AP Chemistry
Chapter 9. Molecular Geometry and Bonding Theories
Chapter 9. Molecular Geometry and Bonding Theories
9.1 Molecular Shapes
Read Sec. 9.1 and 9.2, then complete the Sample and Practice Exercises in these sections.
Sample Exercise 9.1 (p. 347)
Use the VSEPR model to predict the molecular geometries of
a) O3
b)
SnCl3-
Practice Exercise 1 (9.1)
Consider the AB3 molecules and ions: PCl3, SO3, AlCl3, SO32-, and CH3+. How many of these molecules and
ions do you predict to have a trigonal-planar geometry?
a)
1
b) 2
c) 3
d) 4
-1-
e) 5
AP Chemistry
Chapter 9. Molecular Geometry and Bonding Theories
Practice Exercise 2 (9.1)
Predict the electron-domain geometry and the molecular geometry for
a) SeCl2
b)
CO32-
-2-
AP Chemistry
Chapter 9. Molecular Geometry and Bonding Theories
The Effect of Nonbonding Electrons and Multiple Bonds on Bond Angles
•
•
•
•
Consider three molecules with tetrahedral electron domain geometries:
CH4, NH3, and H2O.
By experiment, the H–X–H bond angle decreases from C (109.5o in CH4) to N (107 o in NH3) to O
(104.5 o in H2O).

Since electrons in a bond are attracted by two nuclei, they do not repel as much as lone pairsTherefore,
the bond angle decreases as the number of lone pairs increases.
A bonding pair of electrons is attracted by two nuclei. They do not repel as much as lone pairs which are
primarily attracted by only one nucleus.
• Electron domains for nonbonding electron pairs thus exert greater repulsive forces on adjacent electron
domains.
• They tend to compress the bond angles.
• The bond angle decreases as the number of nonbonding pairs increases.
• Similarly, electrons in multiple bonds repel more than electrons in single bonds. (e.g. in Cl2CO the O–
C–Cl angle is 124.3 o, and the Cl–C–Cl bond angle is 111.4 o).
Cl
111.4o
Cl
C O
124.3o
-3-
AP Chemistry
•
Chapter 9. Molecular Geometry and Bonding Theories
11 basic molecular shapes:
• Three atoms (AB2)
• Linear
• Bent
• Four atoms (AB3):
• Trigonal planar
• Trigonal pyramidal
• T-shaped
• Five atoms (AB4):
• Tetrahedral
• Square planar
• See-saw
• Six atoms (AB5):
• Trigonal bipyramidal
• Square pyramidal
• Seven atoms (AB6):
• Octahedral
Molecules with Expanded Valence Shells
•
Consider a trigonal bipyramid:
• The three electron pairs in the plane are called equatorial;
• The two electron pairs above and below this plane are called axial;
• The axial electron pairs are 180 o apart and 90 o to the equatorial electrons;
• The equatorial electron pairs are 120 o apart;
• To minimize electron–electron repulsion, nonbonding pairs are always placed in equatorial positions
and bonding pairs in either axial or equatorial positions.
-4-
AP Chemistry
Chapter 9. Molecular Geometry and Bonding Theories
-5-
AP Chemistry
Chapter 9. Molecular Geometry and Bonding Theories
Sample Exercise 9.2 (p. 354)
Use the VSEPR model to predict the molecular geometry of
a) SF4
b)
IF5
Practice Exercise 1 (9.2)
A certain AB4 molecule has a square-planar molecular geometry. Which of the following statements about the
molecule is or are true?
(i) The molecule has four electron domains about the central atom
(ii) The B-A-B angles between neighboring B atoms is 90o.
(iii) The molecule has two nonbonding pairs of electrons on atom A.
a)
b)
c)
d)
e)
Only one of the statements is true.
Statements (i) and (ii) are true.
Statements (i) and (iii) are true.
Statements (ii) and (iii) are true.
All three statements are true.
Practice Exercise 2 (9.2)
Predict the electron-domain geometry and molecular geometry of
a)
BrF5+
b) SF5+
Shapes of Larger Molecules
•
In acetic acid, CH3COOH, there are three interior atoms: two C and one O.
•
We assign the molecular (and electron-domain) geometry about each interior atom separately:
• The geometry around the first C is tetrahedral;
• The geometry around the second C is trigonal planar; and
• The geometry around the O is bent (tetrahedral).
-6-
AP Chemistry
Chapter 9. Molecular Geometry and Bonding Theories
Sample Exercise 9.3 (p. 355)
Eyedrops for dry eyes usually contain a water-soluble polymer called polyvinylalcohol, which is based on the
unstable organic molecule called vinyl alcohol. Predict the approximate values for the H-O-C and O-C-C bond
angles in vinyl alcohol.
