```CHEMISTRY 333 – Fall 2006
Exam 3, in class
Name __________________
Mathematica & calculators. You may use Mathematica or a calculator to do arithmetic. Do not use
Mathematica for any higher-level manipulations (algebra, equation solving and/or simplification,
integration, differentiation, eigensystem calculations, etc.).
SHMO, Textbook, Lecture notes, etc.. No SHMO usage is allowed. The exam is closed-book, closedweb page, closed lecture notes, closed homework assignments, etc.
1a.
What values of the n, l, and ml quantum numbers characterize the 3dz2 wave function? (Hint: the
3dz2 wave function is a function of r and θ only; it is not a function of φ).
n = 3, l = 2 (d orbital), ml = 0 (not a function of φ)
1b.
How many hydrogen atom wave functions are degenerate with ψ310? List them.
Hydrogen atom orbital energies depend on n only. There are 9 (n2) orbitals with n = 3 so 8 orbitals are
degenerate with 310:
300, 31-1, 311, 32-2, 32-1, 320, 321, 322.
1c.
What boundary condition(s) are applied to the radial component of the hydrogen atom wave
function? What consequences do these conditions have for the physical properties of a hydrogen
atom?
Boundary condition – wave function approaches zero as distance (r) becomes large. This condition, like
all other boundary conditions that we have seen, results in the energy being quantized.
2
Recall that [ Ĥ , lˆ 2 ] = [ lˆz , lˆ 2 ] = 0 for the hydrogen atom. Also, lˆ 2Yl ml = = 2l (l + 1)Yl ml and lˆzYl ml = =mlYl ml .
2.
Evaluate the following integrals. Explain your reasoning (no integrations are necessary).
∫ψ
310
Ĥ ψ 210 dτ
This integral equals zero. The 210 and 310 functions are eigenfunctions of the Hamiltonian and are
orthogonal. Going a bit farther … the integrand can be rewritten as Ψ310E210Ψ210, and the energy can be
removed from the integrand because it is a constant, leaving the integrand as the product of orthogonal
functions: Ψ310Ψ210
∫ψ
210
lˆz ψ 210 dτ
This integral equals zero. Applying the operator to the 210 function multiplies the function by ml, but ml =
0 so the integrand vanishes.
∫ (Y )
1 *
2
lˆ 2 Y2−1 dτ
This integral equals zero. The spherical harmonics are eigenfunctions of the operator and are orthogonal.
The reasoning is similar to that used in the first integral.
3.
Three quantum chemists approach you for research funds for the same research idea: use the
variation method to develop a model wave function of the hydrogen atom ground state.
Chemist A wants to optimize φ(r) = N exp(-ar2). Chemist B wants to optimize φ(r) = N (ar2 + br4 +
cr6). Chemist C wants to optimize φ(r) = N sin(ar). The variational parameters in these functions
are a, b, and c.
You only have money to fund one project. Which one do you fund and why? (Hint: Explain why
two of the ideas are definitely worthless.)
Fund chemist A. The others are wasting their time.
The variation method provides a way for optimizing model wave functions only when the model observes
the same mathematical and physical constraints observed by the real wave function. Chemist B wants to
optimize a model that blows up when r gets large (the real wave function vanishes). Similarly, Chemist C
wants to optimize a model that fails to decay as r gets large. Only Chemist A has a model that obeys the
constraints on the system.
Note: don’t mistake A’s model for the “real” wave function which is N exp(-ar).
3
4a.
The first ionization potentials (IP) of H, He, and Li are: 13.6 (H), 24.6 (He), and 5.4 (Li) eV. The
nuclear charge (Z) in these atoms, however, increases +1 (H), +2 (He), and +3 (Li). Explain the
up-and-down trend in IP in terms of specific interactions and quantum mechanical principles.
In the independent particle approximation, the IP is the negative of the electron’s orbital energy. The
energy of an orbital in a one-electron atom with nuclear charge Z depends on only two factors: n and Z.
The energy increases with increasing Z and decreases with increasing n.
The ionized electrons in H and He both come from 1s orbitals (n = 1). The dominant factor, then, is Z
which is larger for He. As expected, He has a larger IP.
The ionized electron in Li comes from a 2s orbital (n = 2), which argues for smaller IP smaller. On the
other hand, Z = 3, so this argues for larger IP. How to decide? The core (1s) electrons shield the 2s
electron from the nucleus so, in fact, Zeff is between 1 and 2, and the change in n dominates. In terms of
specific interactions, “shielding” is created by electron-electron repulsions that offset nuclear-electron
attraction.
4b.
