Chapter The Ionic Bond. Why are ionic compounds stable? Spontaneous Processes. Chemical Bonding I: Basic Concepts.

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1 Lewis Dot ymbols hapter 9 hemical Bonding I: Basic oncepts Introduced by G.. Lewis Element symbol plus 1 dot for each valence e Elements tend to form octets, noble gas configurations Useful for representative (sp block) elements only 9.2 The Ionic Bond Typical ionic reactions with Lewis structures Electrostatic force that holds ions together in an ionic compound Very strong force ionic compounds have high melting and boiling points Lewis dot symbols are used to represent formation of ionic compounds 2 + a + a + 2 Li + 2 Li Why are ionic compounds stable? onsider formation of a a(s) + 1/2 2 (g) d a(s) written on mole basis (6.02 x atoms) Ionization: a(g) d a + (g) requires energy I 1 (a) = 496 kj/mol e attachment: (g) d (g) releases energy EA() = 328 kj/mol ( = 328 kj/mol) E released is not enough to make up for E required! What is missing? pontaneous Processes Exothermic processes are more favorable than endothermic processes There s more to it than that! hap. 18 spontaneous, favorable reactions proceed to products 2 2 (g) + 2 (g) d 2 2 (l) = product reactants = 286 kj/mol (vigorous!)

2 9.3 Lattice energy Energy required to completely separate one mole of a solid, ionic compound into gaseous ions Lattice energy makes formation of an ionic compound favorable overcomes unfavorable ionization energy Physical basis of lattice energy is electrostatic attraction between cations and anions Bornaber ycle a + (g) + (g) I 1 = 495 kj e + e EA = 328 kj sub = 108 kj a(g) + (g) a(s) + 1/2 2 (g) 1/2 diss = 79 kj f = 574 kj latt = 928 kj provides driving force! Lattice energy (a) = U = D latt = 928 kj/mol a (s) U = = f + sub + I 1 + 1/2 diss EA Lattice Energy ummary E required to separate 1 mole of a solid, ionic compound into gaseous ions Physical basis of lattice energy is electrostatic attraction between cations and anions ighly charged ions d larger lattice energy (U) U (kj/mol): Li 1017 < Mg 3890 Attraction cation charge X anion charge igh U d high mp, bp mp: Li 845, Mg ovalent Bond Bond in which (2) electrons are shared by (2) atoms Bonds w/ more than 2 centers are possible ovalent ompounds compounds that contain only covalent bonds nly valence electrons are involved in bonding 2 l l l or l l Lone pairs Bonding pairs Usual representation Lone pairs pairs of valence electrons not involved in covalent bond formation Lewis structure representation of covalent bonding w/ lone pairs shown as pairs of dots and bonding pairs (usually) shown as lines ctet Rule an atom (other than ) tends to form bonds until it is surrounded by 8 valence e s full s and p orbitals doesn t work for transition elements l l Each l achieves octet, 6 from lone pairs and 2 from sharing in bond achieves octet, 2 from lone pair and 6 from shared pairs Each achieves "duet" from a shared pair

3 Multiple bonds (bond orders) ingle bond = 2 atoms held together w/ 1 e pair Double bond = 2 atoms held together w/ 2 e pairs Ethene (ethylene) d(=) = 133 pm Ethane d() = 107 pm d () = 154 pm arbon dioxide d(=) = 120 pm Triple bond = 2 atoms held together w/ 3 e pairs Ethyne (acetylene) d( ) = 120 pm arbon monoxide d( ) = 119 pm Bond Length Distance between the nuclei of 2 covalently bonded atoms in a molecule Bond is shorter (and stronger) as bond order increases triple < double < single ~120 pm < 130 pm < 150 pm for bonds Table 9.2 = bond lengths ovalent vs. Ionic ompounds has very small radius d short bonds I has very large radius d long bonds I 2 bond length = 272 pm ovalent compounds contain only covalent bonds Weak attractions between molecules usually leads to low mp and bp Do not conduct electricity if dissolved in water (rarely soluble) or molten Ionic compounds contain ionic bonds trong electrostatic attraction leads to high mp and bp onduct electricity when dissolved in water or molten Ions are mobile charge carriers ontinuum of bond polarity Electron Density Distribution in + a l Ionic bond (early) complete e transfer Low e density igh e density l Polar covalent bond unequal sharing of e pair e s are polarized toward l ow do we describe and predict this tendency to polarize? l l onpolar covalent bond equal sharing of e pair Representation of bond dipole Electrons are more strongly pulled by than Tendency to attract electron in a bond = electronegativity

