CHEMISTRY 161
HW CH#8B
88, 89, 91, 95, 98, 100, 102, 104, 106, 108, 112
8-88
For SN4 there are 26 valence electrons and all the possible resonance structures
with formal charges are
None of these resonance structures have all zero formal charges. Therefore, the most
stable structures are ones that have minimal formal charges with negative charges on
the most electronegative atom (N) and where the formal charges are not separated in
the molecule but are next to each other. Two structures meet these criteria:
8-89
For nitrous oxide, N2O, there are 16 valence
electrons, and the Lewis structures with formal charges assigned to the atoms are
Because oxygen is more electronegative than nitrogen, none of these structures is
likely to be stable because the formal charge on O is positive, when it would be
predicted by electronegativity to be negative.
8-91
(a)
In CH3NO2 there are 24 valence electrons. Completing the octets for all
the atoms (duplet for hydrogen), drawing an alternate resonance structure, and
assigning formal charges to the atoms gives
(b)
In CNNO2 there are 26 valence electrons. Completing the octets for all the
atoms, drawing the alternate resonance structures, and assigning formal charges to
the atoms gives
Formal charges are minimized
in these two structures with the
negative formal charge on the
most electronegative atom
(oxygen), so these are the
preferred structures.
(c) The two structures of CNNO2 given in the text are not resonance structures
because their atoms differ in connectivity. When two molecules have the same
number of atoms, but in a different arrangement, they are called isomers.
8-95
In order for the atom to accommodate more than 8 e– in covalently bonded
molecules, it would require the use of orbitals beyond s and p. The d orbitals are not
available to the small elements in the second period but do become available for the
third period (and subsequent periods!) elements such as P, S, and Cl.
8-98
Not Graded
8-100
(a) POCl3 has 32 valence electrons, and its Lewis structure with formal charges
assigned to its atoms is
We can reduce the formal charges on P and O through a double bond.
There are 10 e– in five covalent bonds around phosphorus in POCl3.
(b) H3PO4 has 32 valence electrons, and its Lewis structure with formal charges
assigned to its atoms is
To reduce the formal charges on P and O, we form a double bond between them.
This gives 10 e– in five covalent bonds around phosphorus in H3PO4.
(c) H3PO3 has 26 valence electrons, and its Lewis structure with formal charges
assigned to its atoms is
All the formal charges on the atoms in this structure are zero. There are 6 e– in three
covalent bonds around phosphorus in H3PO3.
(d) PF6– has 48 valence electrons, and its Lewis structure with formal charges
assigned to its atoms is
All the formal charges on the F atoms in this structure are zero. There are 12 e– in six
covalent bonds around phosphorus in PF6–.
8-102
Both molecules have 32 valence electrons, and the Lewis structures with formal
charges are as follows:
Because oxygen is more electronegative than N, the structures where the O atom has
a –1 formal charge seem reasonable. However, the formal charges on P and O in
PO43– can be reduced because P can expand its octet to form a double bond with O.
In PO43– there is a double bond in a single resonance form, whereas in NO43– all
bonds are single bonds.
8-104
NF4+ has 32 valence electrons, and its Lewis structure is
SbF6– has 48 valence electrons, and its Lewis structure is
8-106
The Lewis structure for Cl2O2 is
In this structure, neither Cl atom needs to expand its octet. However, the formal
charges on the atoms of this structure are fairly high. To reduce this we could form
double bonds between the Cl and O atoms.
In this structure, all of the formal charges of all the atoms are zero, and the central
chlorine atom has an expanded octet.
8-108
(a) NO has (1N  5 e–) + (1O  6 e–) = 11 e–
(b) NO2 has (1N  5 e–) + (2O  6 e–) = 17 e–
(c) NO3 has (1N  5 e–) + (3O  6 e–) = 23 e–
(d) N2O4 has (2N  5 e–) + (4O  6 e–) = 34 e–
(e) N2O5 has (2N  5 e–) + (5O  6 e–) = 40 e–
The odd-electron molecules are NO, NO2, and NO3 (a–c).
8-112
This structure is unfavorable because it has high formal charges on both nitrogen and oxygen, and
the –2 formal charge is on N, which is less electronegative than O.