http://www.chem.ucla.edu/~rgiafe/chemistry.html
Name:
______________________________________________
November
1,
2010
TA:
Robert
20A
Disc
1/2
Quantum
Mechanics
and
Atomic
Structure
Worksheet
2
1. Which
of
the
following
combinations
are
allowed
for
an
electron
in
a
one‐electron
atom?
a. n
=
4,
l
=
3,
m
=
2,
ms
=
1
not
allowed,
ms
=
±½
b. n
=
8,
l
=
6,
m
=
‐1,
ms
=
½
allowed
c. n
=
1,
l
=
2,
m
=
1,
ms
=
‐½
not
allowed,
l
=
n‐1…0
d. n
=
3,
l
=
2,
m
=
0,
ms
=
½
allowed
2. Draw
the
probability
density
function
for
l
=
1
in
three
dimensional
space.
Label
the
axes.
3. Draw
the
probability
density
function
for
l
=
2
in
three
dimensional
space.
Label
the
axes.
4. How
many
radial
nodes
are
present
for
n
=
3,
l
=
1?
How
many
angular
nodes?
Radial
nodes
=
n
–
l
–
1
=
3
–
1
–
1
=
1
radial
node
Angular
nodes
=
1
€
5. The
effective
charges
for
nitrogen
are:
Zeff(1s)
=
6.66,
Zeff(2s)
=
3.85,
and
Zeff(2p)
=
3.83.
Estimate
the
energy
level
diagram
for
the
1s,
2s,
and
2p
orbitals
of
nitrogen.
[Z eff (n)]2
εn = −
(2.18 *10−18 J) n2
[6.66]2
ε1s = −
(2.18 *10−18 J) =
‐9.67
x
10‐17
J
2
1
ε2s
=
‐8.07
x
10‐18
J
ε2p
=
‐7.99
x
10‐18
J
€
109
=
G
106
=
M
103
=
k
10‐1
=
d
10‐2
=
c
10‐3
=
m
10‐6
=
µ
10‐9
=
n
10‐12
=
p
10‐15
=
f
http://www.chem.ucla.edu/~rgiafe/chemistry.html
€
€
€
€
6. The
effective
charges
for
carbon
are:
Zeff(1s)
=
5.67,
Zeff(2s)
=
3.22,
and
Zeff(2p)
=
3.14.
Calculate
the
average
distance
of
the
electron
from
the
nucleus
in
the
1s,
2s,
and
2p
orbitals.
n 2 a0  1  ( + 1) 
r n =
1+ 1−

Z eff (n)  2 
n 2 
12 (5.29 *10−11 m  1  0(0 + 1) 
r10 =
1+ 1−
 =
1.40
x
10‐11
m

2


5.67
1
 2

r 20 =
9.86
x
10‐11
m
r 21 =
8.42
x
10‐11
m
7. Give
the
ground‐state
electron
configurations
for
the
following
elements:
a. He:
1s2
b. B:
[He]2s22p1
c. S:
[Ne]3s23p4
d. Cl:
[Ne]3s23p5
e. Zn:
[Ar]4s23d10
f. Kr:
[Kr]
or
[Ar]4s23d104p6
g. Na:
[Ne]3s1
h. Cr:
[Ar]4s13d5
(exception!!)
i.
Mn:
[Ar]4s23d5
j.
Fe:
[Ar]4s23d6
k. Re:
[Xe]6s24f145d5
l.
Br1‐
:
[Kr]
or
[Ar]4s23d104p6
m. Rb1+
:
[Kr]
or
[Ar]4s23d104p6
n. Zr2+
:
[Kr]5d2
o. Al3+
:
[Ne]
or
[He]2s22p6
p. Te2‐
:
[Xe]
or
[Kr]5s24d105p6
q. Pb4+
:
[Xe]4f145d10
109
=
G
106
=
M
103
=
k
10‐1
=
d
10‐2
=
c
10‐3
=
m
10‐6
=
µ
10‐9
=
n
10‐12
=
p
10‐15
=
f