Chem 1000A Final Examination - Solutions
December 9th, 2003: 9:00 to 12.00 am
Your name _______________
Instructor: Dr. M. Gerken
Student ID _______________
Time: 3h
No. of pages: 3 + 2
Total achievable marks: 114
Report all your answers using the correct number of significant figures. Show all units and their
conversions throughout your calculations. Use SI units for your calculations.
WRITE ALL YOUR ANSWERS INTO THE EXAMINATION BOOKLET!
Question 1 (5 Marks)
Complete the following table.
223 +
Fr
Symbol
Number of electrons
__86_______
Number of neutrons
__136______
Number of protons
__87_______
Overall charge
1+
_219At-_____
86
134
_85________
1-
Question 2 (4 Marks)
What is the maximum number of electrons that can be associated with the following combinations of
quantum numbers? Zero is a possible answer.
(a) n = 3, l =1, ml = -1
This is one of the 3p orbitals: two electrons
(b) n = 2, l = 2, ml = -1, ms = ½
no electrons, since l cannot be the same as n.
(c) n = 2
2s, 2p orbitals: eight electrons
(d) n = 4, l = 3
These are the 4f orbitals: fourteen electrons.
Question 3 (4 Marks)
Draw a 2pz, a 1s, and a 3dz2 orbital.
z
z
z
x
x
x
Question 4 (7 Marks)
Borazine (B3N3H6) is sometimes called “inorganic benzene”. Its Lewis structure (one resonance structure) is
drawn below. (a) Indicate formal charges and partial charges in the Lewis structure. (b) Will a chloride ion
attack the boron or the nitrogen atoms in borazine? Explain your answer.
1
(a)
δ−
H
H
δ+
_
H δ− Bδ+ δ− H δ+
H + B + H
N
N
N
N
Bδ+ Bδ+
_
_
B + B
N
Hδ−
H
δ−
H
N
H
δ−
H
H
δ+
(b) Chloride, Cl-, would attack on an atom with a positive partial charge: boron.
Question 5 (8.5 Marks)
Balance the following redox reaction in a basic aqueous solution and show that the final equation has the
correct electron, material, and charge balance. Give the half-reactions and show as many steps as possible.
What is the name for this kind of redox reaction?
NO2 → NO3-(aq) + NO2-(aq)
N+IVO-II2(g) →
N+VO-II3-(aq) + N+IIIO-II2-(aq)
oxidation half-reaction (unbalanced): NO2 → NO3reduction half-reaction (unbalanced): NO2 → NO2principle element balance:
oxidation half-reaction (unbalanced): NO2 → NO3reduction half-reaction (unbalanced): NO2 → NO2oxygen balance:
oxidation half-reaction (unbalanced): H2O + NO2 → NO3reduction half-reaction (unbalanced): NO2 → NO2hydrogen balance (basic solution):
oxidation half-reaction (unbalanced): H2O + 2OH- + NO2 → NO3- + 2H2O
reduction half-reaction (unbalanced): NO2 → NO2Inserting electrons:
oxidation half-reaction (balanced): H2O + 2OH- + NO2 → NO3- + 2 H2O+ ereduction half-reaction (balanced): NO2 + e- → NO2electron balanced
combination:
H2O + 2OH- + NO2 + NO2 +e-→ NO3- + 2H2O + e- + NO2simplification: 2OH- + 2NO2 → NO3- + NO2- + H2O
material balance:
2H, 6O, 2N|2H, 2N, 6O
charge balance:
(2-) = 2|(-)+(-)=2correct electron, material, and charge balance!
This reaction is a disproportionation reaction.
2
Question 6 (11 Marks)
Hydrogen fluoride is a gas which readily dissolves in water. Aqueous HF is usually stored in plastic bottles
and cannot be handled in glass since it reacts with glass (SiO2), producing the gas, SiF4, and H2O.
SiO2(s) + HF(aq) → SiF4(g) + H2O(l)
Inside a polyethylene beaker, 5.00 g of quartz wool (pure SiO2) is combined with 100. mL of 10.0 M
HF(aq).
(a) Balance the above equation.
SiO2(s) + 4HF(aq) → SiF4(g) + 2H2O(l)
(b) Which reactant is the limiting reagent?
M(SiO2) = 60.0843 g/mol
n(SiO2) = 5.00g/ (60.0843 g/mol) = 0.0832 mol
n(HF) = 0.100 L × 10.0 mol/L = 1.00 mol
mole ratio: SiO2 : HF = 0.0832:1.00 = 1.00:12.0
SiO2 is the limiting reagent, HF is present in excess.
