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Microscopic Physical Chemistry ( CHEM 4420 ) Mid-term Exam 3 ( Wednesday, 30 April 2014 )

This is a closed-book, closed-notes exam administered in a 1-hour class period.

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NAME: ___________________________

RIN Number: ___________________________

Write out all work and answers on these pages.

If extra sheets needed, (1) use provided, (2) refer to extra sheets on exam pages, (3) save & staple to exam when turned in.

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Grading Area: FOR TA USE ONLY,

Problem graded score max value

1 14 2 12 3 4 4 12 5 22 6 4 7 6

total 74

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Useful physical constants.

symbol meaning value

h Planck’s const 6.626 × 10−34 J⋅s

ħ (h / 2π) 1.055 × 10−34 J/(rad/s)

c speed of light 2.998 × 108 m/s

kB Boltzmann’s const 1.38 × 10−23 J / K

= 0.695 cm−1 / K

NA Avogadro’s # 6.022 × 1023 molec / mole

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(1) [ 14 pts ] The ro-vib spectrum of HCl is depicted at right. Provide a brief reason for each of the following observations.

[ Hint: each has to do with the fact that more than one isotope is common for chlorine atoms. ]

(a) [ 3 pts ] Two sets of peaks appear at distinct amplitudes. Explain the ratio of amplitudes for the two sets.

(b) [ 3 pts ] One set is uniformly shifted in frequency

from the other. Explain what is responsible for the shift.

(c) [ 3 pts ] On the plot above, clearly label the regions that correspond to the P, Q and R branches

(d) [ 3 pts ] What values of ΔJ and Δn correspond to the Q branch?

The amplitude ratio is given by the natural isotopic abundance of 35Cl (~75%) and 37Cl (~25%). Thus, the approx 3× higher peaks correspond to 35Cl. The Q branch has ΔJ = 0 and Δn = 1.

The frequency shift is due to the fact that EJ = hcB(J(J+1)), where B ∝ (1/I = 1/µr02). For spectroscopy, the more exact reason is that the energy spacings are in multiples of hcB, and thus also ∝ (1/µr0

2). Because 37Cl has the larger reduced mass, it has smaller energy spacings 35Cl.

(solution provided on plot).

Solution: The Q branch has ΔJ = 0 and Δn = 1.

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(e) [ 2 pts ] Is the above spectrum consistent with a diatomic molecule? Explain your answer in terms of the selection rules for diatomics.

(2) [ 12 pts ] Slater determinants are used to set up n-electron wavefunctions [ Ψ(1,2, …n) ] in the orbital approximation. The determinant guarantees that the resulting product state will be physically acceptable.

a. [ 4 pts ] Name two distinct properties of the wavefunction that are guaranteed by using a Slater determinant?

Solution: (1) Indistinguishability of the electrons (i.e., the wavefunction is antisymmetric)

(2) Satisfaction of the Pauli exclusion principle.

b. [ 2 pts ] What is the electronic configuration for the ground state of C+? Solution: C+ is a five-electron system, and its configuration is 1s22s22px

1

c. [ 6 pts ] Write the Slater determinant for Ψ(1,2,…n) representing the ground state of C+. Use the orbital notation described in lecture notes,

i.e., from lecture, where I defined (electron #)orbital type, spin stateφ , for

example (3)3d,αφ would correspond to the electron

labeled “3” residing in a 3d orbital with spin state α (spin up).

Do not evaluate the determinant.

Solution: Yes. The Q branch is absent because diatomics have the selection rule ΔJ = ±1.

Therefore, ΔJ = 0 (the Q-band location) is disallowed. Features that are present correspond to transitions with ΔJ = −1 (P) and ΔJ = +1 (R).

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Solution: The five-electron system will fill the states !1s,", !1s,#, !2s,",!2s,#and!2px," and the Slater determinant sets up an anti-symmetric wavefunction with indistinguishable representations of the five electrons.

