Chapter 8 Concepts of Chemical Bonding

1. Chapter 8Concepts of Chemical Bonding

2. 8.1 Chemical Bonds, Lewis Symbols, Octet Rule • A chemical bond is a strong attractive force that exists between atoms • Three basic types of bonds: • Ionic • Electrostatic attraction between ions • Ex: Mg2+ + O2- MgO • Covalent • Sharing of electrons between atoms • Ex: Br2, S8, C12H22O11 • Metallic • Metal atoms bonded to several neighboring atoms

3. Lewis Symbols • The electrons involved in chemical bonding are the valence electrons, which usually reside in the outermost occupied shell of an atom. • A Lewis symbol is a simple way of showing these electrons. • Ex: Sulfur [Ne]3s23p4 • See Table 8.1 Octet Rule • Atoms gain, lose or share electrons until they are surrounded by eight valence electrons. • An octet consists of full s and p subshells in an atom. • An octet is a very stable configuration of electrons.

4. 8.2 Ionic Bonding • The chemical reaction between Na(s) and Cl2(g) to produce NaCl(s) is very exothermic: [8.1] • Formation indicates an electron has been lost by sodium and gained by chlorine: • NaCl is a typical ionic compound because it contains a metal of low ionization energy and a nonmetal with high electron affinity. • Ionization energy indicates how easily an electron can be lost by one atom, and electron affinity measures the ease with which another atom gains an electron.

5. Energetics of Ionic Bonding • As we saw in the last chapter, it takes 495 kJ/mol to remove electrons from sodium. • We get 349 kJ/mol back by giving electrons to chlorine. • If the transfer of an electron from Na(g) to Cl(g) were the only factor in forming an ionic bond, the process would rarely be exothermic. 496 -349 = 146 kJ/mol • The positive energy change indicates the ions are not interacting with each other.

6. Yet, we see the reaction as very exothermic. Both light and heat are given off by the reaction.

7. The principal reason that ionic compounds are stable is the electrostatic attraction between ions of opposite charge. • This attraction draws ions together, releasing energy , and causing the ions to form a stable, solid array, or lattice. • A measure of the stabilization that occurs is given by the lattice energy. • Lattice energy is the energy required to completely separate a mole of a solid ionic compound into its gaseous ions.

8. For NaCl, ΔHlattice= +788 kJ/mol • This process is highly endothermic. The reverse process – the formation of NaCl – is highly exothermic: ΔH = -788 kJ/mol • The energy released by the attraction of ions of unlike charge more than makes up for the endothermic nature of ionization energies, making the formation of ionic compounds an exothermic process.

9. Q1Q2 d Eel =  • The energy associated with electrostatic interactions is governed by Coulomb’s law: • Lattice energy, then, increases with the charge on the ions. • It also increases with decreasing ion size.

10. By accounting for all three energies (ionization energy, electron affinity, and lattice energy), we can get a good idea of the energetics involved in such a process.

11. Electron Configurations of Ions of the Representative Elements • The energetics of ionic bond formation explains why so many representative elements tend to have noble-gas electron configurations. • Na, for example, loses its one valence electron to form Na+, which has the noble gas configuration of Ne. • We never find Na2+ ions because the second electron removed would have to come from the inner shell of the Na atom, which would require a large amount if energy. • The increase in lattice energy is not enough to compensate for the energy needed to remove an inner shell electron.

12. Similarly, the addition of electrons to nonmetals is exothermic or slightly • endothermic as long as the electrons are being added to the valence shell. • The Cl atom easily adds an electron to form Cl-, which has the electron configuration of Ar. • To form the Cl-2 ion, the second electron would have to be added to the next higher shell of the Cl atom, which is energetically unfavorable.

13. Transition Metal Ions • The lattice energies of ionic compounds are generally large enough to compensate for the loss of only up to 3 electrons from atoms. • Most transition metals have more than three electrons beyond a noble gas core. • Silver, for example, has a [Kr]4d105s1 electron configuration. In forming Ag+, the 5s electron is lost leaving a full 4d subshell. • In forming ions, transition metals lose the valence shell s electrons first, then as many d electrons as necessary to reach the charge on the ion. • Ex: Fe2+ and Fe3+

14. 8.3 Covalent Bonding • In these bonds atoms share electrons. • There are several electrostatic interactions in these bonds: • Attractions between electrons and nuclei • Repulsions between electrons • Repulsions between nuclei

