Transcript Chapter 9 Molecular Geometries and Bonding Theories
Chemistry, The Central Science
, 10th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten
Chapter 9 Molecular Geometries and Bonding Theories
John D. Bookstaver St. Charles Community College St. Peters, MO 2006, Prentice-Hall, Inc.
Molecular Geometries and Bonding
Molecular Shapes
• The shape of a molecule plays an important role in its reactivity.
• By noting the number of bonding and nonbonding electron pairs we can easily predict the shape of the molecule.
Molecular Geometries and Bonding
What Determines the Shape of a Molecule?
• Simply put, electron pairs, whether they be bonding or nonbonding, repel each other.
• By assuming the electron pairs are placed as far as possible from each other, we can predict the shape of the molecule.
Molecular Geometries and Bonding
Electron Domains
• This molecule has four electron domains.
• We can refer to the electron pairs as electron domains .
• In a double or triple bond, all electrons shared between those two atoms are on the same side of the central atom; therefore, they count as one electron domain.
Molecular Geometries and Bonding
Valence Shell Electron Pair Repulsion Theory (VSEPR)
“The best arrangement of a given number of electron domains is the one that minimizes the repulsions among them.”
Molecular Geometries and Bonding
Electron-Domain Geometries
These are the electron-domain geometries for two through six electron domains around a central atom. (p. 319) Molecular Geometries and Bonding
Electron-Domain Geometries
• All one must do is count the number of electron domains in the Lewis structure.
• The geometry will be that which corresponds to that number of electron domains.
Molecular Geometries and Bonding
Molecular Geometries
• The electron-domain geometry is often
not
the shape of the molecule, however.
• The molecular geometry is that defined by the positions of
only
the atoms in the molecules, not the nonbonding pairs.
Molecular Geometries and Bonding
Molecular Geometries
Within each electron domain, then, there might be more than one molecular geometry.
Molecular Geometries and Bonding
Linear Electron Domain
• In this domain, there is only one molecular geometry: linear.
• NOTE: If there are only two atoms in the molecule, the molecule will be linear no matter what the electron domain is.
Molecular Geometries and Bonding
Trigonal Planar Electron Domain
• There are two molecular geometries: Trigonal planar, if all the electron domains are bonding Bent, if one of the domains is a nonbonding pair.
Molecular Geometries and Bonding
Nonbonding Pairs and Bond Angle
• Nonbonding pairs are physically larger than bonding pairs.
• Therefore, their repulsions are greater; this tends to decrease bond angles in a molecule.
Molecular Geometries and Bonding
SAMPLE EXERCISE 9.1
Using the VSEPR Model
Use the VSEPR model to predict the molecular geometry of
(a)
O 3 ,
(b)
SnCl 3 – .
Solve: (a)
We can draw two resonance structures for O 3 : Because of resonance, the bonds between the central O atom and the outer O atoms are of equal length. In both resonance structures the central O atom is bonded to the two outer O atoms and has one nonbonding pair. Thus, there are three electron domains about the central O atoms. (Remember that a double bond counts as a single electron domain.) The best arrangement of three electron domains is trigonal planar ( Table 9.1
). Two of the domains are from bonds, and one is due to a nonbonding pair, so the molecule has a bent shape with an ideal bond angle of 120 ° ( Table 9.2
).
Molecular Geometries and Bonding
SAMPLE EXERCISE 9.1
continued
(b)
The Lewis structure for the SnCl 3 – ion is The central Sn atom is bonded to the three Cl atoms and has one nonbonding
pair. Therefore, the Sn atom has four electron domains around it. The resulting
electron-domain geometry is tetrahedral (
Table 9.1
) with one of the corners
occupied by a nonbonding pair of electrons. The molecular geometry is thus
trigonal pyramidal ( Table 9.2
), like that of NH 3 .
PRACTICE EXERCISE
Predict the electron-domain geometry and the molecular geometry for
(a) (b)
CO 3 2– .
Answers: (a)
tetrahedral, bent;
(b)
trigonal planar, trigonal planar SeCl 2 , Molecular Geometries and Bonding
Multiple Bonds and Bond Angles
• Double and triple bonds place greater electron density on one side of the central atom than do single bonds.
• Therefore, they also affect bond angles.
Molecular Geometries and Bonding
Tetrahedral Electron Domain
• There are three molecular geometries: Tetrahedral, if all are bonding pairs Trigonal pyramidal if one is a nonbonding pair Bent if there are two nonbonding pairs Molecular Geometries and Bonding
Trigonal Bipyramidal Electron Domain
• There are two distinct positions in this geometry: Axial Equatorial Molecular Geometries and Bonding
Trigonal Bipyramidal Electron Domain
Lower-energy conformations result from having nonbonding electron pairs in equatorial, rather than axial, positions in this geometry.
