Gateway 125,126, 130 Fall 2006 Studio 3a p
1) Define Aufbau principle, Pauli Exclusion Principle, and Hund's Rule.
2) List and describe the four quantum numbers.
3) Count valence electrons
4) Write electron configurations
Reading: 7.57.6 p 288299; 7.7 p 301306
Group roles: A Leader; B Recorder; C Reporter
I. The Atomic Hotel is a special hotel designed for electrons. The hotel has a strict policy called the Aufbau principle that states "ground" floors must be filled first and in order. It costs more to get rooms on higher floors. In the atomic world energy is money so "excited" electrons get rooms on higher floors, thus the exception to hotel policy.
Looking back through the old guest logs the following layouts can be sketched. Determine which electrons had more "money". Label those electrons as excited.
II. Another policy that the hotel enforces is called the Pauli Exclusion Principle. It was determined in 1925 that electrons can occupy the same room, but only if they have opposite spins so they do not interfere with one another.
Which electrons below did not follow the Pauli exclusion principle?

The Pauli Exclusion Principle serves to identify each electron. Four numbers (n, l, m_{l}, and m_{s}) are assigned to each guest in the hotel. The number n, is the principle quantum number which corresponds to the "floor" the electron is on. The letter l describes the room layout or the "floor area" in which the electron is staying (s = 0, p = 1, d = 2, etc.). There is only one s type room on each floor; there are three p type rooms on each floor from the second up; there are five d type rooms on each floor starting with the third floor and going up; and there are seven f type rooms on each floor starting with the fourth floor and moving upwards. The letter m_{l} describes the specific room (most analogous to a room number.) Figure 1 shows a diagram of the hotel rooms available by floor.
Figure 1: The Atomic Hotel by Floor
s
p
d
f
4
Floor n = 1, 2, 3….
Room type l
s = 0
p = 1
d = 2
f = 3
m_{l} specific room
0
1 0 +1
2 1 0 +1 +2
3
3 2 1 0 +1 +2 +3
0
1 0 +1
2 1 0 +1 +2
2
0
1 0 +1
1
0
The preferred hotel diagram is one which shows the hotel rooms by cost (which, as stated in part I is the also the order in which the rooms are filled. Generally, the most inexpensive rooms are on the first floor, with prices increasing with floor. Note that d rooms are more expensive than the s or p rooms on the next floor. Electrons can pair up in a single room, therefore, the hotel has to have a way of identifying each one separately. This is done with m_{s}, the spin quantum number which is +½ or ½.
4f
3  2 1 0 +1 +2 +3
6s
0
5p
1 0 +1
4d
5s
2 1 0 +1 +2
0
4p
1 0 +1
3d
2 1 0 +1 +2
NOTE: d and f suites are more expensive than the s rooms on the next floor/s (3d > 4s).
4s
0
3p
1 0 +1
3s
Example: an electron with the numbers 3, 0, 0, ½ is staying on the 3^{rd} floor, in section s, in room #0, and is in + ½ spin state.
0
2p
1 0 +1
2s
0
1s
0
Using the above pattern, identify the circled electrons using the four quantum numbers.
IV. There is one more policy that helps the hotel run smoothly and keep customers happy. Hund's Rule states that if electrons are being placed in the same section of a floor (rooms that cost the same) then each one gets their own room and has the same spin until the floor is halffilled. If any more electrons want to stay in that same section, then they must pair up with another electron and assume the opposite spin. This does not necessarily apply to electrons that have purchased more expensive rooms.
In the diagrams below identify the areas in which Hund's Rule was broken. Describe how the rule is being broken in each case.
A
B
C
V. OBSERVATIONS:
A. How many "rooms" are there in an "s" suite of a floor? _____
B. How many "rooms" are there in a "p" suite of a floor? _____
C. How many "rooms" are there in a "d" suite of a floor? _____
D. How many electrons can stay on the first floor? _____
second floor? _____
third floor? _____
E. Describe how the rooms in each section are numbered.
F. In diagram C (Section IV above), describe why the electron could go into the 4s
orbital instead of the 3p.
G. Why might section 3d fall between section 4s and 4p?
Draw hotels for and write out the electron configurations for each of the following elements:
B, N, F, Na
Then identify the atoms whose atomic hotels were drawn in IC, IIB, IIIA, IIIB, IIIC, and IVB
34) Valence electrons: Electrons in an atom’s highest occupied shell (s and p electrons) and in partially filled subshells of lower shells d or f electrons).
Example 1: K (potassium) has the electron configuration 1s^{2}2s^{2}2p^{6}3s^{2}3_{p}^{6}4s^{1} Its highest occupied shell is 4 and it has one electron in the 4s orbital so it has one valence electron.
Example 2: S (sulfur) has the electron configuration 1s^{2}2s^{2}2p^{6}3s^{2}3_{p}^{4}. Its highest occupied shell is 3 and it has two electrons in the 3s orbital and four electrons in the 3p orbital. Sulfur has six total valence electrons.
Example 3: Co (cobalt) has the electron configuration 1s^{2}2s^{2}2p^{6}3s^{2}3_{p}^{6}4s^{2}3d^{7}. Its highest occupied shell is 4 and it has two electrons in the 4s orbital. Cobalt also has 7 electrons in a partially filled 3d orbital so it has a total of nine valence electrons.
Example 4: Se (selenium) has the electron configuration 1s^{2}2s^{2}2p^{6}3s^{2}3_{p}^{6}4s^{2}3d^{10}4p^{4}. Its highest occupied shell is 4. Selenium has two electrons in the 4s orbital and four electrons in the 4p orbital. Selenium has a full 3d orbital so these ten electrons are NOT valence electrons. Selenium has six total valence electrons.
How many valence electrons do the following elements have?
F _____ Mo _____
Sb _____ Rb _____
Nd _____ Ag _____
Electron configurations can get cumbersome (think about the 107 electrons in Bohrium). They can be abbreviated using the noble gas (group 8) core notation to abbreviate most of the configuration.
Example1:
Mg as the electron configuration: 1s^{2}2s^{2}2p^{6}3s^{2}; Ne has the electron configuration 1s^{2}2s^{2}2p^{6}
The electron configuration of Mg can be abbreviated: [Ne] 3s^{2}
Example 2:
Ru 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{6} which can be abbreviated: [Kr]5s^{2}4d^{6}
Using the noble gas core notation, write the electron configurations for:
P
Sr
Co
As
