The University of British Columbia Okanagan
CHEM 121 – ATOMIC & MOLECULAR CHEMISTRY
Final Exam, Fall 2011
Wednesday, Dec 14, 9AM
Name:
______________________
Circle your Section:
Student Number:
______________________
McNeil / Neeland
Signature:
______________________
DO NOT TURN THE PAGE UNTIL INSTRUCTED TO DO SO!
Question
Points
1
6
2
5
3
4
4
4
5
2
6
5
7
5
8
4
9
6
• Carefully read each question before answering.
Where appropriate, you must show your work
to receive full credit.
10
2
11
3
12
4
• Include units and the proper significant figures in
all numerical answers.
13
3
14
3
• With the exception of a non-programmable
calculator, no aids or notes of any kind are
permitted or required.
15
3
16
6
17
4
18
5
19
4
20
12
21
6
22
4
TOTAL
100
• Indicate your name, student number, and
section on this front page.
• Make sure you have all ten pages, including 9
pages with 22 questions. You should also have
a periodic table and a page of helpful
information.
• Answer all questions directly on this exam.
Make note of the point value of each question,
and allocate your time accordingly.
Total Points: 100
Total Time: 3 hours
Score
Potentially Helpful Information Constants
Equations
electron mass = 9.109 × 10–31 kg
E = hν
E = ½mv2
E = mc2
proton mass = 1.673 × 10–27 kg
λν = c
λ = h/mv
hν = hν0 + KE
neutron mass = 1.675 × 10
1 u = 1.66054 × 10
–27
–27
kg
kg
h = 6.626 × 10–34 Js
c = 2.998 × 108 m/s
RH = 2.178 × 10–18 J
a0 = 5.29 × 10–11 m
PV = nRT
P=
MM = ρRT/P
e = 1.602 × 10–19 C
NA = 6.022 × 1023 mol–1
k = 1.381 × 10–23 JK–1
Ptot = PA + PB + ⋅⋅⋅
1 eV = 96.5 kJ/mol
absolute zero = –273.15°C
PA = XAPtot
PA = XAP°A
XA = nA/ntot
R = 0.082057 L atm/mol K = 8.3145 J/mol K
ΔP = XsoluteP°solvent Π = MRT
1 atm = 101.325 kPa = 760 torr = 760 mmHg
ΔTf = –Kfm
ΔTb = Kbm
Electronegativity Values
H
2.2
Li
1.0
Na
0.9
K
0.8
Be
1.6
Mg
1.3
Ca
1.0
B
2.0
Al
1.6
Ga
1.8
C
2.6
Si
1.9
Ge
2.0
N
3.0
P
2.2
As
2.2
O
3.4
S
2.6
Se
2.6
F
4.0
Cl
3.2
Br
3.0
I
2.7
Chemistry 121 Final Exam, Fall 2011
(6) 1) Circle the one best answer for each question.
i.
At a given temperature, which compound should exhibit the highest vapour pressure?
a) CH4, because it has the lowest intermolecular forces of attraction
b) I2, because it has the biggest and most polarizable electron clouds
c) H2S, because it has the largest molecular dipole
d) LiF, because it has the smallest molar mass
ii.
SiCl4 (bp = 58°C), AsCl3 (bp = 130°C), and Br2 (bp = 59°C) all have similar molar masses,
between 160 and 181 g/mol. Why is the boiling point of AsCl3 so much greater?
a) AsCl3 is a polar molecule, but the others are non-polar.
b) AsCl3 has a lone pair, but the other compounds have no lone pairs.
c) As-Cl bonds have greater polarity than Si-Cl or Br-Br bonds.
d) As-Cl bonds are stronger than Si-Cl or Br-Br bonds.
e) All of the above factors contribute to the higher boiling point.
iii.
At low pressure and temperature, which gas will exhibit a pressure that is much lower than
expected?
a) CH4, because the molecules are smaller than expected.
b) SiH4, because the molecules move more slowly.
c) HF, because it has strong intermolecular attractive forces.
d) N2, because the molecule has a strong triple bond.
iv.
Which gas has the particles with the highest average molecular speed?
a) NH3 at 25°C
b) CO2 at 25°C
c) NH3 at 75°C
d) CO2 at 75°C
v.
