Download - CHEM1300 fall 2011 with solutions
CHEM 1300 final exam, Fall 2011 December 14, 2011 Version 1
3 hour exam Page 1
Part A
1. Which exam version are you writing (see the header above)? This question is not worth any
marks, but no concessions will be made if you answer it incorrectly.
a. Version 1
b. Version 2
2. The concentration of a solution prepared by dissolving 10.0 g of sodium chloride in sufficient
water to make 750.0 mL of solution is:
a. 0.133 M
b. 0.228 M
c. 0.0133 M
d. 0.456 M
e. 0.0750 M
3. Electrical (Coulomb) energy:
a. Increases in magnitude as the distance between charged particles decreases.
b. Is greater for two Ca2+
cations 10 pm apart than for two Na+ cations 10 pm apart.
c. Is negative for two oppositely charged particles.
d. Is the potential energy associated with the forces of attraction and repulsion between
electrically-charged particles.
e. All of the above.
4. Which radiation has the highest frequency?
a. green visible light
b. radio waves
c. infrared radiation
d. microwave radiation
e. cell phone radiation
5. CANCELLED. A police radar unit is operating at a frequency of 9.527 gigahertz (109 Hz).
What is the wavelength of the radiation being employed?
a. 631.3 nm
b. 6.313×10–24
m
c. 2.858 nm
d. 31.47 cm
e. 2.858×1018
m
CHEM 1300 final exam, Fall 2011 December 14, 2011 Version 1
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6. The figure below represents the energy level diagram for the hydrogen atom, and the arrows labeled
a – f are possible electron transitions. Rank the transitions that involve the emission of a photon
from lowest to highest frequency.
a. a < c < b < e < f < d
b. d < f < e < b < c < a
c. d < e < c
d. c < e < d
e. a < b < f
7. The notation for the subshell with n = 5 and l = 3 is:
a. 3d
b. 3h
c. 5d
d. 5p
e. 5f
8. A possible set of quantum numbers for an electron in the partially filled subshell in a potassium
atom in its ground state configuration would be
n l ml ms
a. 3 0 0 ½
b. 3 1 0 -½
c. 4 0 0 -½
d. 4 1 0 ½
e. 4 2 1 ½
9. Which one of the following pictures shows a 2p orbital?
a. b. c. d. e.
10. Which atom has the smaller 3s orbital?
a. An atom with more protons
b. An atom with fewer protons
c. An atom with more neutrons
d. An atom with fewer neutrons
e. The size of the 3s orbital is the same for all atoms.
n = 1
n = 2
n = 3
n = 4c
d
a b
e
f
Ener
gy
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11. In the ground state electron configuration of gold (Au), how many electrons are there in the 3p
orbitals?
a. 0
b. 1
c. 2
d. 3
e. 6
12. Place the following species in order of decreasing radius.
2Te F 2O
a. F > 2O
> 2Te
b. F > 2Te
> 2O
c. 2Te >
2O
> F
d. 2Te
> F >
2O
e. 2O
> F >
2Te
13. Which one of the following orbital diagrams represents the ground state electron configuration
of a Cr3+
cation?
4s 3d 4s 3d
a. [Ar]
d. [Ar]
b. [Ar]
e. [Ar]
c. [Ar]
14. Choose the diamagnetic species from below.
a. Sn2+
b. Br
c. P
d. Cr
e. None of the above are diamagnetic.
15. Which one of the following is a set of isoelectronic species?
a. C, N, O
b. F, Cl, Br
c. Na+, Mg
2+, Cl
–
d. S2–
, Ar, Ca2+
e. Rb+, Sr
2+, Zr
2+
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16. Place the following in order of increasing first ionization energy:
N F As
a. N < As < F
b. N < F < As
c. F < N < As
d. As < N < F
e. As < F < N
17. Which one of the following species has the highest third ionization energy?
a. Al
b. Ca
c. Mg
d. P
e. P3–
18. Which one of the following has 4 valence electrons?
a. Al
b. Si
c. P
d. As
e. Be
19. The primary reason that an ionic compound of formula LiBr2 does not form is:
a. the electron affinity of Br is endothermic.
b. the lattice energy for LiBr2 is smaller than that of LiBr.
c. the total ionization energy required to produce Li2+
is too high.
d. the second electron affinity of Br is positive.
e. the bond energy of Br2 is prohibitively large.
