1. LEWIS STRUCTURES OBJECTIVES: 1) TO SHOW ELECTRON DOMAINS AROUND THE CENTRAL ATOM. ELECTRON DOMAINS ARE: A) CHEMICAL BONDS B) UNBONDED PAIRS OF ELECTRONS 2) TO SHOW THE CORRECT GEOMETRY AS INDICATED IN THE VSEPR CHART. 3) TO ANALYZE FOR POLARITY AND DIPOLE MOMENT. 4) TO ANALYZE RESONANCE IN DOUBLE BONDS PRESENT

2. WE WILL USE AMMONIA, NH3 AS AN EXAMPLE
WE WILL USE AMMONIA, NH3 AS AN EXAMPLE FOR AMMONIA N: 1 X 5 = H: 3 X 1= TOTAL VALENCE ELECTRONS
STEP 1: CALCULATE THE TOTAL VALENCE ELECTRONS IN THE MOLECULE
ADD ONE ELECTRON FOR EACH NEGATIVE CHARGE. SUBTRACT ONE ELECTRON FOR EACH POSITVE CHARGE.
ALL VALENCE ELECTRONS MUST BE INCLUDED IN LEWIS STRUCTURE, NO MORE, NO LESS

3. FOR AMMONIA N: 1 X 8 = 8 H: 3 X 2= 6 14 TOTAL VALENCE ELECTRONS NEEDED
STEP 2: CALCULATE ELECTRONS NEEDED TO COMPLETE THE OCTET OF ALL NON HYDROGEN ELEMENTS, EACH HYDROGEN NEEDS 2 ELECTRONS.
THE NUMBER OF ELECTRONS YOU WOULD NEED TO STABALIZE EACH OCTET ATOM WITH 8, AND EACH HYDROGEN WITH 2.

4. # SHARED(BONDED) = # NEEDED - # VALENCE 6 = 14 - 8
STEP 3: CALCULATE ELECTRONS SHARED
# SHARED(BONDED) = # NEEDED - # VALENCE

5. # BONDS = # SHARED(BONDED) /2 # BONDS = 6/2 = 3 BONDS
STEP 4: CALCULATE NUMBER OF BONDS (BONDED ELECTRON PAIRS)
# BONDS = # SHARED(BONDED) /2

6. H H N H STEP 5: RENDER YOUR PRELIMINARY STRUCTURE, SHOW BONDS.
NOTE: THE CENTRAL ATOM IS THE LEAST ELECTRONEGATIVE NON HYDROGEN ELEMENT. HYDROGEN CANNOT BE CENTRAL AS IT ONLY MAKES ONE BOND H H N H

7. H H N H STEP 7:CALCULATE THE UNBONDED ELECTRON PAIRS .
# UNSHARED= # VALENCE ELECTRONS – BONDED e- =
ASSIGN UNBONDED ELECTRONS TO LIGANDS FIRST AND THEN ASSIGN TO CENTRAL. H H N H

8. H H N H ∂- ∂+ ∂- ∂+ ∂+ ∂+ ASSIGEN THE MOST ELECTRONEGATIVE ATOM(S)
A (NEGATIVE POLE) SYMBOL. ASSIGN THE LEAST ELECTRONEGATIVE ATOMS A (POSITIVE POLE SYMBOL). DRAW ARROWS TO INDICATE THE POLARITY OF EACH BOND, POINT ARROW TO MOST ELECTRONEGATIVE ATOM. ∂- ∂+ ∂- H H N ∂+ ∂+ H ∂+

9. H H N H STEP 7:CALCULATE THE UNBONDED ELECTRON PAIRS .
# UNSHARED= # VALENCE ELECTRONS – BONDED e- =
ASSIGN UNBONDED ELECTRONS TO LIGANDS FIRST AND THEN ASSIGN TO CENTRAL. H H N H

10. STEP 1: CALCULATE THE TOTAL VALENCE ELECTRONS IN THE MOLECULE
EXAMPLE MOLECULE , IF FOR IODINE TETRA FLOURIDE I:1 X 7= F:4 X 7= VALENCE ELECTRONS e- FOR + CHARGE TOTAL VELANCE e-
STEP 1: CALCULATE THE TOTAL VALENCE ELECTRONS IN THE MOLECULE
ADD ONE ELECTRON FOR EACH NEGATIVE CHARGE. SUBTRACT ONE ELECTRON FOR EACH POSITVE CHARGE.
ALL VALENCE ELECTRONS MUST BE INCLUDED IN LEWIS STRUCTURE, NO MORE, NO LESS

