Name ____________________________
School _______________________
COLLEGE IN HIGH SCHOOL
2010 - 2011
Exam III
Tuesday, March 15 2011
INSTRUCTIONS:
DO NOT TEAR ANY PAGES OFF YOUR EXAM
PRINT YOUR NAME AND SCHOOL AT THE TOP OF THIS PAGE.
The exam contains 14 multiple-choice questions (5.5 points each) to be answered on the machinescored sheet, using a #2 pencil. Be sure to print your name on this sheet (last name, space, first
name); also write and code in your special 7-digit I.D. code.
Do all your working on the exam sheets. The last 23 points will be given for your written answers
for Question 15 and your written answers to the 5 questions marked ***.
TURN IN THE WHOLE EXAM and the MACHINE-SCORED SHEET.
USEFUL INFORMATION
c = 
c = 3.00 x 108 m/s
E = h
h = 6.63 x 10-34 J.s
RH
n2
1 Hz = 1 s-1
En = -
OR
En = RH -
RH
n2
RH = 2.179 x 10-18 J
A periodic chart is on the reverse of this sheet
GOOD LUCK!
1. Light with a wavelength of 458nm and a frequency of 6.55 x 1014 Hz was passed through a
“doubling crystal”, which converts the light into photons with twice the photon energy of the
original light. Which one is definitely correct for the action of the doubling crystal?
A) It causes doubling of the wavelength of the light
B) It does not affect the wavelength of the light
C) The speed, c, of the light wave is doubled
D) It causes doubling of the frequency of the light
E) It has no effect on the wavelength and frequency of the light but it doubles the value of
Planck’s constant.
2. Consider the n = 5 to n = 3, n = 5 to n = 1 and n = 3 to n = 1 transitions of the electron in the
hydrogen atom. Which is the correct set of frequencies in Hz of the light emitted as a result of
these transitions?
*** Give your reasoning or show your work
A)
B)
C)
D)
E)
F)
ni = 5 to nf = 3
2.337 x 1014Hz
2.337 x 1014Hz
2.921 x 1015Hz
2.921 x 1015Hz
3.155 x 1015Hz
3.155 x 1015Hz
ni = 5 to nf = 1
2.921 x 1015Hz
3.155 x 1015Hz
2.337 x 1014Hz
3.155 x 1015Hz
2.337 x 1014Hz
2.921 x 1015Hz
ni = 3 to nf = 1
3.155 x 1015Hz
2.921 x 1015Hz
3.155 x 1015Hz
2.337 x 1014Hz
2.921 x 1015Hz
2.337 x 1014Hz
3. Which of the following statements concerning quantum theory and the periodic table is/are
completely correct?
1. In the periodic table, the 2s subshell of orbitals is filled before the 2p subshell. That is
consistent with the principle from quantum theory that subshells are filled in the order of
increasing energy.
2. The d block of the periodic table consists of 10 groups (e.g. Sc through Zn). This is
consistent with the finding from quantum theory that the d subshell contains 10 orbitals.
3. The first elements to have electrons in a p subshell are B through Ne. We label this
subshell 2p (rather than 1p or 3p) in accordance with the rule that the ℓ quantum number
can have integral values from 0 to n-1 but no higher values.
A) 1 only
F) 2 and 3
B) 2 only
C) 3 only
D) 1 and 2
G) All of them
H) None of them
E) 1 and 3
4. Which one of the following is NOT correct for the 4p subshell of orbitals?
A) The value of the n quantum number is n = 4
B) The value of the ℓ quantum number is ℓ = 1
C) A possible value of the ml quantum number is ml = +3
D) A possible value of the ms quantum number for an electron in this subshell is ms= ½
E) The number of orbitals in this subshell is 3
5. Which is the correct valence electron configuration for the given atom? The correct answer
lists all the valence electrons for that atom and no other electrons. Valence electrons are those
which can participate in the chemistry of the element.
A) S: 2s2 2p4
B) Ti: 4s2 3d2
C) Br: 3d10 4p5
D) Sr: no valence electrons
E) Pb: 6s2 6f2
*** Give the correct valence electron configuration for one of the incorrect responses
6. Which one of the following statements concerning the ground state electronic configurations
of Co and S is NOT correct?
A) The Co atom has 3 unpaired electrons
B) The Co2+ ion has 5 unpaired electrons
C) The S atom has 2 unpaired electrons
D) The S2- ion has no unpaired electrons
E) A species with one or more unpaired electrons is termed paramagnetic
*** For either the Co or the S atom, draw an orbital diagram for the electrons in the two
highest energy occupied subshells for the atom.
7. For which pair of atoms or ions does the atomic radius increase the most in going from the
first to the second species? Example: K and Rb have atomic radii of respectively 201pm and
217pm; the size increases by 16pm in going from K to Rb.
A) O → F
B) F → Ne
C) Ne → Na
D) Na → Mg
E) Mg → Mg2+
8. Aluminum and oxygen have the following ionization energies, I.E., (for losing an electron)
and electron affinities, E.A., (for gaining an electron) – but not necessarily in the given order:
+1300 kJ/mol -140 kJ/mol -40 kJ/mol
+550 kJ/mol
Which is the correct set of values, in kJ/mol?
