AP Chem Norato/Sun
Ch.9
classwork
—
When using Localized Electron Model to describe bonding in a molecule, follow these
three steps,
1.
Draw the Lewis structure
2.
Determine the arrangement of electron pairs using the VSEPR model.
3.
Specify the hybrid orbitals needed to accommodate the electron pairs.
Use Localized Electron Model to describe bonding in the following molecules.
1.
CO
2.
2
CO
3.
BFZI
4.
2
XeF
5.
4
H
3
C
6.
13
7.
2
N
Drawing VESPR Structures
Name:
Date:__________ Period:________
Determine the geometry of each of the following structures or ions.
1. Co
2
2. SO
3. BrF
3
4. Xe0
4
5. 1C1
+
2
Predicted Molecular Structure (VSEPR)
Total Sets of Electronic Regions
Bond Angles
0
2
3
4
5
180°
Linear
1200
Trigonal Planar
109.5°
Tetrahedral
0
a
120°/90°
Trigonal Bipyramidãi
0
Q_4l
—o
I
/
90°
Octahedrãf
0
o——-o
a
Bent
6
Trigonal Pyramidal
Seesaw
Bent
T-shape
Square Pyramidal
0;
DO
0
AE
Linear
/.{q
Square Planar
2
Ci)
a
9
•3
ci
0)
0
Linear
Linear
T-shape
3
0..
•0
-
Linear
Cl)
Al3
,--,
F.
Linear
4
- -
Linear
5
Arrangement of Valence Pairs
(Bonding & Lone)
Total Regions
2
3
4
6
Electonic
Structure
Linear
Trigonal Planar
Tetrahedral
Trigonal
Bypyramidal
Octahedral
*
An electronic region is the shared
electrons between 2 atoms. One region
may include one, two or three pairs of
electrons.
-4
:t
t—
I
%
•
-
-‘F
‘w.
Ch.9
Covalent Bonding
9.1 Hybridization
Problems in Bonding
Example: methane, CH
4
If kept separate, the bonding
between valence orbitals should be
varied:
i.
•
•
one type for is from H and 2s from C
one type for is from H and 2p from C
but all the bonds are equal
(bond energies!)
2. 2p orbitals are perpendicular but C-H
bonds are at 109.5° angles
I
Problems in Bonding
Example: methane, CH
4
H
py
H—C-—H
H
(a)
(b)
y/ê’x
:
2
P
Solutions for Bonding
Example: methane, CH
4
carbon must accept a new set of
orbitals used for bonding other than
its 2s and 2p
> all 4 must be the same
> must form tetrahedral shape (109.5°)
>
hybridization mixing of normal
orbitals to for special bonding orbitals
-
2
3 hybridization of CH
sp
4
four new sp
3 orbitals are created from
one 2s and three 2p orbitals
identical in shape:
• one large lobe
• one small lobe
tetrahedral arrangement (1O95°)
> When a set of equal tetrahedrally
arranged orbitals is needed, the atom
adopts a set of four sp
3 orbitals
3 hybridization of CH
sp
4
IIyhndizal,on
H
Hybridization
3
sp
E
—2s
Orbitaisina
free C atom
Orbitals in C in
the CH
4 molecule
3
4
3 hybridization of CH
sp
3
P
4
H
2
2 hybridization of C
sp
/14
c=c
2 orbitals are created from
three new sp
one 2s and two 2p orbitals
>identicalin shape:
• one large lobe
• one small lobe
> trigonal planar arrangement (120°)
> When a set of equal trigonal planar-
>
arranged orbitals is needed, the atom
2 orbitals
adopts a set of three sp
4
2 hybridization of C
sp
4
H
2
p
2
2p
Hybridization
— —
—
E
—
2
sp
2s
Orbitals in an isolated
carbon atom
Carbon orbitals
in ethylene
2 hybridization of C
sp
4
H
2
FH1s
/
2
r—Sp
2 —ï-—
sp
Jsp2—
2
-sp
1
H
2
sp
2
sp
1
—H
5
4
H
2
2 hybridization of C
sp
What happens to the other 2p orbital
left out of the hybridization in each C?
