Chapter 8 Chemical Bonding

 Na Cl + 1s22s22p63s1 1s22s22p63s23p5 Na+ Cl– + 1s22s22p6

Ionic Bond Cl– electrostatic attraction between oppositely charged ions non-directional Na+

Ionic Bond Cl– electrostatic attraction between oppositely charged ions non-directional Na+ Na+ Cl–

Ionic Bond Cl– electrostatic attraction between oppositely charged ions non-directional Na+ Cl– Na+ Na+ Cl– =

Ionic Bond Cl– electrostatic attraction between oppositely charged ions non-directional Na+ Cl– Cl– Na+ Na+ = Cl– = Na+

Ionic Bond Cl– electrostatic attraction between oppositely charged ions non-directional strength directly proportional to charges of ions and inversely proportional to distance between ion centers Na+

E  Q c a d E = ionic bond strength Qc = charge on cation Qa = charge on anion (absolute value) d = distance between centers of ions

E  Q c a d NaCl Na+ Cl–

E  Q c a d NaCl Na+ Cl– CaCl2 Ca2+ Cl–

E  Q c a d NaCl Na+ Cl– CaCl2 Ca2+ Cl– CaS Ca2+ Cl–

E  Q c a d NaCl Na+ Cl– CaCl2 Ca2+ Cl– CaS Ca2+ S–2 Al2S3 Al3+ S2–

E  Q c a d d Na+ F–

E  Q c a d d d Cl– Na+ F– Na+

E  Q c a d d d Cl– Na+ F– Na+ d Br– Na+

E  Q c a d d d Cl– Na+ F– Na+ d d I– Br– Na+ Na+

Lattice Energy energy required to separate 1 mol of an ionic compound into its gaseous ions eg MgF2(s)  Mg2+(g) + 2F–(g) Hlattice = 2910 kJ/mol for NaF, Hlattice = 911 kJ/mol for KF, Hlattice = 815 kJ/mol WHY???

Ionic Bond Formation Born Haber Cycle Breaks formation of an ionic compound from its elements into a series of theoretical steps and considers the energetics of each Example: consider formation of NaCl Na(s) + 1/2Cl2(g)  NaCl(s) H° = –410.9 kJ

725.4 E (kJ) 376.4 229.4 107.7 Na(s) + 1/2Cl2(g) –410.9

E (kJ) Na(g) + 1/2Cl2(g) Hsub, Na Na(s) + 1/2Cl2(g) 725.4 376.4 229.4 107.7 Hsub, Na Na(s) + 1/2Cl2(g) –410.9

E (kJ) Na(g) + Cl(g) Na(g) + 1/2Cl2(g) Hatom,Cl Hsub, Na 725.4 E (kJ) 376.4 Na(g) + Cl(g) 229.4 Na(g) + 1/2Cl2(g) Hatom,Cl 107.7 Hsub, Na Na(s) + 1/2Cl2(g) –410.9

Na+(g) + Cl(g) + e– E (kJ) IENa Na(g) + Cl(g) Na(g) + 1/2Cl2(g) 725.4 E (kJ) IENa 376.4 Na(g) + Cl(g) 229.4 Na(g) + 1/2Cl2(g) Hatom,Cl 107.7 Hsub, Na Na(s) + 1/2Cl2(g) –410.9

Na+(g) + Cl(g) + e– EACl E (kJ) Na+(g) + Cl–(g) IENa Na(g) + Cl(g) 725.4 EACl E (kJ) Na+(g) + Cl–(g) IENa 376.4 Na(g) + Cl(g) 229.4 Na(g) + 1/2Cl2(g) Hatom,Cl 107.7 Hsub, Na Na(s) + 1/2Cl2(g) –410.9

Na+(g) + Cl(g) + e– EACl E (kJ) Na+(g) + Cl–(g) IENa Na(g) + Cl(g) 725.4 EACl E (kJ) Na+(g) + Cl–(g) IENa 376.4 Na(g) + Cl(g) –Hlattice 229.4 Na(g) + 1/2Cl2(g) Hatom,Cl 107.7 Hsub, Na Na(s) + 1/2Cl2(g) NaCl(s) –410.9

H°f = 107.7kJ + 121.7kJ + 496.0kJ + –349.0kJ – 787.3 kJ H°f = Hsublimation, Na + Hatom, Cl + IE1, Na + EACl – Hlattice, NaCl H°f = 107.7kJ + 121.7kJ + 496.0kJ + –349.0kJ – 787.3 kJ H°f = –410.9 kJ

Lewis Symbols atomic symbol indicates element symbol considered to have 4 sides valence electrons indicated by dots maximum 2 dots per side

