Chapter 8 Basic Concepts of Chemical Bonding

Chapter 8 Basic Concepts of Chemical Bonding

8.1 Lewis Symbols and the Octet Rule

Chemical Bonds Three basic types of bonds Ionic
Electrostatic attraction between ions. Covalent Sharing of electrons. Metallic Metal atoms bonded to several other atoms.

Chemical Bonds If the white powder were sugar, C12H22O11, how would this change the picture?

Are all these Lewis symbols for Cl correct?
Yes, all three are correct. No, the first two are different Lewis structures because of the differing placement of unpaired electrons. No, because the Lewis structure on the right has only six valence electrons and elemental Cl has seven valence electrons. Answer: c

Lewis Symbols G. N. Lewis developed a method to denote potential bonding electrons by using one dot for every valence electron around the element symbol. When forming compounds, atoms tend to gain, lose, or share electrons until they are surrounded by eight valence electrons (the octet rule).

8.2 Ionic Bonding

Ionic Formation Atoms tend to lose (metals) or gain (nonmetals) electrons to make them isoelectronic to the noble gases.

Do you expect a similar reaction between potassium metal and elemental bromine?

Describe the electron transfers that occur in the formation of calcium fluoride from elemental calcium and elemental fluorine. Each calcium atom loses one electron and each fluorine atom gains two electrons. Each calcium atom loses two electrons and each fluorine atom gains one electron. Each calcium atom gains one electron and each fluorine atom loses two electrons. Each calcium atom gains two electrons and each fluorine atom loses one electron. Answer: b

If no color key were provided, how would you know which color ball represented the Na+ and which represented Cl-?

Energetics of Ionic Bonding— Born-Haber Cycle
Many factors affect the energy of ionic bonding. Start with the metal and nonmetal elements: Na(s) and Cl2(g). Make gaseous atoms: Na(g) and Cl(g). Make ions: Na+(g) and Cl–(g). Combine the ions: NaCl(s).

Energetics of Ionic Bonding
We already discussed making ions (ionization energy and electron affinity). It takes energy to convert the elements to atoms. (endothermic) It takes energy to create a cation (endothermic). Energy is released by making the anion (exothermic). The formation of the solid releases a huge amount of energy (exothermic). This makes the formation of salts from the elements exothermic.

If you were to perform the reaction KCl(s) ⟶ K+ (g) + Cl–(g), would energy be released?
Yes, the separation of the ions causes the release of energy. No, the separation of the ions requires energy. No, the separation of ions does not cause any change in energy. Answer: b

Lattice Energy That huge, exothermic transition is the reverse of the lattice energy, the energy required to completely separate a mole of a solid ionic compound into its gaseous ions. The energy associated with electrostatic interactions is governed by Coulomb’s law: Eel =  Q1Q2 d

Lattice Energy Lattice energy increases with:
increasing charge on the ions decreasing size of ions

Energetics of Ionic Bonding
These phenomena also helps explain the “octet rule.” Metals, for instance, tend to stop losing electrons once they attain a noble gas configuration because energy would be expended that cannot be overcome by lattice energies. 17

Sample Exercise 8.1 1) Without using table 8.2, arrange the following ionic compounds in order of increasing lattice energy: NaF, CsI, and CaO.

Practice Exercise 1 Predict which one of the following orderings of lattice energy is correct for these ionic compounds. (a) NaCl > MgO > CsI > ScN (b) ScN > MgO > NaCl > CsI (c) NaCl > CsI > ScN > CaO (d) MgO > NaCl > ScN > CsI (e) ScN > CsI > NaCl > MgO

Practice Exercise 2 Which substance do you expect to have the greatest lattice energy: MgF2, CaF2, or ZrO2?

Using this figure, find the most likely range of values for the lattice energy of KF.

Sample Exercise 8.2 1) Predict the ion generally formed by:
Sr S Al 2) Predict the charges on the ions formed when magnesium reacts with nitrogen.