!(1,2,3,4,5) = 15!

"1s,!(1) "1s,!

(1) "2s,!(1) "2s,!

(1) "2px,!(1)

"1s,!(2) "1s,!

(2) "2s,!(2) "2s,!

(2) "2px,!(2)

"1s,!(3) "1s,!

(3) "2s,!(3) "2s,!

(3) "2px,!(3)

"1s,!(4) "1s,!

(4) "2s,!(4) "2s,!

(4) "2px,!(4)

"1s,!(5) "1s,!

(5) "2s,!(5) "2s,!

(5) "2px,!(5)

Evaluating this determinant gives the final form of the wavefunction, and it is easy to verify that it is antisymmetric.

(3) [ 4 pts ]The Hamiltonian for a one-dimensional harmonic oscillator

is ! = − ħ!

!!!!

!"!+ !

!!!!!!. By using the trial function,

! = !!!!! , we are able to use the Variational Method to get

!! =!∗!"!

!!

!∗!!!!

=ħ!!2! +

!!!

8!

Find the value of α which optimizes the trial function.

d!!/dα = 0 = ħ!

!!− !

!!!!/!!

Therefore, α = !!/2ħ

(4) [ 12 pts ] Given are the plots that correspond to 1s states of row two elements with (Z−ζ) vs Z, and (Z−ζ)/Z vs Z. The plots incorporate ζ1s for row 2 elements in their neutral state, as reported in Table 21.1 of Engel & Reid.

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a. [ 4 pts ] Provide a physical explanation for the observed trend of increasing (Z−ζ) with Z.

Solution: As Z increases across row 2, additional electrons are added in the n = 2 level. As more electrons are added the magnitude of the shielding, characterized by (Z−ζ), increases.

Incidentally, if we utilize language of the orbital approximation, these first fill in the 2s states and then the 2p states. It is interesting that the 1st and 4th additions to p-type orbitals have no apparent effect on ζ1s. However, I haven’t an immediate explanation for these plateaus, and note that they do not occur in plots of (Z−ζ2s).

b. [ 4 pts ] Based on this, do you expect the 2s orbitals for atoms across row 2 to be the same or distinct in the Hartree-Fock limit?

Solution: The 2s orbitals will be distinct from one atom to the next. These orbitals are HF-SCF solutions to one-electron Schrödinger equations, and they change (due to changing Veff) when electrons are added (and when Z is increased).

c. [ 4pts ] Provide a physical explanation for the observed trend of decreasing (Z−ζ)/Z.

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Solution: This plot shows the fractional screening, (Z−ζ)/Z, of the nuclear charge. As we go across the row 2 elements, we add one proton and one electron in each step. Each successive electron is a bit less effective at screening its “accompanying” proton than was the previous one because it moves into higher orbital types with P(r) extending further from the nucleus. Thus, the fractional screening actually decreases.

(5) [ 22 pts ] The following Hamiltonian corresponds to the atomic electronic system at right.

H = !!2

2me

"

#$$

%

&'' (e1

2 +(e22( ) (i)

2

0 1 2

1 1 4Ze

r r⎛ ⎞⎛ ⎞

− +⎜ ⎟⎜ ⎟πε ⎝ ⎠⎝ ⎠ (ii)

2

0 12

1 4e

r⎛ ⎞⎛ ⎞

+ ⎜ ⎟⎜ ⎟πε ⎝ ⎠⎝ ⎠ (iii)

(a) [ 6 pts ] Sets of terms from H above are labeled (i), (ii), and (iii). Briefly define the energy contribution represented by each term.

(i) Kinetic energy of the electrons

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(ii)

(iii)

(b) [ 5 pts ] Write the Hamiltonian for the H2 molecule shown on the right.

[ 3 pts ] Which term or set of terms renders it impossible to analytically solve the corresponding Schrödinger equation?