> they share an electron pair
> occupies the space above and below
2 orbitals
the sharing between sp
> forms pi (rc) bond
>
sigma
band
4
H
2
2 hybridization of C
sp
the sp
2 orbitals are used to share
electrons too
> shared in the area centered on a direct
line between 2 atoms
> form a sigma (a) bond
>
1ti+
+
—aitbond
+
____.___
6
C,,
rD
C
V
C
C,)
C,,
C,)
C
(Th
C
C,,
tr
C,,
\
/
\
\
C
./
\
x
C
C
Tj
C,,
Cii
C C
C
n
C
C
N
crJ
2
sp hybridization of CO
two new sp orbitals are created from
one 2s and one 2p orbitals
> identical in shape:
• one large lobe
• one small lobe
(1800)
> linear arrangement
> When a set of equal linearly-arranged
orbitals is needed, the atom adopts a
set of two sp orbitals
:o=cd:
2
sp hybridization of CO
;514’
HAI
4
Hybridization
2p
—sp
E
—sp
2s
Orbitals in a
free C atom
Orbitals ip the sp
2
hybridized C in CO
8
sp hybridization of CO
2
2
sp
2
sp
0
C
sp
2
sp
2
sp
sp hybridization of CO
2
What happens to the other two 2p
orbitals left out of the hybridization in
each C?
they share electron pairs with the one
extra 2p orbital from sp
2 hybridized 0
sp
1
2
sp
p
9
sp hybridization of CO
2
sigma bond
(1 pair of electrons)
pi bond
(1 pair of
electrons)
d hybridization of PCi
3
sp
5
:C1:..
,C1:
4
I
:C1—..
:C1:
five new sp
d orbilals are created from
3
one 3s, three 3p, and one 3d orbitals
> trigonal bipyramidal arrangement
> When a set of equal
trigonal bipyramidal
dsP3r?
-arranged orbitals
is needed, the atom
dsp
adopts a set of
five sp
d orbitals
3
>
10
3
s
2
d
p hybridization of SF
6 :iJ:
•_*
>
six new sp
2 orbitals are created :F
d
3
%b••
F:
from one 3s, three 3p, and two 3d
orbitals
> octahedral arrangeme
>Whena set of
equal octahedrally
-arranged orbitals
is needed, the atom
adopts a set of six
2 orbitals
d
3
sp
Rq
?
11
Describe the bonding in
ammonia using the LE model
Example 1
steric #: 4
3
5P
PG:tetrahedral
MS: trigonal
pyramidal
I
3
HF’
L
H
is
N
H
H
is
three sigma bonds, no pi bonds
•M
•IN
Example 2
M•
2
Describe the bonding in N
using the LE model
N
N
SI,
lone pair
sigma bond
lone pair
SI,
SI,
4
one sigma
bond
SI,
1:wo pi
(b)
bonds
(dl
12
r..__.-‘._•• 1”
1:1
L
I
Example 3
1:1
J
Describe the bonding in
13 using the LE model
steric #:
> PG: trigonal bipyramidal
> MS: linear
requires 5: sp
d for central I
3
• 3 hold lone pairs
• 2 make sigma bonds with outside I’s sp
3
> requires 4: sp
3 for outside I’s
• 3 hold lone pairs
• 1 makes sigma bond with central I’s sp
d
3
Describe the bonding in
4 usmg the LE model
XeF
p1 4
Ex
••
>steric#:6
> PG: octahedral
> MS: square planar
> requires 6: sp
2 for central Xe
d
3
• 2 hold lone pairs
• 4 make sigma bonds with outside F’s sp
3
> requires 4: sp
3 for outside F’s
• 3 hold lone pairs
• 1 makes sigma bond with central Xe’s
2
d
3
sp
13
AP Chemistry Norato
Chapter 8 Bonding
Study Guide
—
Example 91 A
3 Hybridization
sp
Describe the bonding in the water molecule using the localized electron model.
Strategy
We must establish the VSEPR structure. This is done by drawing the Lewis structure and determining
the number of effective electron pairs around the central atom.
