Chapter 9 Chemical Bonding I lewis Theory

Chapter 9 Chemical Bonding I lewis Theory

Ionic Bonding Edward Wen Copyright 2011 Pearson Education, Inc. Bonding Theories Explain how and why atoms attach together to form molecules Explain why some combinations of atoms are stable and others are not why is water H2O, not HO or H3O Can be used to predict the shapes of molecules Can be used to predict the chemical and physical properties of compounds Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 2 Copyright 2011 Pearson Education, Inc.

Lewis Bonding Theory One of the simplest bonding theories is called Lewis Theory Lewis Theory emphasizes valence electrons to explain bonding Using Lewis theory, we can draw models called Lewis structures aka Electron Dot Structures Lewis structures allow us to predict many properties of molecules such as molecular stability, shape, size, polarity Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 3 G.N. Lewis (1875-1946) Copyright 2011 Pearson Education, Inc.

Why Do Atoms Bond? Chemical bonds form because they lower the potential energy between the ions or atoms The potential energy of the bonded atoms is less than the potential energy of the separate atoms To calculate this potential energy, you need to consider the following interactions: nucleustonucleus repulsions electrontoelectron repulsions nucleustoelectron attractions Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 4 Copyright 2011 Pearson Education, Inc. Types of Bonds

We can classify bonds based on the kinds of atoms that are bonded together Types of Atoms metals to nonmetals nonmetals to nonmetals metals to metals Tro: Chemistry: Edward Wen A Molecular Approach, 2/e Type of Bond Ionic Bond Characteristic electrons transferred Covalent electrons

shared Metallic electrons pooled 5 Copyright 2011 Pearson Education, Inc. Types of Bonding Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 6 Copyright 2011 Pearson Education, Inc. Ionic Bonds When a metal atom loses electrons it becomes a cation metals have low ionization energy (low Zeff), making it relatively easy to remove electrons from them

When a nonmetal atom gains electrons it becomes an anion nonmetals have high electron affinities (high Zeff), making it advantageous to add electrons to these atoms The oppositely charged ions are then attracted to each other, resulting in an ionic bond Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 7 Copyright 2011 Pearson Education, Inc. Valence Electrons & Bonding Because valence electrons are held most loosely, and Because chemical bonding involves the transfer or sharing of electrons between two or more atoms, Valence electrons are most important in bonding

Lewis theory focuses on the behavior of the valence electrons Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 8 Copyright 2011 Pearson Education, Inc. Stable Electron Arrangements and Ion Charge Metals form cations by losing enough electrons to get the same electron configuration as the previous noble gas Nonmetals form anions by gaining enough electrons to get the same electron configuration as the next

noble gas The noble gas electron configuration must be very stable Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 9 Copyright 2011 Pearson Education, Inc. Lewis Bonding Theory Atoms bond because it results in a more stable electron configuration. more stable = lower potential energy Atoms bond together by either transferring or sharing electrons Usually this results in all atoms obtaining an outer shell with eight electrons octet rule

there are some exceptions to this rulethe key to remember is to try to get an electron configuration like a noble gas Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 10 Copyright 2011 Pearson Education, Inc. Octet Rule When atoms bond, they tend to gain, lose, or share electrons to result in eight valence electrons ns2np6 noble gas configuration Many exceptions H, Li, Be, B attain an electron configuration like He He = two valence electrons, a duet Li loses its one valence electron H shares or gains one electron

o though it commonly loses its one electron to become H+ Be loses two electrons to become Be2+ o commonly shares its two electrons in covalent bonds (4 valence electrons) B loses three electrons to become B3+ o commonly shares its three electrons in covalent bonds (6 valence electrons) expanded octets for elements in Period 3 or below using empty valence d orbitals Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 11 Copyright 2011 Pearson Education, Inc. Lewis Theory Predictions for Ionic Bonding The number of electrons a metal atom should lose or a nonmetal atom should gain in order to attain a stable electron arrangement

the octet rule This allows us to predict the formulas of ionic compounds that result also allows us to predict the relative strengths of the resulting ionic bonds from Coulombs Law q q F k r2 Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 12 Copyright 2011 Pearson Education, Inc. Predicting Ionic Formulas Using Lewis Symbols Electrons are transferred until the metal loses all its valence electrons and the nonmetal has an octet Numbers of atoms are adjusted so the electron transfer comes out even

