Chapter Nine. Chemical Bonding I

Transcription

1 Chapter Nine Chemical Bonding I 1

2 The Ionic Bond and Lattice Energies 2

3 Lewis Dot Symbols Consists of atomic symbol surrounded by 1 dot for each valence electron in the atom Only used for main group elements # valence electrons = group number 3

4 Ionic Bonding Electrons are transferred from 1 atom to another Metal atoms: Lose electrons to form cations Nonmetal atoms: Gain electrons to form anions Electrostatic force bonds ions into an ionic compound Form an ionic salt with repeating structure: NaCl, LiF 4 Ionic Bonds follow the octet rule Atoms lose or gain valence e- to make an octet (8e-) 8 valence e- = Noble gas configuration Li Li + + e - e - + F F - Li + + F - Li + F - 1s 2 2s 1 1s 2 1s 2 2s 2 2p 5 1s 2 2s 2 2p 6

5 The Born Haber Cycle Formation of a salt can be tracked by individual steps Enthalpies associated with each step are additive Lattice energies are exothermic: salt is more stable than ions ( H) associated with gaseous atoms, so convert to gas Ionization energies: Electron loss Electron affinity: Electron gain Lattice energy: Energy released when forming a crystal Formation energy ( H f ) not based on gaseous atoms Energy associated with forming 1 mole of a compound from its elements in their normal states Sum of energy transfers called Born-Haber cycle Can use the cycle to the enthalpy of any step in cycle 5

6 Energy Changes in the Born-Haber Cycle Li (s) Li(g) H subl = +155 kj 1/2 F 2 (g) F(g) H diss = +75 kj Li (g) Li + (g)+ e - H IE = + 520kJ F(g) + e- F - (g) H EA = kj Li + (g) + F - (g) LiF (s) H lattice E =? kj Li(s) + 1/2F 2 (g) LiF(s) H f = kj 6 Hess s Law: H f = H subl + H diss + H IE + H EA + H lattice E -594kJ =155kJ + 75kJ + 520kJ - 328kJ + H lattice E H lattice E = -1016kJ

7 Lattice Energies of Common Salts 7

8 The Covalent Bond and Electronegativity 8

9 Lewis Structures Lewis structures represent covalent bond formation Shared pairs of electrons give both atoms an octet F + F F F 1s 2 2s 2 2p 5 1s 2 2s 2 2p 6 9 7e - 7e - 8e - 8e - Bonding Pairs: Shared electrons count for both atoms Represented by a dash (-) between bonding atoms Lone Pairs: Non-shared electrons count for 1 atom Represented by a pair of dots () around atom lone pairs F F lone pairs single covalent bond

10 Multiple Bonds More than one pair of electrons is shared between atoms so each atom can form an octet. 10 Single Bond: 1 shared pair: 1 dash (-) Double bonds: 2 shared pairs: 2 dashes (=) Triple bonds: 3 shared pairs: 3 dashes ( ) Allows atoms in the Lewis structure to share extra electrons if there are not enough for the central atom O C O double bonds N N triple bond

11 Electronegativity and Polar Covalent Bonds Electronegativity The ability of an atom to attract electrons F is the most electronegative atom Nonmetals high electronegativies H F 11 Polar Covalent Bonds Differences in electronegativity result in unequal sharing of electrons between atoms More electronegative atom has a partial negative charge Percent Ionic Character Measure of polarity of bond 100% ionic is full transfer of electron, no sharing 100% covalent is equal sharing, H 2, Cl 2, etc.

12 The Electronegativities of Common Elements

13 Writing Lewis Structures 13

14 Writing Lewis Structures: General Information Electronegativity Central atom usually has the lowest electronegativity (atom lower or to the left in periodic table) Terminal atoms (except H) have higher electronegativities 14 Terminal Atoms Bonded to only one other atom Hydrogen atoms are terminal atoms Bonding Hydrogen atoms are bonded to oxygen atoms in oxoacids Make the molecule as symmetrical as possible

15 Write the Lewis Structure of HNO 1. Add up the valence electrons in the structure 1 (H) + 5 (N) + 6 (O) = 12 valence electrons 2. Arrange the atoms & place bonding electrons H - N O nitrogen less electronegative, put in the center Place e- pairs around terminal atoms O: 4. Place remaining electron pairs on central atom H N O : 5. Add double bond to finish nitrogen octet H H N N = O

16 Formal Charge and Resonance Theory 16

17 Formal Charge Difference between the # of valence electrons in a free atom & the # of electrons assigned to that atom in a Lewis structure. F.C. = Group # - (# of lone e- + # bonds) Molecule is most stable if formal charge is 0 for each atom. 17 H N = O H: = 1-(0 +1)= 0 O: = 6-(6 +1)= -1 N: = 5-(2+3)= 0 S: = 6-(2 +3)= +1 O: = 6-(4+2)= 0 O: = 6-(4 +2)= 0 The most likely Lewis structure has the lowest formal charges Negative formal charge must be on more electronegative atom Sum of formal charges must =0 for molecules or = ionic charge :O S = O

18 Resonance Theory If a molecule or ion can be represented by 2 or more Lewis structures that differ only in electron location, the true structure is a composite of them. 18 : O S = O O = S O: Resonance Structures Equivalent Lewis structures that can be drawn for a molecule Formal charges will usually be present Delocalization Electrons are spread out over several atoms Stabilizes molecule

19 Exceptions to the Octet Rule 19

20 Incomplete Octet 20 Not enough electrons to cover central atom Be 2e - BeH 2 2H 2x1e - H Be H 4e - Terminal atoms unwilling to donate electrons Destabilizes terminal atoms: creates formal charge BF 3 B 3e - 3F 3x7e - 24e - F B F F

21 Coordinate Covalent Bonds 21 1 atom provides both electrons Electrons are then shared between 2 atoms Ex: BF 3 and NH 3 B needs 2 electrons to fill octet N has a lone pair to share F F B F H N H H

22 Free Radicals and Expanded Octets Free Radicals: Molecules with an odd number of valence electrons (N) Extremely reactive, odd electron wants to be part of a pair 22 :O N Compounds with expanded valence shells Central atom has more than eight electrons May have lone pair electrons as well as bonding pairs = O

23 Bond Enthalpy 23

24 Bond Enthalpy Bond Enthalpy Energy required to break a particular bond in a molecule in the gas phase. 24 Enthalpy change for the Reaction ( H) H = H bond breaking + H bond formation Enthalpy change: 2H 2 (g) + O 2 (g) 2 H 2 O(g) H bond breaking = 2 H H-H + H O=O = 2(436kJ) + 499kJ H formation = +1371kJ (endothermic) = 4 H H-O =4(460kJ) = -1840kJ (exothermic) H reaction =+1371kJ-1840kJ = -469kJ

25 Estimate the enthalpy change for the combustion of 1 mole of methane 1. Write the reaction: CH 4 (g)+ 2 O 2 (g) CO 2 (g)+ 2 H 2 O(g) Calculate energy needed to break the bonds in reactants and energy produced when the bonds of products form. Bonds broken H Bonds formed H 4C-H: 4 x 414 kj = +1656kJ 2C=O: 2 x 736 kj = -1472kJ 20=0: 2 x 498 kj= + 996kJ 4H-O: 4 x 464 kj = -1856kJ E dissociation = kj E formation = -3328kJ Bond-breaking:endothermic (+) Bond-forming:exothermic (-) 3. Calculate H for the reaction H approx = +2652kJ kJ = -676kJ

26 Representative Bond Enthalpies 26