Chemistry 20

Chapters

1 & 2

Bonding

G3.G4.G5 CHEMICAL BONDING G2 LEWIS DIAGRAMS

Lewis Diagrams for Atoms The octet rule and the concepts of bonding electrons and lone-pair electrons have ready application in Lewis (electron-dot) diagrams. These diagrams provide a convenient means for keeping track of the origin and distribution of valence electrons involved in the covalent bonding of atoms. Lewis diagrams for chemical bonding were developed by G. N. Lewis in 1916 in order to ease the comprehension of covalent bonding.

Lewis diagrams use:

1. the atomic symbol of the atom to represent both the nucleus and the filled innermost energy levels of the atom which do not participate in the bonding

2. dots (or crosses), one per electron, arranged around the atomic symbol to represent the valence electrons. (Unpaired electrons are usually separated from one another and paired electrons are shown close together. Thus X indicates two unpaired (bonding) electrons, whereas X : indicates a lone electron pair.)

The dots (or crosses) in Lewis diagrams have no significance as far as the actual positions of the electrons are concerned, however, the dots have significance relative to orbital occupancy by electrons. (In this unit Lewis diagrams will be limited to atoms and molecules obeying the octet rule (with the exception of hydrogen). Lewis diagrams are useful for situations beyond the octet rule but are not necessary for secondary high school chemistry.)

The rules for placing electrons in the four available orbitals of an atom are listed below.

1. One electron is placed in each of the four available orbitals before any electron pairing occurs. (This distribution results in a more stable state; i.e., less electron repulsion.)

2. If there are five to eight valence electrons, a second electron may go into the singly occupied orbitals. (Three electrons in an orbital would result in large repulsive forces and an unstable state, and is therefore not allowed.)

Table G1Lewis Diagrams for Atoms

Atom Group # Valence Electrons

# Valence Orbitals # Lone Pairs # Bonding

ElectronsLewis

DiagramH IA 1 1 0 1 H

C IVA 4 4 0 4.

C.

N VA 5 4 1 3. .

N.

O VIA 6 4 2 2. .

: O .

F VII 7 4 3 1. .

: F . .

Complete the following table.

Atom Group # Valence Electrons

# Valence Orbitals Lewis Diagram # Lone

Pairs# Bonding Electrons

1. S

2. Si

3. P

4. Cl

5. Br

6. Se

1

G8 CHEMICAL BONDING G11 ELECTRONEGATIVITY

Molecular State Theory

A restricted version of current molecular state theory is presented below. The evidence for the existence of intermolecular forces is partially explained by progressively uniting the man-made ideas of electronegatives, bond dipoles and molecular dipoles. These ideas have been reorganized for presentation here and do not necessarily represent a historical sequence.

Electronegativity

Experimental evidence indicates that covalently bonded atoms usually exhibit unequal attractions for shared electrons. In fact, different atoms, as shown by a variety of experimental evidence, have different electron-attracting abilities. The relative attraction that an atom has for shared electrons in a covalent bond is known as its electronegativity. A scale of electronegativities, in which the most electronegative (fluorine) is arbitrarily assigned a value of 4.0, was developed by Linus Pauling (1901-1994). Pauling electronegativities are given on the ALCHEMperiodic table.

Examination of the electronegativities of elements given in the periodic table will indicate the following trends.

1. Electronegativities increase from left to right within a period. 2. Electronegativities decrease from top to bottom within a group.3. Relative electronegativities of the nonmetals are high and those of the metals are low.

The latter trend is consistent with the trends given in 1 and 2 and correspond to the trend in variation of non- metallic character; i.e., as nonmetallic character increases, electronegativity increases and vice-versa. The electronegativity scale is a composite of experimental evidence and as such reflects the relative reactivity of metals and nonmetals (i.e., the lowest electronegativity belongs to the most reactive metals, Cs and Fr, while the highest electronegativity belongs to the most reactive nonmetal, F).