The nuclear charge on Be (Z = +4) is larger than that of Li (Z = +3). Describe whether Be’s first IP
is larger or smaller than Li’s. Justify your answer in terms of specific interactions and quantum
mechanical principles.
Li has a 1s22s1 electron configuration and Be has a 1s22s2 electron configuration. n = 2 for the ionized
electron in both cases. Z and also Zeff are larger for Be than for Li, so Be’s IP should be larger.
Remember that shielding, electron-electron repulsion that offsets nuclear-electron attraction, is mainly
created by “inner” shell electrons. Li and Be both have two inner electrons. Electrons in the same orbital
repel, but this repulsion does not offset nuclear-attraction that much.
4
5.
Write the Hamiltonian operator in for H2 assuming fixed nuclei, i.e., do not include operators for
nuclear kinetic energy. To make the operators easier to write, use Cartesian coordinates for the kinetic
energies, but use “interparticle distances” for potential energies.
There are four particles in H2. Thus, there should be 4 kinetic energy operators and 6 potential energy
operators (each particle-particle interaction is described by a separate operator). The kinetic energy
operators for the nuclei can be omitted because we assume fixed nuclei. Particle coordinates are
indicated as follows: 1 = electron 1, 2 = electron 2, A = nucleus A, and B = nucleus B.
1
=2
Hˆ = −
∇12 + ∇ 22 ) +
(
2 me
4πε o
⎛ −1 −1 −1 −1 1
1 ⎞
+
+
+ +
⎜ +
⎟
⎝ r1 A r1B r2 A r2 B r12 rAB ⎠
where
⎛ ∂2
∂2
∂2 ⎞
∇ i2 = ⎜ 2 + 2 + 2 ⎟
⎝ ∂xi ∂yi ∂zi ⎠
A chemist decides to use the following two-determinant model wave function for H2. (Ignore
normalization.) A refers to an atomic orbital on one H and B refers to an atomic orbital on the other H.
Ψ (1, 2) =
6a.
A(1)α (1)
B(1) β (1)
A(2)α (2) B(2) β (2)
+
B(1)α (1)
A(1) β (1)
B(2)α (2)
A(2) β (2)
Rewrite Ψ(1,2) as the product of a two-electron spatial function and a two-electron spin function.
No one had trouble expanding these determinants, but several folks stopped after expanding the function.
They did not factor apart the spatial and spin functions. I wanted the following, a product of two functions,
one function that depends on spatial coordinates only and another that depends on spin coordinates only:
Ψ = [ A(1) B (2) + B (1) A(2)][α (1) β (2) − β (1)α (2) ]
5
6b.
What are the symmetries of the spatial and spin functions in part 6a with respect to electron
interchange? Does Ψ(1,2), the combined function, satisfy the Pauli principle? Explain.
Interchanging the coordinates of electrons 1 and 2 gives:
[ A(2) B(1) + B(2) A(1)][α (2)β (1) − β (2)α (1)]
The spatial function is unchanged (“symmetric”) from 6a. The spin function is the negative of 6a
(“antisymmetric”). Overall, the wave function is the negative of the original (“antisymmetric”), so it satisfies
the Pauli principle which requires a wave function to change sign when the coordinates of any two
electrons are interchanged.
6c.
Is Ψ(1,2) an eigenfunction of Sˆz ? Show your work. If it is an eigenfunction, what is MS? Possibly
Sˆz = sˆ1z + sˆ2 z
sˆzα = ( = / 2)α
sˆz β = (− = / 2) β
In working this problem, it is helpful to notice that a spin operator works on spin functions only, so the
spatial component of Ψ can be ignored. As a further simplification, I will use k to represent = / 2 . The
following algebra shows that the operator makes the wave function vanish. In other words, the wave
function is an eigenfunction with an eigenvalue of zero (MS = 0).
`
S z HaH1L bH2L - bH1L aH2LL =
= Hs`1 z + s`2 zL HaH1L bH2L - bH1L aH2LL
= s`1 z HaH1L bH2L - bH1L aH2LL + s`2 zHaH1L bH2L - bH1L aH2LL
= k a(1)b(2)-(-k)b(1)a(2) + (-k)a(1)b(2)-(k)b(1)a(2) = 0
6
6d.