20. Which one of the following molecules is an alcohol?
a. b. c. d. e.
OH
O O
NH2
O OH
O
O
O
21. Which one of the following bonds is least polar?
a. C–B
b. C–C
c. C–N
d. C–O
e. C–F
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22. Draw the best Lewis structure for the BF‹ anion. What is the formal charge on the boron atom?
a. +2
b. 0
c. +1
d. –1
e. –2
23. C is the central atom in the thiocarbonate ion, CO2S2–
. Which of the following would best reflect
the actual charges on the various atoms based on its Lewis structure(s)? Note: Oa and Ob are the
two oxygen atoms in the thiocarbonate ion.
C Oa Ob S
a. –½ –½ –½ –½
b. –1 0 0 –1
c. 0 0 –1 –1
d. 0 –1 –1 0
e. 0 0 0 –2
24. Which one of the following molecules is nonpolar?
a. CH2Br2
b. SeCl6
c. SO2
d. PCl3
e. CH3OCH3
25. Place the following species in order of decreasing XO bond length, where X represents the central
atom in each species.
CO2 CO⊀‾ SiO⊀‾
a. CO2 > CO⊀‾ > SiO⊀‾
b. CO⊀‾ > SiO⊀‾ > CO2
c. CO2 > SiO⊀‾ > CO⊀‾
d. SiO⊀‾ > CO2 > CO⊀‾
e. SiO⊀‾ > CO⊀‾ > CO2
26. How many - and -bonds are there in the CO molecule?
a. 1 -bonds, 2 -bonds
b. 2 -bonds, 1 -bonds
c. 1 -bonds, 1 -bonds
d. 1 -bonds, 0 -bonds
e. 2 -bonds, 0 -bonds
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27. Based on MO theory, what is the bond order in the LiBe molecule?
a. 0
b. 0.5
c. 1
d. 1.5
e. 2
28. The picture below illustrates the delocalized π-bonding molecular orbital in ozone constructed
from the overlap of the three parallel p-orbitals on adjacent O atoms. How many electrons can
this molecular orbital contain?
a. Up to two electrons having opposite spin.
b. Up to four electrons, two in each lobe.
c. Up to six electrons, two for each of the three O p-orbitals
d. All of the electrons involved in the σ-bonding in ozone.
e. All of the lone pair electrons from the Lewis structure of ozone.
29. How many delocalized -electrons are there in the following molecule?
a. 5
b. 6
c. 8
d. 10
e. 12
30. Which one of the following figures is the best representation of the electronic band structure of
silicon doped with arsenic atoms?
a. b. c. d. e.
O
CHEM 1300 final exam, Fall 2011 December 14, 2011 Version 1
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31. Place the following molecules in order of increasing boiling point for the pure liquid.
a. < <
b. < <
c.
< <
d.
< <
e. <
<
32. Which one of the following liquids is expected to have the highest vapour pressure at 20ºC?
liquid Surface tension at 20ºC (J/m2)
a. hexane 0.0184
b. acetone 0.0237
c. cyclohexanol 0.0344
d. water 0.0728
e. mercury 0.425
33. Solid Kr and solid Cu both have a face-centred cubic unit cell. Why are their physical
properties so different?
a. Kr atoms are bigger than Cu atoms.
b. Cu is more electronegative than Kr.
c. Kr atoms in the solid interact via dispersion forces while Cu atoms in the solid interact via
metallic bonding.
d. Cu atoms have unpaired electrons while all electrons are paired for Kr atoms.
e. Cu is more expensive than Kr.
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34. Which of the following images represents a network covalent solid?
a.
b.
c.
d.
e. All of these images represent network covalent solids.