Comparing Electronegativities of Atoms

The shared electrons in a covalent bond will be displaced (more strongly attracted) to which atom in each of the following chemical bonds? (The first is done as an example. These diagrams represent bonds only - not molecules.)

Example: H - F The arrow indicates the shared electron pair is attracted more strongly by the fluorine atom. 2.1 4.0

1. N - H 2. B - F 3. S - O 4. P - H

5. Si - Cl 6. Cu - Br 7. N - I 8. Br - Cl

9. C - H 10. O - H 11. C - Cl 12. C - O

13. Explain why cesium and francium are the most reactive metals.

14. Explain why fluorine is the most reactive nonmetal.

2

G3 CHEMICAL BONDING G34 IONIC BONDING

Ionic Bonding

There are varying degrees of charge separation, depending on the difference between electronegativities of the atoms bonded. As the difference between electronegativities of bonded atoms increases, charge separation can reach a condition at which the shared electron can be regarded as becoming the exclusive property of the more electronegative atom. At this point electron transfer has taken place. The more electronegative atom has become a negatively charged ion and the less electronegative atom, a positively charged ion. Thus, ions, the structural units of ionic compounds, are formed by the transfer of one or more electrons from one atom to another. Metallic elements, having relatively low electronegativity, tend to lose one or more electrons to form positive ions. Nonmetallic elements, having relatively high electronegativity, tend to gain one or more electrons to form negative ions.

When sodium metal reacts with nonmetallic chlorine, ions are formed by transferof electrons from atoms of sodium to atoms of chlorine.

Figure G8

The process of electron transfer from a metallic atom to a nonmetallic atom involves a fixed number of electrons. Usually, atoms lose or gain an appropriate number of electrons so as to acquire the electronic configuration of the nearest noble gas, which, except for He, is an octet structure. As a general rule, atoms of Groups IA, IIA and IIIA metallic elements lose one, two and three electrons respectively, and atoms of Groups VA, VIA and VII nonmetallic elements tend to gain three, two and one electrons respectively. In the case of transition metals (to which no simplifying general rule can be applied) ionic charges can be obtained from the ALCHEM periodic table. Additionally, it should be recalled that ionic compounds can involve complex ions.

Ions in ionic compounds are held together by simultaneous electrostatic forces of attraction among oppositely charged ions. The overall force of attraction among oppositely charged ions is called an ionic bond. Ionic bonding produces an orderly three dimensional arrangement of ions into ionic crystals as illustrated in the diagram following.

Positive and Negative Ions Arrange to Form Stable Aggregates Called Ionic CrystalsFigure G9

Note that in the NaCl crystal lattice every ion is closest to, and simultaneously attracted by, six ions of opposite charge. Each Na+ ion is simultaneously attracted to six Cl- ions. Each Cl- ion is simultaneously attracted to six Na+ ions.

3

G3 CHEMICAL BONDING G35IONIC BONDING

There are several features associated with ionic bonding that collectively lead to the stable ionic crystal.

1. The formation of ions involves an electron transfer that usually produces ions with the most stable electronic configuration of a noble gas. Note that the process of electron loss is not spontaneous and energy is required to remove an electron from a neutral metallic atom. However, when an atom of a metal closely approaches an atom of a nonmetal, an electron transfer results in a more stable condition. This is because the gain of electrons by an atom of a nonmetal is accompanied by a release of energy greater than the energy absorbed when the electron is removed from the metal.

2. An energy release (lattice energy) occurs when positive and negative ions form the regular three-dimensional arrangement found in ionic crystals.

Because the forces which hold ionic crystals together are not concentrated between the individual atoms, the bonding force involves all of the ions. This gives ionic compounds the general property of high melting and boiling points so that they are generally found as solids at ordinary conditions.

Exercise:

For Questions 1 through 3 assume a metallic Element M with two valence electrons chemically reacts with a nonmetallic Element X with seven valence electrons.

