar Explain.
From part (b), the first MO is a core orbital and does not participate in bonding. It consists of an AO on
only a single atom. An MO would have to depend on AOs centered on at least two atoms and combined in
an additive way in order to be considered as a bonding MO.
From part (b), the second MO consists of a combination of AOs on Li and H. The coefficients in front of
both of these orbitals are positive and since the orbitals are both s-type orbitals, the orbitals add together to
form a bonding orbital. Since there is only one bonding orbital (and no other electrons in antibonding
orbitals), LiH has a single bond. Since the coefficients of the two orbitals that make up the bonding orbital
are not equal (0.493 and 0.820), the bond is polar. In addition, since the coefficient multiplying the H 1s
orbital is larger than the coefficient multiplying the Li 2s orbital, the MO is weighted towards the H atom.
This suggests that more electron density is located on the H atom than on the Li atom. This makes sense
in terms of electronegativities: Li is less electronegative than H; therefore, Li is partially positive and H is
partially negative.
We can construct an energy level diagram for the molecular orbitals to show how they are formed from the
atomic orbitals. This is presented in the figure below. On the left and right sides of the x-axis of the
graph, the atomic orbitals of Li and H are given. In the center, the MOs of LiH are given. We can see how
the 1s orbital on Li is nearly unchanged, while the 1s H orbital and the 2s Li orbital mix to form a bonding
orbital (labeled 1σ) and an antibonding orbital (labeled 1σ*).
0.5
0.0
1σ*
-0.5
E (a.u.)
-1.0
2s Li
1σ
1s H
-1.5
-2.0
-2.5
1s Li
1s
Li LiH H

9
6. A Hartree-Fock SCF calculation on the ground state of the H 2 molecule was carried out at
a fixed internuclear distance of 1.40 a.u. A minimal basis was used, which consisted of a
1s atomic orbital on each hydrogen atom.
The calculation yielded two MO energies,
ε 1 = − 0.619 a.u.
ε 2 = − 0.401 a.u.
The two-electron Coulomb and Exchange integrals that arise in the calculation are listed
below.
€
φ 1 (1) φ 1 (2)
1
r 12
φ 1 (1)φ 1 (2) = 0.566 a.u.
€
φ 1 (1) φ 2 (2)
1
r 12
φ 1 (1)φ 2 (2) = 0.558 a.u.
€
φ 1 (1) φ 2 (2)
1
r 12
φ 2 (1)φ 1 (2) = 0.140 a.u.
€
φ 2 (1) φ 2 (2)
1
r 12
φ 2 (1)φ 2 (2) = 0.582 a.u.
In these integrals, the functions φ 1 and φ 2 represent the molecular orbitals.
€
a.) Write down a Slater determinant for the ground electronic state of H 2 .
€ €
H 2 has two electrons so only one spatial MO will be occupied. For the ground state of H 2 , both electrons
will occupy MO1, φ 1 . The possible spin orbitals are φ 1 α and φ 1 β . Since the spins have to be paired if
the electrons are in the same spatial orbital, the Slater determinant must be:
€
( ) =
Ψ H 2
1,2
€
1
2
φ 1 € ( 1)α 1
φ 1 1
( ) φ 1 ( 2)α( 2)
( ) β( 1) φ 1 ( 2) β( 2) .
b.) Using the molecular € orbital energies and the two-electron integrals given above,
compute the electronic energy of H 2 at an internuclear separation of 1.40 a.u.
The electronic energy is given by
E el = E =
n
n
n
∑ 2ε i − ∑ ∑ ( 2J ij − K ij ) .
i=1 i=1 j=1
For H 2 , there are two electrons so N=2. This means that the number of occupied spatial orbitals, n, equals
N/2 = 1. Writing out € the summations above for H 2 with n=1 yields
E el ( H 2 ) = 2ε 1 − ( 2J 11 − K 11 ).
€

10
6. b.) Continued
From the information given, we know that ε 1 = −0.619 a.u. We also can determine from above that
J 11 = K 11 = φ 1 (1) φ 1 (2)
€
1
r 12
φ 1 (1)φ 1 (2) = 0.566 a.u.
Substituting, the electronic energy of H 2 becomes
€
( ) = 2ε 1 − ( 2J 11 − K 11 )
E el H 2
= 2ε 1 − J 11
= 2 −0.619 a.u.
( ) − ( 0.566 a.u. )
( ) = −1.804 a.u.
E el H 2
c.) Determine the total € energy from the SCF calculation for H 2 at an internuclear
separation of 1.40 a.u. The total energy includes the electronic energy plus the
nuclear-nuclear repulsion energy.
The total energy of the system is just the electronic energy plus the nuclear-nuclear repulsion,
€
E TOT = E el + V nn ,
where the nuclear-nuclear repulsion V nn in atomic units is
€
V nn = Z α Z β
r αβ
.
€
For H 2 ,
Z α = Z β = 1. Substituting, the total energy becomes
€
E TOT = E el +
1
r HH
= −1.804 a.u. +
1
1.40 a.u.
= −1.804 + 0.714 a.u.
E TOT = −1.090 a.u.
€