Practice Exercise 1 Which of these elements is most likely to form ions with a 2+ charge? (a) Li (b) Ca (c) O (d) P (e) Cl

Which element forms a 3+ ion that has the electron configuration [Kr]4d6?
Fe Co Pd Rh Answer: d

8.3 Covalent Bonding

Covalent Bonding In covalent bonds, atoms share electrons.
There are several electrostatic interactions in these bonds: attractions between electrons and nuclei, repulsions between electrons, and repulsions between nuclei. For a bond to form, the attractions must be greater than the repulsions.

Covalent Bonding What would happen to the magnitude of the attractions and repulsions represented in (a) if the nuclei were farther apart?

Ionizing an H2 molecule to H2+ changes the strength of the bond
Ionizing an H2 molecule to H2+ changes the strength of the bond. Based on the description of covalent bonding given previously, do you expect the H—H bond in H2+ to be weaker or stronger than the H—H bond in H2? Stronger, because a H—H covalent bond in H2+ has one less electron than in H2. Stronger, because a H—H covalent bond in H2+ has one more electron than in H2. Weaker, because a H—H covalent bond in H2+ has one less electron than in H2. Weaker, because a H—H covalent bond in H2+ has one more electron than in H2. Answer: c

Lewis Structures Sharing electrons to make covalent bonds can be demonstrated using Lewis structures. We start by trying to give each atom the same number of electrons as the nearest noble gas by sharing electrons. The simplest examples are for hydrogen, H2, and chlorine, Cl2, shown below.

Electrons on Lewis Structures
Lone pairs: electrons located on only one atom in a Lewis structure Bonding pairs: shared electrons in a Lewis structure; they can be represented by two dots or one line

Sample Exercise 8.3 Given the Lewis symbols for nitrogen and fluorine, predict the formula of the stable binary compound (a compound composed of two elements) formed when nitrogen reacts with fluorine and draw its Lewis structure.

Practice Exercise 1 Which of these molecules has the same number of shared electron pairs as unshared electron pairs? (a) HCl (b) H2S (c) PF3 (d) CCl2F2 (e) Br2

Practice Exercise 2 Compare the Lewis symbol for neon with the Lewis structure for methane, CH4. How many valence electrons are in each structure? How many bonding pairs and how many nonbonding pairs does each structure have?

Multiple Bonds Some atoms share only one pair of electrons. These bonds are called single bonds. Sometimes, two pairs need to be shared. These are called double bonds. There are even cases where three bonds are shared between two atoms. These are called triple bonds.

Single covalent bond Double covalent bond Triple covalent bond
The C—O bond length in carbon monoxide, CO, is 1.13 Å, whereas the C—O bond length in CO2 is 1.24 Å. Without drawing a Lewis structure, do you think that CO contains a single, double, or triple bond? Single covalent bond Double covalent bond Triple covalent bond Answer: c

8.4 Bond Polarity and Electronegativity

Polar Covalent Bonds The electrons in a covalent bond are not always shared equally. Fluorine pulls harder on the electrons it shares with hydrogen than hydrogen does. Therefore, the fluorine end of the molecule has more electron density than the hydrogen end.

Electronegativity Electronegativity is the ability of an atom in a molecule to attract electrons to itself. On the periodic table, electronegativity generally increases as you go from left to right across a period. from the bottom to the top of a group.

How does the electronegativity of an element differ from its electron affinity?
EN and EA are the same; they both measure the same characteristic. EA measures the energy released when an isolated atom/ion gains an electron; EN measures the ability of an atom in a molecule to attract electrons to itself. EN values of neutral atoms are just the negative of EA values of neutral atoms. EN measures the energy released when an isolated atom/ion gains an electron; EA measures the ability of an atom in a molecule to attract electrons to itself. Answer: b

For the group 6A elements, what is the trend in electronegativity with increasing atomic number?

Electronegativity and Polar Covalent Bonds
When two atoms share electrons unequally, a polar covalent bond results. Electrons tend to spend more time around the more electronegative atom. The result is a partial negative charge (not a complete transfer of charge). It is represented by δ–. The other atom is “more positive,” or δ+.

Polar Covalent Bonds The greater the difference in electronegativity, the more polar is the bond.