(c) What is the mass and the volume (in m3 and L) of SiF4(g) produced at 25.0 °C and 685 Torr?
M(SiF4) = 104.0791 g/mol
n(SiF4) = 0.0832 mol SiO2 ×(1mol SiF4/1mol SiO2) = 0.0832 mol
m(SiF4) = 0.0832 mol×104.0791 g/mol = 8.66 g
p = 685 Torr ×(101325 Pa/760 Torr) =91300Pa; T = (25.0 + 273.15)K = 298.2K
pV = nRT, V = nRT/p = 0.0832 mol ×8.314 JK-1 mol-1×298.2 K/91300Pa = 2.26× 10-3 J/Pa= 2.26× 10-3
(kg m2 s-2)(kg-1ms2) = 2.26× 10-3 m3 × (1L/0.001 m3 )= 2.26 L
(d) You would like to react the 5.00 g of SiO2 with anhydrous HF, which is a gas at 25.0 °C. What is the
volume (in m3 and L) of stoichiometric amounts of HF gas that you need at 25.0 °C at 685 Torr.
n(HF) = 0.0832 mol SiO2 ×(4mol HF/1mol SiO2) = 0.333 mol HF
pV = nRT, V = nRT/p = 0.333 mol ×8.314 JK-1 mol-1×298.2 K/91300Pa = 6.78× 10-3 J/Pa= 6.78× 10-3 (kg
m2 s-2)(kg-1ms2) = 6.78× 10-3 m3 × (1L/0.001 m3 )= 6.78 L
Question 7 (21 Marks)
(i) Draw Lewis structures for the following compounds. (Central atom is underlined) (ii) What are the
electron-pair geometries and molecular geometries according to the VSEPR model. (iii) Do these molecules
have molecular dipole moments. Indicate the dipole moment if applicable. (iv) Specify the intermolecular
forces present in these compounds.
(a) BrF5
..
..
:
:
F
:
..
.. F: ..
..
:F
:F
..
:
..
..
..:
.. Br ..F
..F
.. :F Br
..
:F.. ..
_
F..:
F..: ..
electron pair geometry: octahedral
molecular geometry: square planar
this molecule has a dipole.
intermolecular forces: dipole-dipole, dipole-induced dipole, London dispersion
3
(b) NOF3
..
:O :..
.. + N F :
..
:..F
:..F:
..
:O :..
.. + N F :
..
:..F
:..F:
_
electron pair geometry: tetrahedral
molecular geometry: tetrahedral
this molecule has a dipole.
intermolecular forces: dipole-dipole, dipole-induced dipole, London dispersion
(c) XeO3F2
..
:F : ..
..
O..
..
O
.. Xe O
..
: ..
F:
electron pair geometry: trigonal bipyramidal
molecular geometry: trigonal bipyramidal
no dipole moment
intermolecular forces: London dispersion
(d) XeOF3..
.. ..- ..F:..
..
.. Xe
..F :
..
.. ..- ..F:.. ..O
:
F
..
..O.. Xe
..F :
: F.. ..
electron pair geometry: octahedral
molecular geometry: square planar (seesaw is an acceptable answer)
dipole moment
intermolecular forces: dipole-dipole, dipole-induced dipole, London dispersion + ion ion/ion dipole/ion
induced dipole
Question 8 (11 Marks)
Draw the Lewis structures and determine the oxidation states for all individual atoms in the following
compounds.
(a) H-O-F (connectivity as indicated)
:O :
H : F.. :
H: +I, O: 0, F: -I
4
(b) XeOF5- (central atom:Xe)
.. :O : ..:
F..
: ..F
.. Xe ..F..:
: F.. ..
..F :
Xe: +VI, O: -II, F: -I
+
(c) NF4 (central atom: N)
..
:F :
.. N+ F..:
:F
..
..
:..F :
N: +V, F: -I
(d) [O3S-S-S-SO3]2- (connectivity as indicated, oxygens bonded to sulfur)
..
..
:O
O:
.. ..
.. -: ..
O
S
S
.. :
.. ..S S O
..
:O
O
..
.. :
+ several resonance structures O-II3S+V-S0-S0-S+VO-II32Question 9 (6 Marks)
Pure iodine (105 g) is dissolved in 325 g of CCl4 at 65 °C.
(a) Given that the vapour pressure of CCl4 at 65 °C is 531 Torr, what is the vapour pressure (in Torr) of this
solution at 65 °C. Assume I2 does not contribute to the vapour pressure.
(b) Calculate the boiling point of CCl4 in this solution. The boiling point of CCl4 is 76.7 °C and Kb.p. = 5.03
K kg mol-1.
