C 9 10 · polar molecular compound with hydrogen bonds e. polar molecular substance without...

CHAPTERS 9 AND 10Covalent Bonding and Molecular Structures

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Objectives

You will be able to:

1. Write a description of the formation of the covalent bond between two hydrogen atoms to form a hydrogen molecule. Your description should include mention of overlapping atomic orbitals to form molecular orbitals. It should also include rough pictures of the atomic orbitals and the molecular orbitals.

2. Write a description of the assumptions made in the Linear Combination of Atomic Orbitals (LCAO) approximation for the molecular orbital approach to covalent bonding.

3. Write an explanation for why bonding molecular orbitals are more stable and why antibonding molecular orbitals are less stable than the separate atomic orbitals that form them.

4. Write a description of how the σ1s, σ*1s, σ2s, σ*2s, σ2p, σ*2p, π*2p, and π*2p are formed from the 1s, 2s and 2p atomic orbitals.

5. Draw sketches of the σ1s, σ*1s, σ2p, σ*2p, π2p, and π*2p molecular orbitals. 6. Write an explanation for why the σ2p molecular orbital of O2 is lower energy

than the π2p molecular orbitals and why the σ*2p is higher energy than the π*2p. 7. Draw molecular orbital diagrams for H2, O2, F2, and Ne2. 8. From a molecular orbital diagram of a molecule, predict its bond order. 9. From a molecular orbital diagram for a molecule, predict whether it is

paramagnetic or diamagnetic. 10. Draw the molecular orbital diagrams for CO and NO and use them to predict

the bond order of each and to explain why each is relatively stable despite their strange Lewis Structures.

11. Describe the advantages and disadvantages of scientific models. 12. Write a list of the basic assumptions of the valence-bond model of covalent

bonding. 13. Write or identify the number of valence electrons for each of the Representative

elements. 14. With reference to the valence bond model, describe how the covalent bond forms

between fluorine atoms in F2. 15. Write or identify a description of the information given in a Lewis structure. 16. With reference to the valence bond model, describe how the covalent bond forms

between hydrogen atoms in H2. 17. Explain how the carbon atom in CH4 is able to form four equivalent covalent

bonds to hydrogen atoms with bond angles of 109.5°.18. Describe (1) how sp3 hybrid orbitals form, (2) their shapes, and (3) how they are

arranged in space. 19. Write or identify a description of the tetrahedral electron group geometry.

174 Chapter 9 & 10 Covalent Bonding and Molecular Structures

20. Explain the following for ethene (ethylene) molecules, C2H4. a. Explain how the carbon atoms are able to form four equivalent covalent

bonds to hydrogen atoms.b. Explain why all the bond angles are about 120°.c. With reference to the valence bond model, explain how the double bond

between carbon atoms forms and explain why one bond in the double bond is weaker than the other.

21. Describe (1) how sp2 hybrid orbitals form, (2) their shapes, and (3) how they are arranged in space.

22. Write or identify a description of the trigonal planar electron group geometry. 23. Explain the following for ethyne (acetylene) molecules, C2H2.

a. Explain how the carbon atoms are able to form two equivalent covalent bonds to hydrogen atoms.

b. Explain why the bond angles are 180°.c. With reference to the valence bond model, explain how the triple bond

between carbon atoms forms and explain why two bonds in the triple bond are weaker than the other one.

24. Describe (1) how sp hybrid orbitals form, (2) their shapes, and (3) how they are arranged in space.

25. Write or identify a description of the linear electron group geometry. 26. Explain how the nitrogen atom in NH3 is able to form three equivalent covalent

bonds to hydrogen atoms with bond angles of 107°.27. Write or identify a description of the trigonal pyramid molecular geometry. 28. Explain how the nitrogen atom in NH4

+ is able to form four equivalent covalent bonds to hydrogen atoms with bond angles of 109.5°.

29. Explain how the oxygen atom in H2O is able to form two equivalent covalent bonds to hydrogen atoms with bond angles of 105°.

30. Write or identify a description of the bent molecular geometry. 31. With reference to the valence bond model, explain how the triple bond forms in

the carbon monoxide, CO, molecule. 32. Explain how the boron atom in BF3 is able to form three equivalent covalent

bonds to fluorine atoms with bond angles of 120°.33. Explain how the phosphorus atom in PF5 is able to form five equivalent covalent

bonds to fluorine atoms with bond angles of 90°, 120°, and 180°.34. Write or identify a description of the trigonal bipyramid electron group geometry. 35. Explain how the sulfur atom in SF4 is able to form four covalent bonds to

fluorine atoms with bond angles of 90°, 120°, and 180°.36. Write or identify a description of the see-saw molecular geometry. 37. Explain how the iodine atom in IF3 is able to form three covalent bonds to

fluorine atoms with bond angles of 90° and 180°.38. Write or identify a description of the T-shaped molecular geometry. 39. Explain how the xenon atom in XeF2 is able to form two covalent bonds to

fluorine atoms with a bond angle of 180°.40. Write or identify a description of the linear molecular geometry. 41. Explain how the sulfur atom in SF6 is able to form six equivalent covalent bonds

to fluorine atoms with bond angles of 90° and 180°.

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42. Write or identify a description of the octahedral electron geometry. 43. Explain how the iodine atom in IF5 is able to form five equivalent covalent bonds

to fluorine atoms with bond angles of 90° and 180°.44. Write or identify a description of the square pyramid molecular group geometry. 45. Explain how the xenon atom in XeF4 is able to form four covalent bonds to

fluorine atoms with bond angles of 90° and 180°.46. Write or identify a description of the square planar molecular geometry. 47. Write an explanation for why the elements listed on the Table 9,10.3 can have

each of the bonding patterns listed. 48. Given a Lewis structure or enough information to get one, do the following

• List the hybridization for each atom in the structure. • Write a description of the formation of each bond in terms of the overlap of

atomic orbitals. • Describe the electron group geometry around each atom that has two or

more atoms attached to it. • Draw the geometric sketch of the molecule, including bond angles. • Describe the molecular geometry around each atom that has two or more

atoms attached to it. 49. Given a Lewis structure or enough information to get one, determine the formal

charge for each atom in the structure. 50. Given a formula for a molecule or polyatomic ion, draw a reasonable Lewis

structure that corresponds to the formula. 51. Given a Lewis structure or enough information to get one, do the following.