Practice Exercise 1 (9.3)
The atoms of the compound methylhydrazine, CH6N2, which is used as a rocket propellant, are connected as
follows (note that lone pairs are not shown): See p. 356 in your textbook.
What do you predict for the ideal values of the C-N-N and H-N-H angles, respectively?
a) 109.5o and 109.5o
b) 109.5o and 120o
c) 120o and 109.5o
d) 120o and 120o
e) None of the above
Practice Exercise 2 (9.3)
Predict the H—C—H and C—C—C bond angles in the molecule shown, called propyne.
-7-
AP Chemistry
Chapter 9. Molecular Geometry and Bonding Theories
9.3 Molecular Shape and Molecular Polarity
•
•
If two charges, equal in magnitude and opposite in sign, are separated by a distance d, then a dipole is
established.
The dipole moment, , is given by  = Qr, where Q is the magnitude of the charge.
•
We can extend this to polyatomic molecules.
• For each bond in a polyatomic molecule, we can consider the bond dipole.
• The dipole moment due only to the two atoms in the bond is the bond dipole.
• Because bond dipoles and dipole moments are vector quantities, the orientation of these individual
dipole moments determines whether the molecule has an overall dipole moment.
•
Examples:
• In CO2 each +CO- dipole is canceled because the molecule is linear.
•
•
In H2O, the +HO- dipoles do not cancel because the molecule is bent.
Molecules with polar bonds can be either polar or nonpolar, depending on the molecular geometry.
Molecules containing polar bonds.
Two of these molecules have a zero dipole moment because their bond dipoles cancel one another, while the other
molecules are polar.
-8-
AP Chemistry
Chapter 9. Molecular Geometry and Bonding Theories
Sample Exercise 9.4 (p. 357)
Predict whether the following molecules are polar or nonpolar:
a)
BrCl
b)
SO2
c)
SF6
Practice Exercise 1 (9.4)
Consider an AB3 molecule in which A and B differ in electronegativity. You are told that the molecule has an
overall dipole moment of zero. Which of the following could be the molecular geometry of the molecule?
a)
b)
c)
d)
e)
Trigonal pyramidal
Trigonal planar
T-shaped
Tetrahedral
More than one of the above
Practice Exercise 2 (9.4)
Determine whether the following molecules are polar or nonpolar:
a)
SF4
b)
SiCl4
-9-
AP Chemistry
Chapter 9. Molecular Geometry and Bonding Theories
9.4 Covalent Bonding and Orbital Overlap
•
valence-bond theory:
9.5 Hybrid Orbitals
sp Hybrid Orbitals
•
According to the valence-bond model, a linear arrangement of electron domains implies sp hybridization.
• Since only one of 2p orbitals of Be has been used in hybridization, there are two unhybridized p orbitals
remaining on Be.
• The electrons in the sp hybrid orbitals form shared electron bonds with the two fluorine atoms.
- 10 -
AP Chemistry
Chapter 9. Molecular Geometry and Bonding Theories
Formation of two equivalent Be—F bonds in BeF2.
Each sp hybrid orbital on Be overlaps with a 2p orbital on F to form a bond. The two bonds are equivalent to each other and
form an angle of 180°.
sp2 and sp3 Hybrid Orbitals
•
Three sp2 hybrid orbitals are formed from hybridization of one s and two p orbitals.
• Thus, there is one unhybridized p orbital remaining.
• The large lobes of the sp2 hybrids lie in a trigonal plane.
• Molecules with trigonal planar electron-pair geometries have sp2 orbitals on the central atom.
•
Four sp3 hybrid orbitals are formed from hybridization of one s and three p orbitals.
• Therefore, there are four large lobes.
• Each lobe points towards the vertex of a tetrahedron.
• The angle between the large lobes is 109.5 o
• Molecules with tetrahedral electron pair geometries are sp3 hybridized.
- 11 -
AP Chemistry
Chapter 9. Molecular Geometry and Bonding Theories
Hybrid Orbital Summary
•
To assign hybridization:
• Draw a Lewis structure.
• Assign the electron-domain geometry using VSEPR theory.
• Specify the hybridization required to accommodate the electron pairs based on their geometric
arrangement.
• Name the geometry by the positions of the atoms.
Sample Exercise 9.5 (p. 365)
Describe the orbital hybridization around the central atom in NH2-.
Practice Exercise 1 (9.5)
For which of the following molecules or ions does the following description apply? “The bonding can be
explained using a set of sp2 hybrid orbitals on the central atom, with one of the hybrid orbitals holding a
nonbonding pair of electrons.”
a) CO2
b) H2S
c) O3
d) CO32e) more than one of the above
Practice Exercise 2 (9.5)
Predict the electron-domain geometry and the hybridization of the central atom in SO32-.
