Chemistry 217
Problem Set 1
Recommended Problems from the Book (1st ed. in parentheses): 1.1, 1.3-1.9, 1.36-1.42,
1.57-1.62 (1.1-1.8, 1.34-1.40, 1.54-1.58)
Supplemental Problems: Lewis structures: Weeks, pp. 1-14. Formal Charge: Weeks,
pp. 20-25; Klein, section 1.5. Lone pairs: Klein, section 1.6. Skeletal structures
(condensed structures): Klein, section 1.1-1.4.
1.
Draw the Lewis structure for each of the following molecules. Show all lone
pairs. All molecules are neutral and all formal charges are zero.
H
H
H
(a)
CH2Cl2
C
(CH3)2O (same as CH3OCH3)
H
H
H
(d)
CH3CH2CO2CH3
H
(b) (CH3)3COH
H
(c)
C
H
Cl
H
C
H
O
C
H
C
O
H
C
H
H
H
C
H
H
C
H
Cl
H
H
H
H
C
O
O
C
O
H
C H
H
H
(e)
CH3CHO
H
H
H
C
H
H C H
H
H
C
H
C
H
C
C
C
H
H
H
N
H
C
H
H
(f)
CH3CH(CH2CH3)CH2NHCH3
2.
Each of the following molecules is neutral. Determine the formal charge for each
atom in the molecules (All lone pairs are shown).
H
H
4-1/2(6)-2=-1
1-1/2(2)-0=0
H
CH3
H
H
C
C
H
P
CH3
C
CH3
H
(a)
4-1/2(4)-2=0
(b)
5-1/2(8)=+1
CH3
H
(c)
P
CH3
CH3
all atoms have f.c.=0
5-1/2(8)-0=+1
O
H3N
CH3
6-1/2(2)-6=-1
H3C
O
(d)
C
all atoms have f.c.=0
CH3
(e)
4-1/2(6)-0=+1
O
O
C
O
O
all atoms have f.c.=0
(f)
6-1/2(2)-6=-1
(g)
4-1/2(6)-2=-1
5-1/2(8)-0=+1
H2C
N
H2C
N
5-1/2(4)-4=-1
(h)
3.
C
N
N
5-1/2(8)-0=+1
(i)
Fill in the lone pairs on the following molecules. All non-zero formal charges are
shown.
N
O
(a)
H3C
(d)
H3C
Si
(b)
(c)
H
O
Cl
CH3
N
(d)
no lone pairs
H
(e)
H
C
H
no lone pairs
4.
In class, we determined the rules of thumb in determining formal charge for
carbon. This information is shown in the first rows of the table below. Fill in the
remaining blank spaces for H, O, and N. Remember that you cannot violate the octet
rule.
Atom
C
H
O
N
# of bonds
4
3
3
1
0
0
2
3
1
3
4
2
# of lone pairs
0
0
1
0
0
1 (2 unshared)
2 (4 unshared)
1 (2 unshared)
3 (6 unshared)
1 (2 unshared)
0
2 (4 unshared)
charge
0
+1
-1
0
+1
-1
0
+1
-1
0
+1
-1
Rationale:
We must fulfill the octet rule for each one.
Remember: F.C. = # electrons on free atom – ½ (shared) – unshared
H:
f.c. = 0 = 1-1 the bonds plus the unshared electrons must total 1. Therefore, there must
be one bond, and no unshared pairs.
f.c. = +1 = 1 – 0 the bonds plus the unshared electrons must equal 0. There cannot be any
bonds or unshared pairs.
f.c. = -1=1-2 the bonds plus the unshared electrons must equal 2. If there are two bonds,
or one bond and one unshared electron, the octet rule is violated. Therefore, there must
be no bonds and one lone pair.
O:
f.c.=0=6-6 the bonds plus the unshared electrons must total 6. Possibilities:
1 bond, 5 unshared electrons: violates octet
2 bonds, 2 lone pairs: o.k.
3 bonds, 3 unshared electrons: violates octet
4 bonds, one lone pair: violates octet
f.c.=+1=6-5 the bonds plus the unshared electrons must total 5. Possibilities:
1 bond, 2 lone pairs: violates octet
2 bonds, 3 unshared pairs: violates octet
3 bonds, 1 lone pair: o.k.
4 bonds, 1 unshared pair: vioates octet
f.c.=-1=6-7 the bonds plus the unshared electrons must total 7. Possibilities:
1 bond, 3 lone pair: o.k.
2 bonds, 5 unshared: violates octet
3 bonds, two lone pair: violates octet
N:
f.c.=0=5-5 the bonds plus the unshared electrons must total 5. Possibilities:
1 bond, two lone pair: violates octet
2 bonds, 3 unshared: violates octet
3 bonds, one lone pair: o.k.