Based on differences in electronegativity, how would you characterize the bonding in sulfur dioxide, SO2? Do you expect the bonds between S and O to be nonpolar, polar covalent, or ionic? Nonpolar, because both S and O are in the same family. Polar covalent, because the difference in electronegativity values is 1.0. Ionic, because the difference in electronegativity values is –1.0. Ionic, because the difference in electronegativity values is >0.9. Answer: b

Sample Exercise 8.4 1) In each case, which bond is more polar? Indicate in each case which atom has the partial negative charge. A) B-Cl or C-Cl B) P-F or P-Cl 2) Which of the following bonds is the most polar: S-Cl, S-Br, Se-Cl, or Se-Br?

Practice Exercise 1 Which of the following bonds is the most polar? (a) H—F (b) H—I (c) Se—F (d) N—P (e) Ga—Cl

Dipoles When two equal, but opposite, charges are separated by a distance, a dipole forms. A dipole moment, , produced by two equal but opposite charges separated by a distance, r, is calculated:  = Qr It is measured in debyes (D).

Dipoles If the charged particles are moved closer together, does μ increase, decrease, or stay the same?

They have the same dipole moment.
Chlorine monofluoride, ClF, and iodine monofluoride, IF, are interhalogen compounds—compounds that contain bonds between different halogen elements. Which of these molecules has the larger dipole moment? IF ClF They have the same dipole moment. Neither has a dipole moment; they are both nonpolar. Answer: a

Dipole Moments Useful relationships: 1 Å = 10-10m
1 D = 3.34 x C-m Electronic charge, e, is equal to x C

Sample Exercise 8.5 The bond length in the HCl molecule is 1.27 Å.
A) Calculate the dipole moment that would result if the charges on the H and Cl atoms were +1 and -1, respectively. B) The experimentally measured dipole moment of HCl(g) is 1.08 D. What magnitude of charge, in units of e, on the H and Cl atoms would lead to this dipole moment?

Practice Exercise 1 Calculate the dipole moment for HF (bond length Å), assuming that the bond is completely ionic. (a) D (b) 1.91 D (c) 2.75 D (d) 4.39 D (e) 7.37 D

Practice Exercise 2 The dipole moment of chlorine monofluoride, ClF(g), is 0.88 D. The bond length of the molecule is 1.63 Å. A) Which atom is expected to have the partical negative charge? B) What is the charge of that atom, in units of e?

The bond between carbon and hydrogen is one of the most important types of bonds in chemistry. The length of an H—C bond is approximately 1.1 Å. Based on this distance and differences in electronegativity, do you expect the dipole moment of an individual H—C bond to be larger or smaller than that of an H—I bond? H—C has the same bond dipole moment as H—I, because the magnitude of Q and r in μ = Qr are about the same for both. H—C has the larger bond dipole moment, because Q and r are larger in H—C than in H—I. H—I has the smaller bond dipole moment, because Q is about the same for both, but r is larger in H—C than in H—I. H—I has the larger bond dipole moment, because Q is about the same for both, but r is larger in H—I than in H—C. Answer: d

How do you interpret the fact that there is no red in the HBr and HI representations?

Is a Compound Ionic or Covalent?
Simplest approach: Metal + nonmetal is ionic; nonmetal + nonmetal is covalent. There are many exceptions: It doesn’t take into account oxidation number of a metal (higher oxidation numbers can give covalent bonding). Electronegativity difference can be used; the table still doesn’t take into account oxidation number. Properties of compounds are often best: Lower melting points mean covalent bonding, for example.

