© 2012 Pearson Education, Inc. Chapter 9 Molecular Geometries and Bonding Theories.

© 2012 Pearson Education, Inc.

Chapter 9

Molecular Geometries

and Bonding Theories

MolecularGeometries

and Bonding© 2012 Pearson Education, Inc.

A. Shape of geometrical structureB. Bond angles between atomsC. Ionization energy of atomsD. Atomic radii

MolecularGeometries


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.

MolecularGeometries


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.

MolecularGeometries


A. OctahedralB. TetrahedralC. Trigonal bipyramidalD. Trigonal planar

MolecularGeometries


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.

• The central atom in this molecule, A, has four electron domains.

MolecularGeometries



Within each electron domain, then, there might be more than one molecular geometry.

MolecularGeometries


A. trigonal planar

B. tetrahedral

C. trigonal bipyramidal

D. octahedral

MolecularGeometries


MolecularGeometries


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.”

MolecularGeometries


Electron-Domain Geometries

Table 9.1 contains the electron-domain geometries for two through six electron domains around a central atom.

MolecularGeometries


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 the number of electron domains.

MolecularGeometries



• 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.

MolecularGeometries


A. There are no common features.

B. Each occurs about the central atom only.

C. Each represents a single electron domain.

D. All exist when a particular Lewis structure is drawn.

MolecularGeometries


Linear Electron Domain

• In the linear 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.

MolecularGeometries


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.

MolecularGeometries


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.

MolecularGeometries


A. No, the domain with the double bond should “push” the other two electron domains resulting in the bond angle between single bonds to less than 120°.

B. Yes, the existence of resonance with three resonance structures equalizes repulsions between electron domains resulting in all bond angles equaling 120°.

MolecularGeometries


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.

MolecularGeometries


A. The bonding electron pair domain is “squeezed” by the repulsion experienced with two nuclei versus the nonbonding electron pair domain is repelled by only one nearby nucleus.B. The bonding electron pair domain is attracted to two nuclei versus the nonbonding electron pair domain is attracted to only one nearby nucleus .C. Bonding electron pair electrons move more slowly near two nuclei and thus expand less in volume compared to the nonbonding electron pairs that move more quickly in the presence of only one nucleus.D. Bonding electron pair electrons move more quickly near two nuclei and thus expand less in volume compared to the nonbonding electron pairs that move more slowly in the presence of only one nucleus.

MolecularGeometries


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.

MolecularGeometries


Sample Exercise 9.1 Using the VSPER ModelUse the VSEPR model to predict the molecular geometry of (a) O3, (b) SnCl3– .

Predict the electron-domain and molecular geometries for (a) SeCl2, (b) CO32– .

Practice Exercise

MolecularGeometries


A. Bond angles are not determined by a particular arrangement of electron domains.

B. A tetrahedral arrangement of electron domains results in greater electron repulsions and a less favorable geometry than electron domains in a square planar geometry.

C. A tetrahedral arrangement of electron domains results in smaller electron repulsions and a more favorable geometry than electron domains in a square planar geometry.

D. A tetrahedral arrangement of electron domains results in no greater electron repulsions and no greater favorable geometry than electron domains in a square planar geometry.

MolecularGeometries


Trigonal Bipyramidal Electron Domain

• There are two distinct positions in this geometry:– Axial– Equatorial

MolecularGeometries


A. 45° B. 90° C. 109.5° D. 120°

MolecularGeometries



Lower-energy conformations result from having nonbonding electron pairs in equatorial, rather than axial, positions in this geometry.

MolecularGeometries



• There are four distinct molecular geometries in this domain:– Trigonal bipyramidal– Seesaw– T-shaped– Linear

MolecularGeometries


Octahedral Electron Domain

• All positions are equivalent in the octahedral domain.

• There are three molecular geometries:– Octahedral– Square pyramidal– Square planar

MolecularGeometries


Use the VSEPR model to predict the molecular geometry of (a) SF4, (b) IF5.

Sample Exercise 9.2 Molecular Geometries of Molecules with Expanded Valence Shells

Predict the electron-domain and molecular geometries of (a) BrF3, (b) ICI4– .

