Chemistry I
Midterm Exam
16 Nov, 2012
1
H
1.0
3
Li
6.9
11
Na
23.0
19
K
39.1
37
Rb
85.5
55
Cs
132.9
Periodic Table of Elements
4
Be
9.0
12
Mg
24.3
20
21
22
23
24
25
26
27
28
29
30
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
40.1 45.0 47.9 50.9 52.0 54.9 55.8 58.9 58.7 63.5 65.4
38
39
40
41
42
43
44
45
46
47
48
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
87.6 88.9 91.2 92.9 95.9 (98) 101.1 102.9 106.4 107.9 112.4
56
57
72
73
74
75
76
77
78
79
80
Ba
La
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
137.3 138.9 178.5 181.0 183.8 186.2 190.2 192.2 195.1 197.0 200.6
Formulae
E = hn
En = n2h2/8mL2
(mev2)/2 = hn - Φ
− Z 2 hR
En =
, n = 1, 2, 3, 
n2
m e4
R = e3 2 = 3.29 ×1015 Hz
8h ε 0
Constants
R = 8.314 J / mol K
= 0.0821 L atm / K mol
= 8.314 L kPa / K mol
5
B
10.8
13
Al
27.0
31
Ga
69.7
49
In
114.8
81
Tl
204.4
6
C
12.0
14
Si
28.1
32
Ge
72.6
50
Sn
118.7
82
Pb
207.2
7
N
14.0
15
P
31.0
33
As
74.9
51
Sb
121.8
83
Bi
209.0
 2 2

∇ + V ( x , y , z ) ψ ( x , y , z )
−
 2m

Hψ = Eψ
1 atm = 760 Torr = 1.01x105 Pa
c = 2.99×108 m/s
h = 6.63×10-34 J⋅s
Mass of e , me = 9.10939 × 10 -31 kg
Nature Exponential Base e = 2.71828
Page 1
8
O
16.0
16
S
32.1
34
Se
79.0
52
Te
127.6
84
Po
(209)
9
F
19.0
17
Cl
35.5
35
Br
79.9
53
I
126.9
85
At
(210)
2
He
4.0
10
Ne
20.2
18
Ar
40.0
36
Kr
83.8
54
Xe
131.3
86
Rn
(222)
1.
The energy level of a particle of mass m (m = 3×10-30 kg) in a two-dimensional rectangular box can be
expressed as
E (nx , n y ) = (
2
n x2 n y h 2
)⋅
+
L 2x L 2y 8m
where Lx and Ly are the length and the width of the box (Lx = 2Ly = 2×10-9 m), h is the Planck’s
constant (h = 6.626×10-34 J．s), and nx and ny are two independent quantum numbers (nx = 1, 2, 3, … ;
ny = 1, 2, 3, …).
(a) If the particle make a transition from (nx, ny) = (4, 1) to (nx, ny) = (2,1), and emit photon. What is
the wavelength of the electromagnetic wave? Express your result in µm (1 µm = 10-6 m, the speed
of light = 3×108 m/s). (5%)
(b) What type of electromagnetic wave is it? Microwave, infrared, visible, or ultraviolet? (5%)
(c) Find the values of the quantum numbers nx and ny that correspond to the lowest degenerate states.
(5%)
2. True and false
(a) For the photoelectric-effect experiment, the Planck constant equals to the slope of a plot of E
vs. n (E: the kinetic energy of the emitted electrons; n : the frequency of the incident light). (3%)
(b) In the double-slit experiment, in order to see the diffraction pattern, electrons must interfere.
Therefore, if the electron is sent one by one to a double-slit, one would never observe diffraction
pattern. (3%)
(c) If an H atom at an excited state emits a photon with a frequency n, the energy of the excited state
is equal to hn. (3%)
(d) There are three orbitals in the n = 3 shell of hydrogen atoms. (3%)
(e) The principal and orbital angular momentum quantum numbers for the 3d orbital are (3, 2). (3%)
Page 2
3. Peter tried to write the Lewis structures of the resonance structures of diazomethane and ozone listed
below. Please determine the formal charge on each atom and then help him to exam whether the
structure is a proper resonance structures or not. If the structure is a proper resonance structure, please
determine whether it is the structure of the lowest energy among the possible resonance structure. Please
explain it briefly.