You have a yellow solid that melts at 41 °C and boils at 131 °C and a green solid that melts at °C. If you are told that one of them is Cr2O3 and the other is OsO4, which one do you expect to be the yellow solid? OsO4 Cr2O3 Melting points are not useful in distinguishing between types of compounds. More information about types of bonds within structures is needed to answer this question. Answer: a

8.5 Drawing Lewis Structures

Writing Lewis Structures (Covalent Molecules) Sample Exercise 8.6
Sum the valence electrons from all atoms, taking into account overall charge. If it is an anion, add one electron for each negative charge. If it is a cation, subtract one electron for each positive charge. PCl3 Keep track of the electrons: (7) = 26

Writing Lewis Structures
Write the symbols for the atoms, show which atoms are attached to which, and connect them with a single bond (a line representing two electrons). Keep track of the electrons: 26 − 6 = 20

Complete the octets around all atoms bonded to the central atom. Keep track of the electrons: 26 − 6 = 20; 20 − 18 = 2

Place any leftover electrons on the central atom. Keep track of the electrons: 26 − 6 = 20; 20 − 18 = 2; 2 − 2 = 0

Practice Exercise 1 Which of these molecules has a Lewis structure with a central atom having no nonbonding electron pairs? (a) CO2 (b) H2S (c) PF3 (d) SiF4 (e) more than one of a, b, c, d

Practice Exercise 2 (a) How many valence electrons should appear in the Lewis structure for CH2Cl2? (b) Draw the Lewis structure.

Writing Lewis Structures Sample Exercise 8.7
5. If there are not enough electrons to give the central atom an octet, try multiple bonds.

Practice Exercise 1 Draw the Lewis structure(s) for the molecule with the chemical formula C2H3N, where the N is connected to only one other atom. How many double bonds are there in the correct Lewis structure? (a) Zero (b) One (c) Two (d) Three (e) Four

Practice Exercise 2 Draw the Lewis structure for: (a) NO+ ion (b) C2H4

Sample Exercise 8.8 Draw the Lewis structure for: BrO3– ClO2– PO43–

Practice Exercise 1 How many nonbonding electron pairs are there in the Lewis structure of the peroxide ion, O22–? (a) 7 (b) 6 (c) 5 (d) 4 (e) 3

Then assign formal charges. Formal charge is the charge an atom would have if all of the electrons in a covalent bond were shared equally. Formal charge = valence electrons – ½ (bonding electrons) – all nonbonding electrons

The dominant Lewis structure is the one in which atoms have formal charges closest to zero. puts a negative formal charge on the most electronegative atom.

The structure actually represents an ion.
Suppose a Lewis structure for a neutral fluorine-containing molecule results in a formal charge of +1 on the fluorine atom. What conclusion would you draw? The structure actually represents an ion. The F atom in the structure must have four covalent bonds attached to it. There must be another F atom in the structure carrying a –1 formal charge, since F is the most electronegative element and it should carry a negative formal charge. There must be a better Lewis structure, since F is the most electronegative element and it should carry a negative formal charge. Answer: d

Sample Exercise 8.9 Lewis Structures and Formal Charges
The following are three possible Lewis structures for the thiocyanate ion, NCS–: (a) Determine the formal charges of the atoms in each structure. (b) Which Lewis structure is the preferred one?

Practice Exercise 1 Phosphorus oxychloride has the chemical formula POCl3, with P as the central atom. To minimize formal charge, how many bonds does phosphorus make to the other atoms in the molecule? (a) 3 (b) 4 (c) 5 (d) 6 (e) 7

The cyanate ion (NCO–), like the thiocyanate ion, has three possible Lewis structures. (a) Draw these three Lewis structures, and assign formal charges to the atoms in each structure. (b) Which Lewis structure is dominant? Practice Exercise 2

8.6 Resonance Structures

What feature of this structure suggests that the two outer O atoms in are in some way equivalent to each other? Start here 6/11/10

The Best Lewis Structure?
Following our rules, this is the Lewis structure we would draw for ozone, O3. However, it doesn’t agree with what is observed in nature: Both O to O connections are the same. Start here 6/11/10

Resonance One Lewis structure cannot accurately depict a molecule like ozone. We use multiple structures, resonance structures, to describe the molecule.

Is the electron density consistent with equal contributions from the two resonance structures for O3? Explain.

