Chem 1000A Final Examination - Solutions
April 15th, 2003: 9:00 to 12.00 am
Your name _______________
Instructor: Dr. M. Gerken
Student ID _______________
Time: 3h
No. of pages: 3 + 1
Report all your answers using significant figures. Show all units and their conversions throughout your
calculations.
Question 1 (5 Marks)
Complete the following table.
140
Symbol
Ce
Number of electrons
__58_______
Number of neutrons
__82_______
Number of protons
__58_______
Overall charge
0
__120Sn2+___
48
70
__50_______
2+
Question 2 (2 Marks)
What is Ψ in Schrödinger’s equation?
In Schrödinger’s equation, Ψ is the symbol for a wavefunction.
Question 3 (5 Marks)
Draw the five 3d orbitals and label them.
x
x
z
y
dxy
dxz
x
y
x
y
z
dyz
z
dx2-y2
dz2
Question 4 (4 Marks)
What is the maximum number of electrons that can be identified with each of the following sets of quantum
numbers? (zero is a possible answer)
zero, l has to be smaller than 2 if n = 2
(a) n = 2, l = 2, ml = -1, ms = ½
(b) n = 3, l = 2
10 electrons (in the five 3d orbitals)
(c) n = 2
8 electrons (2s and three 2p orbitals)
(d) n = 4, l = 3, ml = -½
zero, ml has to be an integer number
Question 5 (4 Marks)
You dissolve 10.0 mg of NaOH and 10.0 mg of HCl in 100. mL of water. (a) Write the balanced reaction
equation. (b) What is the NaCl concentration (neglecting the change in volume during the dissolution and
reaction)?
(a) NaOH + HCl → NaCl + H2O
1
(b) M(NaOH) = 39.9971 g mol-1; M(HCl) = 36.4606 g mol-1
n(NaOH) = 0.0100 g/(39.9971 g mol-1) = 2.50 × 10-4 mol
n(HCl) = 0.0100 g/(36.4606 g mol-1) = 2.74 × 10-4 mol
The reaction has a 1:1 stoichiometry, therefore, NaOH is the limiting reagent.
NaCl produced: n(NaCl) = 2.50 × 10-4 mol
[NaCl] = 2.50 × 10-4 mol/0.100 L = 2.50 × 10-3 mol L-1
Question 6 (15 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.
a) XeF3\
.. - .. .. ..
..
..
.. ..
:F
F:
:F
Xe
F:
.. Xe
..
..
..
..
..
:F:
:F:
..
..
electron-pair geometry: octahedral
molecular geometry: T-shaped
dipole moment: yes
b) SCl2
.. ..
.. S ..
:Cl
Cl
..
.. :
.. ..
.. S ..
:Cl
Cl
..
.. :
_
electron-pair geometry: tetrahedral
molecular geometry: bent
dipole moment: yes
c) ClO2F
_
:O :
:O :
Cl : : Cl :
:O
O
..
..
:..F :
:..F :
electron-pair geometry: tetrahedral
molecular geometry: T-shaped
dipole moment: yes
d) XeO2F4
:O: ..:
F....
..
: ..F Xe F..:
..
:F
.. :O :
electron-pair geometry: octahedral
molecular geometry: octahedral
dipole moment: no
2
e) XeO64:O : .. O
..
....:
-:
O
Xe
O
....
.. : :O..
:O:
+ several resonance structures
electron-pair geometry: octahedral
molecular geometry: octahedral
dipole moment: no
Question 7 (5 Marks)
For extinguishing fires in rooms with expensive electrical equipment a halon has been used (Halon 1301:
CF3Br). Unfortunately, this halon is depleting the ozone layer. As a replacement, a new fire-extinguishing
agent has been introduced last year. This new fire-extinguishing agent contains 22.8 % C, 72.1 % F, and 5.1
% O. (a) Determine the empirical formula for this compound. (b) The molar mass of this agent has been
found to be 316.046 g mol-1. Determine the molecular formula.
In 100 g of sample:
22.8 g C → 1.90 mol
72.1 g F → 3.79 mol
5.1 g O → 0.32 mol
C : F : O = 1.90 mol : 3.79 mol : 0.32 mol = 5.96 mol : 11.9 mol : 1.00 mol
Empirical formula: C6F12O
M(C6F12O) = 31.6046 g mol-1
This calculated molar mass is the same than the experimentally determined one.
