FALL 2009 – VERSION 01 – GREY
PART A: Multiple Choice Section [2 points each]
1. Complete this sentence: Atoms emit visible and ultraviolet light as
A) electrons jump from lower energy levels to higher levels. O
B) the atoms condense from a gas to a liquid. O
C) electrons jump from higher energy levels to lower levels. O
D) they are heated and the solid melts to form a liquid. O
E) the electrons move about the atom within an orbit. O
2. Which one of the following sets of quantum numbers is not possible?
A) n = 4, l = 3, ml = -2, ms = +½ P
B) n = 3, l = 0, ml = 1, ms = -½ O
C) n = 3, l = 0, ml = 0, ms = +½ P
D) n = 2, l = 1, ml = 1, ms = +½ P
E) n = 2, l = 0, ml = 0, ms = -½ P
3. What is the ground-state electron configuration of Ti2+?
A) [Ar] 3d2 4s2
Z for Ti = 22
From periodic table: [Ar]4s 2 3d 2
B) [Ar] 3d1 4s1
Ground state for Fe: [Ar]3d 2 4s 2
C) [Ar] 4s2
To create the cation, remove the two outermost electrons,
2
D) [Ar] 3d
i.e., ground state Ti2+: [Ar]3d 2
E) [Ar] 3d4 4s2
4. How many unpaired electrons are in an atom of vanadium (V; Z=23)?
A) 1
Z for V = 23 From periodic table: [Ar]4s 2 3d 3
B) 2
Ground state for V: [Ar]3d 3 4s 2
C) 3
Energy Diagram:
4s
_↑↓_
D) 5
3d: _↑_ _↑__ _↑_ ___ ___
E) none of these
5. Which two electron configurations represent elements that would have similar chemical properties?
1. 1s2 2s2 2p4
2. 1s2 2s2 2p5
3. [Ar]4s2 3d10 4p3
4. [Ar]4s2 3d10 4p4
A) 1, 2
B) 1, 3
Elements with similar chemical properties are ones from the same group (i.e., np4 ).
C) 1, 4
D) 2, 3
E) 2, 4
1
6. For the elements Mg, Al, P, and S, the first ionization energy increases in the order
A) Mg < S < Al < P
B) Al < Mg < S < P
C) Mg < Al < P < S
D) Al < Mg < P < S
E) Mg < Al < S < P
7. Arrange the following ions in order of increasing ionic radius, K+, P3−, S2−, Cl−.
A) K+ < Cl– < S2− < P3−
B) K+ < P3− < S2− < Cl–
K+: V = 18 e– ; P3– : V = 18 e– ; S2– : V = 18 e– ; Cl– : V = 18 e–
C) P3− < S2− < Cl– < K+
All the ions are isoelectronic!
–
2−
3−
+
+
D) Cl < S < P < K
∴ K < Cl– < S2– < P3– (cations < anions)
E) Cl– < S2− < K+ < P3−
8. For the following Lewis structure, the formal charges on C, N, and S, respectively, are
A)
B)
C)
D)
E)
0, +1, –2
–1, 0, 0
–1, +1, –1
–2, +1, 0
–2, +2, –1
Formal Charge: Number of valence electrons (based on periodic table)
– electrons drawn around atom (i.e., lone pairs + ½ bonded electrons)
C: 4 – 6 = –2;
N: 5 – 4 = +1;
S: 6 – 6 = –1
9. According to the VSEPR theory, the F –As –F bond angles in the AsF4 – ion are predicted
to be
..
:F:
A) 109.5°
..
B) 90° and 120°
V = 5 + (4 × 7) + 1 = 34 electrons
:F:
C) 180°
Framework: AB4 E
:
D) < 109.5°
Shape: Seesaw
:F:
..
E) < 90° & < 120°
Angles: < 90°, < 120° & < 180°
:F:
..
:F:
10. What set of hybrid orbitals is used by Cl in ClF 3 ?
A) sp
B) sp2
C) sp3
D) sp3 d
V = (4 × 7) = 28 electrons
E) sp3 d2
Framework: AB3 E2
Steric Number = 5
:
..
F:
..
:
:F:
11. How many sigma and how many pi bonds are present in H2 NCHO?
A) 4 sigma and 1pi
V = 1+1+5+4+1+6 = 18 electrons
..
