Chem 4502 Prof. Doreen Leopold __________________________________ 12/23/2015 (Wednesday) Name (Optional) Quantum Chemistry and Spectroscopy Final Exam (185 points, 37 questions, 22 pages) • Students have 2 hours to do this exam (1:30 PM - 3:30 PM). • It has 37 multiple choice questions worth 5 points each, for a total of 185 points. Each question will be graded as 0 or 5 points; there is no partial credit. There is just one correct answer for each question, and no penalty for incorrect answers. If more than one answer is bubbled in, no credit will be given for that question. • On the bubble sheets, please bubble in your name (last name first), X.500, student ID number, and sign the back. • You can take the question portion of the exam with you. The answers will be posted later today. • There are 6 - 8 questions on the material potentially covered on midterm Exams 1, 2, 3, and 4, and 8 questions on molecular spectroscopy (Chapter 13 and Homework 10). • You may use a non-programmable, non-graphing calculator. No notes are allowed. The equation sheets (2 pages) are included at the end of the exam (and may be torn off). • Final exam grades, a bar graph with statistics, and a description of the end-of-semester grading curve will be posted by Wednesday, December 30. PLEASE DO NOT TURN THE PAGE UNTIL IT IS ANNOUNCED THAT YOU MAY BEGIN.
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Chem 4502 Prof. Doreen Leopold __________________________________ 12/23/2015 (Wednesday) Name (Optional) Quantum Chemistry and Spectroscopy
Final Exam (185 points, 37 questions, 22 pages)
• Students have 2 hours to do this exam (1:30 PM - 3:30 PM).
• It has 37 multiple choice questions worth 5 points each, for a total of 185 points.
Each question will be graded as 0 or 5 points; there is no partial credit.
There is just one correct answer for each question, and no penalty for incorrect answers.
If more than one answer is bubbled in, no credit will be given for that question.
• On the bubble sheets, please bubble in your name (last name first), X.500, student ID number, and
sign the back.
• You can take the question portion of the exam with you. The answers will be posted later today.
• There are 6 - 8 questions on the material potentially covered on midterm Exams 1, 2, 3, and 4,
and 8 questions on molecular spectroscopy (Chapter 13 and Homework 10).
• You may use a non-programmable, non-graphing calculator.
No notes are allowed.
The equation sheets (2 pages) are included at the end of the exam (and may be torn off).
• Final exam grades, a bar graph with statistics, and a description of the end-of-semester grading curve
will be posted by Wednesday, December 30.
PLEASE DO NOT TURN THE PAGE UNTIL IT IS ANNOUNCED THAT YOU MAY BEGIN.
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Your Exam is Version A. Please check that your bubble sheet is premarked with the correct version.
Part I. (6 questions, 30 points) - Questions selected from among the following topics:
2-Slit Experiment, Dawn of Quantum Mechanics, Classical Wave Equation, Complex Numbers,
22. Slater determinants are used for writing atomic and molecular wave functions because they satisfy which of the following requirements? Choose the best answer.
A. The wave function is zero if any two electron labels are interchanged.
B. The wave function changes sign if any two electron labels are interchanged.
C. The wave function is zero if any two electrons occupy the same spin-orbital.
D. The wave function changes sign if any two electrons occupy the same spin-orbital.
E. both A and C
F. both A and D
G. both B and C
H. both B and D
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23. For homonuclear diatomics like O2, combining one 2s and three 2p atomic orbitals on each atom
yields ________ bonding and ________ antibonding molecular orbitals.