Molecular Geometries and Bonding
Trigonal Bipyramidal Electron Domain
• There are four distinct molecular geometries in this domain: Trigonal bipyramidal Seesaw T-shaped Linear Molecular Geometries and Bonding
Octahedral Electron Domain
• All positions are equivalent in the octahedral domain.
• There are three molecular geometries: Octahedral Square pyramidal Square planar Molecular Geometries and Bonding
SAMPLE EXERCISE 9.2
Molecular Geometries of Molecules with Expanded Valence Shells
Use the VSEPR model to predict the molecular geometry of
(a)
SF 4 ,
(b)
IF 5 .
Solve: (a)
The Lewis structure for SF 4 is The sulfur has five electron domains around it: four from the S—F bonds and one from the nonbonding pair. Each domain points toward a vertex of a trigonal bipyramid. The domain from the nonbonding pair will point toward an equatorial position. The four bonds point toward the remaining four positions, resulting in a molecular geometry that is described as seesaw-shaped: Molecular Geometries and Bonding
SAMPLE EXERCISE 9.2
continued
(b)
The Lewis structure of IF 5 is (There are three lone pairs on each
of the F atoms, but they are not
shown.) The iodine has six electron domains around it, one of which is from a nonbonding pair. The electron-domain geometry is therefore octahedral, with one position occupied by the nonbonding electron pair. The resulting molecular geometry is therefore
square pyramidal
( Table 9.3
): Molecular Geometries and Bonding
SAMPLE EXERCISE 9.2
continued
PRACTICE EXERCISE
Predict the electron-domain geometry and molecular geometry of
(a)
ICl 4 – .
ClF 3 ,
(b) Answers: (a)
trigonal bipyramidal, T-shaped;
(b)
octahedral, square planar Molecular Geometries and Bonding
Larger Molecules
In larger molecules, it makes more sense to talk about the geometry about a particular atom rather than the geometry of the molecule as a whole.
Molecular Geometries and Bonding
Larger Molecules
This approach makes sense, especially because larger molecules tend to react at a particular site in the molecule.
Molecular Geometries and Bonding
SAMPLE EXERCISE 9.3
Predicting Bond Angles
Eyedrops for dry eyes usually contain a water-soluble polymer called
poly(vinyl alcohol),
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.
Solution Analyze:
We are given a molecular structure and asked to determine two bond angles in the structure.
Plan:
To predict a particular bond angle, we consider the middle atom of the angle and determine the number of electron domains surrounding that atom. The ideal angle corresponds to the electron-domain geometry around the atom. The angle will be compressed somewhat by nonbonding electrons or multiple bonds.
Solve:
For the H—O—C bond angle, there are four electron domains around the middle O atom (two bonding and two nonbonding). The electron-domain geometry around O is therefore tetrahedral, which gives an ideal angle of 109.5
° . The H—O—C angle will be compressed somewhat by the nonbonding pairs, so we expect this angle to be slightly less than 109.5
° .
To predict the O—C—C bond angle, we must examine the leftmost C atom, which is the central atom for this angle. There are three atoms bonded to this C atom and no nonbonding pairs, and so it has three electron domains about it. The predicted electron-domain geometry is trigonal planar, resulting in an ideal bond angle of 120 ° . Because of the larger size of the domain, however, the O—C—C bond angle should be slightly greater than 120 ° .
Molecular Geometries and Bonding
SAMPLE EXERCISE 9.3
continued
PRACTICE EXERCISE
Predict the H—C—H and C—C—C bond angles in the following molecule, called
propyne
:
Answers:
109.5
° , 180 ° Molecular Geometries and Bonding
Polarity
• In Chapter 8 we discussed bond dipoles.
• But just because a molecule possesses polar bonds does not mean the molecule
as a whole
will be polar.
Molecular Geometries and Bonding
Polarity
By adding the individual bond dipoles, one can determine the overall dipole moment for the molecule.
Molecular Geometries and Bonding
Polarity
Molecular Geometries and Bonding
SAMPLE EXERCISE 9.4
Polarity of Molecules
Predict whether the following molecules are polar or nonpolar:
(a)
BrCl,
(b)
SO 2 ,
(c)
SF 6 .
Solution Analyze:
We are given the molecular formulas of several substances and asked to predict whether the molecules are polar.
