CHEM 232 PS1 Exam with Complete Solutions
CHEM 232 A Review of General Chemistry Problem Set Chapter 1
1. Rank the items in each set below according to the trends observed for the physical and chemical properties
indicated.
FC = Ve - Bonds - dots electronegativity = EN
A. Rank the atoms in bold in order of increasing value of their formal charge (1 = most negatively charged; 4
= most positively charged). All lone electron pairs are shown for all atoms. Assume no implicit Hs on bold
atoms. 2 bonds /2
3 bonds / 2 dots 4 bonds / 0 3 bonds / 2 dots
LP = 5 -3 -2 = 0 dots
= 3 - 4 = -1 neutral = 4 - 2 -2 = 0
=6-3-2 = +1
EN of P = 2.1 EN of C =
EN of Cl = 3.0 2.5 EN of H
= 2.0
4 1 3 Polarized towards 2
Cl
1 bond / 6
dots 3 bonds / 0 1 bond / 6 3 bonds / 0 dots
B. = 6 - 1- 6 = - dots dots = 4 - 3 = +1
1 =3-3=0 =5-1-6
neutral = -2
2 bonds / 4 dots
= 5 - 2 - 4 = -1
3 bonds / 2 dots 4
2 = 5 - 3 1- 2 = 0 has point charge
3 neutal
2 bonds / 4 4 bonds / 0 dots
dots = 5 - 4 = +1
C. = 5 -2 - 4 = -
1
diffused
charge
N N N N N N N N N P
M CH3
Me Me
e CH3
2 4 3 1
D. The skeletons (atoms and single bonds only) have been drawn for four molecules below. Rank
them in order of increasing number of remaining electrons that must be added to each as
bonded or lone electron pairs to complete the Lewis structure (1 = least remaining electrons;
4 = most remaining electrons). Assume each molecule is neutral unless otherwise indicated.
3–
1
, = 12 electrons O
+P H
O +
Cl Be O O S O H B
Cl O
H
2 4 3 1
= 24 electorns
= 14 electrons =0
2
CHEM 232 A Review of General Chemistry Problem Set Chapter 1
1. Rank the items in each set below according to the trends observed for the physical and chemical properties
indicated.
FC = Ve - Bonds - dots electronegativity = EN
A. Rank the atoms in bold in order of increasing value of their formal charge (1 = most negatively charged; 4
= most positively charged). All lone electron pairs are shown for all atoms. Assume no implicit Hs on bold
atoms. 2 bonds /2
3 bonds / 2 dots 4 bonds / 0 3 bonds / 2 dots
LP = 5 -3 -2 = 0 dots
= 3 - 4 = -1 neutral = 4 - 2 -2 = 0
=6-3-2 = +1
EN of P = 2.1 EN of C =
EN of Cl = 3.0 2.5 EN of H
= 2.0
4 1 3 Polarized towards 2
Cl
1 bond / 6
dots 3 bonds / 0 1 bond / 6 3 bonds / 0 dots
B. = 6 - 1- 6 = - dots dots = 4 - 3 = +1
1 =3-3=0 =5-1-6
neutral = -2
2 bonds / 4 dots
= 5 - 2 - 4 = -1
3 bonds / 2 dots 4
2 = 5 - 3 1- 2 = 0 has point charge
3 neutal
2 bonds / 4 4 bonds / 0 dots
dots = 5 - 4 = +1
C. = 5 -2 - 4 = -
1
diffused
charge
N N N N N N N N N P
M CH3
Me Me
e CH3
2 4 3 1
D. The skeletons (atoms and single bonds only) have been drawn for four molecules below. Rank
them in order of increasing number of remaining electrons that must be added to each as
bonded or lone electron pairs to complete the Lewis structure (1 = least remaining electrons;
4 = most remaining electrons). Assume each molecule is neutral unless otherwise indicated.
3–
1
, = 12 electrons O
+P H
O +
Cl Be O O S O H B
Cl O
H
2 4 3 1
= 24 electorns
= 14 electrons =0
2