1. What kind of bond is most likely to form between M and X? ___________________________________

2. The resulting compound between M and X would form what characteristic kind of solid? ____________________________________________________________________________________

3. Using Lewis diagrams show the electron rearrangement that occurs to form a chemical bond between M and X.

4 Describe the essential difference between a polar covalent bond and an ionic bond.

5. Discuss the statement, "An ionic bond results from the transfer of electrons".

Classify the bonds in the following compounds as predominately covalent or predominately ionic using the staircase line on the periodic table as the dividing line for classification.

6. KCl _____________________________ 9. HI _____________________________

7. LiBr _____________________________ 10. CH4 _____________________________

8. CaS _____________________________ 11. H2S _____________________________

Identify the bond types (ionic, nonpolar covalent or polar covalent) for each of the following substances.

12. BrCl _____________________________ 16. CCl4 _____________________________

13. P4 _____________________________ 17. FeCl3 _____________________________

14. CsF _____________________________ 18. K2S _____________________________

15. CO2 _____________________________ 19. SiF4 _____________________________

20. Can all bonding thus far in this unit be described in terms of simultaneous attractions?

4

Nelson Chemistry

36 IONIC COMPOUNDS

1. What is the key theoretical process used to explain the formation of ionic compounds? What is believed to be the reason for this process?

2. Use electron dot diagrams to represent the formation of lithium oxide from its constituent atoms.

3. Use electron dot diagrams to represent the formation of strontium bromide from its constituent atoms.

5

G5 CHEMICAL BONDING G6LEWIS MOLECULAR THEORY

Lewis Diagrams for Molecules

Consistent with the octet rule, atoms share an appropriate number of electrons to attain eight valence electrons - the stable electron configuration characteristic of the noble gases. (Hydrogen, an important exception to the octet rule, tends to attain two valence electrons - the structure characteristic of the noble gas, helium.)

The rules for drawing Lewis diagrams for molecules in this course are listed below.

1. Draw the Lewis diagrams for each of the atoms in the molecule. 2. Unpaired electrons (called bonding electrons) are available for sharing to form a covalent bond. 3. Paired electrons (called lone pairs) do not partake in the bonding.4. The atom with the most bonding electrons (called the central atom) is placed in the center with the other atoms

bonded to it. 5. In the resulting Lewis diagram all electrons must be paired and each atom except H must be surrounded by an octet of electrons.

These rules are specific to the level of the theory required to cover the molecules studied in this course. In science only the level of theory necessary to the situation is used. When, and only when, a study of more complex species is required should the theory be made more complex: i.e., molecules with unpaired electrons and more valence electrons than an octet.

Example:

The use of dots and crosses to represent valence electrons is not to imply that there are different kinds of electrons. All electrons are identical. The use of dots and crosses is an aid to keep track of the origin of the electrons.

Lewis Diagrams and Structural Formulas Bonding representation is often simplified by omitting the lone electron pairs and by substituting a dash for each bonding pair of electrons. The resulting representation is called a structural formula. (The structural formula does not generally represent the shape of the molecule, it merely represents the bonding that occurs.)

Table G2Lewis Diagram and Structural Formulas - Single Bonds

6

G14 CHEMICAL BONDING G7 LEWIS’ MOLECULAR THEORY

Lewis’ Molecular Theory

Complete the following table. Use Lewis' Molecular Theory to explain the molecular formulas for Questions 5-8, and then to predict the simplest molecular formula formed by the atoms listed in Questions 9-14. In Questions 9-14 the simplest molecular formula is the molecule formed by the least number of atoms.

Molecular Formula or Atoms Combined

Lewis Diagram of Molecule Structural Formula Confirmed or Predicted

Molecular Formula

5. H2O(l)

6. HCl(g)

7. Cl2(g)

8. C2H6(g)

9. H atoms

10. Cl atoms

11. C & H atoms

12. N & I atoms

13. O & F atoms

14. P & Cl atoms

7

NELSON CHEMISTRY

37 MOLECULAR COMPOUNDS

For each of the following molecular compounds, explain the empirically determined formula by drawing a diagram of the structural model.