Practice Exercise

MolecularGeometries


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.

MolecularGeometries


A. The nonbonding electron domains exert a greater repulsive force than the bonding electron domains and this results in the distortion in the C—O—H bond angle.B. The bonding electron domains exert a greater repulsive force than the nonbonding electron domains and this results in the distortion in the C—O—H bond angle.C. The hydrogen atom attached to the carbon atom in the C—O—H interacts with the nonbonding electron domain and this results inthe distortion in bond angle.D. The CH3 group is large and interacts with the C—O—H group and causes the distortion in C—O—H bond angle.

MolecularGeometries


Sample Exercise 9.3 Predicting Bond AnglesEyedrops for dry eyes usually contain a water-soluble polymer called poly(vinyl alcohol), which is based on the unstable organic molecule vinyl alcohol:

Predict the approximate values for the H — O — C and O — C — C bond angles in vinyl alcohol.

Practice ExercisePredict the H — C — H and C — C — C bond angles in propyne:

MolecularGeometries


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.

By adding the individual bond dipoles, one can determine the overall dipole moment for the molecule.

MolecularGeometries


A. The tails of the arrows point toward regions of highest electron density as indicated by the blue color in the electron density picture.B. The tails of the arrows point toward regions of lowest electron density as indicated by the red color in the electron density picture.C. The heads of the arrows point toward regions of highest electron density as indicated by the red color in the electron density picture.D. The heads of the arrows point toward regions of lowest electron density as indicated by the yellow color in the electron density picture.

MolecularGeometries


A. Yes, because COS has different elements than in CO2.

B. No, because COS is linear.C. Yes, because O and S have different

electronegativities, and CO and CS bond dipoles do not cancel each other.

D. No, because O and S have similar electronegativities, and CO and CS bond dipoles cancel each other.

MolecularGeometries


Polarity

MolecularGeometries


Sample Exercise 9.4 Polarity of Molecules

SolutionPredict whether these molecules are polar or nonpolar: (a) BrCl, (b) SO2, (c) SF6.

Practice ExerciseDetermine whether the following molecules are polar or nonpolar: (a) NF3,

(b) BCl3.

MolecularGeometries


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.

MolecularGeometries


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.

MolecularGeometries


A. As the internuclear bonding distance decreases the nucleus-nucleus repulsion increases and the potential energy rises.

B. As the internuclear bonding distance decreases the nucleus-nucleus repulsion decreases and the potential energy rises.

C. As the internuclear bonding distance decreases the electron-nucleus attraction increases and the potential energy rises.

D. As the internuclear bonding distance decreases the electron nucleus repulsion increases and the potential energy rises.

MolecularGeometries


Hybrid Orbitals• Consider beryllium:

– In its ground electronic state, beryllium would not be able to form bonds, because it has no singly occupied orbitals.

– But if it absorbs the small amount of energy needed to promote an electron from the 2s to the 2p orbital, it can form two bonds.

MolecularGeometries


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.

MolecularGeometries


Hybrid Orbitals• These two degenerate orbitals would align

themselves 180 from each other.• This is consistent with the observed geometry of

beryllium compounds: linear.

MolecularGeometries


A. The small lobes of the sp hybrid orbitals of Be can not overlap the lobes of the F orbitals even if oriented properly and cannot form a bond.B. Only large hybrid orbitals of one atom can overlap with large hybrid orbitals of another atom.C. The small lobes of the sp hybrid orbitals of Be can not hold electrons and therefore can not form bonds to F whereas the large lobes can hold electrons and form bonds.D. The small lobes of the sp hybrid orbitals of Be are considerably smaller in spatial extent and form a less effective bond with the lobes of the F orbitals than do the large lobes.

MolecularGeometries


A. Both p orbitals are parallel to the Be-F bonds.B. Both p orbitals intersect the Be-F bonds.C. Both p orbitals are at an angle of 120o to the

Be-F bonds.D. Both p orbitals are perpendicular to the

Be-F bonds.

MolecularGeometries


Hybrid Orbitals

• With hybrid orbitals, the orbital diagram for beryllium would look like this (Fig. 9.15).

