CJE,OSTRU 161
HW CH8A
26, 30, 36, 42, 46, 48, 52, 56, 70, 76, 78, 84,
8-26
No, each atom does not always contribute the same number of valence electrons to form a covalent
bond. A good example of this is nitrogen, specifically, the ammonia molecule. The Lewis structure
for ammonia is
itrogen brought 5 valence electrons to the molecule and each of the hydrogen atoms brought 1
electron. If we write the Lewis symbol of a nitrogen atom,
we can envision that the unpaired electrons on nitrogen pair up with the lone electrons on the
hydrogens. Thus, each bond in ammonia includes one of the nitrogen’s valence electrons. Ammonia,
however, also forms a bond with BF3:
Here the nitrogen atom contributes 2 of its valence electrons (while boron contributes none!) to form
the N—B covalent bond.
8-30
8-36
(a) 5 valence e– (N) + 5 valence e– (N) – 1 e– (positive charge) = 9 valence e–
(b) 4 valence e– (C) + 6 valence e– (S) – 1 e– (positive charge) = 9 valence e–
(c) 4 valence e– (C) + 5 valence e– (N) = 9 valence e–
(d) 4 valence e– (C) + 6 valence e– (O) = 10 valence e–
8-42
(a) CHF2Cl
(Step 1) The number of valence electrons in CHF2Cl is
Element
C
H
2F
Valence electrons
4
1
(2  7)
per atom
+
+
+
Cl
7 = 26
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 CHBr 3 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.
8-46
For NH2Cl
(Step 1) The number of valence electrons in NH2Cl is
Element
N
2H
Cl
Valence electrons per
5
+ (2  1) + 7 = 14
atom
(Step 2) Nitrogen has the most unpaired electrons (3) in its Lewis symbol and
therefore has the highest bonding capacity and will be the central atom in the
structure.
H
H N Cl
(Step 3) We complete the octets on the chlorine atom by adding three lone pairs.
H
H
N
Cl
Step 4) In this structure there are 12 electrons from three lone pairs and three bond pairs. We need
2 more electrons (one pair) to match the valence electrons determined in step 1. We add the lone
pairs to the central nitrogen atom.
H
H
N
Cl
(Step 5) The central nitrogen atom is satisfied with its octet and the hydrogen atoms are satisfied
with their duplet so this Lewis structure is complete.
For N2H4
(Step 1) The number of valence electrons in N2H4 is
Element
2N
4H
Valence electrons per
(2  5)
(4  1)
atom
+
= 14
(Step 2) Nitrogen has the most unpaired electrons (3) in its Lewis symbol. It therefore has the
highest bonding capacity and will be the central atoms in the structure. Also, since hydrogen
atoms must complete a duet, not an octet, the H atoms in hydrazine must be terminal and the N
atoms central in this molecule.
H
H
H N N H
(Step 3) The duplets on the terminal hydrogen atoms are already complete.
(Step 4) In this structure there are 10 electrons from five bond pairs. We need 4 more electrons
(two pairs) to match the valence electrons determined in step 1. We add the lone pairs to the
central nitrogen atoms.
H H
H
N
N
H
(Step 5) The central nitrogen atom is satisfied with its octet, and the hydrogen atoms are satisfied
with their duplets so this Lewis structure is complete.
8-48
Across a period, Zeff increases and so also does electronegativity. Down a group, size
increases so electronegativity decreases.
8-52
When atoms have different electronegativities, they do not equally pull on the
electrons that contribute to the bond and the bond will be polar, with one atom being
slightly negative (– ) and the other slightly positive (+ ).
8-56
Beryllium’s electronegativity is 1.5. The electronegativity values for the halogens
are F = 4.0, Cl = 3.0, Br = 2.8, and I = 2.5. Therefore, only the Be—F bond is ionic
(∆EN = 2.5).
8-70
Lewis structures of all the resonance forms of a molecule or ion have the same
formula, the same arrangement of the atoms in bonds, and the same total number of
electrons.
8-76
For S2O2
(Step 1) The number of valence electrons is
Element
2S
2O
Valence electrons per
(2  6) + (2  6) =
atom
24
(Step 2) Sulfur is less electronegative than oxygen and so the two sulfur atoms are
the central atoms in the structure.
O
S
S
O
(Step 3) We complete the octets on the oxygen atoms by adding three lone pairs to each.
O
S
S
O
(Step 4) In this structure there are 18 electrons from six lone pairs and three bond pairs. We need 6
more electrons (three pairs) to match the valence electrons determined in step 1. The sulfur atoms do
not have octets yet so we will add lone pairs to the S atoms.
O
S
S
O
(Step 5) We can complete the octet for the sulfur atom on the left by forming a double bond between
the left-hand oxygen and that sulfur atom.
O
S
S
O
O
S
S
O
The electrons could also be distributed in two additional resonance forms that also
complete the octet on all the atoms.
For S2O3
(Step 1) The number of valence electrons is
Element
2S
3O
Valence electrons per
(2  6) + (3  6) =
atom
30
(Step 2) Sulfur is less electronegative than oxygen and so the two sulfur atoms are
the central atoms in the structure.
O
O
S
S
O
(Step 3) We complete the octets on the oxygen atoms by adding three lone pairs to each.
O
O
S
S
O
(Step 4) In this structure there are 26 electrons from nine lone pairs and four bond pairs. We need 4
more electrons (two pairs) to match the valence electrons determined in step 1. The sulfur atoms do
not have octets yet so we will add a lone pair to each of the S atoms.
O
O
S
S
O
(Step 5) We can complete the octet for the sulfur atom on the left by forming a double bond between
the left-hand oxygen and that sulfur atom.
O
O
S
S
O
O
O
S
S
O
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.
H
N
N
N
H
N
N
N
H
N
N
N
8-84
The electronegativity of N (3.0) is lower than that of O (3.5), so the positive formal
charge must be on the N atom for the structure that contributes the most to the
bonding in a cation containing N and O.