4. CH3Cl 5. C2H3Cl 6. CH3CN

For each of the following combinations of elements, theoretically predict the molecular formula of the simplest product of the formation reaction. Draw a structural diagram of the reactants and products. Balance the equations.

7. Cl2(g) + Br2(g)

8. N2(g) + Cl2(g)

9. O2(g) + F2(g)

8

G4, G5 CHEMICAL BONDING G8LEWIS DIAGRAMS WITH MULTIPLE BONDS

Multiple Bonds Represented by Lewis Diagrams and Structural Formulas

Atoms which have two or more bonding electrons may share two or three of these bonding electrons with the same atom. If two electron pairs are shared between two atoms, the bond is commonly called a double bond. Three electron pairs shared between two atoms is a triple bond.In the resulting Lewis diagrams all electrons must be paired and each atom except H must be surrounded by an octet of electrons.

Table G3Lewis Diagrams and Structural Formulas – Multiple Bonds

Exercise:

Complete the following table. Use Lewis’ Molecular Theory to confirm the molecular formulas given.

Molecular Substance Molecular Formula Lewis Diagram of

AtomsLewis Diagram of

Molecule Structural Formula

1. carbon dioxide

2. HCN(g)

3. tetrafluoroethene C2F4(g)

4. dichloroethyne C2Cl2(g)

9

G5,G7 CHEMICAL BONDING BETWEEN ATOMS G18 STEREOCHEMISTRY – LAB G1

Purpose:

1. To use VSEPR Theory to predict shapes around central atoms.

2. To construct molecular models from a kit to test the predictions.

Prelab Exercise:

Before going to the lab, complete the following table except for the columns headed Observations (Representation of Actual Shape and Name to Represent Actual Shape).

Procedure:

In the lab, construct models of each molecule using the correct coloured ball for each atom and one spring for each shared electron pair. Draw a diagram of the model under Representation for Actual Shape then write in the name to represent the actual shape. If the actual shape corresponds with the predicted shape, check it off – if not, consult the teacher.

Prelab and Observations:

Prelab Exercise Observations

Molecular substance Lewis Diagram

For Each Central Atom

Name to Represent Predicted

Shape

Representation for Actual

Shape

Name to Represent

Actual Shape# Lone Pairs

# Bonding Pairs

1. NI3(s)

2. C2Cl4(l)

3. CF4(g)

4. OCl2(l)

5. C2F2(g)

6. HOF(l)

10Prelab and Observations:

Prelab Exercise Observations

Molecular substance Lewis Diagram

For Each Central Atom

Name to Represent Predicted

Shape

Representation for Actual

Shape

Name to Represent

Actual Shape# Lone Pairs

# Bonding Pairs

7. NHF2

8. C2IBr

9. C2HF3

10. CHClBr2

11. H2O2(l)

12. CO2(g)

13. N2H3F

14. CH3OH(l)

15. C3H6(g)

(noncyclic)

11

Nelson Chemistry

38 (ENRICHMENT) SHAPES OF MOLECULES

For each of the following molecules, draw an electron dot diagram and use the table on the previous page to describe the shape around each central atom. Then draw a structural diagram for each. Your predictions can be tested using a molecular model kit.

1. hydrogen sulfide, H2S 2. carbon tetrachloride, CCl4

3. phosphine, PH3 4. tetrachloroethylene, C2C14

5. hydrogen peroxide, H2O2 6. carbon dioxide, CO2

7. hydrazine, N2H4 8. propane, C3H8

9. methylcyanide, CH3CN 10. nitrosyl chloride, NOCl

12Answers1. V-shaped 2. Tetrahedral 3. Pyramidal 5. V-shaped 6. Linear 7. Pyramidal8. tetrahedral 9. Tetrahedral and linear 10. pyramidal

G7 (Option 1) CHEMICAL BONDING G21 POLARITY OF BONDS AND MOLECULES

For each of the following molecules draw the diagram to represent the actual shape and use arrows to indicate bond dipoles. Give the name to represent actual shape and indicate whether the molecule is polar or nonpolar. The first is done as an example.