Molecular mass: C6F12O
Question 8 (8 Marks)
Draw the Lewis structures and determine the oxidation states for all atoms in the following compounds.
a) H3C-O-F (connectivity at indicated)
H: +I, C: -II, O: 0, F: -I
2I: +VII, O: -II, F: -I
b) IO2F5 (central atom:I)
c) NF4+ (central atom: N)
N: +V, F: -I
2d) O3S-S-S-SO3 (connectivity as indicated) O-II3S+V-S0-S0-S+VO-II32Question 9 (12 Marks)
Balance the following redox reactions in acidic aqueous solutions (add H+ if necessary). First, write the
balanced half-reactions and combine them. Indicate the (a) electron balance, (b) material balance, and
(c) charge balance of the overall reaction equations.
a) I+VO-II3- + S+IVO-II32- → I02 + S+VIO-II42oxidation half-reaction (unbalanced): S+IVO-II32- → S+VIO-II42- + 2ereduction half-reaction (unbalanced): 2I+VO-II3- + 10e- → I02
electron balance:
oxidation half-reaction (unbalanced): 5S+IVO-II32- → 5S+VIO-II42- + 10ereduction half-reaction (unbalanced): 2I+VO-II3- + 10e- → I02
material balance:
oxidation half-reaction (balanced): 5H2O + 5S+IVO-II32- → 5S+VIO-II42- + 10e- + 10 H+
reduction half-reaction (balanced): 12H+ + 2I+VO-II3- + 10e- → I02 + 6 H2O
3
overall reaction: 2H+ + 2IO3- + 5SO32- → I2 + 5SO42- + H2O
material balance:
2H, 2I, 5S, 21O | 2I, 5S, 2H, 21O
charge balance: (2+) + (2-) +(10-) = 10- | 10correct electron, material, and charge balance!
b) H+ICl+VO-II3 + Cl-I- → Cl+IVO-II2 + Cl02
oxidation half-reaction (unbalanced): 2Cl-I- → Cl02+ 2ereduction half-reaction (unbalanced): H+ICl+VO-II3 + e- → Cl+IVO-II2
electron balance:
oxidation half-reaction (unbalanced): 2Cl-I- → Cl02+ 2ereduction half-reaction (unbalanced): 2H+ICl+VO-II3 + 2 e- → 2Cl+IVO-II2
material balance:
oxidation half-reaction (balanced): 2Cl-I- → Cl02 + 2ereduction half-reaction (balanced): 2H+ + 2H+ICl+VO-II3 + 2 e- → 2Cl+IVO-II2+ 2H2O
overall reaction: 2H+ + 2HClO3 + 2Cl- → 2ClO2+ Cl2 + 2H2O
material balance:
4H, 4Cl, 6O
|
4Cl, 4H, 6O
charge balance: (2+) + (2-) = 0
|
0
correct electron, material, and charge balance!
Question 10 (6 Marks)
Chlorine dioxide, ClO2, (melting point of 11 °C) is a reddish-yellow, highly toxic gas and has a pungent
odor. On gentle heating above 45 °C it decomposes by exploding violently into oxygen and chlorine. (a)
Write the balanced reaction equation for the decomposition. (b) At STP you have 1.505 g of chlorine
dioxide in a 500 mL cylinder. Heating the cylinder to 50 °C results in the decomposition (nicer word for
explosion) of the chlorine dioxide sample. Assuming the cylinder survives the explosion (which may not be
the case in real life – so don’t try it at home), calculate the partial pressures (in Pa and atm) for oxygen and
chlorine and the total resulting pressure (in Pa and atm) in the cylinder.
(a) 2ClO2 → Cl2 + 2O2
(b) M(ClO2) = 67.4515 g mol-1
n(ClO2) = 1.505 g/(67.4515 g mol-1) =0.0223 mol
n(Cl2) = 0.0223 mol ClO2 × (1 Cl2 produced/2 ClO2 consumed) = 0.0112 mol Cl2
n(O2) = 0.0223 mol ClO2 × (1 O2 produced/1 ClO2 consumed) = 0.0223 mol O2
pV = nRT; T = (273.15 + 50) K = 323 K
p(O2) = (nRT)/V = 0.0223 mol × 8.314 J K-1 mol-1 × 323 K/0.0005 m3
= 120000 J m-3 = 120000 kg m2 s-2 m-3 = 120000 kg m-1 s-2 = 120000 Pa (three sign. Fig.)
