1. CH. 9 MOLECULAR SHAPE Bonding patterns Polarity - Dipole
VSEPR, shape, angle
Polarity - Dipole
Bond order
Effects of repulsion on bond :
lone pair - l.p. > l.p. - bonding pair > b.p - b.p.
Look at: Lewis structure, resonance, formal charge, radicals

2. Equations
formal charge (f.c.) = # val. e- - (# unbond e # bond e-)

3. FORMAL CHARGE f.c. = # val. e- - (# unshare e- + 0.5 # share e-)
What does the atom own???
* all unbonded e- pairs
* 0.5 of bonding e-
Have 2+ possible arrangements, which structure imprt??
* smaller f.c. to large
* f.c. not side-by-side
* more -f.c. on more -EN atom

4. . . O Let’s look at O3 Oa: 6 val e- 4 unbonded 4 bonded
6 - [4 + .5(4)]
6 - (4 + 2)
6 - 6 = 0 +1 -1 b
a g
. . O
:O: :O: ..
Ob: 6 val e-
2 unbonded
6 bonded
6 - [2 + .5(6)]
6 - (2 + 3)
6 - 5 = +1
Og: 6 val e-
6 unbonded
2 bonded
6 - [6 + .5(2)]
6 - (6 + 1)
6 - 7 = -1

5. LEWIS DIAGRAMS ^ not show shape
Helps in understanding bonding in cmpds/molecules Y
X Y <---> Y X Y Y
Look at:
1 central atom
2+ central atoms
multiple bonds
resonance

6. BONDING PATTERNS/REQUIREMENTS
Hydrogen (1) Nitrogen (3) Oxygen (2) Halogens (1) carbon (4)
X = F, Cl, Br, I N .. : O .. C H X .. : N +1
P : 3
S : 3
Nitrogen Ion, N+1, (4)

7. LEWIS STRUCTURES
Quick Overview
1. Sum the total number of valence electrons from all
atoms in subst.
2. Identify central atom. Show which atoms are bonded to each other
3. Place 2 e-’s to show a bond between each atom
4. Complete the octet rule for each atom
5. If not enough e-’s to give central atom octet, try multiple bonds

8. LEWIS STRUCTURES 1 central atom, single bonds Determine central atom
* forms 2+ bonds, octet rule
* lower group #, lower EN
N - O N - C P - Cl
N C P
* from same group, the higher period #
N - P S - O Br - F
P S Br
# Val. e-
* add total # e- of all atoms
C O S
* add 1 e- for each “-” charge
C O Cl-1
* substr 1 e- for each “+” charge
C N+1
Halogens
F, Cl, Br, I
F never central atom

9. Bonding central atom
* place single bond to each atom bonded to central atom
* substr 2 e- for each single bond from total val e- count
Remaining e-
* place pairs of e- to complete octet rule to each attached atom
* any remaining e- place pairs around central atom
CBrFI2
bond formed:
# val e-: (2*7) = 32 e-
central atom: C
attached: 1 Br 1 F 2 I

10. 32 - 8 = 24 e-’s left to account for
Place 2 e-’s to
show bonds Br
I:C:I F ..
= 24 e-’s left to account for ..
Complete octet Br
:I:C:I:
: F
= 12
- 8 = 4
- 4 = 0 .. .. :
No e-’s left, no multiple bonds ..
:I C I:
: F:
: Br:
Final step, replace bonding pairs with line
to represent the bond(s) between
PCl3 cen.atom P attach 3 P’s val e- : .. ..
Cl:P:Cl Cl ..
= 20
- 18 = 2
- 2 = 0

11. NHI2
bonds
val e *7 = 20 : .. ..
I :N: I H ..
= 14
- 12 = 2
- 2 = 0
H2S
. .
.S .
H H ..
H--S--H
:F--O--F :
. .
.O .
:F: :F:
OF2

12. LEWIS STRUCTURES .. .. .. .. .. .. 2 central atoms, single bonds CH3OH
Determine central atoms
only C & O can have multiple bonds
H only 1 bond
# Val. e-
= 14 e- C H .. C H .. .. .. O .. O .. H H

13. LEWIS STRUCTURES .. .. .. .. .. 2 central atoms, single bonds NH3O
Determine central atoms
only N & O can have multiple bonds
H only 1 bond
# Val. e-
= 14 e- .. N H .. O .. H N .. O .. H

14. LEWIS STRUCTURES .. .. .. .. .. .. 2+ central atoms, OH bond C2H6O
Determine central atoms
only C & O can have multiple bonds
H only 1 bond
# Val. e-
= 20 e- H .. C O
= 4
bonding e-
w/o OH bond O .. C H H .. .. O .. H C .. C
4 - 4 = 0
nonbonding e-

