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Homework Problems Chapter 9 Homework Problem: 4, 7, 10, 16, 20, 22, 28, 30, 36, 40, 52, 56, 66, 81, 84, 92, 104, 107

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Homework Problems Chapter 9 Homework Problem: 4, 7, 10, 16, 20, 22, 28, 30, 36, 40, 52, 56, 66, 81, 84, 92, 104, 107. CHAPTER 9 Chemical Bonding II: Molecular Geometry and Bonding Theories. Shapes of Molecules and Ions - PowerPoint PPT Presentation

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Page 1: Homework Problems

Homework Problems

Chapter 9 Homework Problem: 4, 7, 10, 16, 20, 22, 28, 30, 36, 40, 52, 56, 66, 81, 84, 92, 104, 107

Page 2: Homework Problems

CHAPTER 9

Chemical Bonding II: Molecular

Geometry and Bonding Theories

Page 3: Homework Problems

Shapes of Molecules and Ions

Molecules and ions have a three dimensional shape. This shape can be important in determining the chemical behavior of a substance.

We will discuss two types of geometries:

1) electron domain geometry - The arrangement of electron containing regions about a central atom. This is sometimes called electron cloud geometry.

2) molecular geometry - The arrangement of atoms about a central atom.

The number of electron containing regions about a central atom is equal to the number of lone pair regions + the number of covalent bonds. Note that in counting electron containing regions a single, double, or triple bond counts as one region.

Page 4: Homework Problems

Examples of Counting Electron Containing Regions

How many electron containing regions are there for central atoms in each of the mole-cules at left?

Page 5: Homework Problems

N 3 regions (two bonds, one lone pair)

C 3 regions (three bonds, no lone pairs)

C at left 4 regions (four covalent bonds)

C at right 3 regions (three covalent bonds)

Note the bond order does not matter in counting electron containing regions.

Page 6: Homework Problems

VSEPR Theory

We may predict both electron cloud geometry and molecular geometry using VSEPR (Valence Shell - Electron Pair Repulsion) theory. The idea behind this theory is that electron containing regions will arrange themselves about a central atom in such a way as to put those regions as far apart as possible. This is due to the repulsive force existing between particles of the same charge (in this case electrons).

The various common cases for electron and molecular geometry are given in Figure 9.2 (page 368) or Table 9.2 (page 369).

Page 7: Homework Problems

Two Regions

If we have two electron containing regions around a central atom, then placing them opposite one another keeps them as far apart as possible.

In this case, both the electron geometry and the molecular geometry are linear.

Page 8: Homework Problems

Three Regions

For three regions about a central atom the electron geometry is trigonal planar (120 angle between the regions). There are two possible molecular geometries:

If all three regions are bonds - trigonal planar

If two of the three regions are bonds, and one is a lone pair of electrons - nonlinear (bent)

Page 9: Homework Problems

Four Regions

For four regions about a central atom the electron geometry is tetrahedral (109 angle between the regions). There are three possible molecular geometries:

4 bonds - tetrahedral 2 bonds - nonlinear (bent)

3 bonds - trigonal pyramid

Page 10: Homework Problems

Five Regions

For five regions about a central atom the electron geometry is trigonal bipyramid. In this case not all of the angles between electron containing regions are the same. Instead, we may divide the regions into equitorial regions and axial regions.

Page 11: Homework Problems

Five Regions

There are four possible molecular geometries:

5 bonds - trigonal bipyramid 3 bonds - T-shape

4 bonds - “see-saw” 2 bonds - linear

Page 12: Homework Problems

Six Regions

For six regions about a central atom the electron geometry is octahedral. The bond angles between adjacent regions are all 90.

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Six Regions

There are three common molecular geometries:

6 bonds - octahedral 5 bonds - square pyramid

4 bonds - square planar

Page 14: Homework Problems

Example: What are the electron geometries and molecular geometries around the central atom in BrF3, POCl3, and H2SO3?

Page 15: Homework Problems

Example: What are the electron geometries and molecular geometries around the central atom in BrF3, POCl3, and H2SO3?

BrF3 5 regions, so electron cloud geometry is trigonal bipyramid.

3 bonds, so molecular geometry is T-shape.

POCl3 4 regions, so electron cloud geometry is tetrahedral.