35. In the unit cell pictured to the right, the solid spheres
represent oxide anions (in all corner and face positions) and
the open spheres with a T inside represent the tetrahedral
holes filled by metal cations (all completely inside the unit
cell). Determine the empirical formula for this ionic
compound, where X represents the metal.
a. XO
b. X2O
c. X3O
d. XO2
e. XO3
36. Order the following crystal structures for increasing number of atoms in the unit cell.
a. simple cubic < face-centred cubic < body-centred cubic
b. face-centred cubic < simple cubic < body-centred cubic
c. face-centred cubic < body-centred cubic < simple cubic
d. simple cubic < body-centred cubic < face-centred cubic
e. body-centred cubic < simple cubic < face-centred cubic
37. For solids with a hexagonal close packed arrangement, what is the coordination number and the
layer packing order?
a. Coordination number = 6, layer packing order = ABABAB
b. Coordination number = 8, layer packing order = ABCABC
c. Coordination number = 8, layer packing order = ABABAB
d. Coordination number = 12, layer packing order = ABABAB
e. Coordination number = 12, layer packing order = ABCABC
CHEM 1300 final exam, Fall 2011 December 14, 2011 Version 1
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Part B
Question 1: /1 point
Balance the combustion reaction for butane:
C4H10 (g) + 13/2 O2 (g) → 4 CO2 (g) + 5 H2O (l)
or
2 C4H10 (g) + 13 O2 (g) → 8 CO2 (g) + 10 H2O (l)
Question 2: /2 points
Write the chemical formula for each compound:
cobalt (III) chloride hexahydrate CoCl3·6H2O
sodium sulfate Na2SO4
dinitrogen pentoxide N2O5
benzene C6H6
Question 3: /1 point
For each set of quantum numbers below, indicate whether the set is valid. If a set is not valid,
provide a new set of quantum numbers that corrects the problem.
Set of quantum numbers Valid (circle) Only if not valid, correct the problem
n = 3, l = 3, ml = 2, ms = –½ No n = 3, l = 2, ml = 2, mS = –½
or
n = 4, l = 3, ml = 2, mS = –½
n = 5, l = 2, ml = –1, ms = ½ Yes
Question 4: /4 points
The metalloid antimony is able to form three different stable ions in its compounds.
Sb [Kr] 5s2 4d
10 5p
3 or 1s
2 2s
2 2p
6 3s
2 3p
6 4s
2 3d
10 4p
6 5s
2 4d
10 5p
3
Sb3–
[Kr] 5s2 4d
10 5p
6 or 1s
2 2s
2 2p
6 3s
2 3p
6 4s
2 3d
10 4p
6 5s
2 4d
10 5p
6 or [Xe]
Sb3+
[Kr] 5s2 4d
10 or 1s
2 2s
2 2p
6 3s
2 3p
6 4s
2 3d
10 4p
6 5s
2 4d
10
Sb5+
[Kr] 4d10
or 1s2 2s
2 2p
6 3s
2 3p
6 4s
2 3d
10 4p
6 4d
10
Rank the four antimony species listed above in
order of increasing size (radius):
Sb5+
< Sb3+
< Sb < Sb3–
Rank the four antimony species listed above in
order of increasing first ionization energy:
Sb3–
< Sb < Sb3+
< Sb5+
CHEM 1300 final exam, Fall 2011 December 14, 2011 Version 1
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Question 5: /2 points
Platinum is a highly inert
metal used in electrical
contacts and electrodes, in
dentistry equipment, and
in jewelry.
The graph to the right
plots the kinetic energy of
ejected electrons as a
function of frequency of
light shone on a platinum
surface.
What is the binding
energy of platinum in
kJ/mol?
From graph, minimum frequency required to eject electrons = 1.3 × 1015
Hz
E = nhν = (6.022×1023
mol–1
)(6.626×10–34
Js)( 1.3 × 1015
s–1
)(1 kJ / 1000 J) = 520 kJ/mol
Question 6: /2 points
One of the less prominent lines in the spectrum of mercury is at 1013.975 nm. How many photons
of this wavelength are required to achieve a total energy of 100.0 kJ?
photons 10104.5photonper J 10959.1
photonsmany for J101.000 photons ofnumber
photon 1for J 10959.1m10 1013.975
m/s 102.998s J 106.626hc nm) E(1013.975
23
19
5
19
9-
8-34
Question 7: /2 points
What mass of FeS is produced when 4.71 g of Fe reacts with 4.25 g S8?