Dipole-dipole forces between molecules are due to
a) overlap of atomic orbitals
b) permanent charge separation within molecules leading to electrostatic attraction
c) electrostatic attraction of protons and electrons within atoms
d) temporary distortion of electron clouds
vi.
Each of the following is decomposed such that hydrogen gas is produced with 100% yield.
Which of the following generates the greatest volume of H2 gas?
a) 13 g C6H6
b) 10 g CH4
c) 18 g H2O
d) 60 g HF
e) 64 g HI
Page 1/9
Chemistry 121 Final Exam, Fall 2011
2) Aluminum tribromide (AlBr3) reacts with hydride (H–) to yield the product [AlBr3H]–.
(1) a) T or F. Five regions of electron density surround the aluminum atom in [AlBr3H]–.
_____
(2) b) Consider the VSEPR-predicted shape of AlBr3. What is the hybridization of atomic orbitals
at the aluminum atom?
(2) c) Why is AlBr3 able to react with hydride and form a new Al-H bond?
(4) 3) Below is a partial molecule which has one mystery atom (in the box). The arrowed bond
consists of the overlap of a 3p orbital with an sp orbital. The C3 atom has bond angles of
120°. Other than the box, all remaining carbon bonds (not shown) are to hydrogen atoms.
Complete the molecule by:
a) showing the identity of the mystery atom in the box
b) drawing all necessary H atoms and the bonds connecting to them, and
c) drawing all remaining C-C sigma/pi bonds consistent with the given information.
(4) 4) Assign formal charges to each central atom. Note that none of these compounds obeys the
octet rule. Explain your reasoning for the last example only.
(2) 5) The ionization energy of gaseous Na atoms is 495.80 kJ/mol. Calculate the lowest possible
frequency of light that can ionize a Na atom.
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Chemistry 121 Final Exam, Fall 2011
(5) 6) Complete the table below.
molecular formula Lewis structure with formal charges on all atoms molecular geometry name hybridization of central atom predicted bond angle(s) overall polarity? (Y / N) [XeF3]+ (3) 7a) In each box, draw a valid Lewis resonance form of the molecule in the centre, clearly
indicating any formal charges. On the centre molecule, draw the electron flow arrows
leading to one of your resonance structures.
(2) 7b) Of the three structures, 1, 2, or 3, which do you think will be the major contributor to the
overall resonance hybrid? Briefly explain your reasoning.
(4) 8) Predict which molecule has the shortest O-O bond length: HOOH, O2, or O3. Note that none
of these compounds is a ring structure. Explain your answer.
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Chemistry 121 Final Exam, Fall 2011
9) Consider the diatomic molecules [NO]2– and BN.
(2) a) Calculate the bond order for each molecule.
(2) b) Indicate the relative predicted bond length and strength of each molecule: which is longer,
and which is stronger?
(2) c) Indicate the paramagnetic or diamagnetic properties for each molecule.
(2) 10) Write the molecular orbital configuration for the diatomic molecule Na2. Project your
knowledge on this one. It’s not hard.
(3) 11) How many electrons in a ground-state xenon atom can have the following quantum numbers?
a) n = 4, l = 2, all of ml = –2 and ml = –1
_______
b) l = 2
_______
c) ms = +½
_______
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Chemistry 121 Final Exam, Fall 2011
(2) 12a) Which of the following sets of four quantum numbers, n, l, ml, ms describes an outermost
electron in a ground state strontium (Sr) atom? Circle all possible correct responses, and
briefly explain your reasoning.
(a) 5, 1, +1, +½
(b) 5, 0, 0, –½
(c) 5, 0, +1, +½
(d) 6, 1, 0, +½
(e) 5, 2, +1, –½
(2) 12b) How could this change when dealing with an excited state of the strontium atom? Circle
all possible correct responses.
(a) 5, 1, +1, +½
(b) 4, 2, 0, –½
(c) 5, 0, +1, +½
(d) 4, 1, 0, +½
(e) 5, 2, +1, –½
(3) 13) Consider a covalent pi bond between two atoms. Compare and contrast how the valence
bond theory and molecular orbital theory account for the pi bond formation?
(1) 14a)
Circle the atom with the largest covalent radius: Si, P, S, Cl, Ar
(1) b) Circle the atom with the highest first ionization potential: O, F, Ne, Na, Mg
(1) c) Circle the atom with the highest electronegativity: Se, Br, Rb, Sr.
