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

1
LEWIS STRUCTURESOBJECTIVES1) TO
SHOW ELECTRON DOMAINS AROUND THE CENTRAL ATOM.
ELECTRON DOMAINS ARE A) CHEMICAL BONDS B)
UNBONDED PAIRS OF ELECTRONS2) 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. FOR AMMONIAN 1 X 5
5 H 3 X 1 3 8 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 AMMONIAN 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

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

STEP 7CALCULATE THE UNBONDED ELECTRON PAIRS .
UNSHARED VALENCE ELECTRONS BONDED e-
2 8
- 6 ASSIGN UNBONDED ELECTRONS TO
LIGANDS FIRST AND THEN ASSIGN TO CENTRAL.
H
H
N
H
8

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

STEP 7CALCULATE THE UNBONDED ELECTRON PAIRS .
UNSHARED VALENCE ELECTRONS BONDED e-
2 8
- 6 ASSIGN UNBONDED ELECTRONS TO
LIGANDS FIRST AND THEN ASSIGN TO CENTRAL.
H
H
N
H
10
EXAMPLE MOLECULE ,
IF4 FOR IODINE TETRA FLOURIDE I1 X
7 7 F4 X 7 28 35
VALENCE ELECTRONS -1 e- FOR
CHARGE 34 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
SHARED(BONDED) NEEDED - VALENCE
6 40
- 34DIVIDE BONDED e- BY 2 TO
CALCULATE BONDS6/2 3 BONDS

STEP 3 CALCULATE ELECTRONS SHARED
SHARED(BONDED) NEEDED - VALENCE
13

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, DONT WORRYWE WILL BOND IT IN THE
NEXT STEP
14

STEP 7CALCULATE THE UNBONDED ELECTRON PAIRS .
UNSHARED VALENCE ELECTRONS BONDED e-
28 (14 PAIR) 34
- 6 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
ionwill 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 ?-.
F
?-
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
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
?-
?-