CH. 9
MOLECULAR SHAPE
Bonding patterns
Molecular shape
Polarity - Dipole
Bond order
VSEPR, shape, angle
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
Equations
formal charge (f.c.) = # val. e- - (# unbond e- + 0.5 # bond e-)
# e - pairs
Bond order :
# bonded pairs
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
Let’s look at O3
. .
O
b
:O: 0
a
Oa: 6 val e- 6 - [4 + .5(4)]
4 unbonded 6 - (4 + 2)
4 bonded 6 - 6 = 0
+1
-1
:O:
..
g
Ob: 6 val e- 6 - [2 + .5(6)]
2 unbonded 6 - (2 + 3)
6 bonded 6 - 5 = +1
Og: 6 val e- 6 - [6 + .5(2)]
6 unbonded 6 - (6 + 1)
2 bonded 6 - 7 = -1
LEWIS DIAGRAMS
^ not show shape
Helps in understanding bonding in cmpds/molecules
Y ..
..
X Y <---> Y X Y
Y
Y
Look at:
1 central atom
2+ central atoms
multiple bonds
resonance
BONDING PATTERNS/REQUIREMENTS
Hydrogen (1)
Nitrogen (3)
Oxygen (2)
Halogens (1)
X = F, Cl, Br, I
carbon (4)
..
N
H
..
O
X
..
:
C
..
N
:
O
C
N
C
+1
Nitrogen Ion, N+1, (4)
N
P:3
S:3
C
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
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
4
6
6
* add 1 e- for each “-” charge
C-4
O-2
Cl-1
8
8
8
* substr 1 e- for each “+” charge
C+4
N+1
2
4
Halogens
F, Cl, Br, I
F never central atom
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-:
4 1 1 1
4 7 7 (2*7) = 32 e-
central atom: C
attached: 1 Br 1 F 2 I
Place 2 e-’s to
show bonds
Br
..
I:..C:I
F
32 - 8 = 24 e-’s left to account for
..
..
.. ..
.. Br
:..I:C:..I:
..
:
:F
..
..
Complete octet
24 - 12 = 12 - 8 = 4 - 4 = 0
No e-’s left, no multiple bonds
..
: Br:
Final step, replace bonding pairs with line
to represent the bond(s) between
..
:I..
..
C I..:
:
PCl3
cen.atom P
.. .. ..
: Cl:P
:
.. .. :Cl
..
: Cl:
..
attach 3 P’s
F
:
..
26 val e-
26 - 6 = 20 - 18 = 2 - 2 = 0
NHI2
bonds 3 1 1
val e- 5 1 2*7 = 20
.. ..
: I :N : I :
.. .. ..
..
20 - 6 = 14 - 12 = 2 - 2 = 0
H
H2S
. .
..
H--S--H
.S .
H
..
H
. .
OF2
..
..
..
..
..
:F--O--F :
.O .
:F:
..
..
:F:
..
LEWIS
STRUCTURE
S
2 central atoms, single bonds
CH3OH
Determine central atoms
only C & O can have multiple bonds
H only 1 bond
..
H ..
..
H ..
C OH
..
H
# Val. e4 + 3 + 6 + 1 = 14 e-
H
H C
H
O
H
LEWIS
STRUCTURES
2 central atoms, single bonds
NH3O
Determine central atoms
only N & O can have multiple bonds
H only 1 bond
..
.. ..
H ..
N OH
..
H
H ..N
H
# Val. e5 + 3 + 6 = 14 e-
O
H
2+ central atoms, OH bond
Determine central atoms
only C & O can have multiple bonds
H only 1 bond
H
..
C O H
..
20 - 16 = 4
bonding e-
..
H
..
H ..
C
H H
H
H C
C
H H
4-4=0
nonbonding e-
O
H
# Val. e8 + 6 + 6 = 20 e-
w/o OH bond
H
H
..
.. ..
H ..
C O ..
C H
H
H
..
H
C2H6O
..
LEWIS
STRUCTURES
LEWIS
STRUCTURES
HCN
multiple bonds
bonds 1 4 3
val e- 1 4 5 = 10
H
..
.... :
: C .. N
..
H--C----N :
EXCEPTION
central atom not octet, move
l.p. to b.p. to central atom
CO2
bonds 4 2
..
..
:O
..
..
4 2*6 = 16
..
C O.. :
C
:O:
val e-
16 - 4 = 12 - 12 = 0
..
:O:
..
O--C--- O..
..
RESONANCE
e- pair “vibrate” back-forth bet atoms,
fills octet rule
dbl bonds next to single bonds
OZONE, O3
O=O=O
Too many “O” bonds
. .
. .
Soooooo
O
:O:
..
O
:O:
:O:
:O:
..
results from “e--pair delocalization”
Can show as
. .
O
:O:
..
:O:
bond order = 3/2 = 1.5
BENZENE -- C6H6
or
Bond order =
9/6 = 1.5
POLYATOMIC IONS
Show as [
]charge
..
: O:




