Chapter 9. Molecular Geometry and Bonding Theoriestamapchemistryhart.wikispaces.com/file/view/Chapter+9+Outline... · AP Chemistry Chapter 9. ... - 1 - Chapter 9. Molecular Geometry

AP Chemistry Chapter 9. Molecular Geometry and Bonding Theories

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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-


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Practice Exercise 9.1 Predict the electron-domain geometry and the molecular geometry for

a) SeCl2

b) CO32-


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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).

C OCl

Cl111.4o

124.3o


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• 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.


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Sample Exercise 9.2 (p. 351) Use the VSEPR model to predict the molecular geometry of

a) SF4

b) IF5

Practice Exercise 9.2 Predict the electron-domain geometry and molecular geometry of

a) ClF3

b) ICl4-

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).


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Sample Exercise 9.3 (p. 352) 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 9.3

Predict the H—C—H and C—C—C bond angles in the molecule shown, called propyne.

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, µ, is given by µ = 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 δ+C−Oδ- dipole is canceled because the molecule is linear.

• In H2O, the δ+H−Oδ- dipoles do not cancel because the molecule is bent.

• Molecules with polar bonds can be either polar or nonpolar, depending on the molecular geometry.


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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.

Sample Exercise 9.4 (p. 354)

Predict whether the following molecules are polar or nonpolar:

a) BrCl

b) SO2

c) SF6


Determine whether the following molecules are polar or nonpolar:

a) NF3

b) BCl3


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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.


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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.


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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. 361) Indicate the hybridization of orbitals employed by the central atom in each of the following:

a) NH2-

b) SF4 (see Sample Exercise 9.2)

Practice Exercise 9.5 Predict the electron-domain geometry and the hybridization of the central atom in

a) SO32-

b) SF6


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9.6 Multiple Bonds • Sigma (σ) bonds: electron density lies on the axis between the nuclei. All single bonds are σ bonds. • Pi (π) bonds: electron density lies above and below the plane of the nuclei. • A double bond consists of one σ bond and one π bond. • A triple bond has one σ bond and two π bonds. • Often, the p orbitals involved in π 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 σ bonds. • There are two unhybridized p orbitals on each C atom. • Both unhybridized p orbitals form the two π bonds; • One π bond is above and below the plane of the nuclei; • One π bond is in front and behind the plane of the nuclei. • When triple bonds form (e.g., N2), one π bond is always above and below and the other is in front and

behind the plane of the nuclei.


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Sample Exercise 9.6 (p. 364)

Describe how the bonds in formaldehyde are formed in terms of overlaps of appropriate hybridized and unhybridized orbitals.


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 σ and π bonds in the molecule.

Resonance Structures, Delocalization and π Bonding

• In the case of benzene: • There are six C–C σ bonds and six C–H σ 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.


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• In benzene there are two options for the three π bonds: • Localized between carbon atoms or • Delocalized over the entire ring (i.e., the π 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. 366) Describe the localized π bonding in the nitrate ion, NO3

-.

Practice Exercise 9.7 Which of the following molecules or ions will exhibit delocalized bonding? SO3, SO3

2-, H2CO, O3, NH4+.

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 σ bonds. • σ 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 π 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?

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Chapter 9. Molecular Geometry and Bonding Theoriestamapchemistryhart.wikispaces.com/file/view/Chapter+9+Outline... · AP Chemistry Chapter 9. ... - 1 - Chapter 9. Molecular Geometry