..
B) 4 sigma and 2 pi
H – N – C = O:
H H
C) 5 sigma and 0 pi
D) 5 sigma and 1pi
Each single bond consists of one σ bond; double bond = σ + π; triple = σ + 2π.
E) 5 sigma and 2 pi
2
12. How many oxygen atoms are present in 1.0 mole of copper nitrate trihydrate, Cu(NO3 )2 ⋅3H2 O?
A) 6.0×1023
1.0 mol × 6.02 ×1023 = 6.02 ×1023 Cu (NO3 )2 ⋅3H2 O
B) 1.8×1024
Cu(NO3 )2 ⋅3H2 O contains 9 oxygen atoms.
24
C) 3.0×10
∴ 6.02 ×1023 Cu(NO3 )2 ⋅3H2 O × 9 H atoms = 5.42 ×1024 H atoms
D) 3.6×1024
1 mol Cu(NO3 )2 ⋅3H2 O
24
E) 5.4×10
13. How many moles of calcium are required to produce 5.50 g of calcium phosphate?
The molar mass of calcium phosphate is 310.18 g/mol
A) 6.28×10–2 mol
3Ca (s) → Ca 3(PO4 )2(s)
–2
B) 5.32×10 mol
5.50 g Ca 3(PO4 )2 × 1 mol/310.18 g = 1.773 × 10-2 mol Ca 3 (PO4 )2
–2
C) 4.07×10 mol
1.773 × 10-2 mol Ca 3(PO4 )2 × 3 mol Ca/1 mol Ca3 (PO4 )2 = 0.0532 mol Ca
–2
D) 3.55×10 mol
E) 1.77×10–2 mol
14. If 0.010 moles phosphoric acid, H3 PO4 is completely neutralised with 0.250 M KOH,
how many mL of potassium hydroxide is required?
A) 8.3 mL
Rxn: H3 PO4 (aq) + 3KOH(aq) → K3 PO4 (aq) + 3H2 O(l)
B) 13 mL
0.010 mol
0.250 mol/L
C) 25 mL
D) 40 mL
# mol KOH = 0.010 mol H3 PO4 × 3 mol KOH/ 1 mol H3 PO4 = 0.030 moles KOH
E) 120 mL
0.030 moles KOH × 1 L/0.250 moles = 0.120 L × 1000 mL/1L = 120 mL
15. Air bags in automobiles are driven by the following reactions:
2NaN 3 (s) → 2Na(l) + 3N2 (g)
10Na(l) + 2KNO3 (s) → K2 O(s) + 5Na2O(s) + N2 (g)
2K2O(s) + 2Na2 O(s) + 2SiO2 (s) → K4 SiO 4 (s) + Na4 SiO 4 (s)
If 2.16×10–4 moles of K4 SiO 4 (s) are produced, how many moles of sodium azide were initially present?
A) 8.64 × 10–3 mol
2.16 × 10–4 moles of K4 SiO4 × 2 mol K2 O × 10 mol Na × 2 mole NaN3
–3
B) 4.32 × 10 mol
1 mol K4 SiO4
1 mol K2 O
2 mol Na
–3
–3
C) 1.08 × 10 mol
= 4.32 × 10 moles of NaN3
D) 4.32 × 10–5 mol
Note: Seeing Na 2 O also appears in equation 2, one can compare K4 SiO4 to Na 2 O, and
E) 1.08 × 10–5 mol
Na 2 O to Na, to arrive at an answer of 8.64 × 10–4 moles of NaN3 (an answer not included
in the selection). This question has therefore been changed to a bonus question. The
multiple choice section is now graded out of 20 questions, instead of 21.
16. Consider the reaction: 2Sb(s) + 3Cl2 (g) ? 2SbCl3 (s)
What are the limiting reactant and the number of moles of the other reactant remaining when 5.0 moles
of antimony and 5.0 moles of chlorine gas react?