= 120000 Pa × 1 atm/ 101325 Pa = 1.18 atm
p(Cl2) = ½ 120000 Pa = 60000 Pa = 0.591 atm
total pressure = 120000 Pa + 60000 Pa = 180000 Pa = 1.18 atm + 0.591 atm = 1.77 atm
Question 11 (5 Marks)
Draw the Lewis structure of chlorine dioxide, ClO2. What class of molecules does ClO2 belong to? Predict
its molecular geometry.
.. .
Cl
:O
O
..
.. :
ClO2 is a radical, since it has an unpaired electron pair. It has a bent geometry.
4
Question 12 (6 Marks)
Write the electron configuration of Europium (Eu) in the orbital box notation.
Eu:
↑↓
↑↓
↑↓ ↑↓ ↑↓
↑↓
↑↓ ↑↓ ↑↓
↑↓
↑↓ ↑↓ ↑↓ ↑↓ ↑↓
1s
2s
2p
3s
3p
4s
3d
↑↓
5s
↑↓ ↑↓ ↑↓↑↓ ↑↓
4d
↑↓↑↓ ↑↓
5p
↑↓
6s
↑ ↑ ↑
4f
↑↓ ↑↓ ↑↓
4p
↑ ↑ ↑ ↑
Question 13 (5 Marks)
Write the electron configurations of Cu and Cu2+ using the noble-gas notation.
Cu: [Ar]4s1 3d10
Cu2+: [Ar]4s0 3d9
Both, Cu and Cu2+ are paramagnetic.
Question 14 (4 Marks)
(a) What is the mass of 1 mole of H2O (in g and in amu)? (b) What is the mass of 1 molecule of H2O (in g
and amu)?
(a) mass of 1 mole H2O: 18.0152 g or 1.085 × 1025 g
(b) mass of 1 H2O molecule: 18.0152 amu or 2.992 × 10-23 g
Question 15 (7 Marks)
Describe the bonding situation in protonated cyanic acid (H-C≡N-H+) using the valence bond theory. (a)
What is the hybridization of the carbon and nitrogen atoms? (b) Draw two energy diagram indicating the
formation of the hybrid orbitals on nitrogen starting from the atomic orbitals on N, going to the hybrid
orbitals on N. (c) Describe which orbitals are involved in forming bonds between C and H, C and N, and N
and H.
(a)
C: sp hybridization, N: sp hybridization
(b) Nitrogen:
E
↑ ↑ ↑
↑↓ ↑
2p
↑ ↑
px py
sp
↓↑
2s
(c)
C-H bond: overlap between the sp hybrid orbital on C and the 1s orbital on H
C≡N bonds: overlap between the sp hybrid orbital on C and the sp hybrid orbital on N, forming the σ bond
Overlap between two p orbitals on C with two p orbitals on N, forming the two π bonds
N-H: overlap between the sp hybrid orbital on N and the 1s orbital on H
5
Question 16 (4 Marks)
Ultraviolet radiation below 290 nm (the ozone cut-off) will be absorbed by the ozone layer and will not
reach the earth’s surface. (a) What range of frequencies will not reach the earth’s surface? (b) Photons of
what energy range will be absorbed by the ozone layer?
(a) ν = c/λ = 2.998 × 108 ms-1/ (2.90 × 10-9 m) = 1.03 × 1015 s-1
Frequencies of 1.03 × 1015 Hz and larger will not reach the earth’s surface.
(b) E=hν = 6.626 × 10-34 Js × 1.03 × 1015 s-1 = 6.85 × 10-19 J
Photons with energies of 6.85 × 10-19 J and larger will not reach the earth’s surface.
Question 17 (3 Marks)
Assuming an intact ozone layer, can sun light/radiation excite an electron of a hydrogen atom from the L
shell (n = 1) to the M shell (n = 2)?