4 bonds, so molecular geometry is also tetrahedral.

H2SO3 4 regions, so electron cloud geometry is tetrahedral.

3 bonds, so molecular geometry is trigonal pyramid.

Page 16: Homework Problems

Summary

electron cloud bonds electron molecular

regions geometry geometry

2 2 linear linear

3 3 trigonal planar trigonal planar

2 nonlinear (bent)

4 4 tetrahedral tetrahedral

3 trigonal pyramid

2 nonlinear (bent)

Page 17: Homework Problems

Summary

electron cloud bonds electron molecular

regions geometry geometry

5 5 trig. bipyramid trig. bipyramid

4 see-saw

3 T-shape

2 linear

6 6 octahedral octahedral

5 square pyramid

4 square planar

Page 18: Homework Problems

Geometry in Large Molecules

For large molecules we can discuss the electron cloud and molecular geometry around each one of the interior atoms.

Example: Give the electron cloud and molecular geometry for each interior atom in the molecule glycine (NH2CH2COOH).

Page 19: Homework Problems

Example: Give the electron and molecular geometry for each interior atom in the molecule glycine (NH2CH2COOH).

Atom electron geometry molecular geometry

N atom tetrahedral trigonal pyramid

Left C atom tetrahedral tetrahedral

Right C atom trigonal planar trigonal planar

O atom tetrahedral nonlinear (bent)

Page 20: Homework Problems

Deviations From “Pure” Geometries

We have assumed that the angles observed in molecular geometries do not depend on how many bonds and lone pair regions there are. In fact, the number of regions containing lone pair electrons has a small effect on the observed bond angles.

In general we may say the repulsive forces between electron containing regions in a molecule or ion are stronger for lone pair electrons than for bonding electrons. So

lone pair - lone pair >

lone pair - bonding >

bonding - bonding

This is due to the fact that bonding electrons are more localized than lone pair electrons, as they must appear between bonded atoms.

Page 21: Homework Problems

In the molecules at left (all of which have four electron containing regions around a central atom) the bond angle decreases from 109.5 in CH4 (pure tetrahedral geometry) to 107.0 in NH3 to 104.5 in H2O.

In CH2O (below) all of the bond angles would be 120. ° in pure trigonal planar geometry.

Page 22: Homework Problems

Representing Three Dimensional Structure

Molecules have a three dimensional structure. We need a systematic method for representing these structures in two dimensions. We do this as follows

For convenience, we will sometimes used a dashed line to replace the hatched wedge for a bond going into the page.

Example: Give the three dimensional structures for CH3Cl and PF5.

Page 23: Homework Problems

Example: Give the three dimensional structures for CH3Cl and PF5.

CH3Cl PF5

tetrahedral trigonal bipyramid

Page 24: Homework Problems

Polar Molecule

A polar molecule is a molecule where the center of positive and negative charge do not coincide. Two things are required for a molecule to be polar

1) The molecule must have at least one polar bond.

2) The contributions from the polar bonds cannot cancel.

Page 25: Homework Problems

For a polar covalent bond the difference in electronegativity between the bonded atoms should be about 0.5 or greater. This means that carbon-carbon bonds are completely nonpolar, and carbon-hydrogen bonds are only slightly polar.

EN(C) = 2.5 EN(H) = 2.1

The table of electronegativities and the molecular geometry for the molecule can be used to determine whether the molecule is polar or nonpolar.

The polarity of a molecule is expressed in terms of the dipole moment of the molecule. The symbol for dipole moment is . Dipole moments are measured in units of Debye (D).

Page 26: Homework Problems

Example

Consider the following molecules: H2O, CF4, NH3, and CH3COCH3. Which of these molecules is expected to have a permanent dipole moment?

Page 27: Homework Problems

H2O = 1.85 D

NH3 = 1.47 D

CF4 = 0. D

CH3COCH3 = 2.88 D

EN Values: H = 2.1; C = 2.5; N = 3.0; O = 3.5; F = 4.0

1 Debye = 3.336 x 10-30 C.m

Page 28: Homework Problems

Theories For Molecular Bonding

The Lewis dot structure method predates quantum mechanics. While it is a good qualitative description of covalent bonding it fails to work well in some cases (resonance structures are one example).