The O—O bonds in ozone are often described as “one-and-a-half” bonds
The O—O bonds in ozone are often described as “one-and-a-half” bonds. Is this description consistent with the idea of resonance? Yes, because in each of the two resonance forms there is one O=O bond and one O—O bond, giving an overall average of 1.5 bonds per O—O bond. No, because a half bond cannot exist in a bonding situation. Answer: a

One-and-a-fifth bonds One-and-a-fourth bonds One-and-a-third bonds
In the same sense that we describe the O—O bonds in O3 as “one-and-a-half” bonds, how would you describe the N—O bonds in NO3–? One-and-a-fifth bonds One-and-a-fourth bonds One-and-a-third bonds One-and-a-half bonds Answer: c

Draw two equivalent resonance structures for the formate ion, HCO2–.
Sample Exercise 8.10 Resonance Structures Which is predicted to have the shorter sulfur–oxygen bonds, SO3or SO32–? Draw two equivalent resonance structures for the formate ion, HCO2–.

Practice Exercise 1 Which of these statements about resonance is true?
(a) When you draw resonance structures, it is permissible to alter the way atoms are connected. (b) The nitrate ion has one long N—O bond and two short N—O bonds. (c) “Resonance” refers to the idea that molecules are resonating rapidly between different bonding patterns. (d) The cyanide ion has only one dominant resonance structure. (e) All of the above are true.

What is the significance of the dashed bonds in this ball and stick model?

Resonance The organic compound benzene, C6H6, has two resonance structures. It is commonly depicted as a hexagon with a circle inside to signify the delocalized electrons in the ring. Localized electrons are specifically on one atom or shared between two atoms; Delocalized electrons are shared by multiple atoms.

Resonance In truth, the electrons that form the second C-O bond in the double bonds below do not always sit between that C and that O, but rather can move among the two oxygens and the carbon. They are not localized; they are delocalized. 86

Yes, because it has multiple Lewis resonance structures.
Each Lewis structure of benzene has three C=C double bonds. Another hydrocarbon containing three C=C double bonds is hexatriene, C6H8. A Lewis structure of hexatriene is Do you expect hexatriene to have multiple resonance structures? If not, why is this molecule different from benzene with respect to resonance? Yes, because it has multiple Lewis resonance structures. Yes, because it has three C=C bonds that can be moved throughout the structure. No, because the carbon chain is linear. No, because we cannot write other reasonable Lewis structures. Answer: d

8.7 Exceptions to the Octet Rule

Exceptions to the Octet Rule
There are three types of ions or molecules that do not follow the octet rule: ions or molecules with an odd number of electrons, ions or molecules with less than an octet, ions or molecules with more than eight valence electrons (an expanded octet).

Odd Number of Electrons
Though relatively rare and usually quite unstable and reactive, there are ions and molecules with an odd number of electrons.

Which of the Lewis structures for NO is dominant based on analysis of the formal charges?
The first NO structure, because all atoms have zero formal charge. The first NO structure, because N and O have equal but opposite formal charges. The second NO structure, because all atoms have zero formal charge. The second NO structure, because N should not have a positive formal charge. Answer: a

Fewer Than Eight Electrons
Elements in the second period before carbon can make stable compounds with fewer than eight electrons. Consider BF3: Giving boron a filled octet places a negative charge on the boron and a positive charge on fluorine. This would not be an accurate picture of the distribution of electrons in BF3.

Fewer Than Eight Electrons
The lesson is: If filling the octet of the central atom results in a negative charge on the central atom and a positive charge on the more electronegative outer atom, don’t fill the octet of the central atom.

More Than Eight Electrons
When an element is in period 3 or below in the periodic table (e.g., periods 3, 4, 5, etc.), it can use d-orbitals to make more than four bonds. Examples: PF5 and phosphate below (Note: Phosphate will actually have four resonance structures with five bonds on the P atom!)

Sample Exercise 8.11 1) Draw the Lewis structure for ICl4-
2) Which of the following atoms is never found with more than an octet of valence electrons around it: S, C, P, or Br? 3) Draw the Lewis structure for XeF2.