A) Cl2 and 1.7 mol Sb remaining
Sb: 5.0 mol × 1 reaction/2 mol = 2.5 reactions
∴ Excess
B) Sb and 2.5 mol Cl2 remaining
Cl2 : 5.0 × 1 reaction/3 mol = 1.667 reactions
∴ Limiting Reagent
C) Sb and 1.7 mol Cl2 remaining
If 1.667 reactions occur, 2 × 1.667 = 3.33 moles of Sb reacts
D) Cl2 and 3.3 mol Sb remaining
5.0 mol Sb – 3.33 moles reacted = 1.667 moles of Sb remain
E) Sb and Cl2 react with no excess
3
17. Sulphur trioxide, SO3 , is made by oxidation of SO2 , i.e., 2SO2 (g) + O2 (g) ? 2SO3 (g)
If a 16-g sample of SO2 gives 16 g of SO3 , the percentage yield of SO3 is
A) 60%
MM (SO2 ) = 32.06 + (2 × 16.00) = 64.06 g/mol; MM (SO3 ) = 80.06 g/mol
B) 75%
16 g SO2 × 1 mol/64.06 g = 0.250 moles SO2
C) 80%
2 moles of SO2 produces 2 moles of SO3 ; ratio 1:1; ∴0.250 moles SO3
D) 100%
0.250 moles SO3 × 80.06 g/1 mol = 20.0 g SO3
E) 125%
% yield = actual yield/ theoretical yield × 100% = 16 g/20.0 g × 100% = 80%
18. What is the concentration of bromide ion in a solution prepared by adding 40.0 mL of water
to 125 mL of 0.96 M aqueous iron(III) bromide?
A) 0.73 M
# moles FeBr3 = 0.125 L × 0.96 mol/L = 0.12 moles
B) 2.2 M
# moles Br– = 0.12 moles FeBr3 × 3 mol Br– /1 mol FeBr3 = 0.36 mol
C) 3.0 M
Volume Total = 40.0 mL + 125 mL = 165 mL = 0.165 L
D) 9.0 M
[Br– ] = 0.36 mol/ 0.165 L = 2.2 M
E) 24 M
19. Consider the reaction: MnO2 (s) + 4HCl(aq) ? MnCl2 (aq) + 2H2 O(l) + Cl2 (g)
How many litres of chlorine gas at 25.0°C and 652 Torr can be produced if 5.00 moles
of HCl is reacted with excess MnO 2 ?
A) 1.25 L
5.00 moles of HCl × 1 mole Cl2 = 1.25 moles of Cl2
B) 24.2 L
4 mol HCl
C) 35.7 L
V = nRT/P = 1.25 moles of Cl2 × 0.0821 atm L/K mol × (25.0 + 273.15K)
D) 88.6 L
(652 Torr × 1 atm/760 Torr)
E) 143 L
V = 35.7 L
20. Given:
P4 (s) + 6Cl2 (g) ? 4PCl3 (l); K
For the reaction, 2PCl3 (l) ? 3Cl2 (g) + ½P4 (s), the equilibrium constant is
A) K−½
B) –K½
C) K2
D) K½
E) K−2
Flip the equation & invert the K:
4PCl3 (l) ? 6Cl2 (g) + P4 (s) ; K-1
Multiply the equation by 2 & square the K:
6Cl2 (g) + P4 (s) ? 4PCl3 (l); K-2
21. At 24°C, the equilibrium constant, Kc, for the reaction: 2NOBr(g) ? 2NO(g) + Br2 (g) is 3.07×10–4 . If
1.5 moles of NOBr, NO & Br2 are mixed in a 1.00 L container at 24°C, which of the following is true?
A) Qc = Kc; the system is at equilibrium and no net change occurs.
B) Qc < Kc; NO(g) and Br2 (g) forms until equilibrium is reached.
Q = [NO]2 [Br2 ] = (1.5)2 (1.5) = 1.5
C) Qc < Kc; NOBr(g) forms until equilibrium is reached.
[NOBr]2
(1.5)2
D) Qc > Kc; NO(g) and Br2 (g) forms until equilibrium is reached.
Q > K, therefore too many products!
E) Qc > Kc; NOBr(g) forms until equilibrium is reached.
Reaction will make more NOBr until K reacted.
4
Part B. [Total of 18 points]
1. Write the balanced net ionic equation for any reaction that occurs when the following aqueous solutions
are mixed. Clearly identify your final answer, including all states. If no reaction occurs, write no reaction.