13

(Option 2) CHEMICAL BONDING G22 POLARITY OF MOLECULES (WITHOUT STEREOCHEMISTRY)

A More Restricted Approach than VSEPR Theory The previous optional section presented a systematic approach to predicting the polarity of molecules from the consideration of shapes of molecules (stereochemistry), electronegitivity, bond dipoles and molecular dipoles. For the majority of molecules encountered in the ALCHEM materials a more restricted approach may be used in which it is not necessary to have as thorough knowledge of stereochemistry as presented in Option 1. The following rules allow the polarity of a molecule to be predicted on the basis of its molecular formula.

Recognizing Nonpolar MoleculesA molecule is nonpolar if its molecular formula has the form: 1. CxRy, where C is carbon and R is any other nonmetallic atom; e.g., CH4(g) and C2F4(l) . 2. Rx, any molecular element; e,g., F2(g), P4(s) and S8(s)

Recognizing Polar Molecules A molecule is polar if its molecular formula has the form: 1. RS, where R and S represent two different nonmetallic atoms (diatomic, binary, molecular compounds);

e,g., HCl(g) and IBr(s) .2. CxRySz, where C is carbon; e.g., CH3Cl(l) and C2H4Cl2(l). (There are a few exceptions to this rule.) 3. NRx or ORx, where N is nitrogen and O is oxygen; usually any molecule containing oxygen and/or nitrogen;

e.g., CH3OH(l) and NH3(g) (but not CO2(g) ).

Rules Versus TheoriesThe rules presented above do not constitute a theory. VSEPR Theory presented previously plus the concept of bond dipoles provides a complete theory for molecular dipoles; i.e., it explains and predicts polarity of molecules. The rules above may be used to predict but not explain molecular dipoles. However, a retroactive explanation of the rules can be made using bond dipoles.

Explanation 1: CH4(g) can be explained as being nonpolar:

The CH4 molecule is nonpolar; i.e., the CH4 molecule does not have a positive end and a negative end, all of the "ends" are partially positive (+).

Explanation 2: H2O can be explained as being polar:

The water molecule is polar if it is assumed to be bent rather than linear. If bent the H2O molecule has a partially positive and a partially negative end and is therefore a polar molecule (a molecular dipole). The rules for recognizing a polar and a nonpolar molecule represent a given way of knowing rather than an empirical or theoretical way of knowing about molecular dipoles. It is important to recognize these different ways of knowing when determining the certainty with which knowledge can be held. Complete the following questions by using the rules presented above to predict the polarity of the molecules. Draw a structural formula and try to explain the polarity or lack of polarity of the molecules by using bond dipoles. (Do the same thing for the molecules on Page G13.) 1. HBr(g) 4. CH3OH(l)

2. O2(g) 5. NI3(s)

3. CCl4(l) 6. C2H3Cl(g)

14

CHEMICAL BONDING G23 BONDING BETWEEN MOLECULES - VAN DER WAALS FORCES

Types of Intermolecular Forces

Several types of intermolecular forces have currently been "identified". It is believed that in most cases, more than one type of intermolecular force act together in contributing to the total attraction between molecules. The most widely accepted theories recognize two main types of intermolecular forces between molecules.

1 . van der Waals forces a. dipole-dipole forces b. London dispersion forces

2. hydrogen bonding

van der Waals Forces

Van der Waals forces are named after the Dutch scientist Johannes Diderik van der Waals (1837-1923), Professor of Physics at Amsterdam University. Van der Waals forces are believed to be present between all chemical species - atoms, molecules and ions. Van der Waals forces are now subdivided into dipole-dipole forces and London dispersion forces.

1. Dipole-Dipole Forces

If the molecules in a sample are polar, the presence of molecular dipoles is believed to give rise to simultaneous intermolecular attraction. The positive side of one molecule should attract the negative side of a neighboring molecule, which attract the next, and so on out to the limits of the sample. Note in the figure below that the central polar molecule is simultaneously attracted to the polar molecules surrounding it.