• Identify whether it is best described as having resonance or not.• If can be seen as having resonance, draw the Lewis Structures for all of the

reasonable resonance forms.• Draw the resonance hybrid for the formula.

52. Given a formula for a molecular compound, predict whether or not it represents a polar molecule.

53. Write a description of how London forces come to be. 54. Write an explanation for why larger molecules have stronger London forces than

smaller molecules. 55. Write a description of how London forces play a part in the attractions between

polar molecules. 56. Given a formula for a pure substance, identify which of the following types of

substances it represents. a. ionic compoundb. metallic substanced. polar molecular compound with hydrogen bondse. polar molecular substance without hydrogen bondsf. nonpolar molecular substance

57. Identify carbon in the diamond form, C(dia), and silicon dioxide, SiO2 as examples of network crystals.


58. Given a formula for a pure substance, identify which of the following attractions is the predominant force holding its particles in the liquid and solid state.

a. ionic bonds b. metallic bondsc. covalent bondsd. dipole-dipole attractionse. hydrogen bondsf. London forces.

59. Given chemical formulas for two substances that are about the same size but have different types of attractions between the particles, predict which substance would have the strongest attractions between the particles in the liquid and solid state.

60. Given chemical formulas for two substances with the same type of attraction between the particles but different sizes, predict which substance would have the strongest attractions between the particles in the liquid and solid state.

61. Convert between the definition and the term for the following words or phrases.

Chapters 9 and 10 Glossary

Molecular orbital A volume that contains a high percentage of the electron charge for an electron in a molecule or a volume within which an electron in a molecule has a high probability of being found.

Bonding molecular orbital Formed from in-phase interaction of two atomic orbitals. is leads to an increase in negative charge between two nuclei where the atomic orbitals overlap and leads to more +/- attraction between the negative charge generated by the electrons and the nuclei.

Antibonding molecular orbital Formed from out-of-phase interaction of two atomic orbitals. is leads to a decrease in negative charge between two nuclei where the atomic orbitals overlap and leads to less +/- attraction between the negative charge generated by the electrons and the nuclei.

Valence bond model e model for covalent bonding that includes the assumptions that (1) only the highest-energy (valence) electrons participate in bonding, (2) covalent bonds arise due to the overlap of atomic orbitals on adjacent atoms, forming molecular orbitals, and (3) covalent bonds often form to pair unpaired electrons.

Valence electrons e highest-energy s and p electrons for an atom. e electrons that are most important in the formation of chemical bonds. e number of valence electrons for the atoms of an element is equal to the element’s A-group number on the periodic table.

Electron-dot symbol A representation of an atom that consists of its elemental symbol surrounded by dots representing its valence electrons.

Lone pair Two electrons that are not involved in the covalent bonds between atoms but are important for explaining the arrangement of atoms in molecules. ey are represented by pairs of dots in Lewis structures.

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Lewis structure A representation of a molecule that consists of the elemental symbol for each atom in the molecule, lines to show covalent bonds, and pairs of dots to indicate lone pairs.

Hybrid orbital An atomic orbital formed from a blend of atomic orbitals, e.g. four sp3 hybrid orbitals are formed from the blend of one s orbital and three p orbitals.

Bond angle e angle formed by straight lines (representing bonds) connecting the nuclei of three adjacent atoms.

Electron group geometry A description of the arrangement of all the electron groups around a central atom in a molecule or polyatomic ion, including the lone pairs.

Molecular geometry e description of the arrangement of all the atoms around a central atom in a molecule or polyatomic ion. is description does not consider lone pairs.

Isomers Compounds that have the same molecular formula but different molecular structures.

Formal charge A tool for evaluating Lewis structures, calculated from the formula:A-group number – number of bonds – number of electrons in lone pairs.

Resonance structures Two or more Lewis structures for a single molecule or polyatomic ion that differ in the positions of lone pairs and multiple bonds but not in the positions of the atoms in the structure. It is as if the molecule or ion were able to shift from one of these structures to another by shifting pairs of electrons from one position to another.

Resonance e hypothetical switching from one resonance structure to another. Resonance hybrid A structure that represents the average of the resonance structures

for a molecule or polyatomic ion. Delocalized pi system A system of overlapping p orbitals used to describe the

bonding in resonance hybrids.Dipole-dipole attraction e intermolecular attraction between the partial negative

end of one polar molecule and the partial positive end of another polar molecule. Hydrogen bond e intermolecular attraction between a nitrogen, oxygen, or

fluorine atom of one molecule and a hydrogen atom bonded to a nitrogen, oxygen, or fluorine atom in another molecule.

London dispersion forces, London forces, or dispersion forces e attractions produced between molecules by instantaneous and induced dipoles.


Molecular Orbital Theory

e goal of molecular orbital theory is to describe molecules in a similar way to

how we describe atoms, that is, in terms of orbitals, orbital diagrams, and electron

configurations. For example, to give you a glimpse at where we are headed, the

following are orbital diagrams for O2 and O.

s1s*

s1s

s2s*

s2p*

s2p

s2s

p2p p2p

p2p p2p* *

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O2 O

Each line in the molecular orbital diagram represents a molecular orbital, which is

the volume within which a high percentage of the negative charge generated by the

electron is found. e molecular orbital volume encompasses the whole molecule. We

assume that the electrons would fill the molecular orbitals of molecules like electrons

fill atomic orbitals in atoms.

