M10 TRO4904 03 SSM C10 PDF

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10

Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory

Review Questions 10.1

The properties of molecules are directly related to their shape. The sensation of taste, immune response, the sense of smell, and many types of drug action all depend on shapespecific interactions between molecules and proteins.

10.3

The five basic electron geometries are (1) Linear, which has two electron groups. (2) Trigonal planar, which has three electron groups. (3) Tetrahedral, which has four electron groups. (4) Trigonal bipyramid, which has five electron groups. (5) Octahedral, which has six electron groups. An electron group is defined as a lone pair of electrons, a single bond, a multiple bond, or even a single electron.

224 10.5

10.7

Chapter 10 Chemical Bonding II (a)

Four electron groups give tetrahedral electron geometry, while three bonding groups and one lone pair give a trigonal pyramidal molecular geometry.

(b)

Four electron groups give a tetrahedral electron geometry, while two bonding groups and two lone pairs give a bent molecular geometry.

(c)

Five electron groups give a trigonal bipyramidal electron geometry, while four bonding groups and one lone pair give a seesaw molecular geometry.

(d)

Five electron groups give a trigonal bipyramidal electron geometry, while three bonding groups and two lone pairs give a T-shaped molecular geometry.

(e)

Five electron groups give a trigonal bipyramidal electron geometry, while two bonding groups and three lone pairs give a linear geometry.

(f)

Six electron groups give an octahedral electron geometry, while five bonding groups and one lone pair give a square pyramidal molecular geometry.

(g)

Six electron groups give an octahedral electron geometry, while four bonding groups and two lone pairs give a square planar molecular geometry.

To determine if a molecule is polar, do the following: 1.

Draw the Lewis structure for the molecule and determine the molecular geometry.

2.

Determine whether the molecule contains polar bonds.

3.

Determine whether the dipole moments for polar bonds add together to form a net molecular dipole moment.

Polarity is important because polar and nonpolar molecules have different properties. Polar molecules interact strongly with other polar molecules, but do not interact with nonpolar molecules, and vice versa. 10.9

According to valence bond theory, the shape of the molecule is determined by the geometry of the overlapping orbitals.

10.11

Hybridization is a mathematical procedure in which the standard atomic orbitals are combined to form new atomic orbitals called hybrid orbitals. Hybrid orbitals are still localized on individual atoms, but they have different shapes and energies from those of standard atomic orbitals. They are necessary in valence bond theory because they correspond more closely to the actual distribution of electrons in chemically bonded atoms.

10.13

The number of standard atomic orbitals added together always equals the number of hybrid orbitals formed. The total number of orbitals is conserved.

10.15

The double bond in Lewis theory is simply two pairs of electrons that are shared between the same two atoms. However, in valence bond theory we see that the double bond is made up of two different kinds of bonds. The double bond in valence bond theory consists of one σ bond and one π bond. Valence bond theory shows us that rotation about a double bond is severely restricted. Because of the side-byside overlap of the p orbitals, the π bond must essentially break for rotation to occur. The single bond consists of overlap that results in a σ bond. Since the overlap is linear, rotation is not restricted.

10.17

In molecular orbital theory, atoms will bond when the electrons in the atoms can lower their energy by occupying the molecular orbitals of the resultant molecule.

Chapter 10 Chemical Bonding II 10.21

The electrons in orbitals behave like waves. The bonding molecular orbital arises from the constr interference between the atomic orbitals and is lower in energy than the atomic orbitals antibonding molecular orbital arises from the destructive interference between the atomic orbitals higher in energy than the atomic orbitals.

10.23

Molecular orbitals can be approximated by a linear combination of atomic orbitals (AOs). The number of MOs formed from a particular set of AOs will always equal the number of AOs used.

10.25

10.27

A paramagnetic species has unpaired electrons in molecular orbitals of equal energy. A parama species is attracted to a magnetic field. The magnetic property is a direct result of the unp electrons. The spin and angular momentum of the electrons generate tiny magnetic A diamagnetic species has all of the electrons paired. The magnetic fields caused by the electron and orbital angular momentum tend to cancel each other. A diamagnetic species is not attracte magnetic field and is, in fact, slightly repelled.