Simultaneous Dipole-Dipole ForcesFigure G4

Note:The ALCHEM authors have defined all bonding in this unit in terms of simultaneous attractions. These definitions are not the original, historical definitions presented by the scientists. The similar definitions given for all of the bond types are typical of the unity which is sought in all scientific fields.

15

G9,G10 CHEMICAL BONDING G26 LONDON DISPERSION FORCES

A fairly valid measure of the comparative strengths of intermolecular forces for substances is a comparison of their boiling points. This works fairly well since the stronger the intermolecular forces holding the molecules together in liquid phase, the more heat will be required to cause them to separate. Table G6 compares the boiling points of the noble gases to the number of electrons they have.

Table G6Variation of the Boiling Point of the Noble Gases

with Number of Electrons

Element Number of Electrons Boiling Point (C)He 2 -269Ne 10 -246Ar 18 -186Kr 36 -152Xe 54 -107Rn 86 -62

Exercise: Further confirm the trend illustrated in Table G6 by comparing the boiling point of another group of nonpolar molecules - the halogens.

Element Number of Electrons Boiling Point (C)1. F2

2. Cl2

3. Br2

4. I2

Limitations of van der Waal Forces

The previous sections on intermolecular forces emphasized that these forces increase with increasing number of electrons in an atom or molecule. Try to predict the boiling point for HF by examining the data in the table below.

Table G7Variations of the Boiling Point of Hydrogen Halides

Hydrogen Halide

Electronegativity of Halogen

Number of Electrons

Boiling Point (C)

HF 4.0 10HCl 3.0 18 -83.7HBr 2.8 36 -67.0HI 2.5 54 -35.4

The four compounds in the preceding table are similar in that they are hydrogen halides. Each compound is molecular in that intermolecular attractions involve van der Waals forces. The generalization about increasing van der Waals forces with increasing number of electrons holds up when HCl, HBr and HI are compared. But hydrogen fluoride, HF, does not fit into the predicted trend. Instead of having a boiling point lower than -83.7C, thus fitting into the trend, HF has the highest boiling point of all four compounds, 19.4C. Examination of the electronegativities of the halogens reveals that the HF molecule is the most polar. However, molecular polarity alone could not account for the magnitude of reversal in trend. The existence of an additional intermolecular force, greater than van der Waals forces, is suggested. This additional force is hydrogen bonding.

16

G10 CHEMICAL BONDING G30BONDING BETWEEN MOLECULES

Intermolecular Forces

1. Using the data from Table G6, plot a graph of boiling point versus number of electrons for the noble gases.

2. State a generalization relating London dispersion forces to the number of electrons in atoms or molecules.

Explain the generalization in terms of Ar, boiling point -186C, and F2, boiling point -188°C.

3. Both Kr (boiling point, -152C) and HBr (boiling point, -67 °C) are isoelectronic (have the same number of electrons). Explain what factors could affect intermolecular bonding to cause the difference in boiling points between Kr and HBr.

4. The boiling point of Cl2 is -35C and the boiling point of C2H5Cl (monochloroethane) is 13°C. Does the explanation proposed for Question 3 apply here? Explain.

17

G10 CHEMICAL BONDING G31BONDING BETWEEN MOLECULES

Refer to the graphs in Figure G7 when answering Questions 5, 6, and 7 following.

Boiling Points for Hydrogen Compounds ofGroup IVA, Group VA, Group VIA and Group VIIA

Figure G7

5. The hydrogen compounds of Groups VA, VIA and VIIA elements have consistently increasing van der Waals forces (except for the first hydrogen compounds) with increasing number of electrons. Explain why the boiling point of the first hydrogen compounds of Groups VA, VIA and VIIA elements display a reversal in trend.

6. Explain why CH4, the first member of the Group IVA hydrogen compounds, does not show the reversal in trend displayed by the first hydrogen compound of the other elements.

7. The boiling points of the hydrogen compounds of the Group IVA elements are consistently lower than the boiling points of the other hydrogen compounds. Give a reason for this effect.