• e molecular orbitals are filled in a way that yields the lowest potential

energy for the molecule.

• e maximum number of electrons in each molecular orbital is two. (We

follow the Pauli exclusion principle.)

• Orbitals of equal energy are half filled with parallel spin before they begin to

pair up. (We follow Hund’s Rule.)

Before we continue with a description of a model used to generate molecular orbital

diagrams, let’s get a review of light and electron waves and how two waves can interact.

e wave description of light describes the effect that the light has on the space around

it. is effect is to generate an oscillating electric and magnetic fields. ese fields can

vary in intensity, which is reflected in varying brightness of light.

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e wave description of the electron describes the variation in the intensity of negative charge generated by the electron.

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Light waves can interact in-phase, which leads to an increase in the intensity of the light (brighter) and out-of-phase, which leads to a decrease in the intensity of the light (less bright).

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Electron waves can also interact in-phase and out-of-phase. In-phase interaction leads to an increase in the intensity of the negative charge. Out-of-phase interaction leads to a decrease in the intensity of the negative charge.

One common approximation that allows us to generate molecular orbital diagrams for some small diatomic molecules is called the Linear Combination of Atomic Orbitals (LCAO) approach. e following assumptions lie at the core of this model.

• Molecular orbitals are formed from the overlap of atomic orbitals. • Only atomic orbitals of about the same energy interact to a significant

degree. • When two atomic orbitals overlap, they interact in two extreme ways

to form two molecular orbitals, a bonding molecular orbital and an antibonding molecular orbital.

Figure 9,10.1Electromagnetic Waves

Figure 9,10.21s Orbital

Figure 9,10.3Wave Interference


For example, our model assumes that two 1s atomic orbitals can overlap in two

extreme ways to form two molecular orbitals. One of the ways the atomic orbitals

interact is in-phase, which leads to wave enhancement similar to the enhancement of

two in-phase light waves. Where the atomic orbitals overlap, the in-phase interaction

leads to an increase in the intensity of the negative charge in the region where they

overlap. is creates an increase in negative charge between the nuclei and an increase

in the plus-minus attraction between the electron charge and the nuclei for the atoms

in the bond. e greater attraction leads to greater stability and lower potential energy.

Because electrons in the molecular orbital are lower potential energy than in separate

atomic orbitals, energy would be required to shift the electrons back into the 1s orbitals

of separate atoms. is keeps the atoms together in the molecule, so we call this orbital

a bonding molecular orbital. e molecular orbital formed is symmetrical about

the axis of the bond. Symmetrical molecular orbitals are called sigma, σ, molecular

orbitals. e symbol σ1s is used to describe the bonding molecular orbital formed from

two 1s atomic orbitals.

e second way that two atomic orbitals interact is out-of-phase. Where the atomic

orbitals overlap, the out-of-phase interaction leads to a decrease in the intensity of the

negative charge. is creates a decrease in negative charge between the nuclei and a

decrease in the plus-minus attraction between the electron charge and the nuclei for

the atoms in the bond. e lesser attraction leads to lower stability and higher potential

energy. e electrons are more stable in the 1s atomic orbitals of separate atoms, so

electrons in this type of molecular orbital destabilize the bond between atoms. We call

this type of molecular orbital an antibonding molecular orbital. e molecular orbital

formed is symmetrical about the axis of the bond, so it is a sigma molecular orbital

with a symbol of σ*1s. e asterisk indicates an antibonding molecular orbital.

e following diagram shows the bonding and antibonding molecular orbitals

formed from the interaction of two 1s atomic orbitals.

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When two larger atoms combine to form a diatomic molecules (like O2, F2, or Ne2), more atomic orbitals interact. e LCAO approximation assumes that only the atomic orbitals of about the same energy interact. For O2, F2, and Ne2, the orbital energies are different enough in energy so only orbitals of the same energy interact to a significant degree.

Like for hydrogen, the 1s from one atom overlaps the 1s from the other atom to form a σ1s bonding molecular orbital and a σ*1s antibonding molecular orbital. e shapes would be similar to those formed from the 1s orbitals for hydrogen. e 2s atomic orbital from one atom overlaps the 2s atomic orbital from the other atom to form a σ2s bonding molecular orbital and a σ*2s antibonding molecular orbital. e shapes of these molecular orbitals would be similar to those for the σ1s and σ*1s

molecular orbitals. Both σ2s and σ*2s molecular orbitals are higher energy and larger than the σ1s and σ*1s molecular orbitals.

e p atomic orbitals of the two atoms can interact in two different ways, parallel or end-on. e molecular orbitals are different for each type of interaction. e end-on interaction between two 2px atomic orbitals yields sigma molecular orbitals, which are symmetrical about the axis of the bond.

Figure 9,10.4Formation of Sigma 1s Molecular Orbitals


2p

nuclei

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s2p

in-phase

out-of-phase

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e two 2py atomic orbitals overlap in parallel and form two pi molecular orbitals. Pi molecular orbitals are asymmetrical about the axis of the bond.

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e 2pz-2pz overlap generates another pair of π2p and π*2p molecular orbitals. e 2pz-2pz overlap is similar to the 2py-2py overlap. To visualize this overlap, picture all of the orbitals in the image above rotated 90°, so that the axes that run through each atomic and molecular orbitals are perpendicular to the paper. e molecular orbitals formed have the same potential energy as the molecular orbitals formed from the 2py-2py overlap.

ere is less overlap for the parallel atomic orbitals. When the interaction is in-phase, less overlap leads to less electron charge enhancement between the nuclei. is leads to less electron charge between the nuclei for the pi bonding molecular orbital than for the sigma bonding molecular orbital. Less electron character between the nuclei means less plus-minus attraction, less stabilization, and higher potential energy for the pi bonding molecular orbital compared to the sigma bonding molecular orbital.