(3) 15) What electronic transition in a hydrogen atom, starting from the n = 6 energy level, will emit
light of wavelength 1094 nm?
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Chemistry 121 Final Exam, Fall 2011
16) Mixtures of benzene (C6H6) and toluene (C7H8)
form ideal solutions. Thermochemical data for
these compounds are shown at right.
normal bp vapour pressure (25°C) benzene 80.1°C toluene 110.6°C 97.3 torr 26.7 torr (4) a) A solution formed from 5.00 g of benzene and 5.00 g of toluene is placed in a sealed flask at
25°C. What is the total vapour pressure of this solution?
(2) b) What is the mole fraction of benzene molecules in the vapour above the solution?
(4) 17) Provide concise rationalizations for the following observation. Note that there are two
observations to explain.
Methanol (CH3OH) is very soluble in water (H2O) but not soluble in carbon disulfide (CS2).
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Chemistry 121 Final Exam, Fall 2011
18) Nicotine (C10H14N2) is a potent insecticide and the
principally addictive component of tobacco. Consider the
effects of dissolving 4.05 g of nicotine into 50.0 g ethanol
(C2H6O). Ethanol has a normal freezing point of
–114.6°C, a density of 0.903 g/mL, a vapour pressure of
43.6 mmHg at 20°C, and a Kf of 1.99 °C/m.
(2) a) What is the vapour pressure of this solution at 20°C?
(3) b) Calculate the new freezing temperature of the ethanol solution.
19) Americium (Am, Z = 95) is a radioactive metallic element used in smoke detectors. Smoke
blocks the passage of emitted alpha particles through an ionization chamber, disrupting an
electric current. Less than one microgram of americium is sufficient to generate about 40000
alpha particles per second. The symbols for elements near americium are:
U (Z=92), Np (Z=93), Pu (Z=94), Cm (Z=96), Bk (Z=97), Cf (Z=98).
(2) a) Write the balanced equation depicting alpha decay from a nuclide of americium-241,
including both mass and charge numbers for all particles.
(2) b) Do you think the presence of radioactive americium in a smoke detector constitutes a health
concern? Briefly explain your reasoning.
Page 7/9
Chemistry 121 Final Exam, Fall 2011
20) Consider the compound hydrazine, N2H4, which has MW = 32.045 g/mol.
(3) a) 48.0 g of hydrazine is placed in a 750.0 mL steel container and heated to 145°C. Hydrazine
is a gas under these conditions. The Van der Waals constants for hydrazine are a = 8.35
atmL2/mol2 and b = 0.0462 L/mol. Calculate the pressure exhibited by the hydrazine.
(3) b) Would you expect equal moles of ethylene (C2H4) to exhibit a higher or lower pressure than
hydrazine under these conditions? Briefly explain your reasoning.
Hydrazine can be used as a rocket fuel, because it reacts exothermically with N2O4 (MW =
92.012 g/mol) to form N2 and H2O:
(3) c) If 3.40 kg of N2H4 reacts with 9.00 kg of N2O4, which material is the limiting reagent?
Clearly show all calculations and necessary reasoning.
2 N2H4(g) + N2O4(g) ⎯→ 3 N2(g) + 4 H2O(g) (3) d) Assuming a 100% yield for the reaction in part c), what volume will the product gases
occupy at 250°C under a pressure of 720 torr?
Page 8/9
Chemistry 121 Final Exam, Fall 2011
21) Consider hydrazine some more, and its phase
diagram, shown at right.
(1) a) At 40 atm pressure and 115°C, in what phase
does hydrazine exist? Circle your response.
solid liquid gas supercritical fluid (2) b) What phase change(s), if any, will occur if the
pressure on the hydrazine in part (a) is reduced to
15 mmHg?
(2) c) Calculate the enthalpy of vaporization of hydrazine, in kJ/mol.
(1) d) What is the name of the point labeled “O” on the diagram?
____________________
(4) 22) A balloon is filled with 0.500 L of gaseous CH4 at initial conditions of 15.0°C and 725 torr.
The balloon rises in the atmosphere to an elevation at which the temperature is –7.5°C and
525 torr. Assume ideal behavior.
a) What is the volume of gas in the balloon at this elevation?
b) What is the density of gas in the balloon at this elevation?
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