4 bonds, one unshared: violates octet
5 bonds, no unshared: violates octet
f.c.=1=5-4 the bonds plus the unshared electrons must total 4. Possibilities:
1 bond, 3 unshared: violates octet
2 bonds, 2 unshared: violates octet
3 bonds, one unshared: violates octet
4 bonds, no unshared: o.k.
f.c.=-1=5-6 the bonds plus the unshared electrons must total 6. Possibilities:
1 bond, 5 unshared: violates octet
2 bonds, 4 unshared: o.k.
3 bonds, 3 unshared: violates octet
4 bonds, two unshared: violates octet
5 bonds, none unshared: violates octet
5.
Fill in the lone pairs and formal charges on the atoms in the following molecules
(all hydrogens are shown).
F
B
O
F
HO
F
(a)
All atoms have
f.c.=0
H
(d)
H
C
(b)
(e)
H
C
O
H
all atoms have
f.c.=0
H
F
C
(c)
H O C H
H3C
F
(f)
O
C
H
O
CH3
O
H3C
(g)
O
C
CH3
-2
O
(h)
H
O
S
O
(i)
N
N
N
6.
Draw valid Lewis structures for the following. For each one, assign all non-zero
formal charges and the overall charge on the molecule.
H
(a)
(1) Arrange atoms in order:
H
H
H
N
C
C
H
H
H
N
(2) Fill in bonds and electrons to fulfill octet.
H
H
H
H
N
C
C
H
N
H H
Add in single bonds first:
The octet is not fulfilled for the carbon and
nitrogen on the right side of the molecule. If we add a double bond between them, the
octet will be fulfilled for the carbon. Then, to fulfill the octet for nitrogen, add a lone pair
(formal charges shown below):
H
H
H
H
N
C
C
H
H
H
N
Overal charge=+1
If we added lone pairs to the carbon and nitrogen to fulfill their octet, we would get:
H
H
H
H
N
C
C
H
N
H H
which has an overall negative charge of -1. This Lewis structure is not
as good as the first, because more atoms contain charge.
(b) CH3CH2CO2
H
(1) Arrange atoms in order:
H
H
O
C
C
C
H
H
O
(2) Fill in bonds and electrons to fulfill octet.
H
H
H
O
C
C
C
O
H H
Add in single bonds first:
The octet is not fulfilled for the carbon on the
right and both oxygens. If we add a double bond between one of the oxygens and the
carbon, it will be fulfilled for the carbon. Then, to fulfill the octet for the oxygens, add in
lone pairs (formal charges shown below):
H
H
H
O
C
C
C
H
H
O
H
or
H
H
O
C
C
C
H
H
O
(both valid Lewis structures) overall charge=-1
(c) CH2CHCH2
(1) Arrange atoms in order:
H
H
H
H
C
C
C
H
(2) Fill in bonds and electrons to fulfill octet.
H
H
H
C
H The octet is not fulfilled for any of the carbons.
Add in single bonds first:
If we add a double bond between two of them and a lone pair to the third, it will be
fullfilled for all three:
H
H
H
H
C
C
C
H
H
or
H
C
H
H
H
C
C
C
C
H
overall charge = -1
Another solution to this problem would be to not fulfill the octet for one of the carbons,
since a carbon with a positive charge is one of the exceptions to the octet rule. In this
case, the answer would be:
H
H
H
H
C
C
C
H
H
or
H
H
H
C
C
C
H
overall charge = +1
(7)
Draw constitutional isomers which fulfill the following criteria:
(a)
Pentane: C5H12 (three different isomers)
(b)
C4H10O
OH
HO
HO
O
(c)
HO
O
C5H10 (Draw at least four. Two should be cyclic molecules. Two should
contain a carbon-carbon double bond.
cyclic:
double bonds:
(d)
C6H10 (Draw at least one isomer of compounds containing the following
types of functionality: 1) a compound with two double bonds; 2) a
compound with a ring and a double bond; 3) a compound with a triple
bond; 4) a compound with two rings
A few examples (there are many others)
two double bonds:
double bond and ring:
triple bond:
two rings:
(e)
C5H10O2 (Draw as many as you can.)
There are many, many constitutional isomers. Here are a few. Just make sure that your compounds
have one double bond or ring in them.
O
HO
O
O
O
O
O
O
OH
O
OH
O
O
OH
OH
OH
OH
O
OH
O
O
(8)
Give the molecular formula of the following compounds:
OH
HO
O
HO
N
O
N
HO
nicotine
(a)
C10H14N2
HO
OH
cholesterol
vitamin C
(b)
C6H8O6
(c)
C27H46O