Chapter 8 Concepts of Chemical Bonding

Chapter 8 Concepts of Chemical Bonding

Lecture Presentation Chapter 8 Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc. Subhash C. Goel South GA State College Douglas, GA 31533 Chemical Bonds Three basic types of bonds Ionic Electrostatic attraction between ions.

Covalent Sharing of electrons. Metallic Metal atoms bonded to several other atoms. Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc. 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). Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc. Ionic Formation Atoms tend to lose (metals) or gain (nonmetals) electrons to make them isoelectronic to the noble gases. Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc.

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). 2015 Pearson Education, Inc.

Basic Concepts of Chemical Bonding 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 Basic Concepts

elements exothermic. of Chemical Bonding 2015 Pearson Education, Inc. 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 Coulombs law: Q1Q2 Eel =

d Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc. Lattice Energy Lattice energy increases with: increasing charge on the ions decreasing size of ions Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc.

Exercise 1 Magnitudes of Lattice Energies Without consulting Table 8.2, arrange the ionic compounds NaF, CsI, and CaO in order of increasing lattice energy. Solution Solve NaF consists of Na+ and F ions, CsI of Cs+ and I ions, and CaO of Ca2+ and O2 ions. Because the product Q1Q2 appears in the numerator of Equation 8.4, the lattice energy increases dramatically when the charges increase. Thus, we expect the lattice energy of CaO, which has 2+ and 2 ions, to be the greatest of the three. The ionic charges are the same in NaF and CsI. As a result, the difference in their lattice energies depends on the difference in the distance between ions in the lattice. Because ionic size increases as we go down a group in the periodic table we know that Cs+ is larger than Na+ and I is larger than F. Therefore, the distance between Na+ and F ions in NaF is less than the distance between the Cs+ and I ions in CsI. As a result, the lattice energy of NaF should be greater than that of CsI. In order of increasing

energy, therefore, we have CsI < NaF < CaO. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. 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. Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc. 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. Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc. 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

Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc. Exercise 2 Lewis Structure of a Compound Given the Lewis symbols for nitrogen and fluorine in Table 8.1, 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. Solve Nitrogen must share a pair of electrons with three fluorine atoms to complete its octet. Thus, the binary compound these two elements form must be NF3: Chemistry, The Central Science, 12th Edition

Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc. 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. Basic Concepts of Chemical Bonding

2015 Pearson Education, Inc. 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. Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc. 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. Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc. 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 . The other atom is more positive, or +. Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc. Polar Covalent Bonds The greater the difference in electronegativity, the more polar is

the bond. Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc. Exercise 3 Bond Polarity In each case, which bond is more polar: (a) BCl or CCl, (b) PF or PCl ? Indicate in each case which atom has the partial negative charge. Solution (a) The chlorine atom is common to both bonds. Therefore, the analysis reduces to a comparison of the electronegativities of B and C. Because boron is to the left of carbon in the periodic table, we predict that boron has the lower electronegativity.

Chlorine, being on the right side of the table, has a higher electronegativity. The more polar bond will be the one between the atoms having the lowest electronegativity (boron) and the highest electronegativity (chlorine). Consequently, the BCl bond is more polar; the chlorine atom carries the partial negative charge because it has a higher electronegativity. (b) In this example phosphorus is common to both bonds, and the analysis reduces to a comparison of the electronegativities of F and Cl. Because fluorine is above chlorine in the periodic table, it should be more electronegative and will form the more polar bond with P. The higher electronegativity of fluorine means that it will carry the partial negative charge. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc.

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) 1 D = 3.34x10-30 C.m 2015 Pearson Education, Inc. Basic Concepts

of Chemical Bonding Exercise 4 Dipole Moments of Diatomic Molecules The bond length in the HCl molecule is 1.27 . (a) Calculate the dipole moment, in debyes, that results if the charges on the H and Cl atoms were and 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 leads to this dipole moment? Solution (a) The charge on each atom is the electronic charge, e = 1.60 1019 C. The separation is 1.27 . The dipole moment is therefore (b) We know the value of , 1.08 D, and the value of r, 1.27 . We want to calculate the value

of Q: We can readily convert this charge to units of e: Basic Concepts of Chemical Bonding 2012 Pearson Education, Inc. Exercise 4 Dipole Moments of Diatomic Molecules Continued Thus, the experimental dipole moment indicates that the charge separation in the HCl molecule is Because the experimental dipole moment is less than that calculated in part (a), the charges on the atoms are much less than a full electronic charge. We could have anticipated this because the HCl bond is polar covalent rather than ionic.

Basic Concepts of Chemical Bonding 2012 Pearson Education, Inc. Is a Compound Ionic or Covalent? Simplest approach: Metal + nonmetal is ionic; nonmetal + nonmetal is covalent. There are many exceptions: It doesnt take into account oxidation number of a metal (higher oxidation numbers can give covalent bonding). Electronegativity difference can be used; the table still doesnt take into account oxidation number. Properties of compounds are often best: Lower melting points mean covalent bonding, for

example. Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc. Writing Lewis Structures (Covalent Molecules) PCl3 1. Sum the valence electrons from all atoms, taking into account overall charge.