There are two general methods that use quantum mechanics to improve on the Lewis dot structure method:

Valence Bond theory - Discusses bond formation in terms of overlap of atomic orbitals (often hybrid orbitals).

Molecular Orbital theory - Uses quantum mechanics to solve the Schrodinger equation for a molecule or ion. This is a very mathematical approach.

Page 29: Homework Problems

Valence Bond Theory For H2

In valence bond theory bond formation is pictured as occurring due to the overlap of atomic orbitals of the atoms that are bonded together. Each of the orbitals contains one electron. When the orbitals overlap, the electron from each atomic orbital is shared by the two atoms. Notice the electrons need to be opposite spin.

The covalent bond forms from the overlap of the two 1s atomic orbitals of the hydrogen atoms that are bonded together. This places the electrons between the two hydrogen nuclei, which holds the molecule together.

Page 30: Homework Problems

Bond Formation and Energy

The formation of a covalent bond between two atoms occurs because it leads to a lower energy than exists for separate atoms. This can be seen in the bonding of two H atoms to form an H2 molecule.

Page 31: Homework Problems

Shortcoming of Simple Valence Bond Theory

The above picture breaks down when you need to form several different valence bonds for the same atom out of orbitals of different types.

Example: Be in BeH2

Be 1s2 2s2 H 1s1

We want the beryllium atom to contribute one electron to each bond between Be and an H atom, but the 2s orbital of beryllium is full, with no unpaired electron.

So what do we do?

Page 32: Homework Problems

Hybrid Orbitals

In this case, before we form the valence bonds we first construct a new set of equivalent orbitals, called hybrid orbitals (in this case sp hybrid orbitals).

We can envision the process taking place as indicated below.

While the process of forming hybrid orbitals raises the energy of the electrons in Be, this is more than made up for when the two Be - H bonds are formed.

Page 33: Homework Problems

Formation of sp Hybrid Orbitals

The sp hybrid orbitals are formed from combinations of the s and px atomic orbitals. Since we begin with two atomic orbitals, we end up with two hybrid orbitals.

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Use of sp Hybrid Orbitals For Bonding (BeH2)

The two sp hybrid orbitals formed in Be can now be used to form the two Be - H valence bonds.

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sp3 Hybrid Orbitals

In methane (CH4) we need the central carbon atom to have four equivalent hybrid orbitals to form the four C - H bonds. This is done using sp3 hybridization.

CH4. C: 1s2 2s2 2p2

As before, we start by “unpairing” the electrons in the carbon atom by promoting one electron from a 2s to a 2p orbital

__ __ __ __ promote __ __ __ __

2s 2p 2s 2p

Page 36: Homework Problems

sp3 Hybrid Orbitals (Continued)

Now that we have one electron in our 2s atomic orbital, and in each of the three 2p atomic orbitals, we combine the four orbitals to form our set of four sp3 hybrid orbitals.

Since we began with four atomic orbitals, we end up with four hybrid orbitals.

Page 37: Homework Problems

These sp3 hybrid orbitals can now be used to form valence bonds or to hold lone pairs of electrons.

Page 38: Homework Problems

Naming and Use of Hybrid Orbitals

We name hybrid orbitals by listing the type of atomic orbitals used to construct the hybrid orbitals (s, p, or d). If we use more than one of a particular type of orbital, we indicate the number of orbitals used by a superscript to the right of the symbol for the orbital.

There is a simple relationship between the number of electron containing regions around a central atom and the type of hybrid orbitals that are required.

Number of regions Hybrid orbitals Electron geometry2 sp linear

3 sp2 trigonal planar

4 sp3 tetrahedral

5* sp3d trigonal bipyramid

6* sp3d2 octahedral * Not possible for atoms from 2nd row of periodic table

Page 39: Homework Problems

Why PF5 Exists and NF5 Does Not

We can now return to an observation we made in the last chapter, that in looking at molecules made from a group 5 nonmetal and fluorine the molecule PF5 exists, but NF5 does not. We can see why this is the case by looking at the hybridization required for the central atom in these two molecules.

N [He] 2s2 2p3

P [Ne] 3s2 3p3 3d0 __ __ __ __ __ __ __ __ __

promote __ __ __ __ __ __ __ __ __

3s 3p 3d

Hybridize to get 5 sp3d hybrid orbitals. This is not possible for nitrogen because there are no 2d orbitals, and so hybrid orbitals requiring use of one or more d orbitals cannot be formed.