(a) M(CCl4) = 153.822 g/mol
M(I2) = 253.810 g/mol
n(CCl4) =325 g/(153.822 g/mol) = 2.11 mol
n(I2) =105 g/(253.810 g/mol) = 0.414 mol
X(CCl4) = 2.11mol/(2.11 mol + 0.414 mol) = 0.836
p(CCl4) = X(CCl4) × p0(CCl4) = 0.836 × 531 Torr = 444 Torr
(b) molality(I2) =0.414 mol/0.325 kg= 1.27 mol/kg
∆Tb.p. = Kb.p. × molality = 5.03 K kg mol-1 ×1.27 mol/kg = 6.41 K = 6.41 °C
Tb.p. = 76.7 °C + 6.41 °C = 83.1 °C
Question 10 (4 Marks)
State which species in each of the following pairs is expected to have the larger radius:
(a) O or O2Oxide is larger.
(b) Si or Cl
Silicone is larger.
(c) S2- or Ar
Sulfide is larger.
5
Question 11 (12.5 Marks)
Oxygen is oxidized by the strong oxidizer, PtF6, to the dioxygenyl cation, O2+ and the PtF6- anion (a) Write
the balanced reaction equation. (b) Determine oxidation states for the atoms that are oxidized and reduced.
(c) Give the half reaction for this redox reaction and show that the electron balance in your reaction equation
from (a) is correct. (d) Draw the MO diagram (correlation diagram) for O2+. (e) Calculate the bond order for
the O2+ cation and compare it to that of O2. (f) What kind of magnetism does the O2+ cation exhibit? (g)
Give the valence electron configuration of the O2+ cation.
(a) O2(g) + PtF6(g) → O2+PtF6-(s)
(b) O02(g) + Pt+VIF6(g) → O+0.52+Pt+VF6-(s)
(c) O2 → O2+ + ePtF6 + e- → PtF6electron balance is correct, since one electron appears on both sides.
d) Draw the MO diagram for O2+.
O
O2+
O+
E
σ2p*
↑
π2p*
↑↓ ↑ ↑
2p
↑
↑ ↑
2p
↑↓ ↑↓ π2p
↑↓
σ2p
↑↓
σ2s*
↑↓
2s
↑↓
2s
↑↓
σ2s
6
(e) Bond order = ½(number of electrons in bonding MO’s – number of electrons in antibonding MO’s)
= ½(8-3) = 2.5
The bond order for the O2+ cation is one, compared to the bond order in O2 of two. The removal
of one electron from O2 strengthens the O-O bond.
(f) The O2+ cation is paramagnetic.
(g) (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)1
Question 12 (3 Marks)
Write the electron configuration of Neodymium (Nd) in the orbital box notation.
Nd:
↑↓
↑↓
↑↓ ↑↓ ↑↓
↑↓
↑↓ ↑↓ ↑↓
↑↓
↑↓ ↑↓ ↑↓ ↑↓ ↑↓
↑↓ ↑↓ ↑↓
1s
2s
2p
3s
3p
4s
3d
4p
↑↓
5s
↑↓ ↑↓ ↑↓↑↓ ↑↓
4d
↑↓↑↓ ↑↓
5p
↑↓
6s
↑ ↑ ↑
4f
Question 13 (3 Marks)
Write the electron configurations of Tc and Tc2+ using the noble-gas notation. What are the magnetic
properties of Tc and Tc2+.
Tc: [Kr]5s1 4d6, but [Kr]5s2 4d5 acceptable
Tc2+: [Kr]5s0 4d5
Both, Tc and Tc2+ are paramagnetic.
Question 14 (4 Marks)
Magnesium is composed of three isotopes, 24Mg, 25Mg, and 26Mg. The isotopes 25Mg and 26Mg have natural
abundances of 10.00 % and 11.01 %, respectively. The isotopic masses of 25Mg and 26Mg are 24.985839 u
and 25.983595 u, respectively. Calculate the natural abundance and the isotopic mass of 24Mg.
Average atomic mass of Mg: 24.3050 u
Abundance of 25Mg: 10.00%
Abundance of 26Mg: 11.01 %
Abundance of 24Mg: 100 % - 10.00% - 11.01 % = 78.99 %
24.3050 u = 0.7899 × x u + 0.1000 × 24.985839 u + 0.1101 × 25.983595 u
= 0.7899 × x u + 2.498 u + 2.861 u
18.946 u = 0.7899 × x u
xu
= 23.99 u
24
The Mg isotope has a natural abundance of 78.99 % and an isotopic mass of 23.99 u.