Solution
0 is (see Sections 8.10 and 8.13)
2
The Lewis structure of H
H\/H
Four effective electron pairs around th oxygen atom indicate a tetrahedral basis leading to a
V-shaped structure. Therefore the oxygen is sp
3 hybridized. Two sp orbitals are occupied with ls
electrons from hydrogen. The other two sp
3 orbitals are occupied by lone pairs.
In molecules or ions with a trigonal planar configuration (3 effective electron pairs around an atom), sp
2
hybridization occurs. With this hybridization, an s orbital and twop orbitals are used. That leaves an
unhybridized (unchanged)p orbital perpendicular to the sp
2 plane.
Look at jgire 9.12 in your textbook.
•
•
•
•
•
Bonds formed from the overlap of orbitals in the plane between two atoms are called sigma (a) bonds.
Bonds formed by the overlap ofunhybridizedp orbitals (above and below the centerp1ane) are called
p1 (sr) bonds.
A single bond is a a bond.
A double bond consists of one a and one r bond.
A triple bond consists of one a bond and two t bonds.
Example 9.1 B
Sigma and P1 Bonds
How many a bonds are there in the commercial inseôticide, “Sevin,” shown below? How many it
bonds?
OH
ii I
H
I
I
II
I
H
I
I
H
O—C—N—C---—H
I
H
H
H
I
I
H
H
Solution
There are 27 a bonds (single bonds and one of the bonds in a double bond).
There are 6 it bonds (the other bond in a double bond).
AP Chemistry Norato
Chapter 8 Bonding
Study Guide
—
Your textbook goes over a variety of different hybridization schemes, all of which center on the idea that
the hybridization of the orbitals of an atom depends on the total number of effective electron pairs
around it. In order to determine the hybridization of an atom, it is essential that you can figure out its VSEPR
structure.
The following table summarizes effective electron pairs around an atom with hybridization. Remember
that, for purposes of the VSEPR model, double and triple bonds count as only one effective electron pair.
Effective Electron
Pairs Around an Atom
2
3
4
5
6
Arrangement
linear
trigonal planar
tetrahedral
trigonal bipyramid
octahedral
Hybridization
sp
2
sp
3
sp
d
3
sp
2
d
3
sp
Keep in mind that a ligand can have a hybridization different from that of the central atom. Each atom in a
molecule must be considered separately based on the Lewis and VSEPR structures of that molecule.
Example 9.1 C
Practice with Hybrid Orbitals
Give the hybridization, and predict the geometry of each of the central atoms in the following
molecules or ions.
a.
b.
c.
d.
+
tF
2
4 (sulfur is the central atom)
OSF
’
2
6
SiF
HCCH (work with both carbons)
Solution
a.
Lewis structure:
: F—I—F:
All three atoms have 4 effective electron pairs. They are all sp
3 hybridized. The VSEPR
structure has a tetrahedral basis. Because the central atom (iodine) has two bonding pairs, it will
take on a V-shape, but the bond angle will be smaller than 109.5°.
b.
Lewis structure:
: 0:
Il-F:
Sulfur has 5 effective electron pairs. The VSEPR structure is a trigonal bipyramid. Sulfur is sp
d
3
hybridized, Each of the fluorines has 4 effective electron pairs.
c.
Lewis structure:
: F:
f__F:
Silicon has 6 effective electron pairs. The VSEPR structure is octahedral. Silicon is sp
d?
3
hybridized. Each of the fluorines has 4 effective electron pairs.
AP Chemistry Norate
Chapter 8 Bonding
Study Guide
—
d.
Lewis structure:
H—CC—H
Each carbon has 2 effective electron pairs. The VSEPR structure is linear. Each carbon is sp
hybridized. Hydrogen atoms bond using1s orbitals. The orbitals are unhybridized.
Example 9J D
Summing it All up
Answer the following questions regarding aspartame ).
m1
(Nutra
sweet
H
H
H—N
\
H
a.
b.
c.
cL
e.
0
How many bonds are in the molecule?
How many t bonds?
What is the hybridization on carbon”a”? Carbon ‘9,”?
What is the hybridization on nitrogen “a”? Oxygen “a”?
What is the CbOliHa bond angle?