Page 40: Homework Problems

Appearance of Hybrid Orbitals

Each particular type of hybrid orbital has its own geometry, which in all cases corresponds to the geometries predicted using VSEPR theory.

Also notice that in all cases the number of atomic orbitals we begin with is equal to the number of hybrid orbitals we end up with.

Page 41: Homework Problems

Sigma () and Pi () Bonds

The above description works well for single bonds between atoms. However, when we have a multiple bond (double or triple bond), the bonds can be divided into two types:

sigma bond - Formed from the overlap of atomic or hybrid orbitals; places electrons directly between the bonded atoms.

pi bond - Formed from the overlap of p type atomic orbitals; places electrons above and below the region between the bonded atoms.

In general, a single bond will always be a sigma bond. For a multiple bond, one of the bonds will be a sigma bond and the other bonds will be pi bonds.

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Example: C2H4 (ethene) (C atoms are sp2 hybridization)

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SummaryTotal number Number of Number of

bonds sigma bonds pi bonds

1 1 0

2 1 1

3 1 2

So the first bond is always a sigma bond, while any additional bonds are pi bonds.

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Example

How many sigma bonds and how many pi bonds are present in the molecule shown below?

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How many sigma bonds and how many pi bonds are present in the molecule shown below?

There are seven sigma bonds.

There are two pi bonds.

Page 46: Homework Problems

Pi Bonds and Free Rotation

Because pi bonds are formed from the overlap of p orbitals, molecules that contain pi bonds (molecules with double or triple bonds) cannot rotate freely around those bonds, since to do so would mean breaking one or more pi bond.

In ethane (C2H6) rotation around the C - C bond does not affect the bond and so is allowed. In ethene (C2H4) rotation around the C = C bond breaks the pi bond, and so will not occur (except at high energy).

Page 47: Homework Problems

Molecular Orbital Theory

In molecular orbital theory we solve the Schrodinger equation to find information about molecules (energies, geometries, electron distribution, and so forth). Just as we find atomic orbitals for atoms, we find molecular orbitals for molecules or ions. This is a very mathematical theory, and so we will only discuss simple cases in a nonmathematical way.

The usual way people do MO theory is called the LCAO-MO method. In this method, linear combinations of atomic orbitals are used to construct molecular orbitals. We also focus on the valence electrons, and generally ignore core electrons.

Page 48: Homework Problems

Molecular Orbital Theory for H2

For H2, we begin with the two 1s atomic orbitals on the two H atoms. There are two ways in which these can be combined, cor-responding to two molecular orbitals. One molecular orbital lowers the energy and therefore corresponds to a bonding orbital, while the other molecular orbital raises the energy and therefore corresponds to a * antibonding orbital (note we use a * to indicate an antibonding orbital). We often also indicate the starting atomic orbitals.

1s atomic orbitals 1s MO *1s MO

Page 49: Homework Problems

MO Diagram

In a molecular orbital diagram we indicate the starting atomic orbitals and the molecular orbitals made from them with the correct relative energies.

The above diagram represents the MO picture for the H2 molecule. As is the case for atoms, we can give an electron configuration for molecules as well. For the above we have H2: (1s)2

Page 50: Homework Problems

Rules for Adding Electrons

The rules for filling molecular orbitals are the same as filling atomic orbitals. They are:

Pauli principle.

Aufbau principle

Hund’s rule

Just as for atoms, we can write electron configurations for molecules by listing the occupied orbitals in order of energy, and indicating the number of electrons in an orbital by a superscript to the right of the orbital name. We usually place each orbital in parentheses.

Page 51: Homework Problems

Example

H2 molecule H2- ion He2 “molecule”

So H2 (1s)2

H2- (1s)2 (1s*)1

He2 (1s)2 (1s*)2

Page 52: Homework Problems

Bond Order

We may define the bond order (BO) as follows

BO = (# bonding e-) - (# antibonding e-)

2

The higher the bond order the stronger the bond. Note that fractional bond orders are possible.

H2 (1s)2 BO = 1

H2- (1s)2 (1s*)1 BO = 1/2

He2 (1s)2 (1s*)2 BO = 0

If the bond order is zero, then we don’t expect the molecule or ion to be stable.