Question 15 (6 Marks)
Phosgene (COCl2) is a highly toxic gas. Describe the bonding situation in phosgene using the valence bond
theory. (a) What is the hybridization of the carbon atom? (b) Draw energy diagram indicating the formation
of the hybrid orbitals on carbon starting from the atomic orbitals on C, generating the hybrid orbitals on C.
(c) Describe the bonding (all bonds) in phosgene in terms of orbital overlap.
7
:O :
.. C ..
:Cl
Cl
..
.. :
(a) C: sp2 hybridization
(b) Carbon:
E
↑ ↑
↑
2p
↑ ↑ ↑
pyz
sp2
↓↑
2s
(c)
C-Cl bonds: overlap between the sp2 hybrid orbital on C and the 3p orbital on Cl
C=O bond: overlap between the sp2 hybrid orbital on C and a 2p orbital on O, forming the σ bond and
overlap between the 2pz orbital on C with a 2p orbital on O, forming the π bond
Question 16 (4 Marks)
What is the longest wavelength (in nm) of electromagnetic radiation that is necessary to ionize Li2+ to give
Li3+?
∆E = (Efinal - Einitial) = Ry(Z2/nfinal2 – Z2/ninitial2)
For Li: Z=3
∆E = En=∞ -En=1 = -Ry(9/∞ - 9/1) = 9Ry =9 × 2.1799 × 10-18 J = 1.9619 × 10-17 J
E = hc/λ
λ = hc/E = 6.626 × 10-34 Js × 2.998× 108 m s-1/ 1.9619 × 10-17 J = 1.013 × 10-8 m = 10.13 nm
8
Prefixes
Pico, p 10-12; nano, n 10-9; micro, µ 10-6 ;milli, m 10-3; centi, c
Fundamental Constants
Planck's constant, h 6.626 × 10-34 J s
Avogadro's number 6.022 × 1023 mol-1
Elementary charge, e 1.6022 × 10-19 C
Electron mass
9.1095 × 10-28 g
Gas constant, R
8.314 J K-1mol-1
n=
10-2; deci, d
Rydberg Constant, RRy
Proton mass
Neutron mass
Speed of light in vacuum, C
1 Ry = 2.1799 × 10-18 J
10-1
1.097 x 107 m-1
1.67252 × 10-24 g
1.6749 × 10-24 g
2.998 x 108 m s-1
m
m
hc
h
c
; λ = ; E = mc 2 ; λ =
; ρ = ; pV = nRT ; Ptot = ∑ p i ; p i = X i ⋅ Ptot ; E = hν =
;
ν
M
V
λ
mv
i
2

n 
p + a   [V − nb] = nRT ;
 V  

u 2 = u rms =
3RT
1
Z2
h
; E kin = mv 2 ; ∆x ⋅ ∆(mv ) >
; E n = −Ry 2 ;
M
2
4π
n
 Z2
 Z2
Z2 
Z2 
∆E = (E final − E initial ) = − R Ryd hc 2 − 2  = −Ry 2 − 2  ; ∆Tb.p. = K b.p. ⋅ molality ;
 n final n initial 
 n final n initial 
∆Tf .p. = K f .p. ⋅ molality ; p i = X i ⋅ Pi0
Physical quantity
Unit
Symbol
Definition
Frequency, f or ν
hertz
Hz
s-1
Energy , W or E
joule
J
kg m2 s-2
Force, F
newton
N
J m-1 = kg m s-2
Pressure, p
pascal
Pa
N m-2 = kg m-1 s-2
Temperature: 0 K = -273.15 °C; 0 °C = 273.15 K
Pressure: 1 atm = 760 Torr = 760 mmHg = 1.01325 bar = 101325 Pa; 1 bar = 105 Pa
Volume: 1 mL = 1 cm3; 1 L = 1000 cm3 = 1 dm3 = 0.001 m3
9
1
Electronegativities
18
2.1
H
1
1.0
Li
3
1.0
Na
11
0.9
K
19
0.9
Rb
37
0.8
Cs
55
0.8
Fr
87
2
13
1.5
2.0
Be
5
1.5
Mg
Ca
20
1.0
3
1.3
Ba
56
1.0
Ra
88
4
1.4
Sc
21
1.2
Sr
38
1.0
2.5
B
4
1.2
12
1.0
14
Y
39
1.1
La
57
1.1
5
1.5
Ti
22
1.3
V
23
1.5
Zr
40
1.3
Nb
41
1.4
Hf
72
Ta
73
6
1.6
Cr
24
1.6
Mo
42
1.5
W
74
7
1.6
Mn
25
1.7
Tc
43
1.7
Re
75
8
1.7
Fe
26
1.8
Ru
44