N


: O: 
 :O
..
.. 
NO3-1
5 3*6
= 5 + 18
= 23 + 1
= 24
..
: O:




N


: 
 : O : O
.. 
Bond order = ???
4/3 = 1.333
..
:O


N

 :O.. : : O :
..
..
:O

 
N
: :O:
 :O
..
..




-1




-1
..
:O:



 
N

 : O : O.. : 
..
: O:


N

: O:
 :O
..
..
-1
..
..
-1
-1




-1
EXCEPTION
S
* e--deficient atoms
* odd # e- atoms
* expanded val shells
deficient atoms
Be
B
2, 4
3
Odd # free radicals
contain unpair eparamagnetic
NO2
..
.
N
N
:O:
:O:
..
:O:
More imprt due to way reacts, as free
radicals react w/ each other to pair e-
:O.
..
EXPANDED
SF6 PCl5
Exothermic
use empty “d” orbitals
period 3+ (nonmetals)
H2SO4
P
5
S
6
I
7
-2H ----> SO4-2
..
: O:
..
..
H O
.. H
.. S O
:O
.. :
: O:
..
..
H O
.. H
.. S O
:O :
More imprt, observed
bond lengths
 : O: 


..
..


O
S
O
.. 
 ..


 :O : 
-2
Bond order:
6/4 = 1.5
Draw most likely structue for:
..
: Cl :
..
..
POCl3
ClO2
..
: Cl :
: O:
*
..
..
P --Cl
:
..
P -- O
-- Cl
:
..
..
:Cl:
..
32 e-
:Cl:
..
P: 5 - [2 + .5(6)]
5 - (2 + 3) = 0
P: 5 - [0 + .5(10)]
5 - (0 + 5) = 0
O: 6 - [4 + .5(4)]
6 - (4 + 2) = 0
O: 7 - [6 + .5(2)]
7 - (6 + 1) = 0
Cl: 7 - [6 + .5(2)]
7 - (6 + 1) = 0
Cl: 6 - [4 + .5(4)]
6 - (4 + 2) = 0
e-
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
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
surrounding atom
m indicates # of
nonbonding en indicates # of
GEOMETRY AROUND CENTRAL ATOM
Possible Bonding Sites
Bonds
Linear 2
Trigonal Planar
(Planar Triangular)
Tetrahedral
Trigonal Bipyramidal
Octahedral
Lone Pair
0
3
2
0
1
4
3
2
0
1
2
5
4
3
2
6
5
4
Geometry
Linear
pg 337
Formula
Bond Angle
AX2
1800
Trigonal Planar
AX3
Angular (Bent or V) AX2E
1200
< 1200
AX4
AX3E
AX2E2
109.50
1070
1050
0
1
2
3
Trigonal Bipyramidal AX5
Seesaw
AX4E
T-shaped
AX3E2
Linear
AX2E3
900, 1200
1730, 1020
87.50
1800
0
1
2
Octahedral
Square Pyramidal
Square Planar
Tetrahedral
Pyramidal
Bent or V
AX6
AX5E
AX4E2
900
850
900
Draw Lewis structure for phosphorus trichloride, PCl3
Total valence e-’s P: 5
Cl: 3*7 =21
total=26
Central atom: P
Attached: 3 Cl’s
..
:Cl
..
..
..
P Cl
.. :
:Cl:
..
GEOMETRY
4 possible bonding sites, tetrahedral
in this case; 3 bonds, 1 lone pair
3-D DRAWING
AX3E Pyramidal
..
..
: Cl
..
P
..
:Cl
..
..
Cl
.. :
Use VSEPR to Predict
Lewis
Structure
assign
e- group
Bond
Angle
Molecular
Shape
Polarity: take shape into acct;
polar bonds present but counterbalanced
results in NP (no dipole moment, m)