• The sp orbitals are higher in energy than the 1s orbital, but lower than the 2p.

MolecularGeometries


Hybrid Orbitals

Using a similar model for boron leads to three degenerate sp2 orbitals.

MolecularGeometries


A. 2B. 3C. 4D. 5

MolecularGeometries


A. The unhybridized p-orbital is 180° from the plane of the sp2 orbitals.

B. The unhybridized p-orbital is coplanar with the plane of the sp2 orbitals.

C. The unhybridized p-orbital is 109.5° from the plane of the sp2 orbitals.

D. The unhybridized p-orbital is perpendicular to the plane of the sp2 orbitals.

MolecularGeometries


Hybrid Orbitals

With carbon, we get four degenerate sp3 orbitals.

MolecularGeometries


Sample Exercise 9.5 Hybridization

SolutionIndicate the orbital hybridization around the central atom in NH2

–.

Practice ExercisePredict the electron-domain geometry and hybridization of the central atom in SO3

2–.

MolecularGeometries


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.

MolecularGeometries


Sigma () Bonds and Pi () Bonds

• Sigma bonds are characterized by

– Head-to-head overlap.

– Cylindrical symmetry of electron density about the internuclear axis.

• Pi bonds are characterized by

– Side-to-side overlap.

– Electron density above and below the internuclear axis.

MolecularGeometries


Single and Multiple BondsSingle bonds are always bonds, because overlap is greater, resulting in a stronger bond and more energy lowering.

In a multiple bond, one of the bonds is a bond and the rest are bonds.

MolecularGeometries


Multiple Bonds

• In a molecule like formaldehyde (shown at left), an sp2 orbital on carbon overlaps in fashion with the corresponding orbital on the oxygen.

• The unhybridized p orbitals overlap in fashion.

MolecularGeometries


Multiple Bonds

In triple bonds, as in acetylene, two sp orbitals form a bond between the carbons, and two pairs of p orbitals overlap in fashion to form the two bonds.

MolecularGeometries


A. To ensure that a sp2 hybrid orbital on one carbon atom can bond to a sp2 hybrid orbital on the other carbon atom or a hydrogen atom.B. To ensure that the H—C—H bond angle is 120°.C. To ensure that the unhybridized p orbitals on the carbon atoms arealigned properly to form a pi bond.D. To ensure that the H—C—C bond angle is 120°.

MolecularGeometries


A. The molecule is both linear and planar.

B. The molecule is not linear, but is planar.

C. The molecule is linear, but not planar.

D. The molecule is neither linear nor planar.

MolecularGeometries


A. Acetylene because it has two carbon-carbon pi bonds and ethylene has only two.B. Acetylene because it has one carbon-carbon pi bond and ethylene has two.C. Ethylene because it has two carbon-carbon pi bonds and acetylene has only two.D. Ethylene because it has one carbon-carbon pi bond and acetylene has two.

MolecularGeometries


Sample Exercise 9.6 Describing σ and π Bonds in a Molecule

Solution

Formaldehyde has the Lewis structure

Describe how the bonds in formaldehyde are formed in terms of overlaps of hybrid and unhybridized oribitals.

Practice Exercise

(a) Predict the bond angles around each carbon atom in acetonitrile:

(b) Describe the hybridization at each carbon atom, and (c) determine the number of σ and π bonds in the molecule.

MolecularGeometries


Delocalized Electrons: ResonanceWhen writing Lewis structures for species like the nitrate ion, we draw resonance structures to more accurately reflect the structure of the molecule or ion.

MolecularGeometries


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.• This means the electrons are not

localized between the nitrogen and one of the oxygens, but rather are delocalized throughout the ion.

MolecularGeometries


ResonanceThe organic molecule benzene has six bonds and a p orbital on each carbon atom.

MolecularGeometries


A. (1) and (4)B. (2) and (3)C. (1) and (2)D. (2) and (4)

(1) C—H(2) C—C(3) C—C—C•C—C—C—C—C—C

MolecularGeometries


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.

MolecularGeometries


Sample Exercise 9.7 Delocalized Bonding

SolutionDescribe the bonding in the nitrate ion, NO3

–. Does this ion have delocalized π bonds?

