1. Chemical Bonding I: The Covalent Bond Chapter 9 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. 9.1 __________________________ are the outer shell electrons of an atom. The valence electrons are the electrons that participate in chemical bonding. 1A 1ns 1 2A 2ns 2 3A 3ns 2 np 1 4A 4ns 2 np 2 5A 5ns 2 np 3 6A 6ns 2 np 4 7A 7ns 2 np 5 Group# of valence e - e - configuration

3. 9.1

4. You must learn to use the ______________ and _____________________to predict: 1.the type of bond that atoms will form (covalent, ionic), 2. the number of bonds an atom of an element can form, and 3. the stability of the product. (266)

5. 9.2 Li + F Li + F - _____________________________ 1s 2 2s 1 1s 2 2s 2 2p 5 1s 2 1s 2 2s 2 2p 6 [He] [Ne] Li Li + + e - e - + FF - F - Li + + Li + F - Atoms are changed to ions when one (or more) electron is lost by one atom and gained by another atom.

6. 9.3 Lattice energy (E) increases as Q increases. cmpd lattice energy MgF 2 MgO LiF LiCl 2957 3938 1036 853 Q= +2,-1 Q= +2,-2 r F < r Cl Electrostatic Energy E = k Q+Q-Q+Q- r Q + is the charge on the cation Q - is the charge on the anion r is the distance between the ions ____________________ (E) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions. Lattice energy (E) increases as as r decreases.

7. 9.3 Born-Haber Cycle for Determining Lattice Energy  H overall =  H 1 +  H 2 +  H 3 +  H 4 +  H 5 oooooo

8. A _________________________ is a chemical bond in which two (or more) electrons are shared by two atoms. Why should two atoms share electrons? FF + 7e - FF 8e - F F F F Lewis structure of F 2 lone pairs single covalent bond 9.4

9. 8e - H H O ++ O HH O HH or 2e - Lewis structure of water _______________ – two atoms share two pairs of electrons single covalent bonds O C O or O C O 8e - double bonds ______________ – two atoms share three pairs of electrons N N 8e - N N triple bond or 9.4

10. Bond Type Bond Length (pm) C-CC-C 154 CCCC 133 CCCC 120 C-NC-N 143 CNCN 138 CNCN 116 Lengths of Covalent Bonds Bond Lengths Triple bond < Double Bond < Single Bond 9.4

11. Comparison of Ionic and Covalent Compounds 9.4

12. H F _________________________ or _______________ is a covalent bond with greater electron density around one of the two atoms electron rich region electron poor region e - rich e - poor F H ++ -- 9.5

13. ______________________ is the ability of an atom to attract toward itself the electrons in a chemical bond. Electron Affinity - _______________, Cl is highest Electronegativity - _______________, F is highest X (g) + e - X - (g) 9.5 The negative of the energy change that occurs when an electron is accepted by an atom of an element in the gaseous state The ability of an atom to attract toward itself the electrons in a chemical bond

14. 9.5 p.270, Fig 9.4 _________________________________________

15. Covalent share e - Polar Covalent partial transfer of e - Ionic transfer e - Increasing difference in electronegativity Classification of bonds by difference in __________________ DifferenceBond Type 0___________  2 ___________ 0 < and <2 ____________ 9.5

16. Classify the following bonds as ionic, polar covalent, or covalent: The bond in CsCl; the bond in H 2 S; and the NN bond in H 2 NNH 2. Cs – 0.7Cl – 3.03.0 – 0.7 = 2.3 H – 2.1S – 2.52.5 – 2.1 = 0.4______________ N – 3.0 3.0 – 3.0 = 0 9.5 ______________

17. 1.Draw skeletal structure of compound showing what atoms are bonded to each other. Put __________ electronegative element in the center. 2.Count total number of valence e -. ____________ 1 for each negative charge. __________________1 for each positive charge. 3.Complete an octet for all atoms except _________ 4.If structure contains too many electrons, form ____________ and ______________bonds on central atom as needed. Writing Lewis Structures 9.6 p.270

18. Write the Lewis structure of nitrogen trifluoride (NF 3 ). Step 1 – N is less electronegative than F, put N in center FNF F Step 2 – Count valence electrons N - 5 (2s 2 2p 3 ) and F - 7 (2s 2 2p 5 ) 5 + (3 x 7) = 26 valence electrons Step 3 – Draw single bonds between N and F atoms and complete octets on N and F atoms. Step 4 - Check, are # of e - in structure equal to number of valence e - ? 3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons 9.6

19. Write the Lewis structure of the carbonate ion (CO 3 2- ). Step 1 – C is less electronegative than O, put C in center Step 2 – Count valence electrons C - 4 (2s 2 2p 2 ) and O - 6 (2s 2 2p 4 ) -2 charge – 2e - 4 + (3 x 6) + 2 = ___________________________________ Step 3 – Draw single bonds between C and O atoms and complete octet on C and O atoms. Step 4 - Check, are # of e - in structure equal to number of valence e - ? 3 single bonds (3x2) + 10 lone pairs (10x2) = ___________________ 9.6 Step 5 - Too many electrons, form double bond and re-check # of e - OCO O 2 single bonds (2x2) = 4 1 double bond = 4 8 lone pairs (8x2) = 16 Total = ______

20. 9.7 Two possible skeletal structures of formaldehyde (CH 2 O) HCOH H CO H An atom’s __________ is the difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure. formal charge on an atom in a Lewis structure = 1 2 total number of bonding electrons () total number of valence electrons in the free atom - total number of nonbonding electrons - The sum of the formal charges of the atoms in a molecule or ion must equal ________________________________.

