NAME _______________________________________ Student No. ___________________________
SECTION (circle one): A01 (Codding)
A02 (Sirk)
A03 (Briggs)
UNIVERSITY OF VICTORIA
Version
A
CHEMISTRY 101
Mid-Term Test 2, November 19
2010
Version
A
This test has two parts and 8 pages, including the cover page. Please count the pages to
ensure you have them all.
PART I is a multiple choice section and is worth 24 marks. The answers for the 13 questions in this
part must be coded on the optical sense form using a SOFT PENCIL.
PART II consists of written answers and is worth 26 marks. Answer these questions on this
examination paper. Answers written partially or completely in pencil cannot be remarked.
- Hand in this entire test paper as well as your optical sense form (i.e. bubble sheet) at the end of the
examination period (60 minutes).
-The basic Sharp EL510 calculator is the only one approved for use in Chemistry 101.
- A DATA sheet accompanies this test.
Marks for Written Answers
Question 1 [6]
Question 2 [6]
Question 3 [4]
Question 4 [5]
Question 5 [5]
TOTAL (/26)
Multiple Choice (/24)
Raw Score (/50)
TOTAL MARK (%)
Chemistry 101, Mid-Term Test 2
Fall 2010
Version A page 2 of 8
PART I – Multiple Choice: Select the BEST response for each question below. [Total marks = 24]
1. This is exam Version A. Mark “A” as the answer to Question 1 on the optical sense form.
2. Choose the CORRECT statement about the SO2 molecule.
A. The sulfur atom has an unshared (i.e. non-bonding) electron pair.
B. The S-O bonds are ionic in character.
C. The two S-O bonds have different lengths since one is a single bond and the other a double bond.
D. The molecule has a linear structure.
E. The oxygen atoms have no unshared (i.e. non-bonding) electron pairs.
Questions 3-6 refer to the molecules in this box. Answer these questions referring to these answers.
3.
i. SiH4
ii. BF3
iii. OF2
vi. BrF5
vii. XeF4
viii. OCCl2
D. ii & iv
E. vii only
B. ii only
C. iii only
D. ii & viii
E. iv only
B. vii
C. vi
D. ii
E. iv
Indicate all of the molecules in the box above that have at least one bond angle of about 90°?
A. ii only
7.
C. v only
Indicate all of the molecules in the box above that have a square pyramidal molecular shape?
A. i
6.
B. ii & v
Indicate all of the molecules in the box above that have both an electron group geometry (electron
domain geometry) and a molecular geometry that are trigonal planar (triangular planar)?
A. i & vii
5.
v. NO
Indicate all of the molecules in the box above that have a molecular structure best described as involving
an ―incomplete octet‖?
A. i only
4.
iv. NF3
B. iv only
C. vii only
D. vi & vii
E. ii & iv
Consider the following three molecules. Which molecule(s) is(are) polar? (That is, which molecules
have a non-zero net molecular dipole moment?)
BF3 NF3 CF4
A. BF3 only
B. NF3 only
C. CF4 only
D. both BF3 and NF3
E. none of them
Chemistry 101, Mid-Term Test 2
8.
Fall 2010
Version A page 3 of 8
The electron-domain geometry of a sulfur-centered compound is trigonal bipyramidal. The
hybridization of the central sulfur atom is __________.
B. sp2
A. sp
C. sp3
D. sp3d
E. sp3d2
9. Which of the following statements is INCORRECT?
A. A linear arrangement of electron domains can often be rationalized using sp hybrid orbitals..
B. When assigning the orbital hybridization, only bonding electrons are considered, since lone pairs do
not participate in hybridization.
C. An octahedral molecular geometry is rationalized by the hybridization of six atomic orbitals.
D. Double bonds take up more space than single bonds.
E. In NF3 there are bond angles that are slightly less than the regular tetrahedral angle of 109.5°.
10. The hybridizations of bromine in BrF5 and of iodine in ICl3 are __________ and __________;
respectively.
A. sp3, sp3d
B. sp3, sp3d2
C. sp3d, sp3
D. sp3d2, sp3d
E. sp3d, sp3d2
11. A Lewis structure of the aspirin molecule is shown below. The approximate bond angles (in degrees, o)
for the bonds labeled ―a‖, ―b‖ and ―c‖ in the drawing are, respectively?
O
H
H
b
C
O
a
H
H
c
O
C
O
A.
B.
C.
D.
E.