10.29

Nonbonding orbitals are atomic orbitals not involved in a bond and will remain localized on the at

10.31

In band theory, bands are the combination of atomic orbitals of the atoms within a solid crysta form molecular orbitals that are not localized on individual atoms, but delocalized over the crystal. In band theory, electrons become mobile when they make a transition from the H Occupied Molecular Orbital into higher-energy empty molecular orbitals known as the L Unoccupied Molecular Orbital. For this reason, the occupied molecular orbitals are often calle valence band and the unoccupied molecular orbitals are called the conduction band.

10.33

Doped semiconductors contain minute amounts of impurities that result in additional electrons conduction band or electron “holes” in the valence band. For example, silicon is a grou semiconductor. Its valence electrons just fill its valence band. When silicon is doped with phospho group 15 element with five valence electrons, its conductivity increases. The phosphorus atom incorporated into the silicon crystal structure, but each phosphorus atom brings with it one addi electron. Since the valence band is completely full, the additional electrons must go into the condu band. These electrons are then mobile and can conduct electrical current. This type of semiconduc called an n-type semiconductor because the charge carriers are negatively charged electrons i conduction band. Silicon can also be doped with a group 13 element, such as gallium, which ha three valence electrons. When gallium is incorporated into the silicon crystal structure, it resu electron “holes,” empty molecular orbitals in the valence band. The presence of holes also allows f

226

Chapter 10 Chemical Bonding II

Problems by Topic VSEPR Theory and Molecular Geometry 10.35

Four electron groups: A trigonal pyramidal molecular geometry has three bonding groups and one lone pair of electrons, so there are four electron pairs on atom A.

10.37

(a)

4 total electron groups, 4 bonding groups, 0 lone pairs A tetrahedral molecular geometry has four bonding groups and no lone pairs. So, there are four total electron groups, four bonding groups, and no lone pairs.

(b)

5 total electron groups, 3 bonding groups, 2 lone pairs A T-shaped molecular geometry has three bonding groups and two lone pairs. So, there are five total electron groups, three bonding groups, and two lone pairs.

(c)

6 total electron groups, 5 bonding groups, 1 lone pair A square pyramidal molecular geometry has five bonding groups and one lone pair. So, there are six total electron groups, five bonding groups, and one lone pair.

(a)

PF3

10.39

Electron geometry–tetrahedral; molecular geometry–trigonal pyramidal; bond angle = 109.5° Because of the lone pair, the bond angle will be less than 109.5°. Draw a Lewis structure for the molecule: PF3 has 26 valence electrons.

Determine the total number of electron groups around the central atom: There are four electron groups on P. Determine the number of bonding groups and the number of lone pairs around the central atom: There are three bonding groups and one lone pair. Use Table 10.1 to determine the electron geometry, molecular geometry, and bond angles: Four electron groups give a tetrahedral electron geometry; three bonding groups and one lone pair give a trigonal pyramidal molecular geometry; the idealized bond angles for tetrahedral geometry are 109.5°. The lone pair will make the bond angle less than idealized. (b)

SBr2

Electron geometry–tetrahedral; molecular geometry–bent; bond angle = 109.5° Because of the lone pairs, the bond angle will be less than 109.5°. Draw a Lewis structure for the molecule: SBr2 has 20 valence electrons.

Determine the total number of electron groups around the central atom: There are four electron groups on S. Determine the number of bonding groups and the number of lone pairs around the central atom: There are two bonding groups and two lone pairs. Use Table 10.1 to determine the electron geometry, molecular geometry, and bond angles:

Chapter 10 Chemical Bonding II (c)

CHCl3 Electron geometry–tetrahedral; molecular geometry–tetrahedral; bond angle = 109.5° Because there are no lone pairs, the bond angle will be 109.5°. Draw a Lewis structure for the molecule: CHCl3 has 26 valence electrons.