21. HCOH C – 4 e - O – 6 e - 2H – 2x1 e - 12 e - 2 single bonds (2x2) = 4 1 double bond = 4 2 lone pairs (2x2) = 4 Total = 12 formal charge on C = 4 -2 -2 -½ x 6 = -1 formal charge on O = 6 -2 -2 -½ x 6 = +1 formal charge on an atom in a Lewis structure = 1 2 total number of bonding electrons () total number of valence electrons in the free atom - total number of nonbonding electrons - +1 9.7

22. C – 4 e - O – 6 e - 2H – 2x1 e - 12 e - 2 single bonds (2x2) = 4 1 double bond = 4 2 lone pairs (2x2) = 4 Total = 12 H CO H formal charge on C = 4 -0 -0 -½ x 8 = 0 formal charge on O = 6 -4 -4 -½ x 4 = 0 formal charge on an atom in a Lewis structure = 1 2 total number of bonding electrons () total number of valence electrons in the free atom - total number of nonbonding electrons - 00 9.7

23. Formal Charge and Lewis Structures 9.7 1.For _________________ molecules, a Lewis structure in which there are no formal charges is preferable to one in which formal charges are present. 2.Lewis structures with large formal charges are _______ plausible than those with small formal charges. 3.Among Lewis structures having similar distributions of formal charges, the most plausible structure is the one in which negative formal charges are placed on the more _________________________ atoms. Which is the more likely Lewis structure for CH 2 O? HCOH +1 H CO H 00

24. A _______________________ is one of two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure. Consider ozone: OOO + - OOO + - OCO O -- OCO O - - OCO O - - 9.8 What are the resonance structures of the carbonate (CO 3 2- ) ion? Q: Which O–O bond is longer?

25. 24.3 C C C CC C H H H H H H C C C CC C H H H H H H Resonance structures of benzene

26. A common misconception about resonance structures: The molecule quickly shifts back and forth from one resonance structure to another. (This is not true.) 9.8 p.277 And remember this fact about resonance structures: The positions of ____________, not _______________ (nuclei), can be rearranged in different resonance structures. Thus, the same atoms must always be bonded to one another in all the resonance structures for a given molecule.

27. Exceptions to the Octet Rule The ________________ Octet HHBe Be – 2e - 2H – 2x1e - 4e - BeH 2 BF 3 B – 3e - 3F – 3x7e - 24e - FBF F 3 single bonds (3x2) = 6 9 lone pairs (9x2) = 18 Total = 24 9.9

28. Exceptions to the Octet Rule _____________________________ N – 5e - O – 6e - 11e - NO N O _________________ (central atom with principal quantum number n > 2) SF 6 S – 6e - 6F – 42e - 48e - S F F F F F F 6 single bonds (6x2) = 12 18 lone pairs (18x2) = 36 Total = 48 9.9

29. The enthalpy change required to break a particular bond in one mole of gaseous molecules is the _______________. H 2 (g) H (g) +  H 0 = 436.4 kJ Cl 2 (g) Cl (g) +  H 0 = 242.7 kJ HCl (g) H (g) +Cl (g)  H 0 = 431.9 kJ O 2 (g) O (g) +  H 0 = 498.7 kJ OO N 2 (g) N (g) +  H 0 = 941.4 kJ N N Bond Energy Bond Energies Single bond < Double bond < Triple bond 9.10 p.280

30. Force is directly proportional to the product of the electrical charge and inversely proportional to the square of the distance. F is the force k is a constant and has the value of 9.00 x 10 9 N  m 2 /C 2 (9.00 x 10 9 newton  meters 2 /coulomb 2 ) q 1 represents the electrical charge of object 1 and q 2 represents the electrical charge of object 2 d is the distance between the two charged particles Excursus: ________________________

31. _________________________ in polyatomic molecules H 2 O (g) H (g) +OH (g)  H 0 = 502 kJ OH (g) H (g) +O (g)  H 0 = 427 kJ Average OH bond energy = 502 + 427 2 = 464 kJ 9.10 Some Bond Energies of Diatomic Molecules and Average Bond Energies for Bonds in Polyatomic Molecules p.282

32. Bond Energies (BE) and Enthalpy changes in reactions  H 0 = total energy input – total energy released =  BE(reactants) –  BE(products) Imagine reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products. 9.10 endothermicexothermic

33. Let’s work through Example 9.9 on page 283

34. Use bond energies to calculate the enthalpy change for: H 2 (g) + F 2 (g) 2HF (g)  H 0 =  BE(reactants) –  BE(products) Type of bonds broken Number of bonds broken Bond energy (kJ/mol) Energy change (kJ) HH1436.4 FF 1156.9 Type of bonds formed Number of bonds formed Bond energy (kJ/mol) Energy change (kJ) HF2568.21136.4  H 0 = __________________________________ = __________ 9.10