Is Ψ(1,2) an eigenfunction of Ŝ 2 ? Show your work. Eigenfunctions of Ŝ 2 , like ψS in the formula
Sˆ 2ψ S = = 2 S ( S + 1)ψ S are characterized by a quantum number S. If Ψ(1,2) is an eigenfunction of Ŝ 2 , what
is its S quantum number? Possibly helpful formulas (see part 6b too):
Sˆ 2 = sˆ12 + sˆ22 + 2( sˆ1x sˆ2 x + sˆ1 y sˆ2 y + sˆ1z sˆ2 z )
sˆxα = (= / 2) β
sˆx β = (= / 2)α
sˆ 2α = (3= 2 / 4)α
sˆyα = (i= / 2) β
sˆ 2 β = (3= 2 / 4) β
sˆy β = (−i= / 2)α
This is only a little more complicated than the previous problem. Once again we can (and should) ignore
the spatial function. I’ll leave some of the math details to you and just point out the following:
3 hbar2
`
2
s1 HaH1L bH2L - bH1L aH2LL =
HaH1L bH2L - bH1L aH2LL
4
3 hbar2
`
2
s2 HaH1L bH2L - bH1L aH2LL =
HaH1L bH2L - bH1L aH2LL
4
2
`s `s HaH1L bH2L - bH1L aH2LL = hbar H bH1L aH2L - aH1L bH2LL
1x 2x
4
2
`s `s HaH1L bH2L - bH1L aH2LL = hbar H bH1L aH2L - aH1L bH2LL
1y 2y
4
2
2
`s `s HaH1L bH2L - bH1L aH2LL = - hbar HaH1L bH2L - bH1L aH2LL = hbar H bH1L aH2L - aH1L bH2LL
1z 2z
4
4
Combining all of these terms shows that the operator makes the wave function vanish. In other words, the
wave function is an eigenfunctions of the operator with an eigenvalue of zero (S = 0).
7
The following graphs are based on a particular hydrogen atomic orbital.
0.35
1.2
0.3
1
0.25
0.8
0.2
0.6
0.15
0.1
0.4
0.05
0.2
Graph #1
7a.
1
2
3
4
5
6
Graph #2
1
2
3
4
5
6
Which graph shows ψ2 and which shows the radial distribution function of ψ. Explain your
reasoning.
I believe that I was asked during the exam whether I really meant Ψ2 and I said to change this to Ψ. Some
of you used this information to answer the question, but others did not. Regardless, the radial distribution
function is the one on the right. The radial distribution function is proportional to r Ψ2, so it rises more
slowly near the nucleus (where r is small) and its maximum occurs at larger r.
7b.
For each atomic orbital listed below, explain why the listed orbital might or might not be the orbital
responsible for the graphs. (Note: Some explanations may be re-usable.)
1s
The graphs cannot refer to this orbital because a 1s orbital has its maximum value at r = 0 (Note: the 1s
orbital is not infinite anywhere.)
2s
The graphs cannot refer to this orbital because a 2s orbital has a non-zero value at r = 0. This is true of
all s orbitals.
2pz
The graphs might refer to this orbital. Its radial function vanishes at the nucleus only.
3pz
The graphs cannot refer to this orbital because the 3pz orbital contains a radial node at r > 0.
3dxy
The graphs might refer to this orbital. Its radial function vanishes at the nucleus only.
8
A chemist decides to model the wave function for H2+ at its equilibrium geometry as Ψ = N(ψ1sA + ψ 1sB)
where ψ 1si is the 1s atomic orbital on atom i. Assume the two nuclei are located on the z axis at -r/2 and
+r/2, respectively.
8a.
Sketch the value of Ψ as a function of z for H2+ for x = 0, y = 0, -r ≤ z ≤ +r
A 1s orbital is a decaying exponential that takes on its maximum value at the nucleus. A linear
combination of these functions looks something like the following where the two maxima are located
approximately at –r/2 and +r/2, respectively.
1
0.8
0.6
0.4
0.2
-2
8b.
-1
1
2
Another chemist suggests a more complicated model for H2+ that combines 1s and 2pz functions.
Write an algebraic formula for Ψcomplicated and identify all variational parameters.
The labeling rule in 8a is to use the subscript to identify the type of atomic orbital and the atom that it is
centered on. The complicated model combines 4 atomic orbitals, 2 from each atom:
Ψcomplicated = c1ψ1sA + c2ψ2pzA + c3ψ1sB + c4ψ2pzB
The complicated model wave function is formed by multiplying each atomic orbital function by a weighting
coefficient, a number, that tells us how much the orbital contributes to the model. The four weighting
coefficients are variational parameters because we adjust them to make the model’s energy as low as
possible.
8c.
How would you use Eoriginal and Ecomplicated, the energies of the two model wave functions, to
decide which model is closer to the *** Quantum Mechanical Truth ***?
Since the complicated model encompasses all of the features of the original model, it must do at least as
well. This does not answer the question, however, as to how energies indicate model quality.
