CHEMISTRY 161
HW 8A
24, 25, 27, 29, 31, 33, 35, 37, 40, 42, 66, 78, 80, 82, 85
8-24
A diatomic molecule that follows the octet rule with a single covalent bond (where “—” represents
two electrons in a shared pair) would have the following Lewis structure:
8-25
For the H—O—H bonding pattern, the oxygen of the central atom forms bonds to the two
hydrogen atoms. This uses 4 of the 8 e– leaving 4 e– left over for the 2 lone pairs. Each hydrogen atom
has a duet of electrons, so the lone pairs reside on oxygen and form an octet on oxygen.
For H—H—O bonding, the two covalent bonds again use 4 of the 8 e –, leaving 4 e– for 2 lone pairs. If
these are placed on the oxygen atom as shown here,
oxygen does not complete its octet and the central hydrogen atom has 4 e–, not a duet. This structure
would violate the Lewis structure formalism.
8-27
8-29
8-31
The ions I– and Ca2+ have complete valence-shell octets. B3+ does not have an octet but rather the
duet of the He atom.
8-33
(a) For a 1+ charge and 1 valence electron:
(b) For a 3+ charge and no valence electrons:
8-35
(a)
3 valence e– (B) + 5 valence e– (N) = 8 valence e–
(b) 1 valence e– (H) + 7 valence e– (F) = 8 valence e–
(c) 6 valence e– (O) + 1 valence e– (H) + 1 e– (negative charge) = 8 valence e–
(d) 4 valence e– (C) + 5 valence e– (N) + 1 e– (negative charge ) = 10 valence e–
8-37
(Step 1) The number of valence electrons in CO is
Element
C
O
Valence electrons per
4
+ 6 = 10
atom
(Step 2) There are only two atoms bonded together so neither is the central atom.
C
O
8-40
(Step 4) In this structure there are 8 electrons from three lone pairs and one bond pair. We
need 6 more electrons (three pairs) to match the valence electrons determined in step 1. We add
the lone pairs to the chlorine atom.
Cl
O
(Step 5) This Lewis structure is complete. To indicate the charge on this ion we add brackets for
the structure and the charge.
Cl
O
(a) For CN–
(Step 1) The number of valence electrons in CN– is
Element
Valence electrons per atom
Gain of electron due to charge
Total valence electrons
C
4
+
N
5 =9
+1
10
(Step 2) There are only two atoms bonded together so neither is the central atom.
C N
(Step 3) We complete the octet on the nitrogen atom by adding three lone pairs.
C
N
(Step 4) In this structure there are 8 electrons from three lone pairs and one bond pair. We need 2
more electrons (one pair) to match the valence electrons determined in step 1. We add the lone
pair to the carbon atom.
C
N
(Step 5) To complete the octet on the carbon atom we convert two lone pairs on the N atom to give
a triple bond between the nitrogen atom and the carbon atom. Finally, we add brackets to the
structure and indicate the charge on this anion.
The Lewis structure is now complete.
C
N
C
N
8-42
(a)
CHF2Cl
(Step 1) The number of valence electrons in CHF2Cl is
Element
C
H
2F
Cl
Valence electrons
4
1
7 = 26
(2  7)
per atom
+
+
+
(Step 2) Carbon has the most unpaired electrons (4) in its Lewis symbol and
therefore has the highest bonding capacity and will be the central atom in the
structure. The hydrogen, fluorine and chlorine atoms will each be bonded to the
carbon.
F
H
C
Cl
F
(Step 3) We complete the octets on the chlorine and fluorine atoms by adding three lone pairs to
each.
F
H
C
Cl
F
(Step 4) In this structure there are 26 electrons from nine lone pairs and four bond pairs. We do not
need any more valence electrons in this structure.
(Step 5) With carbon satisfied with its octet and the hydrogen satisfied with its duplet, the Lewis
structure is complete.
(b) CHBr3
(Step 1) The number of valence electrons in CHBr3 is
Element
C
H
3Br
Valence electrons per
4
1
+ (3  7) = 26
atom
+
(Step 2) Carbon has the most unpaired electrons (4) in its Lewis symbol and
therefore has the highest bonding capacity and will be the central atom in the
structure. The hydrogen and bromine atoms will each be bonded to the carbon.
Br
H
C
Br
Br
(Step 3) We complete the octets on the bromine atoms by adding three lone pairs to each.
Br
H
C
Br
Br
(Step 4) In this structure there are 26 electrons from nine lone pairs and four bond pairs. We do not
need any more valence electrons in this structure.