15. -- -- LEWIS STRUCTURES HCN multiple bonds bonds 1 4 3
val e = 10 .. :
H : C N .. .. : ..
H--C--N : -- --
EXCEPTION
central atom not octet, move
l.p. to b.p. to central atom

16. C .. .. -- .. .. CO2 bonds 4 2 val e- 4 2*6 = 16 O C O 16 - 4 = 12: ..
O C O ..
= 12
- 12 = 0 C
:O: :O:
O--C-- O --

17. Soooooo . . . . O O RESONANCE O=O=O dbl bonds next to single bonds
e- pair “vibrate” back-forth bet atoms,
fills octet rule
OZONE, O3
O=O=O
Too many “O” bonds
Soooooo
. . O
:O: :O: ..
. . O
:O: :O: ..

18. . . O results from “e--pair delocalization” Can show as :O: :O: ..
. . O
:O: :O: ..
Can show as
bond order = 3/2 = 1.5

19. BENZENE -- C6H6 or
Bond order =
9/6 = 1.5

20. POLYATOMIC IONS Show as [ ]charge NO3-1 5 3*6 = 5 + 18 = 23 + 1 = 24: ..
NO3-1 *6 = = .. : .. :
= 24
4/3 = 1.333
Bond order = ??? : ..

21. EXCEPTIONS . . . N N * e--deficient atoms * odd # e- atoms
* expanded val shells
deficient atoms
Be B 2,
Odd # free radicals
contain unpair e-
paramagnetic
NO2 . N
:O: :O: ..
. . N
:O: :O. ..
More imprt due to way reacts, as free
radicals react w/ each other to pair e-

22. EXPANDED SF6 PCl5 Exothermic use empty “d” orbitals
period 3+ (nonmetals)
P S I
H2SO4
-2H ----> SO4-2 ..
H O S O H
:O:
: O: ..
O S O
:O:
: O:
More imprt, observed
bond lengths ..
H O S O H
:O:
: O:
Bond order:
6/4 = 1.5

23. * .. .. . . . . . . . . P --Cl : P -- O -- Cl : POCl3 ClO2
Draw most likely structue for:
POCl3 ClO2
32 e e- * ..
: Cl : : O :
. .
P --Cl :
:Cl: ..
: Cl :
P -- O -- Cl :
:Cl:
P: 5 - [0 + .5(10)]
5 - (0 + 5) = 0
O: 7 - [6 + .5(2)]
7 - (6 + 1) = 0
Cl: 6 - [4 + .5(4)]
6 - (4 + 2) = 0
P: 5 - [2 + .5(6)]
5 - (2 + 3) = 0
O: 6 - [4 + .5(4)]
6 - (4 + 2) = 0
Cl: 7 - [6 + .5(2)]
7 - (6 + 1) = 0

24. ClO2
: O--Cl-- O :
. .
: O--Cl-- O : --
Cl: 7 - [3 + .5(8)]
7 - (3 + 4) = 0
O: 6 - [4 + .5(4)]
6 - (4 + 2) = 0

25. VSEPR pg334 - 43 Valence Shell Electron Pair Repulsion
Lewis: 2-D, shows relative placement of atoms,
the “building” plans, not shape
VSEPR: molecular shape;
minimizes e- repulsion, val e- around central
atom will locate as far away as possible from
other e- to minimize repulsions
assigned designation
AXmEn
central atom
nonbonding e-
n indicates # of
surrounding atom
m indicates # of

26. GEOMETRY AROUND CENTRAL ATOM pg 337
Possible Bonding Sites Bonds Lone Pair Geometry Formula Bond Angle
Linear
Trigonal Planar
(Planar Triangular)
Tetrahedral
Trigonal Bipyramidal
Octahedral
Linear AX
Trigonal Planar AX
Angular (Bent or V) AX2E < 1200
Tetrahedral AX
Pyramidal AX3E
Bent or V AX2E
Trigonal Bipyramidal AX , 1200
Seesaw AX4E , 1020
T-shaped AX3E
Linear AX2E
Octahedral AX
Square Pyramidal AX5E
Square Planar AX4E

27. Draw Lewis structure for phosphorus trichloride, PCl3
Total valence e-’s P: Cl: 3*7 = total=26
Central atom: P Attached: 3 Cl’s ..
:Cl P Cl:
:Cl:
4 possible bonding sites, tetrahedral
in this case; 3 bonds, 1 lone pair
GEOMETRY
AX3E Pyramidal
3-D DRAWING P
. . Cl .. :

28. Use VSEPR to Predict
assign
e- group
Bond
Angle
Molecular
Shape
Lewis
Structure
Polarity: take shape into acct;
polar bonds present but counterbalanced
results in NP (no dipole moment, m)