Determine the total number of electron groups around the central atom: There are four electron groups on C. Determine the number of bonding groups and the number of lone pairs aroun central atom: There are four bonding groups and no lone pairs. Use Table 10.1 to determine the electron geometry, molecular geometry, and bond a Four electron groups give a tetrahedral electron geometry; four bonding groups and n pairs give a tetrahedral molecular geometry; the idealized bond angles for tetrah geometry are 109.5°; however, because the attached atoms have different electronegat the bond angles are less than idealized. (d)

CS2

Electron geometry–linear; molecular geometry–linear; bond angle = 180° Because there are no lone pairs, the bond angle will be 180°. Draw a Lewis structure for the molecule: CS2 has 16 valence electrons.

Determine the total number of electron groups around the central atom: There are two electron groups on C. Determine the number of bonding groups and the number of lone pairs aroun central atom: There are two bonding groups and no lone pairs. Use Table 10.1 to determine the electron geometry, molecular geometry, and bond a Two electron groups give a linear geometry; two bonding groups and no lone pairs linear molecular geometry; the idealized bond angle is 180°. The molecule will not d from this. 10.41

H2O will have the smaller bond angle because lone pair–lone pair repulsions are greater than lone bonding pair repulsions. Draw the Lewis structures for both structures: H3O+ has eight valence electrons. H2O has eight valence electrons.

There are three bonding groups and There are two bonding groups and one lone pair. two lone pairs. Both have four electron groups, but the two lone pairs in H2O will cause the bond angle to be sm

228 10.43

Chapter 10 Chemical Bonding II (a)

SF4

Draw a Lewis structure for the molecule: SF4 has 34 valence electrons.

Determine the total number of electron groups around the central atom: There are five electron groups on S. Determine the number of bonding groups and the number of lone pairs around the central atom: There are four bonding groups and one lone pair. Use Table 10.1 to determine the electron geometry and molecular geometry: The electron geometry is trigonal bipyramidal so the molecular geometry is seesaw. Sketch the molecule:

(b)

ClF3

Draw a Lewis structure for the molecule: ClF3 has 28 valence electrons.

Determine the total number of electron groups around the central atom: There are five electron groups on Cl. Determine the number of bonding groups and the number of lone pairs around the central atom: There are three bonding groups and two lone pairs. Use Table 10.1 to determine the electron geometry and molecular geometry: The electron geometry is trigonal bipyramidal so the molecular geometry is T-shaped. Sketch the molecule:

(c)

IF2 −

Draw a Lewis structure for the ion: IF2− has 22 valence electrons.

Chapter 10 Chemical Bonding II Determine the number of bonding groups and the number of lone pairs aroun central atom: There are two bonding groups and three lone pairs. Use Table 10.1 to determine the electron geometry and molecular geometry: The electron geometry is trigonal bipyramidal so the molecular geometry is linear. Sketch the ion:

(d)

IBr4 −

Draw a Lewis structure for the ion: IBr4 − has 36 valence electrons.

Determine the total number of electron groups around the central atom: There are six electron groups on I. Determine the number of bonding groups and the number of lone pairs aroun central atom: There are four bonding groups and two lone pairs. Use Table 10.1 to determine the electron geometry and molecular geometry: The electron geometry is octahedral so the molecular geometry is square planar. Sketch the ion:

10.45

(a)

C2H2

Draw the Lewis structure:

Atom

Number of Electron Groups

Number of Lone Pairs

Molecular Geometry

Left C

2

0

Linear

Right C

2

0

Linear

Sketch the molecule:

230

Chapter 10 Chemical Bonding II (b)

C2H4

Draw the Lewis structure:

Atom

Number of Electron Groups

Number of Lone Pairs

Molecular Geometry

Left C

3

0

Trigonal planar

Right C

3

0

Trigonal planar

Sketch the molecule:

(c)

C2H6

Draw the Lewis structure:

Atom

Number of Electron Groups

Number of Lone Pairs

Molecular Geometry

Left C

4

0

Tetrahedral

Right C

4

0

Tetrahedral

Sketch the molecule:

10.47

(a)

Four pairs of electrons give a tetrahedral electron geometry. The lone pair would cause lone pair– bonded pair repulsions and would have a trigonal pyramidal molecular geometry.

