HODSON CHE 1301 Petrucci Ch8 Example Problems
The following problems are taken from old exams .
.....
7. The quantum numbers of the last electron of nickel could be:
a)
n
=
3,
ℓ
=
2,
m
ℓ
=
0,
m
s
=
1/2
b)
n
=
3,
ℓ
=
2,
m
ℓ
=
1/2,
m
s
=
1/2
c)
n
=
3,
ℓ
=
2,
m
ℓ
=
0,
m
s
=
0
d)
n
=
4,
ℓ
=
2,
m
ℓ
=
0,
m
s
=
1/2
e)
n
=
3,
ℓ
=
1,
m
ℓ
=
0,
m
s
=
1/2
Answer: A
Explanation: You need to be able to write out the electron configuration of nickel, which
is element number 28, and so we have to arrange 28 electrons. They are entered into
the lowest energy orbitals possible, the sequence of which we can get from the periodic
table. Starting from hydrogen, first period s-block (1s) we work through (like typewriter
reading left to right on each period) the 2s, 2p, 3s, 3p, 4s and finally to the 3d energy
level. Nickel is the eighth element in from the left, and so the highest energy level
present we can represent as 3d
8
. All the levels below this are filled with as many
electrons as they can take:
1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
8
Note that if you add up all the superscripts you get 28
electrons, which is correct for nickel.
You may see the same configuration written as (switched 3d and 4s levels):
1s
2
2s
2
2p
6
3s
2
3p
6
3d
8
4s
2
The Petrucci text uses this, and so we must for consistency. Why ? The 4s sublevel is