Li2O Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 13 Copyright 2011 Pearson Education, Inc. Energetics of Ionic Bond Formation The ionization energy of the metal is endothermic Na(s) Na+(g) + 1 e H = +496 kJ/mol The electron affinity of the nonmetal is exothermic Cl2(g) + 1 e Cl(g) H = 244 kJ/mol Generally, the ionization energy of the metal is larger than the electron affinity of the nonmetal, therefore the formation of the ionic compound should be endothermic (H > 0, unfavorable) But the heat of formation of most ionic compounds is exothermic and generally large. Why? Na(s) + Cl2(g) NaCl(s) Hf = 411 kJ/mol Tro: Chemistry: Edward

Wen A Molecular Approach, 2/e 14 Copyright 2011 Pearson Education, Inc. Ionic Bonding & the Crystal Lattice Crystal lattice: The extra energy is released from the formation of a structure in which every cation is surrounded by anions, and vice versa Formed from the electrostatic attraction of the cations for all the surrounding anions The crystal lattice maximizes the attractions between cations and anions, forming stable arrangement Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 15 Copyright 2011 Pearson Education, Inc.

Crystal Lattice Electrostatic attraction is nondirectional!! no direct anioncation pair Ionic compound does not have molecule the chemical formula is an empirical formula, simply giving the ratio of ions based on charge balance Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 16 Copyright 2011 Pearson Education, Inc. Lattice Energy Lattice energy: Energy required to dissociate the solid crystal into separate ions in the gas state: NaCl(s) Na+(g) + Cl-(g) Hlattice

always largely endothermic Indicates the extra stability that accompanies the formation of the crystal lattice (the reverse of the above equation) hard to measure directly, but can be calculated from knowledge of other processes Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 17 Copyright 2011 Pearson Education, Inc. Determining Lattice Energy The BornHaber Cycle is devised to determine the lattice energy. Hess law! Reactions involved: M(?) M(g) Vaporization of metal: *Formation of nonmetal atom: X(?) X(g) **Ionization of metal atom: M(g) M+(g) + ne **Electron affinity of nonmetal: X(g) + e- X-(g)

Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 18 Copyright 2011 Pearson Education, Inc. The Born-Haber cycle for lithium fluoride. Edward Wen 19 Copyright 2011 Pearson Education, Inc. Example: Use the Born-Haber cycle to find lattice energy for magnesium oxide Given: Electron Affinity Ionization (kJ/mol) energy (kJ/mol) Sublimation

energy (kJ/mol) Hf (kJ/mol) O -141, 844 Mg 148 MgO (-601) O(g) (249.4) Mg 738, 1451 Find: ________ _____ + _____Hlattice = ? First write equations for the above processes: _______________________ H1 = -141 kJ _______________________ H2 = 844 kJ _______________________ H3 = 738 kJ _______________________ H4 = 1451 kJ _______________________ H5 = 148 kJ

_______________________ H6 = -601 kJ _______________________ H7 = 249.4 kJ Edward Wen 20 Copyright 2011 Pearson Education, Inc. O(g) + e- O-(g) H1 = -141 kJ O- (g) + e- O2-(g) H2 = 844 kJ Mg(g) Mg+(g) + e- H3 = 738 kJ Mg+(g) Mg2+(g) + e- H4 = 1451 kJ