(b)

Five pairs of electrons give a trigonal bipyramidal electron geometry. The lone pair occupies an equatorial position in order to minimize lone pair–bonded pair repulsions and the molecule would have a seesaw molecular geometry.

(c)

Six pairs of electrons give an octahedral electron geometry. The two lone pairs would occupy opposite positions in order to minimize lone pair–lone pair repulsions. The molecular geometry would be square planar.

Chapter 10 Chemical Bonding II 10.49

(a)

CH3OH Draw the Lewis structure and determine the geometry about each interior atom:

Atom

Number of Electron Groups

Number of Lone Pairs

Molecular Geometry

C

4

0

Tetrahedral

O

4

2

Bent

Sketch the molecule:

(b)

CH3OCH3 Draw the Lewis structure and determine the geometry about each interior atom:

Atom

Number of Electron Groups

Number of Lone Pairs

Molecular Geometry

C

4

0

Tetrahedral

O

4

2

Bent

C

4

0

Tetrahedral

Sketch the molecule:

(c)

H2O2

Draw the Lewis structure and determine the geometry about each interior atom:

Atom

Number of Electron Groups

Number of Lone Pairs

Molecular Geometry

O

4

2

Bent

232

Chapter 10 Chemical Bonding II Sketch the molecule:

Molecular Shape and Polarity 10.51

Draw the Lewis structure for CO2 and CCl4 and determine the molecular geometry and then the polarity.

Number of electron groups on C 2 4 Number of lone pairs 0 0 Molecular geometry linear tetrahedral Even though each molecule contains polar bonds, the sum of the bond dipoles gives a net dipole of zero for each molecule. The linear molecular geometry of CO2 will have bond vectors that are equal and opposite. The tetrahedral molecular geometry of CCl4 will have bond vectors that are equal and have a net dipole of zero.

10.53

(a)

PF3 – polar Draw the Lewis structure and determine the molecular geometry: The molecular geometry from Exercise 35 is trigonal pyramidal. Determine if the molecule contains polar bonds: The electronegativities of P = 2.1 and F = 4. Therefore, the bonds are polar. Determine whether the polar bonds add together to form a net dipole: Because the molecule is trigonal pyramidal, the three dipole moments sum to a nonzero net dipole moment. The molecule is polar. See Table 10.2 on page 381 in the text to see how dipole moments add to determine polarity.

(b)

SBr2 – nonpolar Draw the Lewis structure and determine the molecular geometry: The molecular geometry from Exercise 35 is bent. Determine if the molecule contains polar bonds: The electronegativities of S = 2.5 and Br = 2.8. Therefore, the bonds are essentially nonpolar. Even though the molecule is bent, since the bonds are very weakly polar, the molecule is essentially nonpolar.

(c)

CHCl3 – polar Draw the Lewis structure and determine the molecular geometry: Th l l t f E i 35 i t t h d l

Chapter 10 Chemical Bonding II Determine whether the polar bonds add together to form a net dipole: Because the bonds have different dipole moments due to the different atoms involve four dipole moments sum to a nonzero net dipole moment. The molecule is pola Table 10.2 on page 381 in the text to see how dipole moments add to determine polari (d)

CS2 – nonpolar Draw the Lewis structure and determine the molecular geometry: The molecular geometry from Exercise 35 is linear. Determine if the molecule contains polar bonds: The electronegativities of C = 2.5 and S = 2.5. Therefore, the bonds are nonpolar. Als molecule is linear, which would result in a zero net dipole even if the bonds were pola The molecule is nonpolar. See Table 10.2 on page 381 in the text to see how d moments add to determine polarity.

10.55

(a)

ClO 3− – polar

Draw the Lewis structure and determine the molecular geometry:

Four electron pairs with one lone pair give a trigonal pyramidal molecular geometry. Determine if the molecule contains polar bonds: The electronegativities of Cl = 3.0 and O = 3.5. Therefore, the bonds are polar. Determine whether the polar bonds add toget...