Plan:
If the molecule contains only two atoms, it will be polar if the atoms differ in electronegativity. If it contains three or more atoms, its polarity depends on both its molecular geometry and the polarity of its bonds. Thus, we must draw a Lewis structure for each molecule containing three or more atoms and determine its molecular geometry. We then use the relative electronegativities of the atoms in each bond to determine the direction of the bond dipoles. Finally, we see if the bond dipoles cancel each other to give a nonpolar molecule or reinforce each other to give a polar one.
Solve: (a)
Chlorine is more electronegative than bromine. All diatomic molecules with polar bonds are polar molecules. Consequently, BrCl will be polar, with chlorine carrying the partial negative charge: The actual dipole moment of BrCl, as determined by experimental measurement, is
µ
= 0.57 D.
(b)
Because oxygen is more electronegative than sulfur, SO 2 written for SO 2 : has polar bonds. Three resonance forms can be Molecular Geometries and Bonding
SAMPLE EXERCISE 9.4
continued For each of these, the VSEPR model predicts a bent geometry. Because the molecule is bent, the bond dipoles do not cancel and the molecule is polar: Experimentally, the dipole moment of SO 2 is
µ
= 1.63 D.
(c)
Fluorine is more electronegative than sulfur, so the bond dipoles point toward fluorine. The six S—F bonds are arranged octahedrally around the central sulfur: Because the octahedral geometry is symmetrical, the bond dipoles cancel, and the molecule is nonpolar, meaning that
µ
= 0.
PRACTICE EXERCISE
Determine whether the following molecules are polar or nonpolar:
(a)
NF 3 ,
(b)
BCl 3 .
Answers: (a)
polar because polar bonds are arranged in a trigonal-pyramidal geometry,
(b)
nonpolar because polar bonds are arranged in a trigonal-planar geometry Molecular Geometries and Bonding
Overlap and Bonding
• We think of covalent bonds forming through the sharing of electrons by adjacent atoms.
• In such an approach this can only occur when orbitals on the two atoms overlap.
Molecular Geometries and Bonding
Overlap and Bonding
• Increased overlap brings the electrons and nuclei closer together while simultaneously decreasing electron electron repulsion.
• However, if atoms get too close, the internuclear repulsion greatly raises the energy.
Molecular Geometries and Bonding
Hybrid Orbitals
But it’s hard to imagine tetrahedral, trigonal bipyramidal, and other geometries arising from the atomic orbitals we recognize.
Molecular Geometries and Bonding
Hybrid Orbitals
• Consider beryllium: In its ground electronic state, it would not be able to form bonds because it has no singly-occupied orbitals.
Molecular Geometries and Bonding
Hybrid Orbitals
But if it absorbs the small amount of energy needed to promote an electron from the 2
s
to the 2
p
orbital, it can form two bonds.
Molecular Geometries and Bonding
Hybrid Orbitals
• Mixing the
s
and
p
orbitals yields two degenerate orbitals that are hybrids of the two orbitals.
These
sp
hybrid orbitals have two lobes like a
p
orbital.
One of the lobes is larger and more rounded as is the
s
orbital.
Molecular Geometries and Bonding
Hybrid Orbitals
• These two degenerate orbitals would align themselves 180 from each other.
• This is consistent with the observed geometry of beryllium compounds: linear.
Molecular Geometries and Bonding
Hybrid Orbitals
• With hybrid orbitals the orbital diagram for beryllium would look like this.
• The
sp
orbitals are higher in energy than the 1
s
orbital but lower than the 2
p
.
Molecular Geometries and Bonding
Hybrid Orbitals
Using a similar model for boron leads to… Molecular Geometries and Bonding
Hybrid Orbitals
…three degenerate
sp 2
orbitals.
Molecular Geometries and Bonding
Hybrid Orbitals
With carbon we get… Molecular Geometries and Bonding
Hybrid Orbitals
…four degenerate
sp 3
orbitals.
Molecular Geometries and Bonding
Hybrid Orbitals
For geometries involving expanded octets on the central atom, we must use our hybrids.
d
orbitals in Molecular Geometries and Bonding
Hybrid Orbitals
This leads to five degenerate
sp 3 d
orbitals… …or six degenerate
sp 3 d 2
orbitals.
Molecular Geometries and Bonding
Hybrid Orbitals
Once you know the electron-domain geometry, you know the hybridization state of the atom.
Molecular Geometries and Bonding
SAMPLE EXERCISE 9.5
Hybridization
Indicate the hybridization of orbitals employed by the central atom in
(a)
NH 2 – ,
(b)
9.2).
SF 4 (see Sample Exercise
Solution Analyze:
We are given two chemical formulas—one for a polyatomic anion and one for a molecular compound—and asked to describe the type of hybrid orbitals surrounding the central atom in each case.
