Wednesday, February 16

We started class today by going over the answers to the 6.5-6.6 reading sheet. The answers are:
1.
a. 7
b. 2
c. 3
2. a. 2s
b. 2s
c. 3p
d. 4s
3. a. 2
b. 6
c. 3s
d. 3p
e. F
g. Cl

4. a. 2p
b. 3s
c. 18th
d. 3d
e. 37, 38
f. 5p
g. 13th, 14th

5. a. S, Ca
b. Cl, Zn
c. Xe, Ba

6. Ar, 4, 3, 4, 5
7. a
8. b
9. a. 2
b. 5
c. 1
10. O, Na, Sc, Cl
11. This is a violation of Hund’s rule.

We then reviewed how to write electron configurations and learned new information about quantum numbers. Mr. H put this chart on the board.

n=1 l=0-----> 1s orbital ml=0
n=2 l=0-----> 2s ml=0
l=1-----> 2p (3 of these) ml=-1, 0, 1
n=3 l=0-----> 3s ml=0
l=1-----> 3p (3 of these) ml=-1, 0, 1
l=2-----> 3d (5 of these) ml=-2, -1, 0, 1, 2

The n value is the number of possible energy levels that atom can have. Since there are 3 possibilities for 2p, there needs to be 3 different ml values to distinguish them from one another. To find the ml value, you start with -l and go to l. For example, since n=3, l=2 has 5 possibilites, you need 5 different ml values. You start with -2 since l=2 and go to 2. The vlaues would be -2, -1, 0, 1, 2. We then talked about how when l=0 it is an s orbital (spherical). When l=1, it is a p orbital (pinched cylinder). When l=2, it is a d orbital (dorky). We then reviewed what each of the quantum numbers represent.
n=energy level (size)
l=orbital type
ml=sperical orientation
ms=spin direction. The ms value is always +1/2 or -1/2

We did a practice problem where you had to find the possible quzntum numbers for a 3p electron in Cl. The answer is n=3, l=1, ml= -1, 0, 1, ms= 1/2, -1/2
To find this you can use the chart from above. Since we were given that it was a 3p electron, that indicates that n=3. We can also see from the chart above that the p orbital for any n value has a corresponding value of 1 for l, so l=1. If you go across the chart for n=3, l=1, you will find the possible ml values are -1, 0, 1. Also, ms is always 1/2, -1,/2, so that is what it is for this problem.

After this we went to the computer lab to work on our new lab, It's Periodic Lab. We went to www.webelements.com to look at trends in the covalent radius and first ionization energy of elements. The covalent radius is the size of the atom that forms part of a covalent bond. By looking at graphs of the periodic table, we saw that going down the periods, the covalent bonds increase and going across the groups from left to right, the covalent bond decreases. We then observed the trends in the ionization energy of elements. Ionization energy is a measure of how much energy it takes to remove an electron from a gaseous atom. We saw that going down the periods, the ionization energy decreased. Going across the groups from left to right, the ionization energy increased. The exception to this was that the last element of each group in the p block was smaller than the first element in each gorup of the p block. The lab is due Tuesday.

The homework was our delicious assignments due Friday and 2 webassigns due Friday and Tuesday. There are 3 assignments on webassign, but only 2 are for credit. If you choose to do all 3, one will count towards an extra homework drop.