8 Fe (s) + S8 (s) → 8 FeS (s)
mol Fe = 4.71 g / 55.85 g/mol = 0.0843 mol Fe
mol FeS if all Fe reacts = 0.0843 mol Fe × (8 mol FeS / 8 mol Fe) = 0.0843 mol FeS if all Fe reacts
mol S8 = 4.25 g / 256.48 g/mol = 0.0166 mol S8
mol FeS if all S8 reacts = 0.0166 mol S8 × (8 mol FeS / 1 mol S8) = 0.133 mol FeS if all S8 reacts
thus, limiting reagent = Fe
mass FeS = 0.0843 mol × 87.91 g/mol = 7.41 g
0
200
400
600
800
0 0.5 1 1.5 2 2.5 3
Kin
etic
En
erg
y (
kJ
/mo
l)
Frequency (1015 Hz)
CHEM 1300 final exam, Fall 2011 December 14, 2011 Version 1
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Question 8: /3 points
Draw two sketches that illustrate how orbitals overlap to form all of the bonds in propene, C3H6.
The first sketch should show the -bonding framework, while the second should show the -
bonding framework. You may want to first draw the Lewis structure. Label the orbitals involved in
the formation of bonds.
C C
CH
H H
H
HH -bond
CH3
p - p
s - sp2s - sp3
s - sp3
s - s
p3
s - sp2
s - sp2
sp2 - sp2
sp2 - sp3
Question 9: /2 points
Draw the best Lewis structure(s) for SO⊀‾, including all equivalent resonance structures, if any.
SO
O
OS
O
O
OS
O
O
O
-2
Question 10: /6 points
Draw the best Lewis structures for Br„ and GaCl3, then fill in the table below.
Br„ GaCl3
Br Br Br-
Ga
Cl
Cl
Cl
Molecule
What is the name
of the electron
group geometry?
What is the name
of the molecular
geometry?
Is the molecule
polar or nonpolar?
Circle one.
List the value(s) of
the bond angles.
Br„ trigonal
bipyramidal
linear nonpolar 180º
GaCl3 trigonal planar trigonal planar nonpolar 120º
CHEM 1300 final exam, Fall 2011 December 14, 2011 Version 1
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Question 11: /3 points
Quinidine, pictured on the
right, is a medication used to
suppress abnormal rhythms
of the heart.
For each of the labeled bond
angles, write the theoretical
(VSEPR) angle value and the
hybridization of the central
atom in the table to the right.
N
O
N
OH
a
bc
d
Hybridization of
central atom
Theoretical
angle
a. sp3 109.5º
b. sp2 120º
c. sp3 109.5º
d. sp3 109.5º
Functional group at position a: ether Functional group at position d: amine
Question 12: /2 points
a. Indicate with an X which intermolecular force(s) must be overcome when the following pure
liquids are vapourized.
b. Circle the compound that has the highest viscosity at room temperature.
Compound Ion-dipole
forces
Dispersion
forces
Dipolar
forces
Hydrogen
bonding
C4H9OH X X X
C4H9F X X
C5H12 X
CH3COCH2CH3 X X
Question 13: Major typo! Final decision pending
Fill in the MO diagram below for O2–
and answer the questions:
EN
ER
GY
σ2p* –––
Bond order of O2– _______
What is the bond order of this species
after it absorbs a photon of light with
sufficient energy to cause ionization? _______
Provide a chemical equation for this ionization:
___________________↓___________________
π2p* ––– –––
π2p ––– –––
σ2p –––
σ2s* –––
σ2s –––
CHEM 1300 final exam, Fall 2011 December 14, 2011 Version 1
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Question 14: /2 points
Draw a line structure (e.g. ) or condensed structure representation (e.g. CH3–CH2–CH3; in
this case you must correctly include all H atoms) of 6,6-dichloro-3-ethyl-4-nonyne (also called:
6,6-dichloro-3-ethyl-non-4-yne).
OR
Question 15: /2 points
Explain the trend in boiling points shown in the graph below (be concise!).
The boiling points of HCl, HBr, and HI increase because the number of electrons and
hence the attractive strength of dispersion interactions increases as: HCl < HBr < HI
The boiling point of HF is much higher because HF molecules can hydrogen bond while
HCl, HBr, and HI molecules cannot.
150
200
250
300
2 3 4 5
Boil
ing P
oin
t (K
)
Period of Halogen
HF
HCl HBr
HI
ClCl
C C CHC
CH2
CH2 CH3
CH3
CH2CH2
CH3
Cl Cl