A. 2 bonding and 2 antibonding
B. 3 bonding and 3 antibonding
C. 4 bonding and 4 antibonding
D. 8 bonding and 8 antibonding
E. 2 bonding and 0 antibonding
F. 4 bonding and 0 antibonding
G. 4 bonding and 2 antibonding
H. 6 bonding and 0 antibonding
I. 6 bonding and 2 antibonding
J. 6 bonding and 4 antibonding
24. For O2, which of the following statements is false regarding the excited 3u "B" state?
A. The transition from the ground state is allowed by the spectroscopic spin selection rule.
B. The bond order is higher in the B state than in the ground state.
C. The B state has a longer bond length and a lower vibrational frequency than the ground state.
D. The B state can be accessed from the ground state by the absorption of ultraviolet light.
E. More than one of the above statements (A-D) is false.
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25. In the simplest LCAO-MO description of H2+, the probability for finding the electron in the
bonding molecular orbital can be calculated as:
Here, the last term (indicated by the bracket)
A. is due to constructive interference
B. is due to destructive interference
C. is due to electron-electron repulsion
D. is zero because the atomic orbitals are orthogonal
E. approaches zero as the internuclear separation decreases
F. more than one of the above statements (A - E) are true
26. The term symbols associated with the 2p3 electron configuration of the nitrogen atom are
2P, 2D, and 4S. According to Hund's Rules,
the lowest energy term is __________ and the highest energy term is __________ .
A. lowest 2P, highest 2D
B. lowest 2P, highest 4S
C. lowest 2D, highest 2P
D. lowest 2D, highest 4S
E. lowest 4S, highest 2P
F. lowest 4S, highest 2D
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27. The variational principle involves the calculation of the quantity:
d
dHE
*
* ˆ
In this expression, in general,
is _______________________________________________ ;
H is _____________________________ ;
and the variational principle states that _________________________ .
A. is the true ground state wave function; H is the exact Hamiltonian operator;
states that E is less than the true ground state energy
B. is the true ground state wave function; H is an approximate Hamiltonian operator;
states that E is less than the true ground state energy
C. is a trial wave function; H is the exact Hamiltonian operator;
states that E is less than the true ground state energy
D. is a trial wave function; H is an approximate Hamiltonian operator;
states that E is less than the true ground state energy
E. is the true ground state wave function; H is the exact Hamiltonian operator;
states that E is not less than the true ground state energy
F. is the true ground state wave function; H is an approximate Hamiltonian operator;
states that E is not less than the true ground state energy
G. is a trial wave function; H is the exact Hamiltonian operator;
states that E is not less than the true ground state energy
H. is a trial wave function; H is an approximate Hamiltonian operator;
states that E is not less than the true ground state energy
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28. Which of the following statements is/are true concerning Hartree-Fock calculations?
I. The instantaneous correlations of the electrons' motions are neglected.
II. The total energy of the He atom is calculated to be about 1 eV more negative than the true energy.
III. The total energy of an atom is calculated as the sum of the energies of the occupied spin-orbitals.
A. I only
B. II only
C. III only
D. I and II
E. I and III
F. II and III
G. I, II and III
H. None of these statements (I, II, III) are true.
29. What is the highest occupied MO in the N2+ molecular ion?
A. u 2pz
B. g
C. g 2pz
D. u
E. u 2s
F. g 2s
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Part V. (8 questions, 40 points)
Topics: Molecular Spectroscopy (Homework 10)
30. Which of the following statements is/are true regarding the harmonic oscillator (HO) as
compared with the Morse potential models for molecular vibrations?
A. In the HO model, the energy levels are equally spaced, but in the Morse potential model,
spacings between consecutive vibrational levels increase with increasing energy.
B. In the HO model, there is only a finite number of vibrational levels, but in the Morse potential
model, there is an infinite number.
C. In the HO model, the average amplitude of the vibration remains the same as the quantum number
v increases, but in the Morse potential model, the amplitude decreases with increasing v.
D. More than one of the above statements are true.
E. None of the above statements (A-C) are true.
31. For a gas phase molecule in a sealed container, which of the following choices correctly ranks
(smallest to largest) the typical energy spacings for various types of quantum states?
A. electronic < translational < vibrational < rotational
B. rotational < vibrational < translational < electronic
C. rotational < translational < vibrational < electronic
D. rotational < electronic < translational < vibrational
E. translational < vibrational < rotational < electronic
F. translational < rotational < vibrational < electronic
G. translational < electronic < rotational < vibrational
H. vibrational < translational < rotational < electronic
I. vibrational < rotational < translational < electronic
J. vibrational < rotational < electronic < translational
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32. Consider a diatomic molecule with rotational constant B = 10 cm-1. At 300 K, what will be the
relative population of molecules in the J = 3 rotational state as compared with J = 0?