So’ution
You must remember to complete the octets in this “shorthand” Lewis structure. Lone pairs are often
“assumed,” so always be on the lookout.
a.
b.
c.
d.
e.
39abonds
67tbonds
, sp
3
sp
2 (3 effective electron pairs)
3 (remember the “assumed” lone pair to complete the octet), sp
sp
2 (3 effective electron pairs,.
including the two to complete the octet)
3 hybridized (complete the octet!); therefore, the angle is based on a tetrahedron, with two
Ob is sp
lone pairs compressing the C—O—H bond angle to 104.5°.
AP Chemistry Noratc
Chapter 8 Bonding
StudyGuide
—
QgIi:i
(chemistry 6th eci page 372/7th ed. page 352)
increases, the bond
As the number of bonds between two atoms
grows shorter and stronger.
EXAMR.E: Arrange the following molecules in order of decreasing
.
6
H
2
,C
H
2
,C
4
H
2
C-C bond length: C
.
H
2
,C
4
H
2
H€, C
2
In order of decreasing C—C bond length: C
First, draw the Lewis structure for each molecule. The C—C triple
bond is th shortest, followed by the double bond, and the single
bond.
Molecular Shapes and Bond Angles
Electron
Pair
No. of
Electron
Pairs
2
Geometry
(Bond
Angle)
Linear
No. of
No. of
Bonding Lone
Electron Electro Molecular
Pairs
n Pairs Geometry Formula
2D Structure
Hybrid
Orbital
2
0
Linear
BeHz
sp
3
0
frigonal
2
3
C0
2
sp
(180°)
3
Trigonal
planar
lanar
(120°)
3
Trigonal
3
1
3ent
2
N0
2
sp
4
0
fetrahedra CH
4
3
sp
planar
(120°)
4
Tetrahedral
(109.5°)
AP Chemistry Noratc..
Chapter 8 Bonding
Study Guide
—
No. of
Electron
Pairs
4
Electron
Pair
Geometry
(Bond
ngle)
2D Structure
No. of
Bonding
Electron
Pairs
rtrahedral
(>109.5°)
3
retahedral
(>>109.5°)
2
No. of
Lone
Electro Molecular
n Pairs Geometry Formula
rrigonal NTh
1
pyramidal
Hybrid
Orbital
3
sp
I
y____
4
2
Bent
3
sp
HzO
@—p
y.
5
Tri011al
5
0
retrahedral
and
>1200)
(900
PCIs
spd
4
SF
d
3
sp
midal
midal
(90°, 120°)
5
rngonal
4
1
Seesaw
unsyrn
metrical
tetrahe
dron
AP Chemistry Norato
Chapter 8 Bonding
Study Guide
—
,
No, of
Electron
Pairs
5
Electron
Pair
Geometry
(Bond
ng1e)
tetrahedral
(900 and
No. of
No.of
Bonding Lone
Electron Electron Molecular
Pairs
Pairs
Geometry Formula
3
2
r’-shaped BrF
3
Hybrid
Orbital
2D Structure
d
3
sp
5
tetrahedral
(180°)
2
3
Linear
2
1C1
d
3
sp
6
Octahedral
(90°)
6
6
Octahedral
6
SF
2
d
3
sp
6
Octahedral
(90°)
5
1
Square
BrFs
pyramidal
2
d
3
sp
24
AP Chemistry Norato’
Chapter 8 Bonding
Study Guide
—
Electron
Pair
No. of
Geometry
Electron (Bond
Pairs
6
No. of
No. of
Bonding Lone
Electron Electron Molecular
Angle)
Pairs
Geometry Formula
Pairs
Octahedral
4
2
Square
ICW
(900)
planar
2D Structure
EXAMPLE: Predict the molecular structure of the carbon dioxide
molecule. Is this molecule expected to have a dipole moment?
First we must draw the Lewis structure for the CO
2 molecule.
O=c=O
In this structure for C0
, there are two effective pairs around the
2
central atom (each double bond is counted as one effective pair).
There are two atoms attached to the central atom,
According to the table, a linear arrangement is required.
Each C—O bond is poiar, but, since the molecule is linear, the
dipoles cancel out and the molecule is nonpolar.