11. FOR IODINE TETRA FLOURIDE I: 1 X 8= 8 F: 4 X 8= 32 40 TOTAL VALENCE ELECTRONS NEEDED
STEP 2: CALCULATE ELECTRONS NEEDED TO COMPLETE THE OCTET OF ALL NON HYDROGEN ELEMENTS, EACH HYDROGEN NEEDS 2 ELECTRONS.
THE NUMBER OF ELECTRONS YOU WOULD NEED TO STABALIZE EACH OCTET ATOM WITH 8, AND EACH HYDROGEN WITH 2.

12. STEP 3: CALCULATE ELECTRONS SHARED
# SHARED(BONDED) = # NEEDED - # VALENCE = DIVIDE BONDED e- BY 2 TO CALCULATE BONDS 6/2 = 3 BONDS
STEP 3: CALCULATE ELECTRONS SHARED
# SHARED(BONDED) = # NEEDED - # VALENCE

13. CENTRAL ATOM IS LEAST ELECTRONEGATIVE NON HYDROGEN ATOM
STEP 5: RENDER YOUR PRELIMINARY STRUCTURE, SHOW BONDS.
NOTE: THE CENTRAL ATOM IS THE LEAST ELECTRONEGATIVE NON HYDROGEN ELEMENT. HYDROGEN CANNOT BE CENTRAL AS IT ONLY MAKES ONE BOND F F I F
CENTRAL ATOM IS LEAST ELECTRONEGATIVE NON HYDROGEN ATOM F
NOTICE WE DO NOT HAVE ENOUGH BONDS FOR ALL THE FLOURINES, DON’T WORRY…WE WILL BOND IT IN THE NEXT STEP

14. F F I F F STEP 7:CALCULATE THE UNBONDED ELECTRON PAIRS .
# UNSHARED= # VALENCE ELECTRONS – BONDED e-
28 (14 PAIR) =
ASSIGN UNBONDED ELECTRONS TO LIGANDS FIRST (to give an octet) AND THEN ASSIGN REMAINNING e- TO CENTRAL. F F I F F
THIS FLOURINE WILL BOND TO THE IODINE BY USING ONE UNSHARED PAIR AS THE NEW BOND. It is not an ion…will have this octet after bonding.
ALL LIGANDS NOW HAVE OCTET, WE HAVE USED 26 OF THE 28 ELECTRONS, THE REMMAINING PAIR GOES ON CENTRAL ATOM

15. THIS MOLECULE HAS 5 DOMAINS, TRIG BIPYRAMIDAL FAMILY: WITH 4 BONDS AND ONE UNSHARED PAIR ON THE CENTRAL ATOM. F F I F F

16. THIS MOLECULE HAS 5 DOMAINS, TRIG BIPYRAMIDAL FAMILY: WITH 4 BONDS AND ONE UNSHARED PAIR ON THE CENTRAL ATOM. F
THIS MOLECULE IS AX4E1 FORM THE VSEPR CHART AND IS DISTORTED TETRAHEDRAL AKA
SEESAW F I F F

17. IDENTIFY THE MOST ELECTRONEGATIVE ELEMENT(S) AND ASSIGN THE NEGATIVE POLE SYMBOL ∂-.
THIS MOLECULE IS AX4E1 FORM THE VSEPR CHART AND IS DISTORTED TETRAHEDRAL AKA
SEESAW F ∂- I F ∂- F ∂-

18. IDENTIFY THE LEAST ELECTRONEGATIVE ELEMENT(S) AND ASSIGN THE POSITIVE POLE SYMBOL ∂+.∂- F F ∂- I ∂+ F ∂- F ∂-

19. ASSIGN POLARITY ARROWS POINTING TO THE MOST ELECTRONEGATIVE ELEMENT IN EACH BOND.∂- F F ∂- I ∂+ F ∂- F ∂-

20. THIS IS A NON-SYMMETRICAL AND POLAR MOLECULE
DRAW A COMPOSIT ARROW TO INDICATE NET DIPOLE MOMENT (POLARITY). ARROW ALWAYS POINTS TO THE MOST ELECTRONEGATIVE ATOM(S) ∂- F
THIS IS A NON-SYMMETRICAL AND POLAR MOLECULE F ∂- I ∂+ F ∂- F ∂-