H
120, 120, 120
120, 109, 180
180, 120, 120
120, 120, 109
109, 109, 120
CH3
Chemistry 101, Mid-Term Test 2
Fall 2010
Version A page 4 of 8
12. Which of the following statements is CORRECT for metallic bonding?
i. Bonding electrons in metals are free to move throughout the metal.
ii. The electron-sea model explains the trends in melting points for the transition metals.
iii. The energy gap between bonding and antibonding molecular orbitals is largest for an
insulator.
iv. There are no antibonding molecular orbitals for a semiconductor.
A)
B)
C)
D)
E)
i
i and ii
ii and iii
i and iii
ii and iv
13. Consider the following molecular orbital energy diagram, which applies to diatomic species
that use only 1s orbitals.
molecular
orbitals
atomic
orbital
atomic
orbital
The MO diagram shown above can represent the electron energy levels for which molecule(s) or
ion(s)?
i.
ii.
iii.
iv.
v.
H2–
He2+
He2–
HHe
H2
A. i only
B. ii only
C. iv only
D. i, ii and iv E. ii, iii and iv
Chemistry 101, Mid-Term Test 2
Fall 2010
Version A page 5 of 8
PART II – Written Answer Questions. [Total Marks = 26]
Write your answers directly on this test paper. Show all your work. This helps us to award part marks
instead of zero where appropriate. Hand in the entire test paper at the end of the test period.
[6 marks]
1. [3] a) Use average bond energies from the datasheet to estimate the enthalpy change for the following
reaction. Note that the carbon atoms are bonded to one another in both C2H4 and C2H6.
H2(g) + C2H4(g)  C2H6(g)
break H-H bond and C=C bond, make C-C bond and two C-H bonds
DeltaHrxn = 432 kJ/mol + 614 kJ/mol – [2(413 kJ/mol) + 348 kJ/mol] = -128 kJ/mol
Datasheet gives:
[1]
H-H* 432, C-H 413, C−C 348, C=C 614
b) Which of the compounds in (a) has the longer carbon-carbon bond? Explain.
C2H6 has a longer C-C bond because it is a single bond.
[2]
c) How many sigma (σ) and pi (π) bonds are present in the entire molecule C2H4?:
____5_____ sigma bonds
____1_____ pi bonds
Chemistry 101, Mid-Term Test 2
Fall 2010
Version A page 6 of 8
[6 marks]
2. [2] a) The ion I3- is known but the ion F3- is not. Draw the Lewis structures for the two ions and explain
why F3- does not form.
I
I
F
F
I
F
Not possible for fluorine. Breaks octet rule
& some comment about F not being able to expand its valence shell since it is in row 2 and doesn’t have
empty d orbitals
b) Use VSEPR theory to predict the shape of the I3- ion and state which set of hybrid orbitals would be
used for the central iodine atom. Draw a three-dimensional representation of the hybrid orbitals
and the molecular shape. What is the value of the I-I-I bond angle?
[4]
VSEPR shape for electron domains is trigonal bipyramidal, sp3d (or dsp3) hybrid, molecular shape is linear,
I
I
I
180 bond angle.
o
4 marks
3. For each of the three oxides CO2, SO2 and XeO3, make a three-dimensional sketch of the molecule
and clearly indicate which one(s) is(are) polar. (That is, state which ones have a non-zero net
molecular dipole moment.) Note: Your sketch does not have to be a complete Lewis structure, but it
does need to correctly represent the shape of the molecule. For each structure, draw an arrow to
indicate the net molecular dipole moment.
Chemistry 101, Mid-Term Test 2
Fall 2010
Version A page 7 of 8
[5 marks]
4. In nitryl chloride, O2NCl (overall neutral molecule), the chlorine atom and the two oxygen atoms are
bonded to the central nitrogen atom.
a) Draw all resonance structure(s) for O2NCl assuming that the octet rule holds for all atoms.
Show each bonding pair of electrons as a line (—) and show non-bonding valence electrons as
dots (:).
b) Show any non-zero formal charges on each atom in all the structures that you draw.
Answer a and b
O
Cl
N
O
Cl
O
N
O
Cl
O
N
O
c) What is the bond order of a nitrogen-oxygen bond in this molecule?
Answer c
Nitrogen-oxygen bond order is 1.5
Chemistry 101, Mid-Term Test 2
Fall 2010
Version A page 8 of 8
[5 marks]
5. Consider the nitroxyl molecule HNO
Draw (i.e. sketch) all of the orbitals used to form ALL sigma (σ) bonds and pi (π) bonds in this
molecule, as well as any orbitals containing lone pairs of electrons. Show clearly how the orbitals
overlap to form bonds. Label all the orbitals and bonds.
(For clarity of presentation you may show the sigma (σ) and pi (π) bonding separately using
two diagrams.)
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