EXAMPLE: Draw the Lewis structure for PF
5 and identify the
molecular geometry.
PF has 40 valence electrons. Connecting 5 F atoms to P. the
5
central atom, uses 10 electrons in five single bonds. The 30
remaining electrons can be used to satisfy the octet rule for each
F atom. There are 5 electron pairs around P and 5 atoms
attached to P. resulting in a trigonal bipyramidal shape.
EXAMPLE: Draw the Lewis
structure for 3
CH
O
H.
H
This molecule has more than one central atom, the carbon and
the oxygen. Hydrogen cannot be in the middle since it can form
25
2
d
3
sp
AP Chemistry Norato
Chapter 8 Bonding
Study Guide
—
only one bond. Draw the correct Lewis structure and then using
the rules for VSEPF{, assign the molecular geometry around each
central atom. The carbon atom has four electron pairs around it
and four atoms attached resulting in tetrahedral geometry.
Oxygen also has four electron pairs around it, but only has two
atoms attached. This leads to bent geometry around oxygen.
EXAMPLE: Arrange the following molecules in order of increasing
.
3
bond angles: H
, NF
4
S, CC1
2
First, draw the correct Lewis structure for each molecule and
then assign an approximate bond angle for each.
Each of the four molecules has four electron pairs around the
central atom, resulting in a tetrahedral arrangement of the
electron pairs about the cetitral atom and approximately a 109.5
bond angle. HS has two lone pairs on the central atom; NF
3 has
one lone pair on the. central atom. These lone pairs repel more
than the bonded electron pairs and will cause the bond angle to
be smaller than the expected 109,5°. The more lone pairs, the
smaller the bond angle.
In order of increasing bond angles, the answer is H
S <NP
2
3
.
4
CC’
<
Exercises
Section 8.1
i.fl
Indicate whether the bonds between the following would be primarily covalent, polar covalent, or ionic:
a. ()—f’{
C.
H-Cl
b. Cs—Cl
d. Br—Br
2.
Calculate the energy of interaction for KCI if the internuclear distance is 0.3 14 nm.
3.
Calculate the energy of interaction between Ag and Bf if the internuclear distance of AgBr is 0.120 rim
(in Id/mole).
Section 8.2
4. )sing a periodic table, order the following from lowest to highest electronegativity:
a. Fr, Mg, Rb
c. P, As, Ga, 0
b. B, Al, C, N
d. Cl, 5, P
5.) Using the periodic chart of elements, place the following in order from the lowest to the highest
electronegativity:
F, Nb, N, Si, Rb, Ca, Pt
6.
Using Figure 8.3 in your textbook, calculate the difference in electronegativity () for each of the
following bonds:
Cl-Cl
b. K-Br
c.
d.
Fe—C)
H—C)
)
S-H
AP Chemistry Norato
Chapter 8 Bonding
Study Guide
—
7.
Place the following in order of increasing polarity:
NaBr,
8.
‘2,
, CN
2
0, MnO
2
H
Which of the following molecules contain polar covalent bonds? List in order of increasing bond
polarity. (Use Fig. 8.3 in your text).
, H
4
S
2
, CH
2
03, Pg, NO, C0
9.
10.
How will the charge be distributed on each of the following molecules: HF, NO, CO. and HCI?
2 is öovalent?
2 is ionic, and BeCI
Why is it that BeF
Section 8.3
II.
Determine the orientation of the dipole of the following, if any.
a.
b.
c.
d.
AICI (planar with aluminum atom at the center)
3
F (tetrahedral with carbon at the center)
3
CH
0 (linear with N-N-O structure)
2
N
4 (planar molecule, silver atom at center, Ag--Cl bonds 900 apart)
AgCI.
12.
Which of the molecules in problem 11 contain one or more polar bonds?
13.
Which of the following molecules would you expect to have a dipole moment of zero? Describe the
dipole orientation of the other two molecules.
a,
b.
KI
4 (tetrahedral structure)
CF
c.
Se (bent structure)
2
H
Section 8.4
14.
Determine the most stable ion for each of the following atoms, and indicate which element they would
be isoelectronic with if they lost or gained electrons:
a.
b.