*** Give your reasoning
A)
B)
C)
D)
E)
F)
I.E. (Al)
+1300
-40
+550
-140
+1300
+550
I.E. (O)
+550
+550
-140
+1300
-40
+1300
E.A. (Al)
-140
+1300
+1300
-40
-140
-40
E.A. (O)
-40
-140
-40
+550
+550
-140
9. Which one statement concerning bonding in stable molecules is NOT correct?
A) If a N atom in a Lewis structure forms more than 3 bonds, it has a negative formal charge
B) In Lewis structures of stable species, each oxygen atom must be surrounded by 8 electrons, in
bonds and lone pairs
C) In Lewis structures, the P atom can be surrounded by up to 10 electrons, in bonds and lone
pairs
D) In the vast majority of organic compounds, each C atom forms 4 bonds and has 0 lone pairs
E) In the Lewis structure of an ion, at least one atom must have a non-zero formal charge
10. *** Draw the Lewis structures of NF3 and IF3. How many lone pairs of electrons are there
on the central atoms, N and I?
A)
B)
C)
D)
E)
NF3
0
0
2
1
1
IF3
1
2
1
0
2
11. For which of the following ions has the charge been given correctly? Note that the rest of
each formula is correct.
1. Phosphate ion:
PO42. Hydrogen carbonate (or bicarbonate) ion: HCO33. Dihydrogen phosphate ion: H2PO42A) 1 only
F) 2 and 3
B) 2 only
C) 3 only
D) 1 and 2
G) All of them
H) None of them
E) 1 and 3
12. Which is the correct set of molecular geometries of the given three species?
A)
B)
C)
D)
E)
CO2
Bent
Linear
Tetrahedral
Linear
Bent
NO2- ion
Trigonal planar
Linear
Trigonal pyramidal
Bent
Trigonal planar
N2H4 (geometry around one N atom)
Trigonal planar
Tetrahedral
Trigonal planar
Trigonal pyramidal
Bent
13. Consider the molecules CF4 and CH2F2. Which one statement concerning polarity is NOT
correct?
A) The molecular geometry of each molecule is tetrahedral
B) The C-F bonds in each molecule are polar
C) The entire CF4 molecule is polar but the CH2F2 molecule is nonpolar
D) A molecule is polar if one end of the molecule has a partial δ+ charge and the other end has a
partial δ- charge.
E) A polar molecule orients (aligns) itself in the presence of a uniform electric field
14. Which statement concerning hybridization is NOT correct?
A) If a central atom in a structure has an electron arrangement which is “tetrahedral” and a
molecular geometry which is “trigonal pyramidal”, its hybridization is sp3.
B) In ethene, C2H4, each carbon atom forms one double bond and two single bonds. Each C atom
in ethene has a hybridization of sp2.
C) The central atom in any molecule with a “bent” molecular geometry has a hybridization of
sp3.
D) If a central atom in a structure has an electron arrangement and a molecular geometry which
are both “trigonal planar”, its hybridization is sp2.
E) sp3d hybridization is shown by an atom, only if it is exceeding an octet of electrons in that
structure.
15. (13 points) Note: there are many connections between different parts of this question.
a) (i) Draw Lewis structures for CH3OH (methanol) and H2CO (formaldehyde)
(ii) Show that the correct Lewis structure for carbon monoxide is :C≡O:
b) The carbon-oxygen bond lengths in the compounds CH3OH, H2CO and CO are, not
necessarily in the correct order, 0.1128, 0.142, 0.1208 nm. Assign each bond length to the correct
compound. Give clear reasoning for your choice.
c) The bond energies in these and similar species are also of interest. It is known that the bond
energy in the CO molecule is 1072 kJ/mol, while in the O2 molecule, it is 494 kJ/mol. Explain
why the bond energy in CO is so much greater than that in O2.
15. d) Continuing the analysis of bond energies. Consider the following thermochemical
equation:
2 CO2(g) → 2 CO(g) + O2(g), ΔH1 = +558 kJ for this reaction
Using this ΔH value and the bond energy information given in part (c), determine the average
carbon-oxygen bond energy in CO2. Hint: remember to consider the bonds which are broken and
those which are formed in the given reaction. Be sure to show all your work!
e) The breaking of the bonds in CO2 probably occurs in two steps: firstly we break one bond:
CO2(g) → CO(g) + O(g), ΔH2
followed by cleavage of the CO bond. Using information from parts (c) and (d), determine the
value of ΔH2, i.e. the energy required to break the first bond in CO2. Again, please show clear
work.
CHS 2010-11
1
D
ANSWER KEY
2
B
3
E
4
C
5
B
6
B
7
C
EXAM 3
8
F
9
A
10
E
11
B
12
D
13
C
14
C
Question #15
a)
(ii) The structure for CO has the correct number of valence electrons, 10, and has 8 electrons
around each atom, as required by the octet rule.
b) The three carbon-oxygen bonds are respectively single, double and triple bonds. Bond length
decreases with increasing bond order, so the bond lengths are respectively 0.142, 0.1208, 0.1128
nm.
c) Bond energy generally increases with bond order. The CO bond is a triple bond, while that in
O2 is a double bond.
d) ΔH of reaction = sum of bond energies of bonds broken MINUS sum of bond energies of
bonds formed.
Bonds broken = 4 x average C=O bond energy in CO2 = 4 BE(C=O)
Bonds formed = 2 x bond energy in CO + bond energy in O2
Thus: +558 = 4 BE(C=O) – (2 x 1072 + 494)
BE(C=O) = (558 + 2638)kJ/4mol = 799 kJ/mol
e) Different approaches possible
(i) Average C=O bond energy = (1st BE + 2nd BE)/2 = (1st BE + 1072)/2
Thus 799kJ = 1st BE/2 + 536kJ
Thus 1st BE = 2 x (799 – 536) kJ = 526kJ/mol
(ii) Analyze the equation:
CO2(g) → CO(g) + O(g)
ΔH2 = 2 x BE(C=O) - BE(CO)
= 2 x 799 – 1072 = 526 kJ