(a) Diazomethane (3% for each structure, 9% total)
(b) Ozone (3% for each structure, 9% total)
4. Please answer the following question briefly.
(a) Are the two ions Ca2+(g) and O2−(g), respectively, more stable than Ca +(g) and O−(g)? Justify your
answer. (4 %)
(b) Why does solid calcium oxide contain Ca2+ and O2− ions rather than Ca+ and O−? (4 %)
5. There are 4 ion pairs including
(1)[Li+][F−]
(2) [Cs+][F−]
(3) [Al3+][O2−]
(4) [Mg2+][O2−]
(a) Which ion pair contains the strongest coulombic interactions? (2%) Please use the properties of
ions and the two tables below to prove the answer. (4%)
(b) Which ion pair has the ionic bond with the most covalent character? (2%) Please use the properties
of ions and the two tables below to prove the answer. (4%)
Electronegativity Table
Ionic Radius Table (unit: pm)
Page 3
6. Treat the π system of a dye molecule containing a conjugated carbon chain of 12 carbon atoms as a 1-D
box, where the molecular orbitals of π system is the linear combination of 12 2p atomic orbitals. Given
that each C atom contributes one electron and that each state of the box can accommodate two electrons.
(a) What is the length of the 1-D box? The average bond lengths between carbon and carbon are
1.44×10-10 m. (2%)
(b) What is the quantum number n of the LUMO? (2%)
(c) Derive an expression for the wavelength of the light absorbed by the transition of an electron from
HOMO to LUMO. (use me as mass of electron, h as Planck’s constant, c as the speed of light) (4%)
7. Please answer the following questions briefly
(a) Write the Lewis structure for N22–. (2%)
(b) According to the valence bond theory, please indicate the hybridization type of N atom in N22–.
(2%)
(c) Please indicate the number of unpaired electrons in N22–. (2%)
8. Please answer the following questions based on the molecular orbital theory
(a) Draw the molecular orbital energy-level diagram for N2 molecule and label the energy levels
according to the type which they are made, whether they are σ- or π-orbitals, and whether they are
bonding or antibonding. (12%)
(b) Please indicate the bond order of ground-state N22–. (3%)
(c) Please indicate the number of unpaired electrons of ground-state N22–. (3%)
(d) Is ground-state N22– paramagnetism or diamagnetism? (3%)
9. For each of the following species, predict the shape and state whether it is polar or nonpolar:
(a) P4 (2%)
(b) SF4 (2%)
(c) SO2 (2%)
(d) XeF4 (2%)
Page 4
Answer:
1. (5% for each question, 15% total)
(a) E (n x , n y ) = (
2
n x2 n y h 2
h2
2
2
n
n
+
⋅
=
+
⋅
)
(
4
)
x
y
L 2x L 2y 8m
32mL 2y
ΔE = E (4,1) - E (2,1) = [(16 + 4) - (4 + 4)] ⋅
ΔE = hn =
hc
l
⇒ l=
hc
=
ΔE
h2
3h 2
=
32mL 2y 8mL 2y
2
8mcL y
hc
=
2
3h
3h
(
)
8mL 2y
8 ⋅ (3 ⋅ 10 -30 ) ⋅ (3 ⋅ 108 ) ⋅ (10 -9 ) 2
24 ⋅ 10 - 40
=
= 3.6 ⋅ 10 -6 (m)
3 ⋅ 6.626 ⋅ 10 -34
6.626 ⋅ 10 -34
l = 3.6 μm
l=
(b) Infrared
(c) E (n x , n y ) = (n x2 + 4n y2 ) ⋅
h2
≡ (n x2 + 4n y2 ) ⋅ Econstant
32mL 2y
E( 1,1 ) = 5,
E( 1,2 ) = 17 ,
E( 2 ,1 ) = 8,
E( 3,1 ) = 13,
E( 4 ,1 ) = 20
E( 2 ,2 ) = 20
E( 1,3 ) = 37
⇒ E( 2 ,2 ) = E( 4 ,1 )
⇒ (2,2) and (4,1) are the quantum numbers corresponding to the lowest degenerate states
2. (3% for each question, 15% total)
(a) True;
(b) False;
(c) False;
(d) False
(e) True
3. (3% for each question, 18% total)
1% for each
correct structure
and formal
charge
1% for each
answer
1% for each
answer
Not corret resonance structure
corret resonance structure
Not corret resonance structure
This structure violates the octet
rule.
