Valenica, Daniel – Exam 1 – Due: Oct 4 2007, 11:00 pm – Inst: Brodbelt
1
This printout should have 27 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
Brodbelt CH301
Exam 1
October 3, 2007
Programmable calculators and cell phones
are prohibited. You may not leave the room
until you turn in your exam.
GOOD LUCK!
001
(part 1 oF 1) 10 points
1
s
2
2
s
2
2
p
6
3
s
2
3
p
5
is the electron confgura
tion For which one oF the Following ions?
1.
P
2+
2.
S
+
3.
P
+
4.
P
2

correct
5.
Cl

Explanation:
We need an ion that has 2 + 2 + 6 + 2 + 5
electrons;
i.e.
, P
2

.
002
(part 1 oF 1) 10 points
IF the Following equation is properly balanced,
what are the coe±cients in Front oF each com
pound?
AlCl
3
+ SCl
2
→
Al
2
S + Cl
2
1.
2; 2; 1; 5
2.
1; 2; 1; 2
3.
2; 2; 2; 5
4.
1; 1; 1; 1
5.
2; 1; 1; 4
correct
6.
1; 2; 1; 4
7.
2; 2; 1; 4
8.
2; 1; 2; 4
Explanation:
AlCl
3
+ SCl
2
→
Al
2
S + Cl
2
I’ll start with Al:
2AlCl
3
+ SCl
2
→
Al
2
S + Cl
2
Al and S are balanced. Balance Cl:
2AlCl
3
+ SCl
2
→
Al
2
+ 4Cl
2
Answer: 2, 1, 1, 4
003
(part 1 oF 1) 10 points
An electron occupies an atomic orbital that
has the shape
∞
. All you can say about its
Four quantum numbers is
1.
n
= 1 or 2 or higher;
‘
= 1;
m
‘
=

1
,
0
,
+1;
m
s
=
±
1
2
.
2.
n
= 2 or higher;
‘
= 1;
m
‘
=

1
,
0
,
+1;
m
s
=
±
1
2
.
correct
3.
n
= 1;
‘
= 1;
m
‘
=

1
,
0
,
+1;
m
s
=
±
1
2
.
4.
n
= 0;
‘
= 0;
m
‘
= 0;
m
s
=
±
1
2
.
5.
n
= 2 or higher;
‘
= 2;
m
‘
=

2
,

1
,
0
,
+1
,
+2;
m
s
=
±
1
2
.
Explanation:
Since
∞
is the shape oF a
p
orbital,
‘
must
be 1.
‘
can only be 1 when
n
= 2 or more, and
the possible values oF
m
‘
are + 1, 0, and

1
when
‘
is 1.
004
(part 1 oF 1) 10 points
An electron is confned to a onedimensional
box oF length L. It Falls From the second en
ergy level to the ground state, and releases a
photon with a wavelength oF 322 nm. What is
the length oF the box?