Solutions 8
1. Only molecules belonging to groups Cn, Cnv, and Cs may have a permanent dipole moment: so,
pyridine (C2v) and nitroethane (Cs) may be polar.
2. The C4v character table:
C4v
E
C2
2C4
2v
2d
h=8
A1
1
1
1
1
1
z, z2, x2 + y2
A2
1
1
1
-1
1
B1
1
1
Final Exam II
CHM 3411, Dr. Mebel, Spring 2006
Each problem is 12.5 pts, problems 9 and 10 are extra credit, also 12.5 pts
each. The total score is 100 pts without extra credit and 125 pts with extra credit.
1. (12.5 pts) Determine in each of the followin
CHEM-443, Fall 2016, Section 010
Quiz 1
Student Name_
09/09/2016
Directions: Please answer each question to the best of your ability. Make sure your response is legible, precise,
includes relevant dimensional units (where appropriate), logically presented
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G
E
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A
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A
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PRESENTED BY MARTY SCHWARTZ DESIGNED BY JON BROMMET
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As you will see from the first diagram below, If you were to place your finger on the 5th fret of the 6th (E) stri
Physical Chemistry
Spring 2006
Prof. Shattuck
Test 1
Name_
1. (5pts) Which of the following particles has a higher kinetic energy?
(a)
(b)
*Answer five (5) of the following seven (7) questions. If you answer more than 5 cross
out the one you wish not to b
1
Physical Chemistry CH342
Spring 2012, Prof. Shattuck
Test 1
Name_
Part 1: Answer 4 of the following 5 questions. If you answer more than 2, cross out the question
that you dont wish to be graded. (8 points each)
1. (a). Give the angular momentum quantum
Physcial Chemistry
Spring 2006
Prof. Shattuck
Test 2
Name_
Part I
Answer 6 of the following 8 questions. If you answer more than 6, cross out the ones you wish
not to be graded. 7 points each.
1. As the internuclear separation, R goes from 0 to , the over
Physical Chemistry
Prof. Shattuck, Spring 2004
Test 1
Name_
Part 1. Answer 6 of the following 7 questions. If you answer more than 6 cross out the one you
wish not to be graded, otherwise only the first 6 will be graded. 10 points each.
1. Name two types
Physical Chemistry
Spring 2011, Prof. Shattuck
Test 2
Name _
Part I: Answer 4 of the following 5 questions. If you answer more than 5 cross out the one you
do not wish to be graded. 7 Points each.
1. Consider the side-to-side overlap of two d orbitals in
nacu prumem Is 11": pts, promems y ana w are exn'a credit, also 1A.: pts
each. The total score is 100 pts without extra credit and 125 pts with extra credit.
1. (12.5 pts) Determine in each of the following cases if the function in the rst column is
an e
Exam II
CHM 3411, Dr. Mebel, Spring 2013
1. (20 pts) The principal line in the emission spectrum of potassium is violet. On close
examination, the line is seen to be a doublet (i.e., two spectral lines close to one another)
with wavelengths of 393.366 and
Solutions 2
2
P = *dx =
a
L
b
1.
b
2 1
x
L
2x
sin2 dx = x
sin
a L L 2 4 L
a
b
a) L = 10 nm, a = 4.95 nm, b = 5.05 nm
P = 0.02
b) a = 1.95 nm, b = 2.05 nm
P = 0.007
c) a = 9.90 nm, b = 10.00 nm
P = 6.610-6
d) a = 5.0 nm, b = 10.0 nm
P = 0.5
b) a =
Solutions 3
1.
1
2.
3.
2
4.
1/2
k
=
2c
= 2c
4 2 c 2 2 m A m B
2
222
k = = 4 c =
m A + mB
H Cl
H Br
HI
CO
, cm
2990
2650
2310
2170
5.
35
= 2 =
81
-1
k, N m-1
516
412
314
The order of stiffness: CO > NO > HCl > HBr > HI
3
1900
NO
1904
1600
6.
7.
4
8.
Solutions 4
1.
h =
hc
1
= I + mv2
2
I=
hc 1 2
mv
2
I = (6.6260810-34 J s) (2.9979108 m s-1) / 58.410-9 m 0.5 (9.1093810-31 kg)
(1.59106 m s-1)2 = 2.2510-18 J = 14.04 eV
2. Please, note the difference between radial wavefunctions given in my notes (p. 11
Solutions 5
1. (a) d electron: l = 2, s = 1/2
j = l + s, , |l s|
j = 5/2, 3/2
(b) f electron: l = 3, s = 1/2
j = l + s, , |l s|
j = 7/2, 5/2
2.
j = l + s, , |l s|
s = 1/2
If j = 3/2, 1/2, then l can be only 1, so we deal with a p electron.
3. The term sym
Solutions 7
1.
1
2.
2
3.
3
4.
4
C3
5.
Symmetry elements:
identity, E;
2C3 clockwise and counter clockwise rotations around
the threefold axis passing through the C-Cl bond;
3v three mirror planes Cl-C-H
Cl
v
v
v
C
H
H
H
6. Follow the chart in Fig. 15.14 (
Solutions 10
1.
2.
3.
A=
A=
2
( )d
2
[max max 2 + 2max max max max ]d =
1
2
1
2
3
2
2
max max + max max max max
1
3
2
We can find from the fact the band starts at 275 nm:
2
1
1
(275nm) = 0 = max 1
275nm 220nm
= 1.2110-8 cm2
A = 2.34108 L mol-1 cm
Exam I
CHM 3411, Dr. Mebel, Spring 2007
1. Determine in each of the following cases if the function in the first column is an
eigenfunction of the operator in the second column. If so, what is the eigenvalue?
Function
Operator
a)
e
b)
sin cos
c)
1d
x dx
x
Exam I
CHM 3411, Dr. Mebel, Spring 2012
1. Determine in each of the following cases if the function in the first column is an
eigenfunction of the operator in the second column. If so, what is the eigenvalue?
Function
Operator
a)
b)
c)
Ae
2 2 2
+ 2+ 2
2
x
Exam I
CHM 3411, Dr. Mebel, Spring 2006
1. (20 pts) Determine in each of the following cases if the function in the first column is
an eigenfunction of the operator in the second column. If so, what is the eigenvalue?
Function
Operator
x2 /2
a)
e
b)
e4 i
Solutions 1
1.
max = hc/(5kT) = 6.6260810-34 J s 2.99792558108 m s-1/ (5 1.3806510-23 J
K-1 5800 K) = 4.9610-7 m = 496 nm - green
2.
= h/p = h/(mv)
v = h/(m) = 6.6260810-34 J s / (9.1093810-31 kg 0.03 m) = 0.0242 m s-1
3.
p = h/ = 6.6260810-34 J s / 3501