1.9
9
1.7
1.8
Co
27
1.8
Ni
28
1.8
Rh
45
1.9
Os
76
10
Pd
46
1.8
Ir
77
Pt
78
11
1.8
Cu
29
1.6
Ag
47
1.9
Au
79
12
1.6
Zn
30
1.6
Ga
31
1.6
Cd
48
1.7
In
49
1.6
Hg
80
Tl
81
3.0
C
6
1.8
Al
13
1.7
15
Ge
32
1.8
Sn
50
1.7
7
2.1
As
33
1.9
Sb
51
1.8
Bi
83
17
8
2.5
F
9
3.0
S
16
2.4
Se
34
2.1
Cl
17
2.8
Po
84
Ar
18
2.9
Br
35
2.5
Te
52
1.9
Ne
10
3.3
Kr
36
2.3
I
53
2.1
At
85
Xe
54
Rn
86
Ac
89
1
Chem 1000 Standard Periodic Table
18
1.0079
1H
He
2
4.0
O
P
15
2.1
Pb
82
3.5
N
Si
14
1.9
16
4.0026
2He
hydrogen
2
13
14
15
16
17
helium
6.941
9.0122
10.811
12.011
14.0067
15.9994
18.9984
20.1797
3Li
4Be
5B
6C
7N
8O
9F
10Ne
11Na
12Mg
lithium berrylium
22.9898 24.3050
boron
26.9815
3
4
5
6
7
8
9
10
11
12
47.88
50.9415
51.9961
54.9380
55.847
58.9332
58.693
63.546
65.39
21Sc
22Ti
23V
24Cr
25Mn
26Fe
29Cu
30Zn
39Y
40Zr
41Nb
42Mo
43Tc
44Ru
72Hf
73Ta
74W
sodium magnesium
39.0983 40.078 44.9559
19K
20Ca
37Rb
38Sr
55Cs
56Ba
La-Lu
88Ra
Ac-Lr
potassium calcium scandium titanium vanadium chromium manganese
iron
85.4678
(98)
101.07
87.62
88.9059 91.224 92.9064
95.94
rubidium strontium
132.905 137.327
cesium
(223)
87Fr
francium
yttrium zirconium niobium
178.49 180.948
barium
226.025
molybdenum technetium
183.85
hafnnium tantalum tungsten
(261)
(263)
(262)
104Rf
rutherfordium
radium
zinc
112.411
45Rh
46Pd
47Ag
48Cd
75Re
rhenium
(262)
76Os
78Pt
79Au
77Ir
osmium
(265)
iridium
(266)
107Bh
hassium meitnerium
108Hs
109Mt
144.24
(145)
150.36
151.965
140.908
58Ce
59Pr
89Ac
90Th
platinum
(269)
15P
phosphorus
31Ga
32Ge
49In
50Sn
16S
fluorine
35.4527
neon
39.948
17Cl
18Ar
74.9216
sulfur
78.96
chlorine
79.904
33As
34Se
35Br
51Sb
52Te
83Bi
gallium germanium arsenic selenium bromine
114.82 118.710 121.757 127.60 126.905
tin
207.19
antimony tellurium
208.980
(210)
53I
85At
86Rn
81Tl
82Pb
158.925
162.50
164.930
167.26
168.934
173.04
174.967
66Dy
67Ho
dysprosium holmium
68Er
69Tm
70Yb
71Lu
65Tb
231.036
238.029
237.048
(240)
(243)
(247)
91Pa
92U
93Np
94Pu
95Am
96Cm
97Bk
uranium neptunium plutonium americium curium
(251)
(252)
erbium
(257)
54Xe
84Po
thallium
bismuth polonium astatine
110Ds
157.25
36Kr
krypton
131.29
xenon
(222)
mercury
lead
argon
83.80
iodine
(210)
gold
terbium
(247)
10
14Si
silicon
72.61
oxygen
32.066
darmstadtium
60Nd
61Pm
62Sm
63Eu
64Gd
praesodymium neodymium promethium samarium europium gadolinium
protactinium
80Hg
nitrogen
30.9738
aluminum
69.723
ruthenium rhodium palladium silver cadmium indium
186.207
190.2
192.22
195.08 196.967 200.59 204.383
106Sg
140.115
actinium9 thorium
copper
107.868
105Db
57La
28Ni
nickel
106.42
dubnium seaborgium bohrium
138.906
lanthanum cerium
227.028 232.038
27Co
cobalt
102.906
13Al
carbon
28.0855
thulium ytterbium lutetium
(259)
(258)
(260)
98Cf
99Es
100Fm 101Md
102No
103Lr
berkelium californium einsteinium fermium mendelevium nobelium lawrencium
radon