4 9.5 Electronegativity (χ) Ability of an atom to attract e s in a chemical bond toward itself Electronegativity determines extent of e transfer in a chemical bond Pauling s relative electronegativity values are scaled to (maximum) χ() = 4.0 an be derived from I 1 and EA of an atom MI igure 9.5 Don t memorize values learn the trend! MAX + a l ontinuum of bond polarity l l l Ionic bond?dc = 2.0 (early) complete e transfer Polar covalent bond 0 < Dc < 2.0 unequal sharing of e pair e s are polarized toward l onpolar covalent bond?dc = 0 equal sharing of e pair Ionic, polar covalent, or nonpolar covalent? KBr χ(k) = 0.8, χ(br) = 2.8, ionic bond χ() = 2.5, χ() = 3.5, polar covalent bond Bl 3 χ(b) = 2.0, χ(l) = 3.0, polar covalent bonds 2 χ() = 3.5, nonpolar covalent bond Remember xidation umber Rules? (ec. 4.4) 3 x # of oxygen is usually 2 Why? χ() = 3.5, exceeded only by (Except 2 2, x # = 1; 2, x # = +2) 4 x # of hydrogen is usually +1 Why? χ() = 2.1, less than most nonmetals Except metal hydrides, e.g. Mg 2, x # () = 1 Why? χ(metals) < luorine has x # 1 in all compounds. Why? χ() = 4.0, largest χ 9.6 Writing Lewis tructures 1. Write skeletal structure Typically: lowest χ atom occupies central position and occupy terminal positions 2. Total valence electrons in molecule anion: add 1 e for every charge cation: subtract 1 e for every + charge

5 Lewis tructures (contd.) Examples 3. atisfy the octet rule for each atom. onnect terminal atoms to central atom w/ single bonds. Use lone pairs to fill octets of terminal atoms. Use all e counted in step Use lone pairs to form double or triple bond to central (or any) atom that does not have octet. 4 Pl P ormal harge Difference between the # valence e in an isolated atom and # e assigned to that atom in a Lewis structure. e s in a bond are divided equally between atoms to obtain formal charges differs from oxidation state, where e s in a bond are assigned to the more electronegative atom eutral molecules, Σ formal charges = 0 Ions, Σ formal charges = charge on ion ormal charges do not represent real, localized charges bookkeeping device ormal charge on atom = # valence e s of free atom # nonbonding e 1/2 (# bonds) [: : ] Examples: x# & ormal harge see next rule ormal charges indicate most plausible Lewis structures or a neutral molecule, structure with no formal charges is preferred. mall formal charges (0,+1, 1) are more favorable than larger formal charges. tructures w/ negative formal charges on most electronegative atoms are favorable. Avoid adjacent formal charges with same sign.

6 3 Lewis structures of nitrite anion Actual structure: each bond is 124 pm single bond, 143 pm Major contributor Major contributor Minor contributor Examples: x# & ormal harge see next rule = double bond, 120 pm (Back to examples) 9.8 Resonance Examples of Resonance Resonance structure 1 of 2 or more Lewis structures for a molecule (ion) that can t be represented w/ a single structure Resonance use of 2 or more Lewis structures to describe a molecule (ion) Each resonance structure contributes to the actual structure no single structure is a complete description positions of atoms must be the same in each actual structure is an average Benzene ( 6 6 ): distances 140 pm single bond, 154 pm = double bond, 133 pm ulfate ( 4 2 ): distances 149 pm 9.9 Exceptions to ctet Rule Incomplete ctet often groups 13, also 2 in gas phase react with electron donors ddelectron Molecules Molecule w/ odd number of electrons cannot satisfy octet rule and 2 form weak bonds B B B 2 Major resonance form w/ zero formal charges 1 of 3 minor resonance forms w/ nonzero formal charges ormation of dative (coordinate covalent) bond with electronpair donor 2

7 Expanded ctet Atoms of secondperiod elements are limited to eight valence e s around atom Atoms of third period and beyond can form compounds w/ > 8 e s around atom used this idea in 4 2 above ontroversial explanations: atoms w/ n = 3 can use d orbitals (not just s and p) atoms w/ n = 3 are large enough to fit more than 4 e pairs around them in multicentered bonds 4 e count : 1 6 = 6 : 4 7 = 28 total e 34 Extra e pair is placed on Expanded ctet Examples Pl 5 Xe 2 I 5 e Bond Dissociation Energy Bond energy = enthalpy change required to break a particular bond in one mole of gaseous molecules 2 (g) d 2 (g) = kj Table 9.4 (pg 356) gives averages exact value depends on molecule ingle bond < double bond < triple bond 347, = 620, o 812 kj/mol Use bond energies to estimate rxn Bonds to lighter elements often stronger Bond Energy (kj /mol) Bond Energy (kj/mol) l l 243 l 432 Br Br 192 Br 366 I I 151 I 298 D = BE(reactants) BE(products)

8 D = BE(reactants) BE(products) Why reactants products rather than products reactants? Bond dissociation energy = enthalpy change required to break a bond d negative sign built into tabulated values (g) + 4(g) d 4 (g) must be exothermic! = ΣBE(reactants) ΣBE(products) = 0 (4 mol)(414 kj/mol) = 1656 kj rxn from Bond Energies Write structures for substances in equation Determine bonds broken and formed Gives approximate answer less accurate than using f doesn t take phases, intermolecular forces into account or the reaction: (g) (g) d 4 2 (g) (g) predict the enthalpy of reaction from average bond energies (Table 9.4) = d 8 = + 12 rxn = 12 mol (414 kj/mol) + 2 mol (347 kj/mol) + 7 mol (498.7 kj/mol) 8 mol (799 kj/mol) 12 mol (460 kj/mol) D rxn = kj Alternative by using enthalpies of formation: rxn = 4 f ( 2 ) + 6 f ( 2 ) 2 f ( 2 6 ) = 4(393.5 kj) + 6(241.8 kj) 2(84.7 kj) D rxn = kj