18

G10 CHEMICAL BONDING G32BONDING BETWEEN MOLECULES

Complete the following table. The first is given as an example. Note that there are six series or groups of molecules.

Table G8Relation of Boiling Point to the number of Electrons

and to the Type of Intermolecular Forces

Molecular Substance with Phase at Room Temperature

Number of Electrons

Boiling Point (C)

Types of Intermolecular Forcesvan der Waals Hydrogen

bondingdipole- dipole London dispersion

e.g. F2(g) 18 -188

1. Cl2(g) -35

2. Br2(l) 59

3. I2(s) 184

4. ClF(g) -101

5. BrF(g) -20

6. BrCl(g) 5

7. ICl(s) 97

8. IBr(s) 116

9. CH4(g) -162

10. C2H6(g) -87

11. C3H8(g) -45

12. C4H10(g) -0.50

13. C5H12(l) 36

14. CF4(g) -129

15. CCl4(l) 77

16. CBr4(s) 189

17. CH3F(g) -78

18. CH3Cl(g) -24

19. CH3Br(g) 3.6

20. CH3I(l) 43

21. CH3OH(l) 65

22. C2H5F(g) -38

23. C2H5Cl(g) 13

24. C2H5Br(l) 38

25. C2H5I(l) 72

26. C2H5OH(l) 78

19G10 CHEMICAL BONDING G33

BONDING BETWEEN MOLECULES

Use the preceding table to answer Questions 27 - 33.

27. Compare the boiling points of BrF(g) and C3H8(g) . 28. Dimethyl ether, CH3OCH3(g) has a boiling point of Account for the difference in boiling points. -24.9C. Compare with the boiling point of ethanol

and account for the difference.

29. The different series of substances given in Table G8, in general, have increasing boiling points with increasing number of electrons. Explain this trend in terms of number of electrons and strength of intermolecular forces.

30. Methanol, CH3OH, and ethanol, C2H5OH, each have the least number of electrons but the highest boiling point of their respective series. Account for this.

31. Explain the difference in boiling point between C2H6 and CH3F.

32. Explain the difference in boiling point between Cl2 and C4H10.

33. Explain the difference in boiling point between BrCl and C2H5Br.

20CHEMICAL BONDING G39

OVERVIEW OF INTERMOLECULAR BONDING

Complete the following table by providing the name of the intermolecular bond type. The bond types (not in order) are metallic, dipole-dipole, hydrogen, ionic, covalent network, and London dispersion.

Bond Type Characteristics of Formulation Some General Properties Examples

1.

The simultaneous attraction by covalent bonds of an atom by adjacent atoms within a 3-D lattice of atoms

Very hard; very high melting point; insoluble in most ordinary solvents; nonconductors of electricity

C(s) (diamond)

SiO2(s) , SiC(s)

(memorize these three)

2.

The simultaneous attraction of an ion by its surrounding ions of opposite charge within an ionic crystal lattice

Crystalline solids under ordinary conditions; high melting and boiling points; dissolve in polar liquids to form conducting solutions; electrical conductors in liquid phase

NaCl(s), Ca(OH)3(s),

CuSO4(s), NH4Cl(s),

NaHCO3(s), KNO3(s)

3.

The simultaneous attraction of free valence electrons by metallic cations

Lustrous, malleable; good electrical conductors; wide range of melting points

Al(s) , Fe(s) , Cu(s) ,Zn(s) , Ca(s) , Na(s) ,

Ag(s) , Pb(s) Hg(l)

4.

The simultaneous attraction of electrons of one molecule by their own nucleus and by the nuclei of adjacent molecules

Relatively low melting solids, gases or liquids because of relatively weak intermolecular forces

H2(g) , CO2 (g) , He(g) , I2 (s) , CH4(g) , S8(s) , CCl4(l) , HBr(g) , CHCl3(l)

5.