When the interaction is out-of-phase, less overlap leads to less shift of electron charge from between the nuclei. is leads to more electron charge between the nuclei for the pi antibonding molecular orbital than for the sigma antibonding molecular orbital. More electron charge between the nuclei means more plus-minus attraction and lower potential energy for the pi antibonding molecular orbital compared to the sigma antibonding molecular orbital.

Figure 9,10.5Formation of Sigma 2p Molecular Orbitals

Figure 9,10.6Formation of Pi 2p Molecular Orbitals

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e expected molecular orbital diagram from the overlap of 1s, 2s and 2p atomic orbitals is as follows. We will use this diagram to describe O2, F2, Ne2, CO, and NO.

s1s*

s1s

s2s*

s2p*

s2p

s2s

p2p p2p

p2p p2p* *

We assume that the electrons would fill the molecular orbitals of molecules like electrons fill atomic orbitals in atoms. We use the following procedure when drawing molecular orbital diagrams.

• Determine the number of electrons in the molecule. We get the number of electrons per atom from their atomic number on the periodic table. (Remember to determine the total number of electrons, not just the valence electrons.)

• Fill the molecular orbitals from bottom to top until all the electrons are added. Describe the electrons with arrows. Put two arrows in each molecular orbital, with the first arrow pointing up and the second pointing down.

• Orbitals of equal energy are half filled with parallel spin before they begin

to pair up.

We describe the stability of the molecule with bond order.

bond order = 1/2 (#e− in bonding MO’s − #e− in antibonding MO’s)

We use bond orders to predict the stability of molecules. • If the bond order for a molecule is equal to zero, the molecule is unstable. • A bond order of greater than zero suggests a stable molecule. • e higher the bond order is, the more stable the bond.

We can use the molecular orbital diagram to predict whether the molecule is paramagnetic or diamagnetic. If all the electrons are paired, the molecule is diamagnetic.

If one or more electrons are unpaired, the molecule is paramagnetic.


EXAMPLES:

1. e molecular orbital diagram for a diatomic hydrogen molecule, H2, is

s1s*

s1s

• e bond order is 1. Bond Order = 1/2(2 - 0) = 1

• e bond order above zero suggests that H2 is stable.

• Because there are no unpaired electrons, H2 is diamagnetic.

2. e molecular orbital diagram for a diatomic helium molecule, He2, shows the

following.

s1s*

s1s

• e bond order is 0 for He2. Bond Order = 1/2(2 - 2) = 0

• e zero bond order for He2 suggests that He2 is unstable.

• If He2 did form, it would be diamagnetic.

3. e molecular orbital diagram for a diatomic oxygen molecule, O2, is

s1s*

s1s

s2s*

s2p*

s2p

s2s

p2p p2p

p2p p2p* *

• O2 has a bond order of 2. Bond Order = 1/2(10 - 6) = 2

• e bond order of two suggests that the oxygen molecule is stable.

• e two unpaired electrons show that O2 is paramagnetic.

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4. e molecular orbital diagram for a diatomic fluorine molecule, F2, is

s1s*

s1s

s2s*

s2p*

s2p

s2s

p2p p2p

p2p p2p* *

• F2 has a bond order of 1. Bond Order = 1/2(10 - 8) = 1

• e bond order of one suggests that the fluorine molecule is stable.

• Because all of the electrons are paired F2 is diamagnetic.

5. e molecular orbital diagram for a diatomic neon molecule, Ne2, is

s1s*

s1s

s2s*

s2p*

s2p

s2s

p2p p2p

p2p p2p* *

• Ne2 has a bond order of 0. Bond Order = 1/2(10 - 10) = 0

• e bond order zero for Ne2 suggests that Ne2 is unstable.

• If Ne2 did form, it would be diamagnetic.


We can describe diatomic molecules composed of atoms of different elements in a similar way. e bond between the carbon and oxygen in carbon monoxide is very strong despite what looks like a strange and perhaps unstable Lewis Structure.

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e plus formal charge on the more electronegative oxygen and the minus formal charge on the less electronegative carbon would suggest instability. e molecular orbital diagram predicts CO to be very stable with a bond order of three.

s1s*

s1s

s2s*

s2p*

s2p

s2s

p2p p2p

p2p p2p* *

We predict the nitrogen monoxide molecule to be unstable according to the Lewis approach to bonding.

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e unpaired electron and the lack of an octet of electrons around nitrogen would suggest an unstable molecule. NO is actually quite stable. e molecular orbital diagram predicts this by showing the molecule to have a bond order of 2.5.

s1s*

s1s

s2s*

s2p*

s2p

s2s

p2p p2p

p2p p2p* *

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109.5°

109.5°109.5° 109.5° 109.5° sp3

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sp3

sp3

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Figure 9,10.7Bonding in F2

Figure 9,10.8Bonding in H2

Figure 9,10.9Formation of sp3 Hybrid Atomic Orbitals


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Figure 9,10.11Formation of sp2 Hybrid Atomic Orbitals

Figure 9,10.10Bonding in CH4

Overlap of atomic orbitals in methane, CH4

Four C-H sigma bonds due to sp3-1s overlap

109.5°

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1sH C

109.5°

1sH

C

109.5°

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Figure 9,10.13Bonding in NH3

Figure 9,10.12Bonding in C2H4


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Figure 9,10.14Bonding in C2H2


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Figure 9,10.17Bonding in H2O


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Figure 9,10.19Formation of sp3dHybrid Atomic Orbitals

Figure 9,10.18Bonding in BF3

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Figure 9,10.20Bonding in PF5

Figure 9,10.21Bonding in SF4


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Figure 9,10.22Bonding in IF3

Figure 9,10.23Bonding in XeF2

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Figure 9,10.24Formation of sp3d2

Hybrid Atomic Orbitals

Figure 9,10.25Bonding in SF6


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Figure 9,10.26Bonding in IF5

Figure 9,10.27Bonding in XeF4

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e following assumptions lie at the core of the valence-bond model. • Only the highest energy electrons participate in bonding. • Covalent bonds usually form to pair unpaired electrons.• Covalent bonds arise due to the overlap of atomic orbitals on adjacent

atoms. When the columns on the periodic table are numbered by the A group convention,

the number of valence electrons in each atom of a representative element is equal to the element’s group number on the periodic table.