Plan:
To determine the hybrid orbitals used by an atom in bonding, we must know the electron-domain geometry around the atom. Thus, we first draw the Lewis structure to determine the number of electron domains around the central atom. The hybridization conforms to the number and geometry of electron domains around the central atom as predicted by the VSEPR model.
Solve: (a)
The Lewis structure of NH 2 – is Because there are four electron domains around N, the electron-domain geometry is tetrahedral. The hybridization that gives a tetrahedral electron-domain geometry is
sp
3 hybrid orbitals contain nonbonding pairs of electrons, and the other two are used to make two-electron bonds with the hydrogen atoms.
(b)
The Lewis structure and electron-domain geometry of SF 4 are shown in Sample Exercise 9.2. There are five electron domains around S, giving rise to a trigonal-bipyramidal electron-domain geometry. With an expanded octet of ten electrons, a
d
orbital on the sulfur must be used. The trigonal-bipyramidal electron-domain
geometry corresponds to
sp
3
d
hybridization (
). One of the hybrid orbitals that points in an equatorial
direction contains a nonbonding pair of electrons; the other four are used to form the S—F bonds.
PRACTICE EXERCISE
Predict the electron-domain geometry and the hybridization of the central atom in
(a)
SO 3 2– ,
(b) Answers: (a)
tetrahedral,
sp
3 ;
(b)
octahedral,
sp
3
d
2 SF 6 .
Molecular Geometries and Bonding
Valence Bond Theory • Hybridization is a major player in this approach to bonding.
• There are two ways orbitals can overlap to form bonds between atoms.
Molecular Geometries and Bonding
Sigma (
) Bonds
• Sigma bonds are characterized by Head-to-head overlap.
Cylindrical symmetry of electron density about the internuclear axis.
Molecular Geometries and Bonding
Pi (
) Bonds
• Pi bonds are characterized by Side-to-side overlap.
Electron density above and below the internuclear axis.
Molecular Geometries and Bonding
Single Bonds
Single bonds are always bonds, because overlap is greater, resulting in a stronger bond and more energy lowering.
Molecular Geometries and Bonding
Multiple Bonds
In a multiple bond one of the bonds is a and the rest are bonds.
bond Molecular Geometries and Bonding
Multiple Bonds
• In a molecule like formaldehyde (shown at left) an
sp 2
orbital on carbon overlaps in fashion with the corresponding orbital on the oxygen.
• The unhybridized
p
orbitals overlap in fashion.
Molecular Geometries and Bonding
Multiple Bonds
In triple bonds, as in acetylene, two
sp
orbitals form a bond between the carbons, and two pairs of
p
overlap in orbitals fashion to form the two bonds.
Molecular Geometries and Bonding
SAMPLE EXERCISE 9.6
Describing
and
Bonds in a Molecule
Formaldehyde has the Lewis structure Describe how the bonds in formaldehyde are formed in terms of overlaps of appropriate hybridized and unhybridized orbitals.
Solution Analyze: Plan:
We are asked to describe the bonding in formaldehyde in terms of orbital overlaps.
Single bonds will be of the type, whereas double bonds will consist of one bond and one bond. The ways in which these bonds form can be deduced from the geometry of the molecule, which we predict using the VSEPR model.
Solve:
The C atom has three electron domains around it, which suggests a trigonal-planar geometry with bond angles of about 120 ° . This geometry implies
sp
2 make the two C—H and one C—O
hybrid orbitals on C ( Table 9.4
bonds to C. There remains an unhybridized 2
p
orbital on carbon, perpendicular to the plane of the three
sp
2 hybrids.
Molecular Geometries and Bonding
SAMPLE EXERCISE 9.6
continued The O atom also has three electron domains around it, and so we will assume that it has
sp
2 well. One of these hybrids participates in the C—O hybridization as bond, while the other two hybrids hold the two nonbonding electron pairs of the O atom. Like the C atom, therefore, the O atom has an unhybridized 2
p
orbital that is perpendicular to the plane of the molecule. The unhybridized 2
p
form a C—O
π
bond, as illustrated in Figure 9.27.
orbitals on the C and O atoms overlap to
Figure 9.27 Formation of
and
bonds in formaldehyde, H 2 CO.
PRACTICE EXERCISE
Consider the acetonitrile molecule:
(a) (c)
Predict the bond angles around each carbon atom;
(b)
determine the total number of and bonds in the molecule.
