1. 1 Resonance and Formal Charge

2. 2 Resonance and Formal Charge: At the conclusion of our time together, you should be able to: 1. Define resonance 2. Determine resonance structures for a molecule 3. Calculate the formal charge for an atom 4. Determine the resonance structure that contributes the most to a compound by using formal charge

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4. 4 SAMPLE PROBLEM:Writing Resonance Structures PLAN: SOLUTION: PROBLEM:Write resonance structures for the nitrate ion, NO 3 -. After Steps 1-4, go to 5 and then see if other structures can be drawn in which the electrons can be delocalized over more than two atoms. Nitrate has 1(5) + 3(6) + 1 = 24 valence e - N does not have an octet; a pair of e - will move in to form a double bond.

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6. 6 1. Smaller formal charges (either positive or negative) are preferable to larger charges; 2. A more negative formal charge should exist on an atom with a larger EN value. 3. Get unlike charges as close together as possible 4. Avoid like charges (+ + or - - ) on adjacent atoms Four criteria for choosing the more important resonance structure:

7. 7 Resonance and Formal Charge Formal charge of atom = # valence e - = (# unshared electrons + 1/2 the # shared electrons)

8. 8 Nitric acid.. :.. H O O O N : :.... ► We will calculate the formal charge for each atom in this Lewis structure.

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10. 10 Nitric acid.. :.. H O O O N : :.... ► Hydrogen shares 2 electrons with oxygen. ► Assign 1 electron to H and 1 to O. ► A neutral hydrogen atom has 1 electron. ► Therefore, the formal charge of H in nitric acid is 0. Formal charge of H

11. 11 Nitric acid.. :.. H O O O N : :.... ► Oxygen has 4 electrons in covalent bonds. ► Assign 2 of these 4 electrons to O. ► Oxygen has 2 unshared pairs. Assign all 4 of these electrons to O. ► Therefore, the total number of electrons assigned to O is 2 + 4 = 6. Formal charge of O

12. 12 Nitric acid.. :.. H O O O N : :.... ► Electron count of O is 6. ► A neutral oxygen has 6 electrons. ► Therefore, the formal charge of oxygen is 0. Formal charge of O

13. 13 Nitric acid.. :.. H O O O N : :.... ► Electron count of O is 6 (4 electrons from unshared pairs + half of 4 bonded electrons). ► A neutral oxygen has 6 electrons. ► Therefore, the formal charge of oxygen is 0. Formal charge of O

14. 14 Nitric acid.. :.. H O O O N : :.... ► Electron count of O is 7 (6 electrons from unshared pairs + half of 2 bonded electrons). ► A neutral oxygen has 6 electrons. ► Therefore, the formal charge of oxygen is -1. Formal charge of O

15. 15 Nitric acid.. :.. H O O O N : :.... ► Electron count of N is 4 (half of 8 electrons in covalent bonds). ► A neutral nitrogen has 5 electrons. ► Therefore, the formal charge of N is +1. Formal charge of N –

16. 16 Nitric acid.. :.. H O O O N : :.... ► A Lewis structure is not complete unless formal charges (if any) are shown. Formal charges – +

17. 17 Sure!! And I suppose the Pope is Jewish???

18. 18 Formal Charge Formal charge = Number of valenceelectrons number of bonds number of unshared electrons –– An arithmetic formula for calculating formal charge.

19. 19 – "Electron Counts" and Formal Charges in NH 4 + and BF 4 - 1 4 N H H H H + 7 4..B F F F F........ : : : : : :..

20. 20 Formal Charge: Selecting the Best Resonance Structure An atom “owns” all of its nonbonding electrons and half of its bonding electrons. Formal charge of atom = # valence e - - # unshared electrons - 1/2 # shared electrons

21. 21 For O A # valence e - = 6 # nonbonding e - = 4 # bonding e - = 4 X 1/2 = 2 Formal charge = 0 For O B # valence e - = 6 # nonbonding e - = 2 # bonding e - = 6 X 1/2 = 3 Formal charge = +1 For O C # valence e - = 6 # nonbonding e - = 6 # bonding e - = 2 X 1/2 = 1 Formal charge = -1

22. 22 EXAMPLE: NCO - has 3 possible resonance forms - Resonance and Formal Charge AB C

23. 23 Now Determine Formal Charges -20+10000 Forms B and C have smaller formal charges; this makes them more important than form A. (rule 1) Form C has a negative charge on O which is the more electronegative element, therefore C contributes the most to the resonance hybrid. (rule 2)

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25. 25 Exceptions to the Octet Rule

26. 26 Exceptions to the Octet Rule ► 4b. Expanded Octets – only on period 3 and higher  Expanded octets form when an atom can decrease (or maintain at 0) it’s formal charge  Ex: SF 6, PCl 5, SO 2, SO 3, SO 4 ► 5a. Electron deficient – have fewer than 8  Ex: BeCl 2, BF 3  may attain an octet by coordinate covalent bond ► Odd number of electrons – aka free radicals  Ex: NO 2  May attain an octet by pairing with another free radical

27. 27 SAMPLE PROBLEM: Writing Lewis Structures for Exceptions to the Octet Rule. PLAN: SOLUTION: PROBLEM: Write the Lewis structure for BFCl 2. Draw the Lewis structures for the molecule and determine if there is an element which can be an exception to the octet rule. BFCl 2 will have only 1 Lewis structure.

28. 28 Resonance and Formal Charge: Let’s see if you can: 1. Define resonance 2. Determine resonance structures for a molecule 3. Calculate the formal charge for an atom 4. Determine the resonance structure that contributes the most to a compound by using formal charge

29. 29 Now determine the formal charges and best structure for the 2 examples at the bottom of page 11. Your Turn

30. 30 Now determine the formal charges and best structure for the middle example on page 12. Your Turn

31. 31 1 0 11 00 Is there a better structure?? No!!

32. 32 The base is under assault!!!