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Chemistry Study Guide: Valence Electrons, Lewis Structures, and Molecular Geometry - Prof., Assignments of Chemistry

Central Oregon Community CollegeChemistry
Prof. Zelda Ziegler

Material Type: Assignment; Professor: Ziegler; Class: General Chemistry I; Subject: Chemistry ; University: Central Oregon Community College; Term: Unknown 1989;

Typology: Assignments

Pre 2010

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Ch 221-Ziegler KEY CONCEPTS Chapters 8 & 9
10th Edition
Class period 1 -- Reading pp. 300- 304
1. State the number of valence electrons for a main-group element. Problems 8.1, 8.7
2. Draw the electron-dot symbol for a main-group element. Problems 8.11, 8.17
3. State the “octet rule.” Problem 8.8
Class period 2 -- Reading pp. 386 – 318
4. Identify periodic trends in electronegativity. Problem 8.37
5. Predict the polarity of a bond. Problem 8.39, 8.40
6. Draw Lewis structures for molecules and polyatomic ions. Problems 8.45, 8.46.
Class period 3 -- Reading pp. 320 – 324; 344 – 353; 357 - 358
7. Use calculated Formal Charges to evaluate Lewis structures. Problem 8.47, 8.50
8. Write resonance forms. Problems 8.51, draw the resonance forms for 8.50 a & b
9. Use the Lewis structure to predict molecular geometry and bond angles. (Note: you will
be required to do this only for molecules and ions in which the octet rule is satisfied for all
atoms.) Problems 9.21 a-d., 9.25, 9.26
10. From the geometry of a polyatomic ion or molecule, predict whether it is polar or
nonpolar. Problems 9.32, 9.35, 9.37.
Class period 4 -- Reading pp. 360 – 365(t); 365 (m) – 369
11. State the hybridization in a covalently bonded species. Problem 9.54 a.
12. State the number of pi and sigma bonds in a molecule or polyatomic ion. Problem 9.54 b-
e; and for the molecules in Problems 9.25, and 9.26, state the number of sigma and pi
bonds in each molecule.
Class Period Key Concepts
1 1, 2, 3, 4
2 5, 6, 7, 8, 9
3 10, 11, 12
Monday - Thursday Section
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) H2CO
H
C
H
O
b) H2O2
OO
H
H
pf2

Partial preview of the text

Download Chemistry Study Guide: Valence Electrons, Lewis Structures, and Molecular Geometry - Prof. and more Assignments Chemistry in PDF only on Docsity!

Ch 221-Ziegler KEY CONCEPTS Chapters 8 & 9

10 th^ Edition

Class period 1 -- Reading pp. 300- 304

1. State the number of valence electrons for a main-group element. Problems 8.1, 8.

2. Draw the electron-dot symbol for a main-group element. Problems 8.11, 8.

3. State the “octet rule.” Problem 8.

Class period 2 -- Reading pp. 386 – 318

4. Identify periodic trends in electronegativity. Problem 8.

5. Predict the polarity of a bond. Problem 8.39, 8.

6. Draw Lewis structures for molecules and polyatomic ions. Problems 8.45, 8.46.

Class period 3 -- Reading pp. 320 – 324; 344 – 353; 357 - 358

7. Use calculated Formal Charges to evaluate Lewis structures. Problem 8.47, 8.

8. Write resonance forms. Problems 8.51, draw the resonance forms for 8.50 a & b

9. Use the Lewis structure to predict molecular geometry and bond angles. (Note: you will

be required to do this only for molecules and ions in which the octet rule is satisfied for all

atoms.) Problems 9.21 a-d., 9.25, 9.

10. From the geometry of a polyatomic ion or molecule, predict whether it is polar or

nonpolar. Problems 9.32, 9.35, 9.37.

Class period 4 -- Reading pp. 360 – 365(t); 365 (m) – 369

11. State the hybridization in a covalently bonded species. Problem 9.54 a.

12. State the number of pi and sigma bonds in a molecule or polyatomic ion. Problem 9.54 b-

e; and for the molecules in Problems 9.25, and 9.26, state the number of sigma and pi

bonds in each molecule.

Class Period Key Concepts

Monday - Thursday Section

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

Ch 221-Ziegler KEY CONCEPTS Chapters 8 & 9

10 th^ Edition

c) C 2 F 6

C C

F F F F

F

F

d) AsO 3 3-

As O O

O

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

O S O

v.e- 6 6 6 -(e on each atom) 6 5 7 Formal Chg: 0 +1 -

S O

(as per ch 4. )

b)SO 3

There are 3 resonance forms.

O S O O

O O S O

v.e- 6 6 6 6 -(e on ea. atom) 7 7 4 6 Formal Chg: -1 -1 +2 0

S O

c) SO 3 -

O S O

O 2

O O S O

v.e- 6 6 6 6 -(e on ea.atom) 7 7 5 7 Formal Chg: -1 -1 +1 -

S O

d) SO 4 -

O S

O

O

O

2 -

O O S O O

v.e- 6 6 6 6 6 -(e on ea. atom) 7 7 5 7 7 Formal Chg: -1 -1 +2 -1 -

S O

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.

9.54 a) Carbons labeled in order from left to right in the book’s figure:

C1: sp^3 ; C2: sp^2 C3: sp^3 ; C4: sp^3

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