HW 5D Answers
1. Write the Lewis structures for the following. Show all resonance structures where applicable.
a.
O N
O
O
N
O
O
d.
N
O
O
O
O
N
O
N
O
c.
O
N
O
N
O
O
O
N
b.
O N
O
N
O
O
O C N
O C N
O C N
N N N
N N
N N
N
N
e.
f.
g.
O
O
O
S
S
S
O
O
O
H
2.
H
C
H
H
C
C
C
H
H
C
H
C
H
H
H
H
C
C
H
3.
H
C
C
C
H
O
H
C
C
O
O
H
N
C
O
H
C
H
H
N
C
O
H
C
H
N
C
O
4.
The Lewis structures for the various species are:
CO (10 e -):
C
CO (10 e ):
e -e):): O
CO
CO
2 (16
2 (16
O
Triple bond between C and O
Triple bond between C and O.
C
CO32 (24 e -):
DoubleDouble
bond between
C and OC and O.
bond between
O
2-
O
C
O
2-
O
C
O
O
2-
O
C
O
O
O
Average of 1 1/3 bond between C and O in CO32.
CH3OH (14 e):
Single bond between C and O.
As the number of bonds increases between two atoms, bond strength increases, and bond length decreases.
With this in mind, then:
Longest  shortest C – O bond: CH3OH > CO32 > CO2 > CO
Weakest  strongest C – O bond: CH3OH < CO32 < CO2 < CO
5. BF3 has 3 + 3(7) = 24 valence electrons. The two Lewis structures to consider are:
0
F
+1
F
B
-1
B
F
F
0
0
F
0
0
F
0
The formal charges for the various atoms are assigned in the Lewis structures. Formal charge = number of
valence electrons on free atom  number of lone pair electrons on atoms 
1/2 (number of shared electrons of atom). For B in the first Lewis structure, formal charge (FC) = 3  0  1/2(8)
= 1. For F in the first structure with the double bond, FC = 7  4  1/2(4) = +1. The others all have a formal
charge equal to zero [FC = 7 – 6 – 1/2(2) = 0].
The first Lewis structure obeys the octet rule but has a +1 formal charge on the most electronegative element
there is, fluorine, and a negative formal charge on a much less electronegative element, boron. This is just the
opposite of what we expect: negative formal charge on F and positive formal charge on B. The other Lewis
structure does not obey the octet rule for B but has a zero formal charge on each element in BF 3. Because
structures generally want to minimize formal charge, then BF3 with only single bonds is best from a formal
charge point of view.
6.
C
O
Carbon: FC = 4  2  1/2(6) = 1; oxygen: FC = 6  2  1/2(6) = +1
Electronegativity predicts the opposite polarization. The two opposing effects seem to partially cancel to give a
much less polar molecule than expected.
7. OCN has 6 + 4 + 5 + 1 = 16 valence electrons.
Formal
charge
O
C
N
O
C
N
O
C
N
0
0
-1
-1
0
0
+1
0
-2
Only the first two resonance structures should be important. The third places a positive formal charge on the
most electronegative atom in the ion and a –2 formal charge on N.
CNO will also have 16 valence electrons.
Formal
charge
C
N
O
C
N
O
C
N
O
-2
+1
0
-1
+1
-1
-3
+1
+1
All the resonance structures for fulminate (CNO) involve greater formal charges than in cyanate (OCN),
making fulminate more reactive (less stable).
8. SCl, 6 + 7 = 13; the formula could be SCl (13 valence electrons), S2Cl2 (26 valence electrons), S3Cl3 (39
valence electrons), etc. For a formal charge of zero on S, we will need each sulfur in the Lewis structure to have
two bonds to it and two lone pairs [FC = 6 – 4 – 1/2(4) = 0]. Cl will need one bond and three lone pairs for a
formal charge of zero [FC = 7 – 6 – 1/2(2) = 0]. Since chlorine wants only one bond to it, it will not be a central
atom here. With this in mind, only S2Cl2 can have a Lewis structure with a formal charge of zero on all atoms.
The structure is:
9. The nitrogen-nitrogen bond length of 112 pm is between a double (120 pm) and a triple (110 pm) bond. The
nitrogen-oxygen bond length of 119 pm is between a single (147 pm) and a double bond (115 pm). The third
resonance structure shown below doesn’t appear to be as important as the other two since there is no evidence
from bond lengths for a nitrogen-oxygen triple bond or a nitrogen-nitrogen single bond as in the third resonance
form. We can adequately describe the structure of N2O using the resonance forms:
N
N
O
N
N
O
10. Assigning formal charges for all three resonance forms:
N
N
O
N
N
O
N
N
O
-1
+1
0
0
+1
-1
-2
+1
+1
For:
, FC = 5 - 4 - 1/2(4) = -1
N
N
, FC = 5 - 1/2(8) = +1 , Same for
, FC = 5 - 6 - 1/2(2) = -2 ;
N
O ,
FC = 6 - 4 - 1/2(4) = 0 ;
O ,
FC = 6 - 2 - 1/2(6) = +1
N
O
N
and
,
FC = 5 - 2 - 1/2(6) = 0
,
FC = 6 - 6 - 1/2(2) = -1
N
We should eliminate N‒N≡O because it has a formal charge of +1 on the most electronegative element (O).
This is consistent with the observation that the N‒N bond is between a double and triple bond and that the N‒O
bond is between a single and double bond.
11. For formal charge values of zero:
(1) each carbon in the structure has 4 bonding pairs of electrons and no lone pairs;
(2) each N has 3 bonding pairs of electrons and 1 lone pair of electrons;
(3) each O has 2 bonding pairs of electrons and 2 lone pairs of electrons;
(4) each H is attached by only a single bond (1 bonding pair of electrons).
Following these guidelines, the Lewis structure is:
12. For a formal charge of zero, carbon atoms in the structure will all satisfy the octet rule by forming four bonds
(with no lone pairs). Oxygen atoms have a formal charge of zero by forming two bonds and having two lone
pairs of electrons. Hydrogen atoms have a formal charge of zero by forming a single bond (with no lone pairs).
Following these guidelines, two resonance structures can be drawn for benzoic acid (see next page).