Examples Na 

Examples Na   Na

Examples  Na   Na Na

Examples  Na   Na Na Na 

Examples  Na   Na Na Na     Ca Ca    Ca   Ca Ca 

Examples  Na   Na Na Na     Ca Ca    Ca   Ca Ca      Al      Ca 

  Na  + Cl   

–  +      Na  + Na + Cl Cl       

–  +      Na  + Na + Cl Cl          Cl    Ca +     Cl   

–  +      Na  + Na + Cl Cl        –      Cl    2+  Ca + Ca +   –       Cl  Cl      

1s22s22p5 1s22s22p5       + F F      

1s22s22p5 1s22s22p5       + F F      

1s22s22p5 1s22s22p5 1s22s22p4 1s22s22p6 –        +       + + F F F F             

1s22s22p5 1s22s22p5 1s22s22p4 1s22s22p6 –        +       + + F F F  F             –        +      + +  F F F F             

1s22s22p5 1s22s22p5 1s22s22p4 1s22s22p6 –        +       + + F F F F              –        +      + +  F F F F                     F F        shared e– pair

y y x x 2px 2px y y x

       F F        shared e– pair       F F      

Covalent Bond attractive force between atoms resulting from sharing electron pair(s) very strong highly directional

single bond: sharing 1 e– pair double bond: sharing 2 e– pairs triple bond: sharing 3 e– pairs

         O   O      O  O       

         O    O     O    O            O O  

         O   O      O O               O O        N   N       N N       

         O   O      O O               O O        N   N       N N          N N  

+  H  H  H H 

+  H  H  H H  y y x        F F       

Pure Covalent Bond +  H  H  H H  y y x        F F       

y y x        Br F   

y y x + –        Br F   

y y x + –        Br F    Polar Covalent Bond Unequal Sharing of Electron Pair(s)

Bonding Continuum pure ionic polar covalent pure covalent BrF CsF H2

Electronegativity The ability of an atom to attract electron density towards itself increases from L to R across a period and up a group

Bonding Continuum pure covalent pure ionic polar covalent nonpolar covalent ionic 2.0 0.5 CsF (3.9) BrF (1.3) H2 (0)  E’neg

KCl  E,neg. = 2.9 – 0.9 = 2.0 ionic HCl  E,neg. = 2.9 – 2.1 = 0.8 polar covalent BrCl  E,neg. = 2.9 – 2.8 = 0.1 nonpolar covalent Br2  E,neg. = 2.9 – 2.9 = 0 pure covalent

Dipole Moment results from separation of charge polar bonds have increases as  E’neg. increases

Dipole Moment results from separation of charge polar bonds have increases as  E’neg. increases indicated by crossed arrow pointing from positive end to negative end of dipole

H I 2.1 2.2

+ – H I 2.1 2.2

+ – H I  E’neg. = 0.1

+ – H I  E’neg. = 0.1 H Cl 2.1 2.9

+ – H I  E’neg. = 0.1 + – H Cl 2.1 2.9

+ – H I  E’neg. = 0.1 + – H Cl  E’neg. = 0.8

+ – H I  E’neg. = 0.1 + – H Cl  E’neg. = 0.8 H F 2.1 4.1

+ – H I  E’neg. = 0.1 + – H Cl  E’neg. = 0.8 + – H F 2.1 4.1

+ – H I  E’neg. = 0.1 + – H Cl  E’neg. = 0.8 + – H F  E’neg. = 2.0

Lewis Structures shorthand notation for placement of atoms and valence electrons in molecules

Drawing Lewis Structures arrange atoms in formula such that element with only 1 atom in formula is in center surrounded by other atoms in formula connect central atom to outer atoms by single covalent bonds total valence electrons in molecule add valence e– of all atoms add 1 e– for each – charge subtract 1 e– for each + charge subtract 2 e– for each bond

use remaining e– to complete octets of outer atoms place remaining e– on central atom, in pairs when possible check form multiple bonds if necessary

Octet Rule all atoms require 8 valence electrons EXCEPTIONS H, He 2 e– B, Al 6 e– Z  15 8 or more e–

Examples draw Lewis structures for each of the following

Methane, CH4

Methane, CH4 H H C H H

Methane, CH4 H H C H H

Methane, CH4 C 4 e– H H 4 H 4 e– C 8 e– H H

Methane, CH4 C 4 e– H H 4 H 4 e– C 8 e– H H 4 bonds – 8 e– 0 e–

SnCl3– Cl Sn Cl Cl

SnCl3– Cl Sn Cl Cl

SnCl3– Cl Sn 4 e– 3 Cl 21 e– Sn 25 e– Cl Cl

SnCl3– Cl Sn 4 e– 3 Cl 21 e– Sn 25 e– Cl Cl 1 – 1 e– 26 e–

SnCl3– Cl Sn 4 e– 3 Cl 21 e– Sn 25 e– Cl Cl 1 – 1 e– 26 e– 3 bonds – 6 e– 20 e–

SnCl3–     Cl   Sn 4 e– 3 Cl 21 e– Sn 25 e–     Cl Cl         1 – 1 e– 26 e– 3 bonds – 6 e– 20 e– – 18 e– 2 e–

