You must answer all questions 1-9. You may chose any 3 questions from 10-15 to answer.
1. (12 points) Give the electron configuration for the following atoms or ion
Ge
[Ar] 4s 2 3d 10 4p 2
Cu
[Ar] 4s 1 3d 10
(this is one of the oddballs you have to memorize)
S 2-
[ Ne] 3s 2 3p 6 -or- [Ar]
Mn 2+
[Ar] 4s 2 3d 3
2 A (5 points) Arrange the atoms N, F, Ga, As from smallest to largest ionization energy.
Smallest: Ga, As, N, F : Largest
B (5 points) Arrange the atoms P, Cl, Rb, Sb in size from smallest to largest.
Smallest: Cl, P, Sb, Rb : Largest
3A (4 points) When we talked about the reactivity of the groups IA metals, I said for gas or solid reactions, the reactivity of K>Na>Li. Why?
Reactivity of group 1A metals in gas or solid form correlates with 1 st ionization energy; that is, the lower the ionization energy, the greater the reactivity. Since the lower ionization energy is lower on the period table, the reactivity goes in the order shown.
B (4 points) On the other hand, in solution chemistry when you rate these three metals for their reducing power Li>K>Na. Why?
When these metals react in solution, you have an additional interaction between the metal atoms and the water. This interaction is strongest for the smallest atom, Li and smallest for the larger atom,
Na. This interaction dominates the reduction potential, so you get the reactivity given in this problem .
C (4 points) On yet the other hand, when you react the pure metal with water K >Na >> Li. Why?
Seeing as how this is a reduction of H
2
O in solution you would expect it to correlate with the reduction potential order given in 3B. However there is yet one more additional factor. This reaction gets HOT. Enough heat is generated to melt both the K and the Na. Once these metals melt, they form a liquid, and the liquid has a larger surface to react with the water, hence they become much more reactive. The reaction is not hot enough to melt the Li, so it stays in a solid chink, and react more slowly.

4. (10 points) Rank the ions Na + , Al 3+ , S 2, Cl by size from smallest to largest.
of these ions Cl and S 2 have the larger [Ar] electron configuration while
Na + and Al 3+ have the smaller [Ne] configuration.
For ions with the same electron configuration, the smaller ions are the ones with the larger atomic number because they have the most positive charge in the nucleus and have the largest
‘pull’ on the electrons.
Smallest: Al 3+ , Na 1+ , Cl , S 2 : largest
5. (10 points) Rank the compounds KBr, CaSe, and RbI from smallest to largest lattice energy.
smallest: RbI, KBr, CaSe: largest lattice energy has the general form
Lattice Energy = k


Q Q
1 2 r


Lets look at r, the distance between the atoms
K, Br, Ca, and Se are in the 4 th period and are smaller than Rb and I that are in the 5 th period, this gives them a stronger lattice energy. For KBr vs CaSe, the charge for K and Br is
+/- 1, while the charge for Ca and Se is +/- 2 so the CaSe has the larger lattice energy
6. (15 points) Using the following bond energy information:
H-H 432
H-F 565
H-Cl 427
C-H
C-C
C-F
413
347
485
C-Cl
F-F
339
Cl-Cl 239
154
Calculate the energy for the reaction:
CH
4
(g) + 2Cl
2
(g) + 2F
2
(g) 6 CF
2
Cl
2
(g)+ 2HF(g) + 2 HCl(g)
(CF
2
Cl
2
is Freon-12 a common refrigerant)
) H = energy required to break bonds in reactants - energy released when bonds form in products
= 4(C-H) + 2(Cl-Cl) + 2(F-F) - [2(C-F) + 2(C-Cl) + 2(H-F) + 2(H-Cl)]
= 4(413) + 2(239) + 2(154) - [2(485) + 2(339) + 2(565) + 2(427)]
= 1652 + 478 + 308 - [970 + 678 + 1130 + 854]
= 2,428-3,632
= -1,194 kJ/mol
2

7. (16 points) Draw Lewis Structures for the following compounds
C
2
H
2
H C C H
N
2
N N
ICl
4
-
Cl
Cl
I
Cl
Cl
-1
BeCl
2
Cl Be Cl
3

8. Draw at least two nonequivalent Lewis structures for SCN and evaluate your structures using formal charge. Which is the best structure?
4
-1
-1
-1
S C N S C N
S C N
Formal Charge
I.
S: 6-5 = 1
C: 4-4 = 0
N: 5-7 = -2
II.
S: 6-6 = 0
C: 4-4 = 0
N: 5-6 = -1
III.
S: 6-7 = -1
C: 4-4 = 0
N: 5-5 = 0
Which is the ‘best’ structure?
I has two atoms with nonzero formal charge, while II and III have only 1 atom with nonzero formal charge, so I is worse than II or III. Based on your knowledge of the periodic table you cannot tell whether N or S is the more electronegative, so I would accept either II or III as being better, if you took a guess which was more electronegative. When you actually look up the numbers, N (3) is more electronegative than S (2.5) so II should be the best structure.