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Material Type: Assignment; Professor: Ziegler; Class: General Chemistry I; Subject: Chemistry ; University: Central Oregon Community College; Term: Unknown 1989;
Typology: Assignments
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Fall 06 10 th^ Ed.
Answers to Problems not in the appendix:
8.1 a) Group IV (AKA 14) b) Group II (AKA 2) c) Group V (AKA 15) 8.8 a) The “octet rule” is the observation that atoms will gain lose or share electrons to achieve the stable configuration (full shell like the noble gases) of 8 valence electrons. The reason it is in quote marks is that it has so many exceptions, specifically, H, He, (which have a full shell of 2) and pretty much everything in the lower periods can have more than 8. b) S needs 2 more electrons. c) This configuration needs only 3 more electrons.
8.40 a) O-F < C-F < Be-F b) C-P< S-Br < O-Cl c) C-S < N-O < B-F
8.46 a) H 2 CO
H C H
O b) H 2 O 2
O O H H
c) C 2 F 6
C C
F F F F
F
F
d) AsO 3 3-
As O O
O
Fall 06 10 th^ Ed.
e) H 2 SO 3
H O (^) S (^) O
H
O f) C 2 H 2 H^ C^ C H
Lewis Structure Formal Charge calculation Oxidation #s a) SO 2
there are 2 resonance forms
O S O
v.e- 6 6 6 -(e on each atom) 6 5 7 Formal Chg: 0 +1 -
(as per ch 4. )
b)SO 3
There are 3 resonance forms.
O S O O
v.e- 6 6 6 6 -(e on ea. atom) 7 7 4 6 Formal Chg: -1 -1 +2 0
c) SO 3 -
O S O
O 2
v.e- 6 6 6 6 -(e on ea.atom) 7 7 5 7 Formal Chg: -1 -1 +1 -
d) SO 4 -
O S
O
O
O
2 -
v.e- 6 6 6 6 6 -(e on ea. atom) 7 7 5 7 7 Formal Chg: -1 -1 +2 -1 -
9.32 The 3 polar bonds in PH 3 would completely cancel each other out if the molecule were flat (planar), and then the net dipole moment would be zero. Since PH 3 is polar, the molecule can’t be flat.
b) total v.e-: 36 c) 26 e- in sigma bonds, 2 e- in pi bonds, 8 e- in nonbonding orbitals
problem sigma bonds pi bonds problem sigma bonds pi bonds 9.25 a 4 0 9.26 a 3 1 b 5 0 b 6 1 c 3 2 c 4 0 d 7 1 d 5 2