∆E = RRyd h c (1/22 – 1/12) = 1.097 × 107 m-1 × 6.626 × 10-34 Js × 2.998 × 108 ms-1× (-0.75)
= -1.634 ×10-18 J
This energy is larger than the energy calculated for 290 nm (6.85 × 10-19 J). Since energy larger than 6.85
× 10-19 J is cut off, photons on earth’s surface do not have sufficient energy for the excitation of an electron
in a hydrogen atom from the L to the M shell.
Electronegativities
1
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
2.0
1.5
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
Zr
40
1.3
Hf
72
V
23
1.5
Nb
41
1.4
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
Ac
89
6
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
Po
Cl
Ar
18
Br
35
2.5
Kr
36
I
53
2.1
At
85
Ne
10
17
2.8
Te
52
1.9
84
He
2
4.0
O
P
15
2.1
Pb
82
3.5
N
Si
14
1.9
16
Xe
54
Rn
86
Prefixes
Pico, p 10-12; nano, n 10-9; micro, µ 10-6 ;milli, m 10-3; centi, c 10-2; deci, d 10-1
Fundamental Constants
Rydberg Constant
1.097 x 107 m-1
Planck's constant, h 6.626 × 10-34 J s
Avogadro's number 6.022 × 1023 mol-1
Proton mass
1.67252 × 10-24 g
Neutron mass
1.6749 × 10-24 g
Elementary charge, e 1.6022 × 10-19 C
-28
Electron mass
9.1095 × 10 g
Speed of light in vacuum, C 2.998 x 108 m s-1
Gas constant, R
8.314 J K-1mol-1 = 0.082057 L atm K-1 mol-1
n=
m
M
ρ=
m
V
pV = nRT
PT = ∑ pi
i
 1
1
∆E = (E final − Einitial ) = − RRyd hc 2 − 2
n
 final ninitial
E = mc 2
Physical quantity
Unit




λ=
Symbol
hc
E = hν =
pi = X i ⋅ PT
h
mv
λ
c
λ =
ν
∆x ⋅ ∆(mv ) >
h
4π
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
1
Chem 1000 Standard Periodic Table
18
1.0079
1H
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
(263)
(261)
(262)
104Rf
rutherfordium
radium
47Ag
48Cd
76Os
78Pt
79Au
gold
mercury
thallium
81Tl
82Pb
157.25
158.925
162.50
164.930
167.26
66Dy
67Ho
dysprosium holmium
68Er
77Ir
osmium
(265)
iridium
(266)
109Mt
144.24
(145)
150.36
151.965
89Ac
90Th
platinum
60Nd
61Pm
62Sm
63Eu
64Gd
praesodymium neodymium promethium samarium europium gadolinium
65Tb
231.036
238.029
237.048
(240)
(243)
(247)
terbium
(247)
91Pa
92U
93Np
94Pu
95Am
96Cm
97Bk
protactinium
50Sn
46Pd
108Hs
59Pr
49In
45Rh
hassium meitnerium
140.908
32Ge
zinc
112.411
107Bh
58Ce
31Ga
uranium neptunium plutonium americium curium
7
80Hg
(251)
oxygen
32.066
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
copper
107.868
106Sg
140.115
actinium9 thorium
75Re
15P
phosphorus
silicon
72.61
ruthenium rhodium palladium silver cadmium indium
186.207
190.2
192.22
195.08 196.967 200.59 204.383
rhenium
(262)
14Si
aluminum
69.723
nickel
106.42
105Db
57La
28Ni
nitrogen
30.9738
cobalt
102.906
dubnium seaborgium bohrium
138.906
lanthanum cerium
227.028 232.038
27Co
13Al
carbon
28.0855
(252)
tin
207.19
lead
erbium
(257)
antimony tellurium
208.980
(210)
53I
argon
83.80
36Kr
krypton
131.29
54Xe
iodine
(210)
xenon
(222)
84Po
85At
88Rn
168.934
173.04
174.967
69Tm
70Yb
71Lu
bismuth polonium astatine
thulium ytterbium lutetium
(259)
(258)
(260)
98Cf
99Es
100Fm 101Md
102No
103Lr
berkelium californium einsteinium fermium mendelevium nobelium lawrencium
radon