15. Lewis Structures Lewis structures are representations of molecules showing all electrons, bonding and nonbonding.

16. Multiple Bonds • Single bonds share one pair of electrons. • Octets can also be obtained by sharing two pairs of electrons to form a double bond, or three pairs of electrons to form a triple bond. • As a general rule, the distance between bonded atoms decreases as the number of shared electron pairs increases

17. 8.4 Bond Polarity and Electronegativity • The concept of bond polarity helps describe the sharing of electrons between atoms. • A nonpolar bond is one in which the electrons are shared equally between two atoms, as in Cl2 and N2. • In a polar covalent bond, one of the atoms exerts a greater attraction for the bonding electrons than the other. • If the difference in relative ability to attract electrons is large enough, an ionic bond is formed.

18. Electronegativity • The ability of an atom in a molecule to attract electrons to itself. • The greater an atom’s electronegativity, the greater its ability to attract electrons to itself. • Electronegativity is related to ionization energy and electron affinity. An atom with very negative electron affinity and a high ionization energy will both attract electrons from other atoms and resist having its electrons attracted away. • On the periodic table, electronegativity increases as you go from left to right across a row and from the bottom to the top of a column.

19. Electronegativity and Bond Polarity • When electrons are shared equally between two atoms, the bond is nonpolar, as in diatomic molecules. • Although atoms often form compounds by sharing electrons, the electrons are not always shared equally. • In HF, for example, fluorine has a higher electronegativity and 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. • We can represent the charge distribution as • The δ+ and δ- (read “delta plus”, delta minus) symbolize the partial charges.

20. Dipole Moments • When two atoms share electrons unequally, a bond dipole results. • The dipole moment, , produced by two equal but opposite charges separated by a distance, r, is calculated:  = Qr • It is measured in debyes (D).

21. Bond Length and Dipole Moments • The greater the difference in electronegativity, the more polar is the bond. • The bond length increases as the electronegativity differences get smaller.

22. Bond Types and Nomenclature • Ionic compounds (metal + nonmetal) are given names based on their component ions, to include the charge on the cation if that is variable (as in transition metals). • Molecular compounds use prefixes to indicate the number of atoms present. • The dividing line between the two ways to name compounds is not always clear. Ex: TiO2 is called titanium(IV) oxide but more commonly titanium dioxide. • Many compounds of metals with high oxidation states (usually above 3+) have properties more similar to covalent compound.

23. 8.5 Drawing Lewis Structures • Lewis structures are representations of covalent bonding in molecules. Though simple, they are useful for better understanding of the bonds between atoms in many molecules. • Follow the procedures below. PCl3 • Find the sum of valence electrons of all atoms in the polyatomic ion or molecule. • If it is an anion, add one electron for each negative charge. • If it is a cation, subtract one electron for each positive charge. 5 + 3(7) = 26

24. The central atom is the least electronegative element that isn’t hydrogen. Connect the outer atoms to it by single bonds of shared electrons. Each single bond contains two electrons. Keep track of the electrons: 26  6 = 20

25. 3. Complete the octets of the outer atoms with pairs of unshared electrons. 4. Place any remaining electrons around the central atom even though it violates the octet rule. The octet rule does not always apply for certain atoms. Keep track of the electrons: 26  6 = 20  18 = 2  2 = 0

26. If you run out of electrons before the central atom has an octet, form multiple bonds by taking unshared electrons and making them shared (bonding) electrons. Remember, shared electrons count towards fulfilling the octet for both bonded atoms. In the above structure, both C and N share the electrons in the triple bond between them to achieve an octet of electrons. The single pair between C and H satisfies the requirement for H to have a duet of electrons. Also remember that H can only have a duet (one shared pair) of electrons. It can never have more than one shared bond and can, therefore, never be a central atom.

27. Lewis Structures for Polyatomic Ions • Add one electron for each negative charge • Subtract one electron for each positive charge • Draw the structure following the rules for neutral compounds. • Put the Lewis structure in brackets with the charge outside the brackets at the upper right. • See sample exercise 8.8 and the practice exercise

28. Formal Charges • Sometimes, it is possible to draw two or more Lewis structures for a particular molecule or ion. • Calculating formal charges for each atom in a molecule allows you to determine which structure is the most reasonable. • The formal charge is the number of valence electrons in an isolated atom minus the number of electrons assigned to it in a Lewis structure. • The formal charge for an atom in a molecule is the charge the atom would have if all the atoms in the molecule had the same electronegativity (i.e., they were equally shared). • Formal charges do not represent real charges on atoms. They are “bookkeeping” conventions.