Keep track of the electrons: 5 + 3(7) = 26 2015 Pearson Education, Inc. If it is an anion, add one electron for each negative charge. If it is a cation, subtract one electron Basic Concepts Chemical for each positive ofBonding charge. Writing Lewis Structures

2. 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 2015 Pearson Education, Inc. Basic Concepts of Chemical Bonding

Writing Lewis Structures 3. Complete the octets around all atoms bonded to the central atom. Keep track of the electrons: 26 6 = 20; 20 18 = 2 2015 Pearson Education, Inc. Basic Concepts of Chemical Bonding Writing Lewis Structures 4. Place any leftover

electrons on the central atom. Keep track of the electrons: 26 6 = 20; 20 18 = 2; 2 2 = 0 2015 Pearson Education, Inc. Basic Concepts of Chemical Bonding Writing Lewis Structures 5. If there are not enough electrons to give the central atom an octet, try multiple bonds. Basic Concepts

of Chemical Bonding 2015 Pearson Education, Inc. Writing Lewis Structures 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 (bonding electrons) all nonbonding electrons Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc.

Writing Lewis Structures 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. Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc. Exercise 5 a) How many valence electrons should appear in the Lewis structure for CH2Cl2?

(b) Draw the Lewis structure. Answers: (a) 20 (b) Basic Concepts of Chemical Bonding 2012 Pearson Education, Inc. Exercise 6 Lewis Structure for a Polyatomic Ion Draw the Lewis structure for the BrO3 ion. Solution

Bromine (group 7A) has seven valence electrons, and oxygen (group 6A) has six. We must add one more electron to our sum to account for the 1 charge of the ion. The total number of valence electrons is, therefore, 7 + (3 6) + 1 = 26. For oxyanions BrO3, SO42, NO3, CO32, and so forththe oxygen atoms surround the central nonmetal atom. After following this format and then putting in the single bonds and distributing the unshared electron pairs, we have Notice that the Lewis structure for an ion is written in brackets and the charge is shown outside the brackets at the upper right. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc.

Exercise 7 Lewis Structures and Formal Charges Three possible Lewis structures for the thiocyanate ion, NCS , are (a) Determine the formal charges in each structure. (b) Based on the formal charges, which Lewis structure is the dominant one? Solution (a) Neutral N, C, and S atoms have five, four, and six valence electrons, respectively. We can determine the formal charges in the three structures by using the rules we just discussed: As they must, the formal charges in all three structures sum to 1, the overall charge of the ion. (b) The dominant Lewis structure generally produces formal charges of the smallest magnitude (guideline 1). That rules out the left structure as the dominant one. Further, as discussed in Section 8.4, N is more electronegative than C or S. Therefore, we expect any negative formal charge to reside on the N atom

(guideline 2). For these two reasons, the middle Lewis structure is the dominant one for NCS. Basic Concepts of Chemical Bonding 2012 Pearson Education, Inc. The Best Lewis Structure? Following our rules, this is the Lewis structure we would draw for ozone, O3. However, it doesnt agree with what is observed in nature: Both O to O

connections are the same. 2015 Pearson Education, Inc. Basic Concepts of Chemical Bonding Resonance One Lewis structure cannot accurately depict a molecule like ozone. We use multiple structures, resonance structures, to describe

the molecule. Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc. 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. 2015 Pearson Education, Inc. Basic Concepts of Chemical Bonding 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). Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc. Odd Number of Electrons Though relatively rare and usually quite unstable and reactive, there are ions and molecules with an odd number of electrons. Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc.

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. Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc.

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, dont fill the octet of the central atom. Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc. 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!) Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc. 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 ClCl bond, D(Cl Cl), is measured to be 242 kJ/mol.

We write out reactions for breaking one mole of those bonds: ClCl 2 Cl Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc. 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. 2015 Pearson Education, Inc. Basic Concepts of Chemical Bonding 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). 2015 Pearson Education, Inc. Basic Concepts of Chemical Bonding Example From the figure on the last slide

CH4(g) + Cl2(g) CH3Cl(g) + HCl(g) In this example, one CH bond and one Cl Cl bond are broken; one CCl and one HCl bond are formed. Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc. Answer H = [D(CH) + D(ClCl)] [D(CCl) + D(HCl)] = [(413 kJ) + (242 kJ)] [(328 kJ) + (431 kJ)] = (655 kJ) (759 kJ) = 104 kJ

Basic Concepts of Chemical Bonding 2015 Pearson Education, Inc. Exercise 8 Using Average Bond Enthalpies Using data from Table 8.4, estimate H for the reaction Solution Analyze We are asked to estimate the enthalpy change for a chemical reaction by using average bond enthalpies for the bonds broken and formed. Plan In the reactants, we must break twelve CH bonds and two CC bonds in the two molecules of C2H6 and seven O2 bonds in the seven O2 molecules. In the products, we form eight C==O bonds (two in each CO 2) and twelve OH bonds (two in each H2O). Solve Using Equation 8.12 and data from Table 8.4, we have

H = [12D(CH) + 2D(CC) + 7D(O2)] [8D(C==O) + 12D(OH)] = [12(413 kJ) + 2(348 kJ) + 7(495 kJ)] [8(799 kJ) + 12(463 kJ)] = 9117 kJ 11948 k J = 2831 kJ Basic Concepts of Chemical Bonding 2012 Pearson Education, Inc. 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. Basic Concepts

of Chemical Bonding 2015 Pearson Education, Inc.

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