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Figure 9,10.28Valence Electrons

Table 9,10.1 Most Common Covalent Bonding Patterns

Element Frequency of pattern

Number of bonds

Number of lone pairs

Example

H always 1 0 �

B usually 3 0 �

C usually 4 0� � ��

rarely 3 1 �

N, P, & As usually 3 1 �

commonly 4 0�

O, S, & Se usually 2 2 ��

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F, Cl, Br, & I usually 1 3 �

Sample Study Sheet 9,10.1 Explanation of bonding Patterns - Valence Bond Model


Formal Charge = Group Number − Number of lines − Number of dots

Formal Charge = (e number of valence electrons necessary to be uncharged)

− (the number of electrons from bonds assuming equal sharing)

− (the number of electrons from lone pairs)

T-: You will be given a bonding pattern for a nonmetal atoms and asked to explain how it is possible in terms of the valence bond model.

G PS Write, “Only the highest energy electrons participate in bonding.”

S Draw the orbital diagram for the valence electrons. Include the empty “d” orbitals for the third period nonmetals and below.

S If there is a formal charge, add or subtract electrons.

a. Add one electron for -1.

b. Subtract one electron for +1.

S If necessary, promote one or more electrons from a pair to an empty orbital to get the number of unpaired electrons equal to the number of bonds to be explained.

S Rewrite the orbital diagram, showing the predicted hybrid orbitals.

S Write, “Covalent bonds form in order to pair unpaired electrons.”

S Indicate that the unpaired electrons form the bonds.

S Indicate that the paired electrons are lone pairs.

EXERCISE 9,10.1 - Explanation of Bonding Patterns

Explain the following bonding patterns

a. C – 4 bonds, no lone pairs, and no formal charge

b. N – 4 bonds, no lone pairs, and a +1 formal charge

c. O – 1 bond, 3 lone pairs, and a −1 formal charge

d. S – 4 bonds, 1 lone pair, and no formal charge

e. I – 5 bonds, 1 lone pair, and no formal charge

199

Sample Study Sheet 9,10.2: From Lewis Structures

T-: You will be given a Lewis structure, and you will be asked the following. • What is the hybridization for each atom in the structure? • Write a description of the formation of each bond in terms of the overlap of

atomic orbitals. • Describe the electron group geometry around each atom that has two or

more atoms attached to it. • Draw the geometric sketch of the molecule, including bond angles. • Describe the molecular geometry around each atom that has two or more

atoms attached to it. G S for predicting hybridization for each atom in a structure.

S Count the number of electron groups around each atom.An electron group is any one of the following.

• single bond• double or triple bond (counts as one group)• lone pair

S Apply the following guidelines. See Table 9,10.1.

1 group - No hybridization

(Hydrogen atoms are the only atoms that have only one group around them in a Lewis structure. e 1s orbital is the only occupied atomic orbital for hydrogen atoms. Because there is no other orbital with which to blend, there is no hybridization. Hydrogen atoms use the 1s atomic orbital to form bonds.)

2 groups - sp hybridization3 groups - sp2 hybridization4 groups - sp3 hybridization5 groups - sp3d hybridization6 groups - sp3d2 hybridization

G S for describing the bonds in terms of the overlap of atomic orbitals. S Describe single bonds that do not involve hydrogen as due to the overlapping of two hybrid orbitals, one from each atom.

S Describe single bonds that involve hydrogen as due to the overlapping a hybrid orbital with a 1s orbital for the hydrogen atom.

S Describe double bonds as one bond forming from the overlap of two hybrid orbitals, one from each atom. Describe the second bond as due to the overlap of two p orbitals, one from each atom.

S Describe triple bonds as one bond forming from the overlap of two hybrid orbitals, one from each atom. Describe each of the other two bonds as due to the overlap of two p orbitals, one from each atom.


To determine the electron group geometry around each atom that is attached to two or more atoms, use the number of electron groups around each atom and the guidelines found on Table 9,10.2.

Use one or more of the following geometric sketches shown on Table 9,10.2 for the geometric sketch of you molecule.

To describe the molecular geometry around each atom that has two or more atoms attached to it, count the number of bond groups and apply the guidelines found on Table 9,10.2. (Single, double, and triple bonds all count as one bond group.)

Note that if all of the electron groups attached to the atom are bond groups (no lone pairs), the molecular geometry is the same as the electron group geometry.