Answers: (a)
two
π
bonds approximately 109 ° around the left C and 180 ° on the right C; (b) sp 3 , sp; (c) five Geometries
Delocalized Electrons: Resonance
When writing Lewis structures for species like the nitrate ion, we draw resonance structures to more accurately reflect the structure of the molecule or ion.
Molecular Geometries and Bonding
Delocalized Electrons: Resonance
• In reality, each of the four atoms in the nitrate ion has a
p
orbital.
• The
p
orbitals on all three oxygens overlap with the
p
orbital on the central nitrogen.
Molecular Geometries and Bonding
Delocalized Electrons: Resonance
This means the electrons are not localized between the nitrogen and one of the oxygens, but rather are delocalized throughout the ion.
Molecular Geometries and Bonding
Resonance
The organic molecule benzene has six bonds and a
p
orbital on each carbon atom.
Molecular Geometries and Bonding
Resonance
• In reality the electrons in benzene are not localized, but delocalized.
• The even distribution of the electrons in benzene makes the molecule unusually stable.
Molecular Geometries and Bonding
SAMPLE EXERCISE 9.7
Delocalized Bonding
Describe the bonding in the nitrate ion, NO 3 – . Does this ion have delocalized bonds?
Solution Analyze:
Given the chemical formula for a polyatomic anion, we are asked to describe the bonding and determine whether the ion has delocalized bonds.
Plan:
Our first step in describing the bonding in NO 3 – is to construct appropriate Lewis structures. If there are multiple resonance structures that involve the placement of the double bonds in different locations, that suggests that the component of the double bonds is delocalized.
Solve:
In Section 8.6 we saw that NO 3 – has three resonance structures: In each of these structures the electron-domain geometry at nitrogen is trigonal planar, which implies
sp
2 hybridization of the N atom. The
sp
2 hybrid orbitals are used to construct the three N—O bonds that are present in each of the resonance structures.
Molecular Geometries and Bonding
SAMPLE EXERCISE 9.7
continued The unhybridized 2
p
orbital on the N atom can be used to make structures shown, we might imagine a single localized N––O bonds. For any one of the three resonance bond formed by the overlap of the unhybridized 2
p
orbital on N and a 2
p
orbital on one of the O atoms, as shown in Figure 9.30(a). Because each resonance structure contributes equally to the observed structure of NO 3 – , however, we represent the bonding as spread out, or delocalized, over the three N—O bonds, as shown in Figure 9.30(b).
Figure 9.30
Localized and delocalized
bonds in NO 3 – .
Molecular Geometries and Bonding
SAMPLE EXERCISE 9.7
continued
PRACTICE EXERCISE
Which of the following molecules or ions will exhibit delocalized bonding: SO 3 , SO 3 2– , H 2 CO, O 3 , NH 4 + ?
Answer:
SO 3 and O 3 , as indicated by the presence of two or more resonance structures involving each of these molecules bonding for Molecular Geometries and Bonding
Molecular Orbital (MO) Theory
Though valence bond theory effectively conveys most observed properties of ions and molecules, there are some concepts better represented by molecular orbitals. Molecular Geometries and Bonding
Molecular Orbital (MO) Theory
• In MO theory, we invoke the wave nature of electrons.
• If waves interact constructively, the resulting orbital is lower in energy: a bonding molecular orbital.
Molecular Geometries and Bonding
Molecular Orbital (MO) Theory
If waves interact destructively, the resulting orbital is higher in energy: an antibonding molecular orbital.
Molecular Geometries and Bonding
MO Theory
• In H 2 the two electrons go into the bonding molecular orbital.
• The bond order is one half the difference between the number of bonding and antibonding electrons.
Molecular Geometries and Bonding
MO Theory
For hydrogen, with two electrons in the bonding MO and none in the antibonding MO, the bond order is 1 2 (2 - 0) = 1 Molecular Geometries and Bonding
MO Theory
• In the case of He 2 , the bond order would be 1 2 (2 - 2) = 0 • Therefore, He 2 does not exist.
Molecular Geometries and Bonding
MO Theory
• For atoms with both
s
and
p
orbitals, there are two types of interactions: The
s
and the
p
orbitals that face each other overlap in fashion.
The other two sets of
p
orbitals overlap in fashion.
Molecular Geometries and Bonding
MO Theory
• The resulting MO diagram looks like this.
• There are both and bonding molecular orbitals and * and * antibonding molecular orbitals.
Molecular Geometries and Bonding
MO Theory
• The smaller
p
-block elements in the second period have a sizeable interaction between the
s
and
p
orbitals.
• This flips the order of the s and p molecular orbitals in these elements.
Molecular Geometries and Bonding
Second-Row MO Diagrams
Molecular Geometries and Bonding