0
Be
c.
d.
I
Te
e.
Na
15.
List four ions that are isoelectronic with argon and have charges from —2 to +2. Arrange these in order
of increasing ionic radius.
16.
List four ions that are isoelectronic to Kr. Arrange these in order of increasing radius.
17.
Determine the formula of the binary compound formed from the following sets of atoms.
a.
b.
c.
d.
18.
Caand0
KandCl
RbandS
BaandP
V
Predict formulas for the following binary ionic compounds.
a.
b.
c.
d.
MgandN
NaandF
CaandS
SrandTe
V
V
V7
AP Chemistry Norato,
Chapter 8 Bonding
Study Guide
—
19.
Using shorthand notation, list the core electron configurations for the ions rn the compo
unds in
problem 18.
20.
Place the following in an order of increasing ionic size. Use the shorthand notatio
n to list the core
electron configuration for each of the ions.
a.
b.
21.
, Te
2
Ba
, Cs, f
2
Cs, S
. 2— K
2
Place the following in an order of increasing ionic size. Use shorthand notation
to list the core electron
configuration for each of the ions.
a.
C1, F, Sr, Ca
2
b.
Na, Mg
, Li, Be
2
2
Section 8.8
22.
Using the bond energy values listed in Table 8.4 of your text, calculate the All
for the following
reactions:
a.
b.
c.
(g) + 0
2
2H
(g) — 2H
2
0(g)
2
2C
g
6
H
2
) + 70
(g) -+ 4C0
2
(g) + 6H
2
0(g)
2
HCN(g) + 2H
(g) - 2
2
N
3
CH
(
g)
N
23,
Use bond energy values from
1L4 jny Jcipck to calculate AN for the following reactio
ns:
a. H-CC-H(g) + H
(g) -* 2
2
C
CH
(
g)
H
b. 2CH
(g) + 302(g) — 2C0(g) + 4H
4
0(g)
2
c. N
(g) + 2H
2
(g) - N
2
(g) 2
4
H
2
(M
)
NH
1
24.
Compare the values obtained in parts “a” and “b” of Proble
m 23 to AN values calculated from AR° data
in Appendix 4 in your textbook.
25.
Calculate the enthalpy of reaction Aff for the following
reaction. Use the enthalpies of formation
found in Appendix 4 of your textbook.
H
(
2
g)
+
C
(
4
H
2
g)
-
C
(
6
H
2
g)
Section 8.10
26.
27.
28.
Draw Lewis dot structures for the following atoms,
ions, or molecules:
a. Sr
d. Ga
b. 3r
e, GaC1
’
4
c. ICN
f. P’
Draw Lewis structures for the following:
a.
C.
P
b. C
d. P
’’
5
Draw Lewis- structures for the following:
a. AsF
3
c. 3
H
0
’
b. 03
d. BH.
g.
h.
2
NH
CSe
2
e.Cr
e.
f.
NH
0
AP Chemistry Norato
Chapter 8 Bonding
Study Guide
—
Section 8.11
29.
Draw Lewis dot structures for the following:
a,
b.
30.
(i’)
3
BCI
AsF
5
()
Br0
3
10 (contains a S—S bond)
F
2
S
Draw Lewis dot structures for the following:
a.
C.
3
SbCI
AIF
6
5
PCI
Section 8.12
Assign formal charges to each of the labeled atoms, a-e.
e
2—
•Br.•
:S—O:
0C—Br
d
a
c:):
b
c1=C—0:
:
ci:
.
4
0
2
Draw the remaining resonance forms for N
/0
N—N
How many reasonable resonance structures are there for carbon monoxide, CO. What are the formal
charges?
, is used as a propellant on the Space Shuttle. Draw all reasonable structures for N
4
H
2
Hydrazin, N
,
4
H
2
charges.
formal
and assign
Draw a Lewis structure and any resonance forms of benzene, C
. (Benzene consists of a ring of six
H
6
carbon atoms with one hydrogen bonded to each carbon.)
Section 8.13
Predict the structure of each of the following molecules or ions:
a.
b.
6
SeF
N
0
2
C.
d.
CIF
C10
e.
