Resonance and Formal Charge

1. Resonance and Formal Charge

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

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4. SAMPLE PROBLEM: Writing Resonance Structures PROBLEM: Write resonance structures for the nitrate ion, NO3-. PLAN: 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. SOLUTION: 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.

6. Four criteria for choosing the more important resonance structure: 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

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

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

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

11. Nitric acid 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 in the structure is 2 + 4 = 6. .. : O .. N H O .. : : O Formal charge of O ..

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

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

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

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

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

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18. .. 1 H : : F .. .. : : N B F F H H .. .. : : F H .. 4 "Electron Counts"and Formal Charges in NH4+ and BF4- 7 + – 4

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

20. Formal Charge An arithmetic formula for calculating formal charge. Formal charge = Number of valence electrons number ofbonds number ofunshared electrons – –

21. For OC For OA # valence e- = 6 # valence e- = 6 # nonbonding e- = 6 # nonbonding e- = 4 # bonding e- = 2 X 1/2 = 1 # bonding e- = 4 X 1/2 = 2 For OB Formal charge = -1 Formal charge = 0 # valence e- = 6 # nonbonding e- = 2 # bonding e- = 6 X 1/2 = 3 Formal charge = +1

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

23. Now Determine Formal Charges -2 0 +1 -1 0 0 0 0 -1 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. Exceptions to the Octet Rule

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: SF6, PCl5, SO2, SO3, SO4 5a. Electron deficient – have fewer than 8 Ex: BeCl2, BF3 may attain an octet by coordinate covalent bond Odd number of electrons – aka free radicals Ex: NO2 May attain an octet by pairing with another free radical

27. PROBLEM: Write the Lewis structure for BFCl2. SAMPLE PROBLEM: Writing Lewis Structures for Exceptions to the Octet Rule. PLAN: Draw the Lewis structures for the molecule and determine if there is an element which can be an exception to the octet rule. SOLUTION: BFCl2 will have only 1 Lewis structure.

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

29. Your Turn Now determine the formal charges and best structure for the 2 examples at the bottom of page 9.

30. Your Turn Now determine the formal charges and best structure for the middle example on page 13.

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

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