- 12 -
AP Chemistry
Chapter 9. Molecular Geometry and Bonding Theories
Ignore
hybridizations that
include d orbitals as
current evidence
suggests they do not
exist.
- 13 -
AP Chemistry
Chapter 9. Molecular Geometry and Bonding Theories
9.6 Multiple Bonds
•
Sigma () bonds: electron density lies on the axis between the nuclei. All single bonds are bonds.
•
Pi () bonds: electron density lies above and below the plane of the nuclei.
• A double bond consists of one  bond and one  bond.
• A triple bond has one  bond and two  bonds.
•
Often, the p orbitals involved in  bonding come from unhybridized orbitals.
•
For example: acetylene, C2H2:
• The electron-domain geometry of each C is linear.
• Therefore, the C atoms are sp hybridized.
• The sp hybrid orbitals form the C–C and C–H  bonds.
• There are two unhybridized p orbitals on each C atom.
• Both unhybridized p orbitals form the two  bonds;
• One  bond is above and below the plane of the nuclei;
• One  bond is in front and behind the plane of the nuclei.
•
When triple bonds form (e.g., N2), one  bond is always above and below and the other is in front and
behind the plane of the nuclei.
- 14 -
AP Chemistry
Chapter 9. Molecular Geometry and Bonding Theories
Sample Exercise 9.6 (p. 368)
Formaldehyde has the Lewis structure
Describe how the bonds in formaldehyde are formed in terms of overlaps of hybridized and unhybridized
orbitals.
Practice Exercise 1 (9.6)
We have just arrived at a bonding description for the formaldehyde molecule. Which of the following
statements about the molecule is or are true?
(i) Two of the electrons in the molecule are used to make the  bond in the molecule.
(ii) Six of the electrons in the molecule are used to make the  bonds in the molecule.
(iii) The C-O bond length in formaldehyde should be shorter than that in methanol, H3COH.
a)
b)
c)
d)
e)
Only one of the statements is true.
Statements (i) and (ii) are true.
Statements (i) and (iii) are true.
Statements (ii) and (iii) are true.
All three statements are true.
Practice Exercise 2 (9.6)
Consider the acetonitrile molecule:
a) Predict the bond angles around each carbon atom
b) Describe the hybridization at each of the carbon atoms
c) Determine the total number  and  bonds in the molecule.
- 15 -
AP Chemistry
Chapter 9. Molecular Geometry and Bonding Theories
Resonance Structures, Delocalization and  Bonding
•
In the case of benzene:
• There are six C–C bonds and six C–H  bonds
• Each C atom is sp2 hybridized.
• There is one unhybridized p orbital on each carbon atom, resulting in six unhybridized carbon p orbitals
in a ring.
•
In benzene there are two options for the three  bonds:
• Localized between carbon atoms or
• Delocalized over the entire ring (i.e., the  electrons are shared by all six carbon atoms).
•
Experimentally, all C–C bonds are the same length in benzene.
• Therefore, all C–C bonds are of the same type (recall single bonds are longer than double bonds).
Sample Exercise 9.7 (p. 371)
Describe the localized  bonding in the nitrate ion, NO3-.
Practice Exercise 1 (9.7)
How many electrons are in the  system of the ozone molecule, O3?
a) 2
b) 4
c) 6
d) 14
e) 18
Practice Exercise 2 (9.7)
Which of the following molecules or ions will exhibit delocalized bonding? SO2, SO3, SO32-, H2CO, NH4+?
- 16 -
AP Chemistry
Chapter 9. Molecular Geometry and Bonding Theories
General Conclusions
•
•
•
•
•
Every pair of bonded atoms shares one or more pairs of electrons.
Two electrons shared between atoms on the same axis as the nuclei are  bonds.
 Bonds are always localized in the region between two bonded atoms.
If two atoms share more than one pair of electrons, the additional pairs form bonds.
When resonance structures are possible, delocalization is also possible.
Sample Integrative Exercise (p. 381)
Elemental sulfur is a yellow solid that consists of S8 molecules. The structure of the S8 molecule is a puckered,
eight-membered ring.
Heating elemental sulfur to high temperatures produces gaseous S2 molecules.:
S8(s)  4 S2(g)
a) With respect to electronic structure, which element in the second row of the periodic table is most
similar to sulfur?
b) Use the VSEPR model to predict the S-S-S bond angles in S8 and the hybridization at S in S8.
d) Use bond enthalpies (Table 8.4) to estimate the enthalpy change for the reaction just described.
Is the reaction exothermic or endothermic?
- 17 -