CF
C
3
I (carbon is central atom)
Predict the structure of HCN.
Using the VSEPR model, determine the molecular geometry for each of the following molecules:
a.
b.
4
SCI
H
S
2
e
c.
d.
IF
4
SnClj
e.
2
TlCl
Which of the molecules or ions in Problem 36 contain polar covalent bonds? e polar?
2
AP Chemistry Norato
Chapter 8 Bonding
Study Guide
—
SectIon 9,1
9
What geometry do the following hybrid bonds possess?
sp
2
sp
a.
b.
3
sp
d. 3
sp
d
C.
e. sp
2
d
3
Predict the type of hybrid orbital that the central atoms of each of the following
compounds display:
a. S1H
4
d. NC1
3
f. SiF
6
b. H
0
3
e. AsH
3
g. CH
3
5
PC1
C.
Predict the geometries of the following compounds:
a.
2
SF
b. SF
4
c. XeF
2
d. XeF
4
e. IF
5
f. CIF
3
Predict the geometry about the indicated atom, and identify the hybridization of
each atom.
a. the two carbon atoms and the nitrogen atom of glycine
H
o
H—C—C
at
\
0
b.
c.
d.
the carbon atom in 2
CI
CF
the phosphorous atom in PCi
5
the nitrogen atom in NH
2
Section 9.2
() Determine the number of sigma and pi bonds in each of the following:
H
HO
)c=—c—oH
I1
b.
H
H—CN
H
I
O.H
II
I
H—N—C—N—H
a.
b.
c.
d.
e.
f.
H
HCC—C<
H
H
The structure of urea is
H
c.
How many a bonds are there?
How many it bonds?
What is the hybridization at the carbon?
How are the nitrogen atoms hybridized?
What is the N—C—N bond angle expected to be?
How many lone pairs of electrons are there?
30
AP Chemistry Norato
Chapter 8 Bonding
Study Guid
—
Answers to Exercises
a.
b.
polar covalent
ionic
2.
—736 x io J
3.
E
—1160 Id/mole
4.
a.
b.
Fr,Rb,Mg
A1,B,C,N
5.
c.
d.
polar covalent
covalent
c.
d.
Ga,As,P,O
P,S,C1
Rb<Ca<Nb<Si<Pt<N<F
0.8
6.
1.0
a.
b,
7.
1.6
1.8
2.2
3.0
0
2.0
4.0
c.
d.
1.7
1.4
e.
0.4
<
1
0
2
<N
C
aB
W
r<M
<H
n0
0
8.
0.5
03, P
8
9.
1.4
1:9
H
S
2
, CH
4
<
2.0
<
NO
<
2
CO
HF;NO;CO;HC1
+—+--. +—+—
10.
The difference in the electronegativity between Be and F is higher
than the one between Be and Cl.
3.0— 1.5 1.5 (covalent bond)
4.0— 1.5 2.5 (ionic)
11.
a.
b.
12.
all
13.
b. The opposing bond polarities in a tetrahedral structure cancel out.
Thus CF
4 has no dipole moment.
KI is a binary ionic compound that has a negative dipole toward I.
Selenium will have a partial negative
charge as its electronegativity is greater than that of hydrogen. Thus
the resulting dipole moment of
Se would be orientated as shown:
2
H
no dipole
negative toward F
c.
d.
negative toward 0
no dipole
H
14.
a.
b.
c.
d.
e.
15.
,K
24
Ca
, CF. S
4
2
16.
Sr
<
17.
a.
b.
CaO
KCI
H
O2,
isoelectronic with neon
Be isoelectronic with helium
,
2
F, isoelectronic with xenon
, isoelectronIc with xenon
2
Te
, isoelectronic with neon
4
Na
4
Rb
<
Br <Se
2
c.
d.
S
2
Rb
Ba
2
P
3
31
AP Chemistry
—
Noratc,
Chapter 8 Bonding
Study Guide
18.
a, Mg
2
N
3
b.NaF
c.
d.
CaS
SrTe
19.
a.
b.
[Ne], [Ne]
c.
[Ne], [Ne]
d.
[Ar], [Ar]
[Kr], [Xe]
a.
b.