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Period 2 Homonuclear Diatomics

We may apply the above procedure to the period 2 homonuclear diatomic molecules and ions. Note that we only look at he molecular orbitals involving atomic orbitals in the valence shell.

The 2s atomic orbitals present in the valence shell of second row atoms can be used to form a bonding and an antibonding sigma orbital, as was the case with the 1s atomic orbitals.

*2s

2s 2s

2s

AO MO AO

Page 54: Homework Problems

Molecular Orbitals From 2pAtomic Orbitals

The 2p atomic orbitals present in second row atoms can be used to form two types of molecular orbitals - sigma orbitals (bonding and antibonding) and pi orbitals (bonding and antibonding).

*2p

*2p

2p 2p 2p

2p

AO MO AO

Page 55: Homework Problems

Formation of Molecular Orbitals From px Atomic Orbitals

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Formation of Molecular Orbitals From py and pz Atomic Orbitals

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Period 2 Homonuclear Diatomics

The molecular orbitals formed from the 2s and 2p atomic orbitals of second period atoms are shown below. The order of these orbitals in terms of energy depends on the particular type of homonuclear diatomic species being discussed.

Order for Li2 to N2 Order for O2, F2

Page 58: Homework Problems

Example: Give the electron configuration and bond order for C2, C2+,

and C2-.

C2 (8 e-) (2s)2 (2s*)2 (2p)4 BO = (6-2)/2 = 2

C2+ (7 e-) (2s)2 (2s*)2 (2p)3 BO = (5-2)/2 = 1 1/2

C2- (9 e-) (2s)2 (2s*)2 (2p)4 (2p)1 BO = (7-2)/2 = 2 1/2

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MOs For Period 2 Homonuclear Diatomics

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Summary of Bonding TheoriesLewis theory - Simple to use, useful in making qualitative

predictions about bond lengths and bond strengths. Not directly useful for three dimensional structure of molecules, sometimes makes incorrect predictions concerning bond order, unpaired electron spins. Difficulties in use for some molecules (requiring resonance structures).

VSEPR- Based on Lewis theory, and allows predictions of molecular geometries. However, some of the same weaknesses of Lewis theory.

Valence Bond theory (with hybridization) - Explains the formation of covalent bonds in more detail than Lewis theory, and also why atoms in the third row and below can violate the octet rule. Fails in the prediction of some properties of molecules, such as paramagnetism in O2.

Molecular Orbital theory - Most rigorous theory for molecules, but also the most complicated and mathematical theory. Difficult to apply in a simple way for large molecules.

Page 61: Homework Problems

Multicenter Pi Bonding

We may combine valence bond theory and MO theory to account for resonance structures in terms of multicenter pi bonds, that is, pi bonds among more than two atoms.

NO2- 24 valence electrons (and so 4 covalent bonds)

When there are resonance structures that is usually evidence that the bonding needs to be described in terms of multicentered pi bonds. A correct description of bonding in this case requires use of both valence bond theory and molecular orbital theory.

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The N atom and three O atoms can each form sp2 hybrid orbitals. These form the N - O sigma bonds, and also contain two sets of lone pair electrons for each oxygen atom.

Formation of Sigma Bonds

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The remaining p orbitals on each atom that are perpendicular to the plane of the molecule can be used to form a final, multicentered pi bond, along with two other multicentered molecular orbitals (not shown) for the additional two sets of lone pair electrons.

Formation of Multicenter Pi Bonds

4 pz

antibonding MO

lone pair MOs

bonding MOs (multicentered pi bond)

Page 64: Homework Problems

Benzene

Benzene (C6H6) is an important example of a molecule that gains stability due to the presence of multicentered pi bonds.

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End of Chapter 9“Physicists are notoriously scornful of scientists from other

fields. When the wife of the great Austrian physicist Wolfgang Pauli left him for a chemist, he was staggered with disbelief. ‘Had she taken a bullfighter I would have understood,’ he remarked in wonder to a friend. ‘But a chemist…’” Bill Bryson, A Short History of Nearly Everything

“The physicist's greatest tool is his wastebasket.” - Albert Einstein

“Hofstadter's law: It always takes longer than you expect, even when you take into account Hofstadter's law.” - Douglas Hofstadter

“Love hides in molecular structures.” - Jim Morrison