Petrucci 8th ed Problem set #4: Chemical bonding 1 of 6
Petrucci 8th Edition problem set #4: Chemical
Bonding
Chapter 11
11.6
a)
b)
d)
e)
a)
b)
c)
d)
11.7
11.10
a)
FC(H) = 1 – 0 – 1 = 0
FC(central C) = 4 – 0 – 4 = 0
FC (terminal C) = 4 – 2 – 3 = -1
b)
FC(C) = 4 – 0 – 4 = 0
FC(-O) = 6 – 6 – 1 = -1
FC(=O) = 6 – 4 –2 = 0
e)
FC(S) = 6 – 2 - 3 = +1
FC(-O) = 6 – 6 - 1 = -1
FC(=O) = 6 – 4 - 2 = 0
c)
Petrucci 8th ed Problem set #4: Chemical bonding 2 of 6
11.14
b)
a)
bent
trigonal pyramidal
c)
tetrahedral
11.15
a)
linear
b)
tetrahedral
c)
trigonal pyramidal
11.27 a) Incomplete octet around C
b) incomplete octet around left C and not enough valence electrons in the
structure
c) this Lewis structure is correct
d) there are too many valence electrons in this structure
11.33 a) FC(O) = 6 - 2 - 3 = +1
b) FC(B) = 3 - 0 - 4 = -1
c) FC(N) = 5 - 0 - 4 = +1
Petrucci 8th ed Problem set #4: Chemical bonding 3 of 6
11.34 b)
(II)
FC(C) = 4 - 4 - 2 = -2
FC(central S) = 6 - 0 - 4 = +2
FC(terminal S) = 6 - 4 - 2 = 0
Since the formal charges on structure I are smaller, it is the more plausible one.
(I)
FC(S) = 6 - 4 - 2 = 0
FC(C) = 4 - 4 - 0 = 0
11.57
a)
b)
linear
linear
c)
d)
tetrahedral
trigonal planar
Petrucci 8th ed Problem set #4: Chemical bonding 4 of 6
Chapter 12
12.12
a)
is a planar molecule.
The hybridization on C is sp2 (one bond to each of the
three attached atoms).
b)
is a linear molecule.
The hybridization on each C is sp (one bond to each
of the two nitrogens).
c)
is neither linear nor planar.
The shape around the left-hand side C (F3C-) is
tetrahedral and that C has sp3 hybridization. The
shape around the right-hand carbon (-C≡N) is linear
and that C has sp hybridization.
d)
is a linear molecule.
The hybridization on C is sp.
Petrucci 8th ed Problem set #4: Chemical bonding 5 of 6
12.25
For each species, we first draw the Lewis structure, as an aid to explaining bonding.
Plausible Lewis structure
Geometry of the
molecule
Hybridization
The molecule is linear
C is sp hybridized
N is the central atom.
N is sp2 hybridized.
The part of the molecule
around N is trigonal
planar and the N-O-H
part of the molecule is
bent.
The O on the left is bent with sp3
hybridization.
Cl is the central atom.
Cl is sp3 hybridized.
a)
b)
In HONO2, there are a total of
1 + 5 + (3 x 6) = 24 valence
electrons, or 12 pairs.
c)
The molecule is trigonal
pyramidal.
In ClO3-, there are a total of
7 + (3 x 6) + 1 = 26 valence
electrons, or 13 pairs.
d)
B is the central atom.
The molecule is
tetrahedral.
In BF4-, there are a total of
3 + 4 (4 x 7) + 1 = 32 valence
electrons, or 16 pairs.
B is sp3 hybridized.
Petrucci 8th ed Problem set #4: Chemical bonding 6 of 6
12.29
Plausible Lewis structure
Geometry of the
molecule
Hybridization
C is the central atom.
The geometry is
tetrahedral.
C is sp3 hybridized.
a)
The Cl-C-Cl bond angles
are 109.5º
In CCl4, there is a total of
4 + (4 x 7) = 32 valence
electrons, or 8 pairs.
b)
The geometry around N
is angular or bent.
In ONCl, there is a total of 6 + 5
+ 7 = 18 valence electrons, or 9
pairs.
The O-N-Cl bond angle
is about 120º
There are four bonds in the
molecule:
Each C-Cl bond is represented
by a σ bond. The carbon is
hybridized sp3.
N is sp2 hybridized.
There are three bonds in the
molecule:
σ bonds between N-Cl and
N-O (two σ bonds).
One π bond between O and N.
c)
In HONO, there are a total of
1 + (2 x 6) + 5 = 18 valence
electrons, or 9 pairs.
The geometry of the O
O on the left is sp3 hybridized.
on the left is angular or
N is hybridized sp2
bent.
There are four bonds in the
The geometry of the N is molecule:
angular or bent.
σ: H - O
σ: O -- N
The O-N-O bond angle
σ: N -- O
is 120º and the H-O-N
π: N -- O
bond angle is 109.5º.
d)
The geometry around C
is trigonal planar, all
bond angles around that
atom are 120º
In COCl2, there are a total of
4 + 6 + (2 x 7) = 24 valence
electrons, or 12 pairs.
The hybridization of C is sp2.
There are four bonds in the
molecule:
2σ: Cl -- C
σ: O -- C
π: O - C