’’<Cs<F<Te all can be written as [Xe]
Ba
;
2
K < Cs <O2’ <
K” [Ar], Cs [Xe], O2
a.
b.
<
2
Ca
<
F”<CF
Sr
2 < Li < Mg
Be
2 < Na
22.
a.
—509kJ
b.
23.
a.
—169kJ
b.. —1091 kJ
24.
a.
6kJ difference
b.
25.
iH
26.
a.
•Sr.
b.
[:Br:]
20.
21.
=
”
2
Ca
Be
2
=
=
[Ne], S
”
2
=
[Ar]
[Ar], Sr
2 = [Kr], F = [Ne]. CV [Ar]
[He], Li” = [He], Mg
” = [Ne], Na = [Ne]
2
—2881kJ
C.
—lS8kJ
C, I
+81 Id
7kJdifference
—136.7 Id/mole
d,.Ga.
ci:
—
g.
[H——Hj’
h.
SeC=Se
:C1—Ga—C1:
ci:
27.
28.
:P:j13—
c.
: I—CN:
f.
a.
[H]
c.
: P.
C.
d.
5
[P]
b.
•
a.
: F—As—F:
c.
: F:
[:ci:J
H—O—H
+
e.
H
H
H—N—H
H
b.
0=0—0:
d.
H
H—B—H
H
f.
00
+
AP Chemistry Norato
Chapter 8 Bonding
Study Guide
—
29.
: Cl—B—Cl:
a.
c.
:O—Br—O:
ci:
0:
:F: :F: :F: :F:
:
b.
\/
../\
F.L_..F:
As
/\
:F: :F:
30.
a.
\/
S—F:
/\..
d : F—S
:F:
: Cl—Sb—Cl:
F: :F: :F:
..
..
\/
b.
:F::F:
Ci:
3—
c.
F—Al—F:
..
/ \
..
:c’a:
:F::F:
31.
a.
b.
32.
0
0
N—N
/
0
-4
0\
N—N
/
\
o
33.
—1
+1
C.
d.
0
:ci:..
:ci.ja:
e. +1
£0
0\
/0
0\
4
0
: oc: is the only reasonable structure. The
carbon has a formal charge of—i, and
oxygen has a +1
formal char
ge. It is possible to draw a structure such
as
o=c: in wbich both C and
0 have a zero
formal charge, but the carbon would be elect
ron deficient (lacks an octet).
34.
The only reasonable structure has an N—N
single bond. All formal charges =0.
H
H
\
../
N—N
/..
H
\
H
Each nitrogen has a formal charge of+1 and
the two hydrogens that are single have
formal charges
of—i.
H
35.
H
H
II
H
I
II
H
H
H
36.
37.
a.
b.
octahedral
linear
H
c.
d.
see-saw
linear
e.
HCN is linear (H—CN:)
33
tetrahedral
AP Chemistry Norato
Chapter 8 Bonding
Swdy Guide
—
38.
39.
a.
see-saw
b.
bent
c.
d.
square planar
trigonal bipyramidal
e.
linear
All contain polar covalent bonds; b, c, d, and e are polar. Negative is toward: 0 in b; equatorial fluorine
inc;Oind;fluorineine,
c.
d.
tetrahedral
trigonal bipyramidal
e.
a,
b.
linear
trigonal planar
a.
b.
sp
3
3
sp
3,
a,
b.
angular (like water)
see-saw
4.
a.
b.
c.
d.
, based on tetrahedron
3
, trigonal planar; nitrogen-sp
2
carbon a-sp
, tetrahedral; carbon b-sp
3
trigonal pyramid
sp-tetrahedral
-trigonal bipyramid
3
dsp
-based on tetrahedron-bent’
3
sp
5,
a.
b.
c.
8sigrna,2pi
Ssigrna,2pi
7sigma,3pi
6.
a.
b.
7
l
1.
2.
c,
d.
e.
3
dsp
sp
3
f,
c, linear
ci square planar
octahedral
sp
3
sp
2
d
3
g.
e, square pyramidal
f. T-shaped
e. 1200
f.4
c sp
2
3
d.sp
34
2
sp