SnCl3–     Cl   Sn 4 e–  3 Cl 21 e– Sn  25 e–     Cl Cl         1 – 1 e– 26 e– 3 bonds – 6 e– 20 e– – 18 e– 2 e– – 2 e–

SnCl3–     Cl   Sn 4 e–  3 Cl 21 e– Sn  25 e–     Cl Cl         1 – 1 e– 26 e– 3 bonds – 6 e– 20 e– – 18 e– 2 e– – 2 e–

XeF4 F F Xe F F

XeF4 Xe 8 e– F F 4 F 28 e– Xe 36 e– F F 4 bonds – 8 e– 28 e–

XeF4        Xe 8 e–  F    F  4 F 28 e– Xe 36 e–     F F         4 bonds – 8 e– 28 e– – 24 e– 4 e–

XeF4        Xe 8 e–  F   F     4 F 28 e– Xe   36 e–     F F         4 bonds – 8 e– 28 e– – 24 e– 4 e– – 4 e– 0 e–

SF6

SF6         S 6 e– F   F        6 F 42 e–  F S F       48 e–     F F         6 bonds –12 e– 36 e– – 36 e– 0 e–

O2 O O 2 O 12 e– 1 bond – 2 e– 10 e–

O2       O O     2 O 12 e– 1 bond – 2 e– 10 e– – 10 e– 0 e–

      O O           O O           O O    

Form Multiple Bond       O O    

Form Multiple Bond       O O    

Form Multiple Bond       O O           O O  

N2 N N 2 N 10 e– 1 bond – 2 e– 8 e–

N2       N N   2 N 10 e– 1 bond – 2 e– 8 e– – 8 e– 0 e–

Form Multiple Bonds       N N  

Form Multiple Bonds       N N       N N    

Form Multiple Bonds       N N       N N       N N  

CO2       O C O   C 4 e–     2 O 12 e– 16 e– 2 bonds – 4 e– 12 e– – 12 e– 0 e–

      O C O       O C O         O C O    

Resonance more than one valid Lewis structure each structure called resonance structure

              O C O O C O O C O          

Resonance Hybrid average of resonance structures written in square brackets

Bond Order 1, single bond 2, double bond 3, triple bond

              O C O O C O O C O           2 + 1 + 3 B.O.left bond = = 2 3

              O C O O C O O C O           2 + 1 + 3 B.O.left bond = = 2 3 2 + 3 + 1 B.O.right bond = = 2 3

              O C O O C O O C O           2 + 1 + 3 B.O.left bond = = 2 3 2 + 3 + 1 B.O.right bond = = 2 3       O C O  

              O C O O C O O C O           2 + 1 + 3 B.O.left bond = = 2 3 2 + 3 + 1 B.O.right bond = = 2 3       O C O  

CO32–

CO32–           O O O       C C C          

CO32–           O O O       C C C           2 + 1 + 1 B.O.each bond = = 1.3 3

CO32–           O O O       C C C           2 + 1 + 1 B.O.each bond = = 1.3 3 2– O C O O

Bond Strength and Bond Length triple < double < single bond strength single < double < triple

Length (Å) C C 1.54 C C 1.34 C C 1.20

Length (Å) Strength (kJ/mol) C C 1.54 348 C C 1.34 614 C C 1.20 839

Formal Charge an accounting tool for electron ownership (# valance e– in free atom) – (# e– in lone pairs on atom) – 1/2(# bonded e– on atom)

Best structure has zero FC on all atoms lowest FC possible –FC on most electronegative atoms and +FC on least electronegative atoms

For example, consider thiocyanate ion

For example, consider thiocyanate ion SCN–

For example, consider thiocyanate ion SCN–             N C S C S N S N C            

For example, consider thiocyanate ion SCN–             N C S C S S N C N             FCN = 5 – 4 – 2 = –1

For example, consider thiocyanate ion SCN–             N C S C S C N S N             FCN = 5 – 4 – 2 = –1 FCC = 4 – 0 – 4 = 0

For example, consider thiocyanate ion SCN–             N C C S C S S N N             FCN = 5 – 4 – 2 = –1 FCC = 4 – 0 – 4 = 0 FCS = 6 – 4 – 2 = 0

For example, consider thiocyanate ion SCN– –1             N C C S C S S N N             FCN = 5 – 4 – 2 = –1 FCC = 4 – 4 – 2 = –2 FCS = 6 – 0 – 4 = +2

For example, consider thiocyanate ion SCN– –1 –2 +2 –1             N C S C S N S N C             FCN = 5 – 0 – 4 = +1 FCC = 4 – 4 – 2 = –2 FCS = 6 – 4 – 2 = 0

For example, consider thiocyanate ion SCN– –1 –2 +2 –1 +1 –2             N C S C S N S N C            

For example, consider thiocyanate ion SCN– –1 –2 +2 –1 +1 –2             N C S C S N S N C             best structure because lowest FC and –FC on most electronegative atom