(i) silver nitrate & sodium chloride:
AgNO3 (aq) + NaCl(aq) → AgCl(s) + NaNO3 (aq)
NIE: Ag (aq) + Cl (aq) → AgCl (s)
+
–
2 pts per section.
–1 for each mistake
(states, charges, balance)
(ii) calcium hydroxide & iron (III) perchlorate:
3Ca(OH)2 (aq) + 2FeClO 4 (aq) → 2Fe(OH)3 (s) + 3Ca(ClO 4 )2 (aq)
NIE: Fe 3+(aq) + 3OH– (aq) → Fe(OH)3 (s)
2. Ozone, O3 , in a sample of air can be determined by reacting the sample with KI(aq) :
O3 (s) + 3I– (aq) + 2H+(aq) → O2 (g) + I3 – (aq) + H2 O(l)
/4
The I3 – formed is then titrated with sodium thiosulphate, Na2 S2 O3 , according to the net ionic equation:
2S2 O32– (aq) + I3– (aq) → S4O6 2– (aq) + 3I– (aq)
An 8.32-g sample of air is reacted with excess KI and acid. If 18.40 mL of 0.0100 M Na2 S2O3 is required to
titrate the I3 – produced, what is the mass percentage of O 3 in the sample? (MM (O 3 ) = 48.00 g/mol)
# moles of S2 O3 2– = 0.01840 L × 0.0100 mol/L = 1.84 × 10–4 moles
(1 pt)
# moles of I3– = 1.84 × 10–4 moles × 1 mol I3 – /2 mol S2 O3 2– = 9.20 × 10–5 moles
(1 pt)
# moles of O3 = 9.20 × 10–5 moles × 1 mol O3 /1 mol I3 – = 9.20 × 10–5 moles
(1 pt)
Mass of O3 = 9.20 × 10–5 moles × 48.00 g/1 mol = 4.416 × 10–3 g
(1 pt)
Mass percent of O3 = 4.416 × 10–3 g/8.32 g × 100% = 0.0531%
/5
(1 pt)
(Note: Break down in grading key is only when students go wrong; If student has 0.0531% - 5/5; no penalty for wrong sig figs)
3. The bonds of oxygen molecules are broken by sunlight. The minimum energy required to break the
oxygen-oxygen bond is 495 kJ/mol. What is the wavelength (in nanometres) of sunlight that can cause
this bond breakage?
E = ε × NA; ∴ε = E (Joules)/N A = 495 kJ/mol × 1000 J/1 kJ × 1 mol/6.022 × 1023 photons
ε = 8.220 × 10–19 J/photon
(1 pt - convert moles to photons)
ε = hc/λ; ∴ λ= hc/ε = (6.626×10–34 J s × 2.998×108 m/s)/ 8.220 ×10–19 J/photon
λ = 2.417 × 10–7 m
(1 pt - convert energy to λ in m)
–7
9
λ = 2.417 × 10 m × 1 × 10 nm/1 m = 242 nm (1 pt convert to nm)
BONUS: What type of electromagnetic radiation is this? _____ ultra-violet ______
/3
Bonus
/2
4. Complete the following table. For dipole moment (i.e., polar) circle yes or no.
Molecule
ClF4
Lewis Structure ?
..
:F:
–
V = (5×7) +1
V = 36 e –
AsF3
V = 5 + (3×7)
V = 26 e –
(1 pt)
. . ..
..
:F
–
Cl
–
F:
..
. . ..
:F:
..
..
:F
..
(1 pt)
..
..
– As – .F:
.
:F:
..
Note electrons must be shown in diagram!
VSEPR Shape ?
AB4 E2
Shape:
Square planar
(1 pt)
..
..
F:
..
:F:
Yes
:F:
..
. . . . :F:
AB3 E
Shape: Trigonal . .
pyramidal :F. .
(1 pt)
Dipole Moment?
..
:F:
Name must be given for point. . .
/6
No
(1 pt)
..
Total:
Yes
:F:
..
No
(1 pt)
/18
Note: Grades were assigned according to what appeared in each box and how it related to each section.
(i.e., if an incorrect ClF4− Lewis structure had only one lone pair & the shape was identified as seesaw, then the shape point was awarded.)
5