Or
The total formal charge is not
equal to zero.
Or
The total number of electrons is
not correct.
This is one of the structures of the
lowest energy.
This structure violates the octet
rule.
Or
The total formal charge is not
equal to zero.
Or
The total number of electrons is
not correct.
Page 5
1% for each
correct
structure and
formal charge
1% for each
answer
1% for each
answer
Not corret resonance structure
corret resonance structure
corret resonance structure
This structure violates the octet
rule.
Or
The total formal charge is not
equal to zero.
Or
The total number of electrons is
not correct.
This is one of the structures of the
lowest energy.
The central oxygen atom violates
the octet rule.
4. (a) Ca2+(g) and O2-(g), respectively, are less stable than Ca+(g) and O-(g), because the formation of
Ca2+(g) and O2-(g) from Mg+(g) and O-(g), respectively, requires a lot of energies. (4%)
(b) Lattice energy of Ca2+O2- is much greater than that of Ca+O-. (4%)
5. (a) Al2O3 (2%),
𝑞 ×𝑞
The strength of ionic bonds is proportional to �(𝑟1+𝑟2) �
For LiF :
𝑞 ×𝑞
�(𝑟1+𝑟 2)
1 2
𝑞 ×𝑞
+1×−1
�＝�(76+133)
+1×−1
�=0.004785
1
2
For CsF : �(𝑟1+𝑟 2) �＝�(167+133) �=0.003333
For Al2O3:
1
2
𝑞 ×𝑞
�(𝑟1 +𝑟2)
1 2
𝑞 ×𝑞
+3×−2
�＝�(54+140) �=0.030928
+2×−2
For MgO: �(𝑟1+𝑟 2) �＝�(72+140) �=0.018868
Because
1
𝑞 ×𝑞
�(𝑟1+𝑟 2)
1 2
2
� of Al2O3 is the largest one, the strongest ionic bond is Al2O3。(4%)
(b) Al2O3(2%),
The smaller difference of the electronegativity between a cation and a anion, the stronger covalent
character of the ionic bond。
The difference of the electronegativity between Li and F＝4-1＝3
The difference of the electronegativity between Cs and F＝4-0.8＝3.2
The difference of the electronegativity between Al and O＝3.4-1.6＝1.8
The difference of the electronegativity between Mg and O 的電負度差＝3.4-1.3＝2.1
Because the difference of the electronegativity between Al and O is the smallest one, the strongest
Page 6
covalent character of the ionic bond is Al2O3.(4%)
6. (2+2+4%)
(a) this conjugate system has 11 CC bonds ; the length of this box = 11×1.44×10-10 = 1.58×10-9 m
(b) there are 12 π-electrons ; 6 energy levels are occupied ; LUMO: n = 7
(c) 𝐸 𝑜𝑜 𝐻𝐻𝐻𝐻 =
62 ∙ℎ2
8∙𝑚𝑒 ∙(1.58×10−9 )2
2
; 𝐸 𝑜𝑜 𝐿𝐿𝐿𝐿 =
72 ∙ℎ2
8∙𝑚𝑒 ∙(1.58×10−9 )2
13 ∙ ℎ
𝑐 𝑐 ∙ ℎ 8 ∙ 𝑐 ∙ 𝑚𝑒 ∙ (1.58 × 10−9 )2
∆𝐸 =
; 𝜆= =
=
𝜈
∆𝐸
8 ∙ 𝑚𝑒 ∙ (1.58 × 10−9 )2
13 ∙ ℎ
7. (2+2+2%)
(a)
(b) sp2
(c) 0
8. (a)(12%)
energy levels: 6%
labels: 6%
(b) 2 ; (3%)
(c) 2 ; (3%)
(d) paramagnetic (3%)
9. (a) tetrahedral, nonpolar ; (2%)
(b) see-saw, polar ; (2%)
(c) angular, polar ; (2%)
(d) square planar, nonpolar. (2%)
Page 7