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Name:
________________________
Student Number:
___________________
CHM 2311
Final Exam
April 29 2009
Professor Darrin Richeson
There are 15 pages in this exam (including supporting materials and tables).
Please count the pages to make sure none are missing.
You may
carefully remove
the last three pages of the exam and use them
for scratch work.
Please write legibly and show your work to receive credit for your answers.
Partial marks
may in some cases
be awarded for partially correct work.
Question
Mark
Question
Mark
1
/5
8
/4
2
/ 9
9
/13
3
/ 7
10
/16
4
/ 16
11
/5
5
/ 10
12
/12
6
/9
7
/4
Total
/110
Good luck and have a great summer!
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1. (
5 points
) A hydrogen orbital has the following
complete
wave function (both
angular and radial portions of the wave function are included and the angular part is
written in Cartesian coordinate form):
{ }
( )
−
=
=
−
2
2
2
3
/
2
2
/
3
15
4
/
1
2
15
81
1
)
,
(
)
(
)
,
,
(
r
y
x
e
a
Z
Y
r
R
r
o
π
ρ
ϕ
θ
ψ
where
o
a
Zr
=
and Z = 1 for a hydrogen atom
a
o
= 52.9 pm
a.
For this orbital, determine the distance from the nucleus in picometers at which
all
radial
nodes occur.
The radial component becomes zero (has a node) when
2
= 0 or
= 0 = Zr/a
o
This is only true at the r = 0 – the nucleus
no radial nodes
b.
Determine the quantum numbers n and l for this orbital.
Clearly
show the process
by which you determined these values. Identify the orbital (1s, 2s, 2p
x
, etc.
..).
In part (a) we determined that there are
no radial node
s.
The number of angular nodes can be determined from the angular part of
the wavefunction. From the equation,
= 0 for x
2
y
2
= 0 (and all values for
z). Rearranging gives a node for x = y and for x = y
There are 2 angular nodes. This corresponds to l = 2 or a d orbital.
The equation for the number of radial nodes is given by nl1 = number of
radial nodes. Since nl1= 0 then n = 2+1 = 3
Therefore this is a 3d orbital and specifically the 3d
x2y2
orbital.
3
2.
(9 points)
a.
Calculate Z
eff
for a 2p electron on each of the following species.
O
2
, F

, Na
+
, Mg
2+
.
Each of these ions has the same electronic configuration:
(1s
2
) (2s
2
, 2p
6
)
This means that they all have the same s value
s =(2 x 0.85) +
(7 x 0.35) = 4.15
Z
eff
= Z
actual
 s
Ion
Z
Z
eff
O
2
8
3.85
F

9
4.85
Na
+
11
6.85
Mg
2+
12
7.85
Match the ion with the radii in the following table.
Crystal (Ionic) Radius
Ion
119pm
86pm
116pm
126pm
Calculate the Z
eff
for a 4d electron of antimony (symbol Sb)
(1s2)
(2s2, 2p6)
(3s2, 3p6)
(3d10)
(4s2, 4p6)
(4d10) (5s2, 5p3)
Z
eff
= Zs = 51 9x0.35 + 36x1 = 5139.15 = 11.85
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3.
(7 points)
There are potentially three isomers for the anion NSO

. The two known
species are NSO
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This note was uploaded on 01/22/2011 for the course BCH 2333 taught by Professor Mezl during the Winter '08 term at University of Ottawa.
 Winter '08
 Mezl
 Biochemistry

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