(c) [ 4 pts ] For the diatomic system above the terms for the Hamiltonian that relate only to nuclear coordinates. [ Label

Repulsive Coulombic potential between the electrons

Coulombic attraction potential between the electrons and the nucleus

The repulsive Coulombic potential energy between electrons 1 and 2.

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these terms for reference in part (e). ]

Solution: Terms for nuclear KE: (iv) → !!2

2mp

"

#$$

%

&'' (A

2 +(B2( )

and Coulombic nuc-nuc repulsion: (v) → e2ZAZB4!"0

#

$%%

&

'((1R#

$%

&

'(

(d) [ 4 pts ] What is the effect of the Born-Oppenheimer approximation on each of the terms in (d)? [ Be clear by referring to your labels in (d).]

Solution: With B.O., we assume nuclei are fixed (i.e., no motion, constant R). So, term (iv) vanishes,

and is used only in the vibrational Sch.Eqn, which is treated as a separate problem. Additionally, term (v) becomes a constant, allowing independent solution of the electronic problem at a given R. Repetition vs. R specifies the vibrational potential.

(6) [ 4 pts ] Why does the effective nuclear charge for the 1s orbital increase by 0.99 in going from

oxygen to fluorine but only increases by 0.65 for the 2p orbital?

Because the radial probability distribution of the 1s electron is quite localized, there is no orbital space between the 1s and the nucleus and since there are no electrons that are added to the 1s level in

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going from oxygen to fluorine, the 1s electron feels the full increase in the nuclear charge in going from oxygen to fluorine. By contrast, the 2p electrons are more delocalized. As an additional p electron is added in going from oxygen to fluorine, only a part of the increase in the nuclear charge is negated by shielding due to the additional 2p electron. Adding an electron in the same shell is not as effective in shielding as adding an electron in an inner shell.

(7) [ 6 pts ] The neutral configuration of Lithium (Li0) is 1s22s1 with nuclear

charge Z = 3, whereas Beryllium has the configuration 1s22s2 and Z = 4. The plot at right shows realistic representations of 1s & 2s radial distribution functions. These are qualitatively correct for both Li0 and Be0, and may be conceptually useful in considering your answers here.

(a) [ 2 pts ] How many electrons contribute to screening a 2s electron in Be0? ____________ 3 (2 from 1s-type orbitals plus 1 from the 2s)

(b) [ 2 pts ] The total screening experienced by a 2s electron in a given atom is (Z−ζ2s). In comparing this value for Be0 and Li0, circle which of the following is correct.

(i) 0 02 2Be Li( ) ( )s sZ Z−ζ < −ζ

(ii) 0 02 2Be Li( ) ( )s sZ Z−ζ = −ζ

(iii) 0 02 2Be Li( ) ( )s sZ Z−ζ > −ζ

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(c) [ 2 pts ] Consider instead the value ξ2s ≡ (Z−ζ2s)/Z, which is a marker of the effectiveness of screening in a 2s orbital. Circle which of the following is correct.

(i) 0 02 2Be Li( ) ( )s sξ < ξ

(ii) 0 02 2Be Li( ) ( )s sξ = ξ

(iii) 0 02 2Be Li( ) ( )s sξ > ξ

Explanations: (a) A 2s electron is strongly screened (~<1 charge unit contributed to screening) by each 1s electron since the latter reside almost completely between the 2s and the nucleus. Additionally, each 2s electron in Be0 will screen the other, but only weakly, contributing <<1 charge unit to screening because it spends so little fractional time within the other 2s. Thus, the 2s experiences total screening ~<2 charge units, meaning (Z − ζ2s) >~2 so 1.5 < ζ1s < 2.5 The actual value is ζ2s = 1.91. (b) As more electrons are added, there is always more screening. This explains why total screening is bigger for Be0. (c) However, each added electron is less effective than the previous. So as you go across the periodic table, each step increases Z by 1, but increases screening by a bit less than 1. Thus, screening efficacy is less for Be0.

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