(Step 5) With carbon satisfied with its octet and the hydrogen satisfied with its duplet, the Lewis
structure is complete.
(c) CH2Cl2
(Step 1) The number of valence electrons in CH2Cl2 is
Element
C
2H
2Cl
Valence electrons per
4
(2  1)
(2  7) = 20
atom
+
+
(Step 2) Carbon has the most unpaired electrons (4) in its Lewis symbol and
therefore has the highest bonding capacity and will be the central atom in the
structure. The hydrogen and chlorine atoms will each be bonded to the carbon.
H
H
C
Cl
Cl
(Step 3) We complete the octets on the chlorine atoms by adding three lone pairs to
each.
H
H
C
Cl
Cl
(Step 4) In this structure there are 20 electrons from six lone pairs and four bond pairs. We do not
need any more valence electrons in this structure.
(Step 5) With carbon satisfied with its octet and the hydrogen satisfied with its duplet, the Lewis
structure is complete.
8-66
Because the double bond in NO is stronger than the 1.5 bond in NO2, the energy
required to vibrate the N–O bond in NO is higher.
8-78
For HN3, hydroazoic acid
(Step 1) The number of valence electrons is
Element
H
3N
Valence electrons per
1
(3  5) = 16
atom
+
(Step 2) We are given that hydroazoic acid is a linear molecule with the connectivity
of the atoms as
H N N N
(Step 3) We complete the octets on the rightmost nitrogen atom by adding three lone pairs. The duplet
on the terminal H atom is already satisfied.
H
N
N
N
(Step 4) In this structure there are 12 electrons from three lone pairs and three bond pairs. We need 4
more electrons (two pairs) to match the valence electrons determined in step 1. The middle nitrogen
atoms do not have octets yet so we will add lone pairs.
H
N
N
N
(Step 5) We can complete the octet for the nitrogen atoms by forming a triple bond between them.
H
N
N
N
H
N
N
N
The electrons could also be distributed in two additional resonance forms that also
complete the octet on all the atoms.
The following resonance form is not valid because it has more than a duet for the H
atom and has less than an octet for the N atom.
H
N
N
N
H
N
N
N
H
N
N
N
8-80
For urea, H2NC(O)NH2, we are given that there is a carbon–oxygen double bond.
(Step 1) The number of valence electrons is
Element
C
4H
O
2N
Valence electrons per
(1  4)
(4  1)
6
+
(2  5) =
atom
+
+
24
(Step 2) We are given that urea has a carbon–oxygen double bond, and from the
formula given, the connectivity of the atoms is
O
H
N
C
N
H
H
H
(Step 3) We complete the octets on the nitrogen atoms by adding one lone pair to each, and we
complete the octet on oxygen by adding three lone pairs. The duplet on the terminal H atom is already
satisfied.
O
H
N
C
N
H
H
H
(Step 4) In this structure there are 24 electrons from five lone pairs and seven bond pairs. This
matched the number of electrons required for the structure.
(Step 5) We can complete the octet for the carbon atom by forming a double bond between it and the
oxygen atom.
O
H
N
O
C
N
H
H
H
N
H
H
C
N
H
H
The electrons could also be distributed in two additional resonance forms that also
complete the octet on all the atoms in which a double bond forms between nitrogen
and carbon.
O
H
N
H
C
O
N
H
H
H
N
H
C
O
N
H
H
H
N
H
C
N
H
H
8-82
For nitrite ion, NO2–
(Step 1) The number of valence electrons is
Element
Valence electrons per atom
N
5
17
2O
+ (2  6) =
Additional electrons for charge on
+ 1
anion
Total valence electrons
18
(Step 2) Nitrogen has the most bonding capacity and is the least electronegative
atom and so it is the central atom in the structure
O N O
(Step 3) We complete the octets on the oxygen atoms by adding three lone pairs to each.
O
N
O
(Step 4) In this structure there are 16 electrons from six lone pairs and two bond pairs. We need 2
more electrons (one pair) to match the number of electrons determined in step 1. These we add to the
nitrogen atom.
O
N
O
(Step 5) We can complete the octet for the nitrogen atom by forming a double bond between the
nitrogen and the oxygen atom.
O
N
O
O
N
O
The electrons could also be distributed in one additional resonance form that also
completes the octet on all the atoms by forming a double bond between nitrogen and
the other oxygen atom.
O
N
O
O
N
O
8-85
For both HNC and HCN there are 10 valence electrons and the Lewis structures with
formal charges are
The formal charges are zero for all the atoms in HCN, whereas in HNC the carbon
atom, with a lower electronegativity than N, has a –1 formal charge.