29. Assigning formal charges • For each atom, count the electrons in lone pairs and halve the electrons it shares with other atoms. • Subtract that from the number of valence electrons for that atom: The difference is its formal charge.

30. The best Lewis structure… • …is the one with the fewest charges. • …puts a negative charge on the most electronegative atom. • Formal charge is a tool for estimating the distribution of electric charge within a molecule. • The concept of oxidation states constitutes a competing method to assess the distribution of electrons in molecules. • If the formal charges and oxidation states of the atoms in carbon are compared, the following values are arrived at:

31. Neither oxidation number nor formal charge gives an accurate description of actual charges on atoms. • Oxidation numbers assign all shared electrons to the most electronegative atom, whereas formal charges divides all shared electrons equally between atoms. • The images below shows the difference. The image on the far right shows the calculated distribution of electrons in HCl. Regions of more negative charge are red.

32. 8.6 Resonance Structures • Sometimes the experimentally determined arrangement of atoms is not adequately described by a single Lewis structure • This is the Lewis structure we would draw for ozone, O3. • We would expect the double bond to be shorter than the single bond. • The negative charge is assigned to one of the oxygens. + -

33. But this is at odds with the true, observed structure of ozone, in which… • …both O—O bonds are the same length. • …both outer oxygens have a charge of 1/2.

34. One Lewis structure cannot accurately depict a molecule such as ozone. • The double bond could just as easily been put between the other O=O bond. • We use multiple structures, called resonance structures, to describe the molecule.

35. Just as green is a synthesis of blue and yellow, ozone is a synthesis of these two resonance structures. • The real molecule is described as an average, or blend, of the two Lewis structures. • Ozone cannot be depicted by a single Lewis structure in which the electrons are locked into position. • A single Lewis structure would have to show the electrons not localized, as shown to the left, but delocalized, i.e., spread over the two bonds.

36. Consider the Lewis structures for the formate ion, below • 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, but rather are delocalized.

37. 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.

38. 8.7 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).

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

40. Less than an Octet of Valence 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.

41. Therefore, structures that put a double bond between boron and fluorine are much less important than the one that leaves boron with only 6 valence electrons.

42. 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.

43. More than an Octet of Valence Electrons • The only way PCl5 can exist is if phosphorus has 10 electrons around it. • It is allowed to expand the octet of atoms on the 3rd row or below. • Presumably d orbitals in these atoms participate in bonding.

44. Sometimes an expanded valence shell is shown even if a structure can be drawn with an octet. • Consider the case with the PO43- ion. • Even though we can draw a Lewis structure for the phosphate ion that has only 8 electrons around the central phosphorus, formal charges show that the better structure puts a double bond between the phosphorus and one of the oxygens, thus eliminating the charge on the phosphorus and one of the oxygens. • The best representation is a series of Lewis structures in resonance.

45. 8.8 Covalent Bond Strength • The strength of a bond is measured by determining how much energy is required to break the bond. • The bond enthalpy is the enthalpy change, ΔH, for the breaking of a particular bond in one mole of a gaseous substance. • The bond enthalpy for a Cl—Cl bond, D(Cl—Cl), is measured to be 242 kJ/mol. • This is the enthalpy change when 1 mol of Cl2 is dissociated into Cl atoms.

46. Average Bond Enthalpies • This table lists the average bond enthalpies for many different types of bonds. • Average bond enthalpies are positive, because bond breaking is an endothermic process. • These are average bond enthalpies, not absolute bond enthalpies. • The C—H bonds in methane, CH4, will be a bit different than the C—H bond in chloroform, CHCl3.

47. Enthalpies of Reaction • Yet another way to estimate H for a reaction is to compare the bond enthalpies of bonds broken to the bond enthalpies of the new bonds formed. • In other words, Hrxn = (bond enthalpies of bonds broken)  (bond enthalpies of bonds formed)

48. CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g) • In this example, one C—H bond and one Cl—Cl bond are broken. • One C—Cl and one H—Cl bond are formed.

49. So, Hrxn = [D(C—H) + D(Cl—Cl)  [D(C—Cl) + D(H—Cl) = [(413 kJ) + (242 kJ)]  [(328 kJ) + (431 kJ)] = (655 kJ)  (759 kJ) = 104 kJ

50. 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.