The variation principle guarantees that model energies are always greater than the true energy, Emodel >
EQMT. The closer Emodel comes to EQMT, the closer the model comes to QM truth. If the complicated model
is really an improvement on the original model, we should observe Eoriginal > Ecomplicated > EQMT.
9
9.
The table shown above was discussed in class in connection with bonding in H2+. Explain how
these data were generated and how they shed light on the mechanism of covalent bonding in this
ion.
The energy values in the “Total” column were obtained using a model wave function that combines two
1s orbitals in a bonding fashion (“original” wave function in problem 8). The energy of this model was
calculated at two geometries – well-separated nuclei and nuclei at the equilibrium bond distance – and
the energy difference between these geometries is shown in the “Total” column. In other words, when the
two nuclei are pushed together, the kinetic energy rises, the potential energy falls by twice as much
(satisfying the virial theorem), and the total energy falls too.
Each value in the “Total” column is obtained using a particular energy operator (or operators) to calculate
the energy. ΔT uses the electron kinetic energy operator. ΔV combines three interaction operators: two
electron-nucleus interactions (the electron interacts independently with each nucleus) and one nucleusnucleus interaction. ΔE = ΔT + ΔV. (ΔE can also be obtained using the total energy operator, the
Hamiltonian, for H2+.)
As discussed in class, the 1s orbitals use the same exponent, but this exponent is optimized for each
geometry. The exponent’s value when the nuclei are well-separated is identical to its value in the
hydrogen atom. The exponent’s value increases when the nuclei are separated by the bond distance.
This compresses the orbital around each nucleus, and this affects the kinetic and potential energies.
Since changes in orbital exponent and geometry both affect energy, a “nonbonding” model was
constructed to separate the two effects. The nonbonding model, defined as ψ1sA + iψ1sB, combined the 1s
orbitals in a fashion that delocalizes the electron, but doesn’t produce bonding. Important: the model uses
the same orbital exponents used above. The energy changes for this model are shown in the
“Nonbonding” column. The kinetic energy rises strongly and the potential energy falls strongly. These
effects are due mainly to orbital compression because the model is nonbonding (notice that the total
energy change is small).
The energy values in the “Bonding” column were obtained by subtracting the values in the “Nonbonding”
column from the values in the “Total” column. The idea behind these data is that they show the “leftover”
energy changes that occur when the nuclei approach, i.e., the energy changes separate from the
changes caused by orbital compression. Supposedly, these leftovers are the changes responsible for
bonding.
10
A five carbon cation and the properties of its two occupied Huckel MO are shown below (the orbital
coefficients have been normalized).
MO
energy
c1
c2
c3
c4
c5
10a.
α + 2.14 β
α + 0.66 β
0.261
0.557
0.465
0.435
0.465
-0.657
-0.435
0.185
0.557
0.185
1
2
3
5
4
Write the Huckel MO secular matrix for this cation (use my numbering scheme).
The “secular matrix” means the matrix of coefficients generated by the secular equations. This matrix is
shown below. Many of you, however, wrote one of the matrices that can be obtained by transforming the
secular matrix and I allowed varying amounts of credit for these depending on their similarity to the
secular matrix.
α
i
j
j
j
j
j
j
j
j
j
j
j
j
k
− E
β
0
0
β
α − E
β
0
0
β
α − E
β
β
α − E
0
0
0
β
0
β
α
0 y
z
β z
z
z
z
0 z
z
z
z
β z
z
z
− E{
10b.
Why do we insist that the determinant of the secular matrix equals zero?
Because this provides the only physically meaningful solution to the secular equations. The equations all
have a similar form, such as 0 = c1(α – E) + c2(β) + c30 + c40 + c50 (this is first secular equation based on
the matrix in 10a). One solution to these equations is to set c1 = c2 = … = c5 = 0, but this is not physically
meaningful. The only other solution is to require that the determinant of the secular matrix equal zero.
10c.
Calculate Eπ for this cation.
In Huckel theory, Eπ is the sum of the electron energies. The ion contains four electrons and these occupy
the two lowest energy MOs (E1 = α + 2.14 β, E2 = α + 0.66 β).
Eπ = 2 E1 + 2 E2 = 4 α + 5.6 β
11
10d.
Calculate the number of pi electrons on each carbon (see previous page for coefficients and
numbering). Based on these results, assign a charge to each carbon.
The spatial distribution of each electron is determined by the square of its MO. In Huckel theory, we
ignore terms containing pipj where i and j are different atoms. In other words, we say ψi2 ≈ c12p12 + c22p22
+ c32p32 + c42p42 + c52p52. The electron density distribution for this electron puts ci2 electrons on atom i.