201

Table 9,10.2 Hybridization and Geometry

e− groups

Hybrid Orbitals

General Geometric Sketch e− group geometry

bond angles

bond groups

lone pairs

molecular geometry

2 sp 180° linear 180° 2 0 linear

3 sp2

120°triangular

planar120° 3 0 triangular

planar

2 1 bent4 sp3

109.5°tetrahedral 109.5°

4 0 tetrahedral 3 1 trigonal

pyramid 2 2 bent5 sp3d

120°

90°

120°

180°

trigonal bipyramid

90°,120°,180°

5 0 trigonal bipyramid

4 1 see-saw 3 2 T-shaped 2 3 linear

6 sp3d2

90°180°

180°

90°

octahedral 90°, 180°

6 0 octahedral 5 1 square

pyramid 4 2 square

planar


Table 9,10.3 Common Covalent Bonding Patterns for the Nonmetal Elements

Element # of bonds # of lone pairs Formal charge

H always 1 0 0C usually 4 0 0 rarely 3 1 -1N usually 3 1 0 commonly 4 0 +1O usually 2 2 0 commonly 1 3 -1 rarely 3 1 +1F always 1 3 0

P and As usually 3 1 0 commonly 5 0 0 commonly 4 0 +1

S and Se usually 2 2 0 commonly 4 1 0 commonly 6 0 0 commonly 1 3 -1 possibly 3 1 +1

halogens except F usually 1 3 0 possibly 3 2 0 possibly 5 1 0 possibly 7 0 0

Xe usually nonerarely 2 3 0rarely 4 2 0

Table 9,10.4 The Most Common Bonding Pattern for Each Nonmetal Element

Group Number of Covalent Bonds Number of Lone Pairs

4A (C) 4 0

5A (N, P, As) 3 1

6A (O, S, Se) 2 2

7A (halogens) 1 3

8A (noble gases) none 4

Sample Study Sheet 9,10.3: Resonance Hybrids

203

T-: You are asked to draw a resonance hybrid.

G S

• Draw the skeletal structure with solid lines for the bonds that are found in all of the resonance forms.

• Where there is sometimes a multiple bond and sometimes not, draw a dotted line.

• Draw in all of the lone pairs that are found on every one of the resonance forms. (Leave off the lone pairs that are on one or more resonance form but not on all of them.)

• Put on full formal charges on those atoms that have the formal charge in all of the resonance forms.

• Put on partial formal charges on the atoms that have formal charges in one or more of the resonance forms but not in all of them.

e resonance hybrid for the nitrate polyatomic ion is below.

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�Figure 9,10.29Resonance Hybrid

Sample Study Sheet 9,10.4 Resonance and Resonance forms


T- - You are asked to draw a Lewis structure for a molecule or a polyatomic ion, and the structure drawn has the following form.

� � �

Z can have more than one lone pair. X and Y can have lone pairs. e X-Y bond can be a triple bond. e Y-Z bond can be a double bond. Z cannot be F with one bond and three lone pairs or O with two bonds and two lone pairs.

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G S: Follow these steps when writing the resonance forms.

• Shift one of the lone pairs on one adjacent atom down to form a multiple bond.

• Shift one of the multiple bonds up to form a lone pair. You might find it useful to draw the arrows to indicate the hypothetical shift of electrons.

• Repeat this process for each adjacent atom with a lone pair.

Sample Study

Sheet 9,10.5

Drawing Lewis

Structures

T-: You are asked to draw a Lewis structure from a chemical formula for a molecule or a polyatomic ion.

G S (As you become more experienced with the process of drawing Lewis structures, you will see shortcuts that will allow you to bypass some or all of the following steps.)

S Determine the number of valence electrons for the molecule or polyatomic ion.

• For neutral molecules, the total number of valence electrons is the sum of the valence electrons of each atom.

Remember that the number of valence electrons for any one of the representative elements is equal to its group number, using the A group convention for numbering the groups. For instance, chlorine, Cl, is in group 7A, so it has seven valence electrons.

Hydrogen has one valence electron.

• For polyatomic cations, the total number of valence electrons is the sum of the valence electrons for each atom minus the charge.

• For polyatomic anions, the total number of valence electrons is the sum of the valence electrons for each atom plus the charge.

205

S Draw a reasonable skeletal structure joining the atoms with single bonds. One or more of the following guidelines might help you with this step.

• Try to set the atoms in space to gain the most common number of bonds for each atom. Table 9,10.3 lists the most common bonding patterns for the nonmetal elements.

• Apply the following guidelines in deciding what goes in the center of your structure.

∗ Hydrogen and fluorine atoms are never in the center. ∗ Oxygen atoms are rarely in the center. ∗ e element with the fewest atoms in the formula is usually in

the center. ∗ e atom that is capable of making the most bonds is usually

in the center. See Table 9,10.3. ∗ e atom with the lowest electronegativity is often in the

center. • Oxygen atoms rarely bond to other oxygen atoms. • e molecular formula often reflects the structural formula.• Carbon atoms commonly bond to other carbon atoms.

S Subtract 2 electrons from the total for each of the single bonds (lines) described in Step 2 above.

S Try to distribute the remaining electrons as lone pairs to get eight total electrons around each atom except hydrogen, beryllium, and boron.

When an atom has a total of eight electrons around it, we say it has an octet of electrons. Atoms of the noble gases other than helium have an octet of electrons around them. e orbital diagram and electron-dot structure for the valence electrons of neon show this.

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e stability of the octet is reflected in the fact that atoms in reasonable Lewis structures often have this octet.

∗ Carbon, nitrogen, oxygen and fluorine always have 8 electrons around them in a reasonable Lewis structure.

∗ Beryllium and boron can have less than eight electrons but never more than eight.

∗ e nonmetal elements below the second period (P, S, Cl, Se, Br, and I) usually have eight electrons around them but often have more.

∗ e metalloids arsenic, As, and tellurium, Te, form covalent bonds in similar ways to phosphorus, P, and sulfur, S, so they usually have eight electrons around them but often have more.

∗ Hydrogen will always have two electrons total from its one bond.


S Do one of the following.

• If in Step 4 you were able to get eight electrons around each atom other than those of hydrogen, beryllium, and boron and if you used all of the remaining valence electrons, go to Step 6.

• If you have electrons remaining after each of the atoms other than those of hydrogen, beryllium, and boron have their octet, you can put more than eight electrons around the elements in the third period and below.

• If you do not have enough electrons to get octets of electrons around each atom (other than hydrogen, beryllium, and boron), convert one lone pair into a multiple bond for each 2 electrons that you are short.