Practice ExerciseWhich of these species have delocalized bonding: SO3, SO3

2– ,H2CO, O3, NH4+?

MolecularGeometries


A. sp hybrid atomic orbitals

B. sp2 hybrid atomic orbitals

C. sp3 hybrid atomic orbitals

D. sp3d hybrid atomic orbitals

MolecularGeometries


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.•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.•If waves interact destructively, the resulting orbital is higher in energy: an antibonding molecular orbital.

MolecularGeometries


A. 1s B. 1s* C. D. s1 *1s

MolecularGeometries


MO Theory• In H2 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.

• For hydrogen, with two electrons in the bonding MO and none in the antibonding MO, the bond order is

12

(2 − 0) = 1

MolecularGeometries


MO Theory

• In the case of He2, the bond order would be

12

(2 − 2) = 0

• Therefore, He2 does not exist.

MolecularGeometries


A. Remain bonded, as the bond order remains the same.

B. Fall apart, as the bond order changes from 1 to 1.5.

C. Remain bonded, as the bond order changes from 1 to 1.5.

D. Fall apart, as the bond order changes from 1 to zero.

MolecularGeometries


A. The two electrons in the σ MO

B. The one electron in the σ MO

C. The two electrons in the σ1s MO

D. The one electron in the σ1s MO

*1s

*1s

MolecularGeometries


What is the bond order of the He2+ ion? Would you expect this ion to be stable relative to the separated

He atom and He+ ion?

Sample Exercise 9.8 Bond Order

Solution

Practice ExerciseDetermine the bond order of the H2

– ion.

MolecularGeometries


A. The 1s orbital of each Li atom is large in spatial extent and they do not overlap effectively to form a bond between two Li atoms.B. The 1s orbital of each Li atom overlaps the domain of the 2s orbital and this prevents forming an effective bond between two Li atoms.C. The net bond order for the σ1s and σ MOs is zero and no effective bond occurs.D. The net bond order for the σ1s and σ MOs is 1 and no effective bond occurs.

*1s*1s

MolecularGeometries


A. No, because the bond order is 0.

B. Yes, because the bond order is 0.5.

C. Yes, because the bond order is 1.

D. Yes, because the bond order is 1.5.

MolecularGeometries


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.

MolecularGeometries


MO Theory

• The resulting MO diagram looks like this (Fig. 9.41).

• There are both and bonding molecular orbitals and * and * antibonding molecular orbitals.

MolecularGeometries


A. Bonding MOsB. Antibonding MOs

MolecularGeometries


MO Theory

• The smaller p-block elements in the second period have a sizable interaction between the s and p orbitals.

• This flips the order of the and molecular orbitals in these elements.

MolecularGeometries


A. σ2p and π2p

B. σ and π

C. σ2p and σ2p

D. σ and π2p *2s

*2p

*2p

MolecularGeometries


Second-Row MO Diagrams

MolecularGeometries


A. Yes, because the reversal of MOs does not change the electron occupation of each MO.

B. No, because the reversal of MOs results in the two MOs containing one electron each.

2 p

MolecularGeometries


A. F2 contains more electrons than N2 and these electrons go into theantibonding π orbital and the bond order decreases.

B. The change in the order of π2p and σ2p MOs from N2 to F2 reducesthe bond order.

C. The increase in number of occupied MOs in F2 reduces electronelectron repulsions and decreases bond enthalpy.

D. The reduction in unoccupied antibonding N2 and that of F2g MOsdecreases the bond enthalpy.

*2p

MolecularGeometries


Sample Exercise 9.9 Molecular Orbitals of a Period 2 Diatomic Ion

Solution

For the O2+ ion, predict (a) number of unpaired electrons, (b) bond order, (c) bond enthalpy and bond

length.

Practice ExercisePredict the magnetic properties and bond orders of (a) the peroxide ion, O2

2– (b) the acetylide ion, C22–

© 2012 Pearson Education, Inc. Chapter 9 Molecular Geometries and Bonding Theories.

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Transcript of © 2012 Pearson Education, Inc. Chapter 9 Molecular Geometries and Bonding Theories.