Two electrons are described by ψ1 and two electrons are described by ψ2, where the MOs are:
ψ1 = .26 p1 + .56 p2 + .46 p3 + .44 p4 + .46 p5
ψ2 = .66 p1 + .44 p2 - .18 p3 - .56 p4 - .18 p5
The number of pi electrons on carbon 1 is 2(.262) + 2(.662) = 1. Carbons 2 and 4 also hold 1 pi electron
each. The charge on each of these three carbons is zero.
The number of pi electrons on carbon 3 (and also carbon 4) is 2(.462) + 2(-.182) = 0.5. The charge on
each of these two carbons is +0.5.
10e.
Calculate the pi bond orders between each pair of neighboring carbons (skip bonds that are
symmetry-related to other bonds)
Huckel theory says the bond order created by one electron between atoms i and j is equal to cicj where
these are the coefficients from the electron’s MO. To get the total bond order between atoms i and j we
must add up the contributions from each electron.
Bond order 1-2 = 2(.26 x .56) + 2(.66 x .44) = 0.86
Bond order 2-3 = 2(.56 x .46) + 2(.44 x (-.18)) = 0.36 = Bond order 2-5 (by symmetry)
Bond order 3-4 = 2(.44 x .46) + 2((-.56) x (-.18)) = 0.61 = Bond order 4-5 (by symmetry)
12
The relative pi electron energies of anthracene and phenanthrene can be compared using Dewar PMO
vs.
anthracene
11a.
phenanthrene
Calculate numerical values for the orbital coefficients in benzyl radical’s NBMO using Dewar’s
rules. (You may not use SHMO for any part of problems 11 and 12; you may use Mathematica to
perform arithmetic operations like multiplication-addition-square root, but nothing more
complicated).
You probably should have obtained the coefficients in three stages. First stage: non-zero coefficients are
located on “starred” atoms only and the sums of the coefficients around any “blank” atom must equal
zero. Second stage: in a normalized NBMO, the sums of the squares of the coefficients equal one. Third
stage: numerical values for the coefficients.
-1
1
11b.
(-1/7)1/2
2
-1
(1/7)1/2
-.38
(4/7)1/2
(-1/7)1/2
.38
.76
-.38
Calculate the change in NBMO energy that occurs when two benzyl radicals unite to form
anthracene by using Dewar’s PMO method.
This could have been worded better. The NBMO energy does not change. Rather, combining the radicals
produces new MO with different energies from the NBMO. What I want, of course, is the difference
between the NBMO and new MO energies.
As shown in the drawing at the top of the page, uniting two radicals to make anthracene brings together a
benzylic carbon in one radical with an ortho carbon in another radical. The product of these coefficients is
-.38 x .76 = -.29. Since this happens in two locations simultaneously, the new MO energy is lower than
the NBMO energy by 2 x .29 β ≈ 0.6 β. The overall change in pi energy is twice this because two
electrons occupy the new MO.
Some of you were confused by the opposite signs of the coefficients and predicted that the NBMO
interaction would be destabilizing. In fact, it is both stabilizing and destabilizing. One new MO is a
bonding combination of the NBMO; the other is an antibonding combination. I am interested in the
bonding combination because this is the MO occupied by the two electrons.
11c.
Repeat part 11b, but for 2 benzyl Æ phenanthrene.
Phenanthrene is obtained by bringing the benzyl carbons together (.76 x .76) and the ortho carbons
together (-.38 x -.38), so the change in NBMO energy is [.76 x .76 + (-.38 x -.38)] β = .72 β. The overall
change in pi energy is twice this because two electrons occupy the new MO.
11d.
Which molecule should have a more stable pi system, anthracene or phenanthrene, and by how
much?
Phenanthrene is expected to have a more stable pi system by 2 (.72 - .6) β = .29 β (if you use lots of sig
figs)
13
12XC. Find another way to use Dewar’s PMO method to compare the energies of anthracene and
phenanthrene and carry out the necessary calculations (show your work for each step: draw your
fragments, calculate their NBMO coefficients, calculate the change in NBMO energy as the fragments
unite, determine the relative pi energies of anthracene and phenanthrene). Do you obtain the same
relative energies?
Just in case I am not being clear, I want you to draw two fragments A and B that will make
anthracene when united one way, and will make phenanthrene when united in another. A and B
are identical benzyl radicals in problem 11, but it is not necessary (or even convenient) for A and
B to be identical in this problem.
This can be done many ways. You just have to make sure that your fragments are odd-alternant
hydrocarbons and that the same fragments can give both anthracene and phenanthrene.
Strange but true fact: some fragment choices predict that anthracene is more stable than phenanthrene.
(This is very annoying, but come see me before you decide Dewar PMO is worthless b/c there may be a