A. about 2 times more molecules in J = 3 than in J = 0
B. about 4 times more molecules in J = 3 than in J = 0
C. about 7 times more molecules in J = 3 than in J = 0
D. about 12 times more molecules in J = 3 than in J = 0
E. about 2 times more molecules in J = 0 than in J = 3
F. about 4 times more molecules in J = 0 than in J = 3
G. about 7 times more molecules in J = 0 than in J = 3
H. about 12 times more molecules in J = 0 than in J = 3
33. For benzene (C6H6), what are the numbers of translational, rotational, and vibrational degrees
of freedom?
A. 1 translational, 2 rotational and 12 vibrational
B. 1 translational, 3 rotational and 32 vibrational
C. 2 translational, 3 rotational and 31 vibrational
D. 3 translational, 2 rotational and 31 vibrational
E. 3 translational, 2 rotational and 36 vibrational
F. 3 translational, 3 rotational and 12 vibrational
G. 3 translational, 3 rotational and 30 vibrational
H. 12 translational, 12 rotational and 12 vibrational
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34. In the rotational-vibrational spectrum of a gas phase diatomic molecule, the absorption lines are
spaced by about 20 cm-1. If the vibrational frequency is 2,000 cm-1, what is the "energy" (in cm-1) of
the transition from J = 3 in the lower vibrational state to J = 2 in the upper vibrational state?
A. 2120 cm-1
B. 2060 cm-1
C. 2000 cm-1
D. 1960 cm-1
E. 1940 cm-1
F. 1880 cm-1
G. 240 cm-1
H. 180 cm-1
I. 120 cm-1
J. 60 cm-1
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35. In the infrared spectrum of carbon monoxide (12C16O), the fundamental line is observed at 2143
cm-1 and the first overtone occurs at 4260 cm-1. What are the values of the harmonic frequency (e)
and the anharmonicity constant (exe) for this molecule?
A. e = 2117 cm-1, exe = 13 cm-1
B. e = 2117 cm-1, exe = 26 cm-1
C. e = 2130 cm-1, exe = 13 cm-1
D. e = 2143 cm-1, exe = 26 cm-1
E. e = 2143 cm-1, exe = 2117 cm-1
F. e = 2156 cm-1, exe = 7.5 cm-1
G. e = 2156 cm-1, exe = 13 cm-1
H. e = 2169 cm-1, exe = 13 cm-1
I. e = 2169 cm-1, exe = 26 cm-1
J. e = 4260 cm-1, exe = 2117 cm-1
36. The visible emission spectrum of gas phase I2 following laser excitation at 514 nm shows a long
vibrational progression. Which of the following statements is/are true concerning this spectrum?
A. The ground and excited electronic states have similar equilibrium bond lengths.
B. Most of the emitted light has wavelengths shorter than 514 nm.
C. I2 relaxes to the zero point level of the excited electronic state before emitting visible light.
D. I2 relaxes to a wide range of vibrational levels in the ground electronic state.
E. More than one of the above statements are true.
37. In this course, we have applied the Schrodinger equation to various systems. For which system(s)
did we set the potential energy term equal to zero when setting up the Schrödinger equation? Choose
the best answer.
A. particle-in-a-box
B. harmonic oscillator
C. rigid rotator
D. hydrogen atom
E. helium atom
F. H2 molecule
G. two of the above
H. three of the above (among A - F)
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Possibly Useful Equations, Conversions and Constants Chem 4502
c = 2.998 x 108 m/s k B = 0.695 cm-1 / K
h = 6.626 x 10−34 J·s = h/(2π) = 1.055 x 10−34 J·s
e = 1.602 x 10−19 C
1 eV = 1.602 x 10−19 J (corresponds to 8066 cm-1)
me = 9.109 x 10−31 kg mp = 1.673 x 10−27 kg amu = 1.661 x 10-27 kg
Normal modes of a vibrating string of length : un(x,t) = An cos (ωnt + n) sin (n π x / )
Schrödinger equation: (−2 / (2m)) d
2Ψ / dx2 + V(x)Ψ(x) = EΨ(x)
Momentum operator: = -i ∂/∂x
--------------------------------------------------------------------------------------------------------------------- PIB ψn(x) = (2/a)½ sin (nπx / a) En = n2h2 / (8ma2) HO where a = (μk)½
/
Ev = (v+½) ho where o = (1 / (2π)) (k/μ)½ and μ = m1m2 / (m1+m2) (diatomic)