1. Lewis Structures and Formal Charge

2. Rules Governing Formal Charge Calculate Formal Charge Add up the lone pair electrons on the atom, and half of the shared electrons around the atom (those in bonds) Subtract the assigned electrons from the element’s valence electrons The sum of the formal charges on all atoms in a given molecule or ion must equal the overall charge on the species Molecules = zero Ions = ionic charge Which is the preferred Lewis Structure? Formal charges closest to zero Negative formal charges on the most electronegative atoms

3. Example #1 – Formaldehyde, CH 2 O Hydrogens  No unshared electrons  ½ of shared pair = 1 electron  1 valence electron – 1 electron = FC of 0 Hydrogens  No unshared electrons  ½ of shared pair = 1 electron  1 valence electron – 1 electron = FC of 0 Carbon  No unshared electrons  ½ of shared pairs = 4 electrons  4 valence electrons – 4 electrons = FC of 0 Carbon  No unshared electrons  ½ of shared pairs = 4 electrons  4 valence electrons – 4 electrons = FC of 0 Check: The sum of the formal charges is equal to the charge (0) Oxygen  4 unshared electrons  ½ of shared pairs = 2 electrons  6 valence electrons – 6 electrons = FC of 0 Oxygen  4 unshared electrons  ½ of shared pairs = 2 electrons  6 valence electrons – 6 electrons = FC of 0 0 0 0 0

4. Example #2 – Ammonium ion Hydrogens  No unshared electrons  ½ of shared pair = 1 electron  1 valence electron – 1 electron = FC of 0 Hydrogens  No unshared electrons  ½ of shared pair = 1 electron  1 valence electron – 1 electron = FC of 0 0 00 0 Nitrogen  No unshared electrons  ½ of shared pairs = 4 electrons  5 valence electrons – 4 electrons = FC of +1 Nitrogen  No unshared electrons  ½ of shared pairs = 4 electrons  5 valence electrons – 4 electrons = FC of +1 +1 Check: The sum of the formal charges is equal to the ionic charge (1+)

5. Evaluating Competing Structures: PO 4 3- In drawing a Lewis structure, the first consideration is the number of valence electrons available.  Phosphorus: 5 valence electrons 5e-  Oxygen: 6 valence electrons each 24e-  Charge: 3- (3 electrons in excess) 3e-  Total: 32e-  Phosphorus: 5 valence electrons 5e-  Oxygen: 6 valence electrons each 24e-  Charge: 3- (3 electrons in excess) 3e-  Total: 32e-

6. Potential Lewis Candidates All of these use exactly our 32e- “budget” All of these use exactly our 32e- “budget”

7. Potential Lewis Candidates Assign formal charges +1 0 0 0 0 0 0 0 -2 -3 0 0 0 0

8. Evaluate Using Formal Charges -3 0 0 0 0 Oxygen is more electronegative than phosphorus. Negative formal charge should be on oxygen.

9. Evaluate Using Formal Charges Oxygen is more electronegative than phosphorus. Negative formal charge should be on oxygen. 0 0 0 -2

10. Evaluate Using Formal Charges Oxygen is more electronegative than phosphorus. Negative formal charge should be on oxygen. 0 0

11. Evaluate Using Formal Charges Oxygen is more electronegative than phosphorus. Negative formal charge IS on oxygen. 0 0 +1  Oxygen is more electronegative than phosphorus. Negative formal charge IS on oxygen.  However, this has a greater set of formal charges assigned than the structure above.  Oxygen is more electronegative than phosphorus. Negative formal charge IS on oxygen.  However, this has a greater set of formal charges assigned than the structure above.