The simultaneous attraction of a hydrogen ion (proton) by the electron pairs of adjacent N, O or F atoms

Relatively high melting solids, gases or liquids because of relatively strong intermolecular attraction

H2O(l) , HF(g) , NH3(g) , C2 H5OH(l) , H2O2(l) , CH3NH2(g) , CH3COOH(l)

6.

The simultaneous attraction of a molecular dipole by the surrounding molecular dipoles

A weak intermolecular force which exists in addition to the stronger dispersion forces. Low melting solids, liquids and gases

H2S(g) , C2 H3Cl(l), C2 H5F(g) , IBr(s) , CH3I(l)

Complete the following table by providing the name of the intermolecular bond type. The bond types represented (not in order) are metallic, dipole-dipole, hydrogen, ionic, covalent network, and London dispersion.

13. Write the type of intermolecular bonding under each compound below. List the compounds in order of increasing boiling point.

NaCl Na C3H8 C2H5OH SiO2 C2H3Cl

21CHEMICAL BONDING G40

OVERVIEW EXERCISE

For each of the following substances, draw the Lewis dot and structural diagrams for the molecule and identify the molecule as polar or nonpolar. (Optional: Draw a shape diagram of the molecule.)

1. BrF(g)

2. CH3Cl(g)

3. C2Br4(s)

4: NCl3(l)

5. List the intermolecular bonding forces present in ethanol.

Consider the substances C2H3Cl and C2H3I when answering the next three questions.

6. Which of the substances. C2H3Cl and C2H3I, would probably be more polar? _____________________

7. Which would probably have the higher boiling point? _______________________

8. What specific type of bonding probably contributes most to the intermolecular attractions in these substances?

___________________________________________________________________________________________

9. (Optional) In which of the following does repulsion from a lone pair not influence molecular shape?

A. H2S B. NI3 C. HBr D. OF2 E. PH3

22 CHEMICAL BONDING G41

OVERVIEW EXERCISE

10. The ALCHEM Organic Chemistry unit will provide information about the molecular compound benzene, C6H6(l) The boiling points for some benzene derivative compounds are:

C6H5F(l): 85C C6H5Cl (l): 132C C6H5I (l): 188C

Which of the following is the boiling point for C6H5Br (l) ?

A. 88C B. 122 C C. 156C D. 249C E. 337C _________________

11. Phenol, C6H5OH(s) (182C) has a boiling point ,very close to that of iodobenzene, C6H5I(l) (188C)

List the bond types contributing to the intermolecular attractions in each substance. Explain why it is not possible to confidently predict which has the higher boiling point.

12. Benzoic acid, C6H5COOH(s), has the structure shown below. Can a confident prediction be made as to whether its boiling point will be greater or less than that of phenol? Explain.

13. Given that phenol melts at 43C, what would be the most likely physical state of benzoic acid at room temperature? Explain. _________________________________

14. Arrange the following substances in a list in order of increasing boiling points. List beside each substance the type of bonding present in the solid state. For the molecular substances, also list the number of electrons per molecule, and note whether any of the substances are isoelectronic. Predict the probable state at room temperature.

C8H18 C3H5(OH)3 C6H14 SiC C4H9Cl NaF Mg

15. (Optional) Molecules are nonpolar when all of the atoms have equal electronegativity. Outline briefly, in point form, the other conditions that result in a molecule being nonpolar (for cases where there is only one central atom).

16. Complete the following statement.

All chemical bonds result from. . .

17. In terms of electron mobility and electronegativity, explain why Na(s) is a very good conductor of electric current while NaCl(s) is a nonconductor.

18. (Optional) Draw a Lewis dot diagram for the ammonium ion, NH4+. Remember that in this ion one of the

original valence electrons is missing. Put large brackets around the diagram with the + sign outside the brackets, superscripted to the right.

19. (Optional) Draw Lewis dot diagrams for the ions - ClO-, ClO2-, ClO3

-, and ClO4-. Begin by considering the dot

diagram for a chloride ion, and remembering that a covalent bond is two nuclei attracting the same pair of electrons. Hint: Consider that by overcoming a small amount of repulsion, the electrons around an oxygen atom would be arranged like this

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