∗ If you would need two more electrons to get octets, convert one lone pair in your structure to a second bond between two atoms.

∗ If you would need four more electrons to get octets, convert two lone pairs into bonds. is could mean creating a second bond in two different places or creating a triple bond in one place.

∗ If you would need six more electrons to get octets, convert three lone pairs into bonds.

∗ Etc.S Check your structure.

• First check to see if all of the atoms have their most common bonding pattern.

If each atom has its most common bonding pattern, your structure is a reasonable structure. Skip Step 7, and proceed to Step 8.

• Determine the formal charge for each atom that has other than its most common bonding pattern. (Atoms with their most common bonding pattern always have a zero formal charge.)

Formal Charge = Group Number − Number of lines − Number of dots

∗ If all of the formal charges are zero, skip Step 7, and go to Step 8.

∗ If you have formal charges, continue with Step 7.

S Try to rearrange your structure to eliminate or at least reduce any formal charges. (is step is unnecessary if there are no formal charges.)

• One way to eliminate formal charges is to return to Step 2 and try another skeleton.

• Formal charges can also be eliminated by converting lone pairs into bonds. is process leads to placing more than eight electrons around some atoms. Here is where we hit controversy. Chemists are not agreed about whether it is better to emphasize the importance of octets in reasonable Lewis structures or better to minimize formal charges.

207

Step 8: Once you have a reasonable Lewis structure, consider the possibility of resonance.

• If resonance is possible, write all of the reasonable resonance forms for the original Lewis structure.

• Draw the resonance hybrid from the resonance forms.It is possible to generate more than one reasonable Lewis structure for a given formula. When this happens, the following rules will help you decide which is most likely.

• e more usual or common the bond is, the more stable the structure. See Table 9,10.4.

• A structure with formal charges is less stable than one without formal charges.

• “+” formal charges are more likely on the less electronegative elements and “-” formal charges are more likely on the more electronegative elements.

It is possible to have two Lewis structures for the same molecular formula, which are considered equally reasonable according to the above criteria. e substances represented by the Lewis structures are isomers, molecules that have the same molecular formula but different structural formulas. e dimethyl ether and ethanol are isomers of C2H6O.

CH

H

H

O

H

H

C H

H

H

C

H

H

CH HO

dimethyl ether ethanol


THE FOLLOWING IS A SUMMARY OF THE STEPS FOR DRAWING

LEWIS STRUCTURES FROM CHEMICAL FORMULAS.

S Determine the number of valence electrons for the molecule or polyatomic ion.

S Draw a reasonable skeletal structure joining the atoms with single bonds.

S Subtract 2 electrons from the total for each of the single bonds described in Step 2 above.

S Try to distribute the remaining electrons as lone pairs to get eight total electrons around each atom except hydrogen, beryllium, and boron.

S Do one of the following.

• If in Step 4 you were able to get eight electrons around each atom other than those of hydrogen, beryllium, and boron and if you used all of the remaining valence electrons, go to Step 6.

• If you have electrons remaining after the atoms other than those of hydrogen, beryllium, and boron have their octet, you can put more than eight electrons around the elements in the third period and below.

• If you do not have enough electrons to get octets of electrons around each atom (other than those of hydrogen, beryllium, and boron), convert one lone pair into a multiple bond for each 2 electrons that you are short.

S Check your structure.

S Try to rearrange your structure to eliminate or at least reduce any formal charges.

S Once you have a reasonable Lewis structure, consider the possibility of resonance. If resonance is possible, draw the reasonable resonance forms and the resonance hybrid for the structure.

EXERCISE 9,10.2 - Drawing Lewis Structures

Draw a reasonable Lewis structure for each of the following. If the structure has resonance, draw all of the reasonable resonance structures and the resonance hybrid.

a. CH3Br h. C3H7OH

b. ClF3 i. HCO2-

c. CH2O j. CH2CHF

d. CN− k. HCO2H

e. CF3CHCl2 l. NH2COCH3

f. C2H6O m. SO42−

g. C2H4F2 n. PO42−

209

Figure 9,10.30Connections Between Skills This figure shows the connections between the skills covered in earlier chapters and the skills presented in this chapter. To master the new skills, you need to have mastered the old ones. Start at the top of the sequence and work your way down, convincing yourself as you go along that you can do each task.

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For a molecule to be polar, it must meet two criteria:

• It must contain polar bonds.

• It must have an asymmetrical distribution of its polar bonds.

H Cl ��

d-d+

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Figure 9,10.31The HCl molecule. The polar covalent H-Cl bond (or bond dipole) leads to an overall separation of positive and negative charge in the molecule. Therefore, the HCl molecule is a polar molecule, or a molecular dipole.

BFF

F

120°

OH

H105°

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Sample Study Sheet 9,10.6 Predicting Molecular Polarity

T- – You are asked to predict whether a molecule is polar or nonpolar; or you are asked a question that cannot be answered unless you know whether a molecule is polar or nonpolar. (For example, you are asked to predict the type of attraction holding the particles together in a given liquid or solid.)

G S

S Draw a reasonable Lewis structure for the substance. S Identify each bond as either polar or nonpolar. (If the difference in electronegativity for the atoms in a bond is greater than 0.4, we consider the bond polar. If the difference in electronegativity is less than 0.4, the bond is essentially nonpolar.)

• If there are no polar bonds, the molecule is nonpolar. • If the molecule has polar bonds, move on to Step 3.

S If there is only one central atom, examine the electron groups around it.• If there are no lone pairs on the central atom, and if all the bonds to the

central atom are the same, the molecule is nonpolar. • If the central atom has at least one polar bond and if the groups bonded to

the central atom are not all identical, the molecule is probably polar. Move on to Step 4.

S Draw a geometric sketch of the molecule. S Determine the symmetry of the molecule using the following steps.