Mg(s) Mg(g) H5 = 148 kJ Mg(s) + O2 (g) MgO(s) O2(g) O(g) find: H6 = -601 kJ H7 = 249.4 kJ MgO(s) O2-(g) + Mg2+(g) Hlattice = ? Hlattice = 3890 kJ Edward Wen 21 Copyright 2011 Pearson Education, Inc. q q F k

r2 Trends in Lattice Energy Ion Size The force of attraction between charged particles is inversely proportional to the distance between them Larger ions mean the center of positive charge (nucleus of the cation) is farther away from the negative charge (electrons of the anion) larger ion = weaker attraction weaker attraction = smaller lattice energy Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 22 Copyright 2011 Pearson Education, Inc. Lattice Energy vs. Ion Size Metal

chloride Lattice energy (kJ/mol) LiCl 834 NaCl 787 KCl 701 CsCl 657 Tro: Chemistry: Edward Wen A Molecular Approach, 2/e

23 Copyright 2011 Pearson Education, Inc. q q F k r2 Trends in Lattice Energy Ion Charge The force of attraction between oppositely charged particles is directly proportional to the product of the charges Larger charge means the ions are more strongly attracted Lattice Energy = 910 kJ/mol larger charge = stronger attraction stronger attraction = larger lattice energy

Of the two factors, ion charge is generally more important Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 24 Lattice Energy = 3414 kJ/mol Copyright 2011 Pearson Education, Inc. Rank the following ionic compounds in order of increasing magnitude of lattice energy: CaO, KBr, KCl, SrO First examine the ion charges and order by product of the charges Ca2+& O2-, K+ & Br, K+ & Cl, Sr2+ & O2 (KBr, KCl) ___ (CaO, SrO) Then examine the ion sizes of each group and order by radius; larger < smaller

(KBr, KCl) same cation, Br > Cl (same Group) (CaO, SrO) same anion, Sr2+ > Ca2+ (same Group) KBr ___ KCl ___ (CaO, SrO) KBr ___ KCl ___ SrO ___ CaO Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 25 Copyright 2011 Pearson Education, Inc. Rank the following ionic compounds in order of increasing magnitude of lattice energy: MgS, NaBr, LiBr, SrS First examine the ion charges and Mg2+& S2-, Na+ & Br, order by product of the charges Li+ & Br, Sr2+ & S2 (NaBr, LiBr) __ (MgS, SrS) Then examine the ion sizes of

each group and order by radius; larger radius, ___ lattice energy (NaBr, LiBr) same anion, Na+ > Li+ (same Group) (MgS, SrS) same anion, Sr2+ > Mg2+ (same Group) NaBr __ LiBr __ (MgS, SrS) NaBr __ LiBr __ SrS __ MgS Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 26 Copyright 2011 Pearson Education, Inc. Ionic Bonding: Model vs. Reality Lewis theory implies: The attractions between ions are strong Ionic compounds should have high melting points (m.p.) and boiling points (b.p.) because breaking down the crystal should require a lot of energy the stronger the attraction (larger the lattice energy), the higher the m.p.

Ionic compounds do have high m.p. and b.p. M.p. generally > 300 C Most ionic compounds are solids at room temperature (except ionic liquids, (C2H5)NH3+NO3 m.p. 12C) Edward Wen 27 Copyright 2011 Pearson Education, Inc. Properties of Ionic Compounds Melting an ionic solid Hard and brittle crystalline solids all are solids at room temperature Melting points generally > 300 C The liquid state conducts electricity the solid state does not conduct electricity Many are soluble in water the solution conducts electricity well Tro: Chemistry:

Edward Wen A Molecular Approach, 2/e 28 Copyright 2011 Pearson Education, Inc. Practice Which ionic compound below has the highest melting point? KBr (734 C) CaCl2 (772 C) MgF2 (1261 C) More online resource Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 29 Copyright 2011 Pearson Education, Inc. Ionic Bonding Model vs. Reality Lewis theory:

The positions of the ions in the crystal lattice are critical to the stability of the structure moving ions out of position should therefore be difficult, and ionic solids should be hard hardness is measured by rubbing two materials together and seeing which streaks or cuts the other the harder material is the one that cuts or doesnt streak Ionic solids are relatively hard compared to most molecular solids Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 30 Copyright 2011 Pearson Education, Inc. Ionic Bonding Model vs. Reality Lewis theory implies that if the ions are displaced

from their position in the crystal lattice, that repulsive forces should occur This predicts the crystal will become unstable and break apart. Lewis theory predicts ionic solids will be brittle. Ionic solids are brittle. When struck they shatter. -+ +- -+ +- -+ +- -+ + - -+ +- -+ +- -+ +- -+ +- + - + - + - + - + - + - + - + - + + + - Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 31 Copyright 2011 Pearson Education, Inc. Ionic Bonding Model vs. Reality To conduct electricity, a material must have

charged particles that are able to flow through the material Lewis theory implies that, in the ionic solid, the ions are locked in position and cannot move around Lewis theory predicts that ionic solids should not conduct electricity Ionic solids do not conduct electricity Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 32 Copyright 2011 Pearson Education, Inc. Ionic Bonding Model vs. Reality Lewis theory implies in the liquid state or when dissolved in water, the ions will have the ability to move around Both a liquid ionic compound and an ionic compound dissolved in water should conduct electricity

Ionic compounds conduct electricity in the liquid state or when dissolved in water Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 33 Copyright 2011 Pearson Education, Inc. Conductivity of NaCl: Solid, Solution, Molten in NaCl(s), the ions are stuck in position and not allowed to move to the charged rods in NaCl(l), the ions are mobile and in NaCl(aq), the ions are separated and allowed to move to the

charged rods allowed to move to the charged rods Tro: Chemistry: Edward Wen A Molecular Approach, 2/e 34 Copyright 2011 Pearson Education, Inc. Example: Use the Born-Haber cycle to find lattice energy for magnesium oxide Given: Electron Affinity Ionization (kJ/mol) energy (kJ/mol) Sublimation energy (kJ/mol) Hf (kJ/mol)

O -141, 844 Mg 148 MgO (-601) O(g) (249.4) Mg 738, 1451 Find: MgO(s) O2-(g) + Mg2+(g) Hlattice = ? First write equations for the above processes: O(g) + e- O-(g) H1 = -141 kJ O- (g) + e- O2-(g) H2 = 844 kJ Mg(g) Mg+(g) + eH3 = 738 kJ Mg+(g) Mg2+(g) + eH4 = 1451 kJ Mg(s) Mg(g) H5 = 148 kJ Mg(s) + O2 (g) MgO(s) H6 = -601 kJ O2 (g) O(g) H7 = 249.4 kJ Edward Wen 35

Copyright 2011 Pearson Education, Inc. O(g) + e- O-(g) H1 = -141 kJ O- (g) + e- O2-(g) H2 = 844 kJ Mg(g) Mg+(g) + e- H3 = 738 kJ Mg+(g) Mg2+(g) + e- H4 = 1451 kJ Mg(s) Mg(g) H5 = 148 kJ Mg(s) + O2 (g) MgO(s) O2(g) O(g)

How to find: H6 = -601 kJ H7 = 249.4 kJ MgO(s) O2-(g) + Mg2+(g) Hlattice = ? MgO(s) Mg(s) + O2 (g) -H6 = 601 kJ O- (g) + e- O2-(g) H2 = 844 kJ Mg+(g) Mg2+(g) + e- H4 = 1451 kJ O(g) + e- O-(g) H1 = -141 kJ

Mg(g) Mg+(g) + e- H3 = 738 kJ Mg(s) Mg(g) H5 = 148 kJ O2 (g) O(g) H7 = 249.4 kJ Hlattice =-H6 + (H1 + H2 + H3 + H4 + H4 + H5 + H7) = 3890 kJ Edward Wen 36 Copyright 2011 Pearson Education, Inc.

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