Practice Exercise 1 In which of these molecules or ions is there only one lone pair of electrons on the central sulfur atom? (a) SF4 (b) SF6 (c) SOF4 (d) SF2 (e) SO42–

Even though we can draw a Lewis structure for the phosphate ion that has only 8 electrons around the central phosphorus, the better structure puts a double bond between the phosphorus and one of the oxygens. 97

This eliminates the charge on the phosphorus and the charge on one of the oxygen atoms. The lesson is: when the central atom in on the 3rd row or below and expanding its octet eliminates some formal charges, do so. 98

8.8 Strengths and Lengths of Covalent Bonds

Covalent Bond Strength
Most simply, the strength of a bond is measured by determining how much energy is required to break the bond. This is called the bond enthalpy. The bond enthalpy for a Cl—Cl bond, D(Cl— Cl), is measured to be 242 kJ/mol. We write out reactions for breaking one mole of those bonds: Cl—Cl → 2 Cl•

The enthalpy of atomization – 6 [D(C—H)] = a good estimate of D(C—C).
How can you use the enthalpy of atomization of the hydrocarbon ethane, C2H6(g), along with the value D(C—H) = 413 kJ/mol to estimate the value for D(C—C)? The enthalpy of atomization l 7 bonds broken = a good estimate of D(C—C). The enthalpy of atomization – 6 [D(C—H)] = a good estimate of D(C—C). The enthalpy of atomization + 6 [D(C—H)] = a good estimate of D(C—C). The enthalpy of atomization l 7 bonds broken – 6[D(C—H)] = a good estimate of D(C—C). Answer: b

Average Bond Enthalpies
Average bond enthalpies are positive, because bond breaking is an endothermic process. Note that these are averages over many different compounds; not every bond in nature for a pair of atoms has exactly the same bond energy.

Based on bond enthalpies, which do you expect to be more reactive, oxygen, O2, or hydrogen peroxide, H2O2? O2 is more reactive, because the O=O bond enthalpy is less than that of the O—O bond enthalpy in hydrogen peroxide. O2 is more reactive, because the O=O bond enthalpy is greater than that of the O—O bond enthalpy in hydrogen peroxide. H2O2 is more reactive, because the O—O bond enthalpy is less than that of the O=O bond enthalpy in O2. H2O2 is more reactive, because the O—O bond enthalpy is greater than that of the O=O bond enthalpy in O2. Answer: c

Using Bond Enthalpies to Estimate Enthalpy of Reaction
One way to estimate H for a reaction is to use the bond enthalpies of bonds broken and the new bonds formed. Energy is added to break bonds and released when making bonds. In other words, Hrxn = (bond enthalpies of all bonds broken) − (bond enthalpies of all bonds formed).

Is this reaction endothermic or exothermic?

Example From the figure on the last slide
CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g)

Sample Exercise 8.12 Using Average Bond Enthalpies
Using Table 8.4, estimate ΔH for the following reaction (where we explicitly show the bonds involved in the reactants and products): © 2009, Prentice-Hall, Inc. 107

Practice Exercise 1 Using Table 8.4, estimate ΔH for the “water splitting reaction”: H2O(g) → H2(g) + O2(g). (a) 242 kJ (b) 417 kJ (c) 5 kJ (d) –5 kJ (e) –468 kJ

Bond Enthalpy and Bond Length
We can also measure an average bond length for different bond types. As the number of bonds between two atoms increases, the bond length decreases.

Predict the N-N bond enthalpy for an N-N bond that has resonance forms that include equal contributions from single and double N-N bonds.

Integrative Exercise Phosgene, a substance used in poisonous gas warfare in World War I, is so named because it was first prepared by the action of sunlight on a mixture of carbon monoxide and chlorine gases. Phosgene is 12.14% C, 16.17% O, and 71.69% Cl by mass. Its molar mass is 98.9 g/mol. A) Determine the molecular formula B) Draw three Lewis structures for the molecule (C is the central atom) C) Using formal charges determine which Lewis structure is the most important one D) Using average bond enthalpies, estimate ΔH for the formation of gaseous phosgene from CO(g) and Cl2(g)

Chapter 8 Basic Concepts of Chemical Bonding

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