• Describe the polar bonds with arrows pointing toward the more electronegative element. Use the length of the arrow to show the relative polarities of the different bonds. (A greater difference in electronegativity suggests a more polar bond, which is described with a longer arrow.)

• Decide whether the arrangement of arrows is symmetrical or asymmetrical ∗ If the arrangement is symmetrical and the arrows are of equal length,

the molecule is nonpolar. ∗ If the arrows are of different lengths, and if they do not balance each

other, the molecule is polar.∗ If the arrangement is asymmetrical, the molecule is polar.

EXERCISE 9,10.3 - Predicting Molecular Polarity

Decide whether the molecules represented by the following formulas are polar or nonpolar. (You may need to draw Lewis structures and geometric sketches in order to do this.)

a. BCl3 e. CBr4

b. C3H8 f. CH3CO2H

c. CS2 g. Br2O

d. NH3 h. IF3

211

Sample Study Sheet 9,10.7: Predicting Types of Attractions and Relative Strengths of Attractions Between Particles

T- – You are asked to predict the relative strengths of attractions between particles of two substances, or you are asked a question that cannot be answered unless you know those relative strengths. (For example, you are asked to compare certain properties of substances, such as their relative condensation temperatures.)

G S -

S Determine the type of attraction between the particles using the following steps.

S () Classify each substance as either a metallic element, an element composed of atoms, an element composed of molecules, an ionic compound, or a molecular compound.

• Metallic elements have metallic bonds.

• e particles in elements that are composed of atoms (the noble gases) and molecules (the other nonmetallic elements) are held together by London forces.

• e ions in ionic compounds are held together by ionic bonds.

• Continue to Step 1(b) for molecular compounds.

S () For molecular compounds, draw the Lewis structure for the molecule.

• If the Lewis structure contains an O-H, N-H, or H-F bond, the attractions that form when the substance condenses are hydrogen bonds enhanced by London forces.

• For other molecular compounds, do Step 1(c).

Step 1(c): If there are no O-H, N-H, or H-F bonds, determine the polarity of the bonds.

• If there are no polar bonds, the molecules are nonpolar and form London forces when they condense.

• If there is at least one polar bond, do Step 1(d).

Step 1(d): Predict whether the polar bonds are symmetrically or asymmetrically arranged.

• If the distribution of polar bonds is symmetrical and their dipoles equal, the molecules are nonpolar and form London forces when they condense.

• If the distribution of polar bonds is asymmetrical, or symmetrical with unequal dipoles, the molecules are polar and form dipole-dipole attractions enhanced by London forces when they condense.


Step 2 Although we cannot predict the relative strengths of attractions between all particles, we can apply one of the following guidelines to predict the relative strengths of attractions between some particles.

• For substances that contain particles of about the same size, the substances with chemical bonds (ionic, covalent, or metallic) have stronger attractions between particles than substances with intermolecular attractions (hydrogen bonds, dipole-dipole attractions, or London forces).

Chemical bonds are generally stronger than intermolecular attractions.

• For molecular substances that contain molecules of about the same size, substances with hydrogen bonds have stronger attractions between the particles than substances with either dipole-dipole attractions or London forces, and substances with dipole-dipole attractions have stronger attractions between the particles than London forces.

Hydrogen bonds are generally stronger than dipole-dipole attractions, which are generally stronger than London forces.

• For molecular substances that have the same type of intermolecular attraction, larger molecules form stronger mutual attractions.

Larger molecules tend to have stronger attractions.

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YesNo

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SymmetricalAsymmetrical

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Figure 9,10.32 Scheme for Predicting Types of Attractions

Table 3.25 Solid and Liquid Form

Type of substance Particles to visualize

Examples Type of attraction between particles in

solid or liquidElements Metal Cations in a

sea of electronsAu Metallic bond

Noble Gases Atoms Xe London forcesCarbon (diamond) Carbon atoms C(dia) Covalent bondsOther Nonmetal Elements

Molecules H2, N2, O2, F2, Cl2, Br2, I2, S8, Se8, P4

London forces

Ionic Compounds

Cations and Anions

NaCl Ionic bond

Molecular Compounds

Nonpolar Molecular

Molecules Hydrocarbons London forces

Polar molecules without H-F, O-H, or N-H bond

Molecules HCl Dipole-Dipole forces

Molecules with H-F, O-H or N-H bond

Molecules HF, H2O, alcohols, NH3

Hydrogen bonds

Network crystals Atoms SiO2 Covalent bonds

EXERCISE 9,10.4 - Predicting Type of Attraction

What is the primary type of attraction that holds the particles of each of these substances in the liquid or solid form.

a. Ir e. NH4OH

b. AlF3 f. BF3

c. C2H6 g. CH2O

d. CH3NH2 h. C3H7OH

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EXERCISE 9,10.5 - Predicting Types and Strengths of attractions Between Particles

For each of the following substances, write the name for the type of particle that forms its basic structure and the name of the primary type of attraction that can form between these particles. From each pair of substances, choose the one that you expect would have stronger interparticle attractions.

a. ethanol, C2H5OH (in alcoholic beverages)dimethyl ether, CH3OCH3 (refrigerant and solvent for propellant sprays)

b. manganese, Mn (improves corrosion resistance and hardness when alloyed with other metals)carbon tetrachloride, CCl4 (used to add chlorine atoms to organic compounds)

c. propane, C3H8 (fuel and aerosol propellant)n-heptane, C7H16 (standard for octane rating – 100% n-hexane is zero octane)

d. acetone, CH3COCH3 (solvent for paints and varnishes)lithium nitrate, LiNO3 (used to make ceramics and in rocket propellants)

e. antimony tribromide, SbBr3 (a molecular compound used as a mordant, which fixes dyes to textiles)antimony trichloride, SbCl3 (a molecular compound used for fireproofing textiles)


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