CH. 9 MOLECULAR SHAPE
Bonding patterns
Molecular shape
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
EquationsEquations
formal charge (f.c.) = # val. e- - (# unbond e- + 0.5 # bond e-)
pairs bonded #
pairs e # :order Bond
-
FORMAL CHARGEFORMAL 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: 6 val e-
4 unbonded 4 bonded . .
O
:O: :O:
..
6 - [4 + .5(4)] 6 - (4 + 2) 6 - 6 = 0
O: 6 val e-
2 unbonded 6 bonded
6 - [2 + .5(6)] 6 - (2 + 3) 6 - 5 = +1
O: 6 val e-
6 unbonded 2 bonded
6 - [6 + .5(2)] 6 - (6 + 1) 6 - 7 = -1
LEWIS DIAGRAMS ^ not show shape
Y .. .. X Y <---> Y X YY Y
Look at: 1 central atom 2+ central atoms multiple bonds resonance
Helps in understanding bonding in cmpds/molecules
BONDING PATTERNS/REQUIREMENTSBONDING PATTERNS/REQUIREMENTS
Hydrogen (1) Nitrogen (3) Oxygen (2) Halogens (1) carbon (4) X = F, Cl, Br, I
H N..
N..
N:
O.. ..
O..
..
X ..: ..
C
C
C
C
Nitrogen Ion, N+1, (4) N
+1P : 3S : 3
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 Determine centralcentral atom 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# 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, IF never central atom
Bonding Bonding centralcentral atom atom * place single bond to each atom bonded to central atom * substr 2 e- for each single bond from total val e- count Remaining eRemaining e-
* place pairs of e- to complete octet rule to each attached atom * any remaining e- place pairs around central atom
CBrFI2
bond formed: 4 1 1 1
# val e-: 4 7 7 (2*7) = 32 e-
centralcentral atom atom: Cattached: 1 Br 1 F 2 I
Place 2 e-’s to show bonds
32 - 8 = 24 e-’s left to account for
- 8 = 4
Final step, replace bonding pairs with lineto represent the bond(s) between
No e-’s left, no multiple bonds..
..
..:
Br I:C:I F
..
..
24 - 12 = 12..
..
..
Complete octet Br :I:C:I:
: F
..
..
.. .. - 4 = 0
.. .. :I C I:
: F:
..
..
..
..: Br:
PCl3 cen.atom P attach 3 P’s 26 val e-
Cl:P:Cl Cl
.. 26 - 6 = 20..
- 18 = 2:
:
:
:....
..
....
- 2 = 0
NHI2
bonds 3 1 1
OFOF22
H2S
val e- 5 1 2*7 = 20
::....
.... I :N: I H
.. 20 - 6 = 14 - 12 = 2
..
- 2 = 0
. .
.S .
H H
..
H--S--H
..
. .
.O .
:F: :F:
..
..
.. .. ..
:F--O--F : .. ..
..
LEWIS STRUCTURES
2 central atoms, single bonds
Determine Determine centralcentral atoms atomsonly C & O can have multiple bondsH only 1 bond
CH3OH
CH
H
H
O....
..
.. ....H
..
# Val. e# Val. e- 4 + 3 + 6 + 1 = 14 e-
C
H
H
H
O
....H
LEWIS STRUCTURES2 central atoms, single bonds
Determine Determine centralcentral atoms atomsonly N & O can have multiple bondsH only 1 bond
NH3O
NH
H
O....
..
.. ....H
..
# Val. e# Val. e- 5 + 3 + 6 = 14 e-
O
....
H
..
NH
H
LEWIS STRUCTURES2+ central atoms, OH bond
Determine Determine centralcentral atoms atomsonly C & O can have multiple bondsH only 1 bond
C2H6O
# Val. e# Val. e- 8 + 6 + 6 = 20 e-
H
..CH
H....
..
..
H
C O
..
H
H
....
20 - 16 = 4bonding e-
4 - 4 = 0nonbonding e-O
....
H
CH
H
H
H
H
C
w/o OH bond
..
..
H
H....
H
..O
..CH
H....
..
H .. C
HCNbonds 1 4 3
val e- 1 4 5 = 10
....H : C N..
LEWIS STRUCTURESmultiple bonds
H--C--N :----
:
EXCEPTIONcentral atom not octet, move l.p. to b.p. to central atom
..
..:
bonds 4 2
COCO22
val e- 4 2*6 = 16
::....
....
16 - 4 = 12 - 12 = 0 O C O.. ..
C
:O: :O:
.. ..
O--C-- O ..
..
-- --
RESONANCE
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:
..
results from “e--pair delocalization”
Can show as
bond order = 3/2 = 1.5
. .
O
:O: :O:
..
BENZENE -- C6H6
or
Bond order = 9/6 = 1.5
POLYATOMIC IONS
Show as [ ]charge
NO3-1
5 3*6= 5 + 18= 23 + 1 = 24
1-
O O
N
O
1-
O O
N
O
1-
O O
N
O
::
....
....
::.... ..
..:
:..
: :
: :
: : : ::
1-
O O
N
O
..
..
..::
:
::
1-
O O
N
O
::
::....
..
:
1-
O O
N
O
..
..:
..
:
: : :
4/3 = 1.333Bond order = ???
EXCEPTIONS
deficientdeficient atoms atoms Be B 2, 4 3
* e--deficient atoms* odd # e- atoms* expanded val shells
Odd #Odd # free radicals contain unpair e-
paramagnetic
NO2
.
N
:O: :O:
..
. .
N
:O: :O.
..
More imprt due to way reacts, as freeradicals react w/ each other to pair e-
EXPANDED SF6 PCl5
Bond order: 6/4 = 1.5
Exothermicuse empty “d” orbitalsperiod 3+ (nonmetals)
PP SS II 5 6 7
H2SO4
.. ..H O S O H
:O:
..
..
..
..: O:
-2H ----> SO4-2
.. ..H O S O H
:O:
....
: O:More imprt, observedbond lengths
.. .. O S O
:O:
....
: O:2
PP: 5 - [2 + .5(6)] 5 - (2 + 3) = 0
OO: 6 - [4 + .5(4)] 6 - (4 + 2) = 0
ClCl: 7 - [6 + .5(2)] 7 - (6 + 1) = 0
POCl3 ClO2
..
: Cl : . . . . . .
P -- O -- Cl :
. . . .
:Cl:
..
..
: Cl : : O :
. .
P --Cl :
. .
:Cl:
..
Draw most likely structue for:
*32 e- e-
PP: 5 - [0 + .5(10)] 5 - (0 + 5) = 0
OO: 7 - [6 + .5(2)] 7 - (6 + 1) = 0
ClCl: 6 - [4 + .5(4)] 6 - (4 + 2) = 0
ClCl: 7 - [3 + .5(8)] 7 - (3 + 4) = 0
OO: 6 - [4 + .5(4)] 6 - (4 + 2) = 0
ClO2
. . . . .
: O--Cl-- O : . . . .
. .
. .
: O--Cl-- O : . . .
. .
-- --
VSEPR pg334 - 43
Valence Shell Electron Pair Repulsion
Lewis: 2-D, shows relative placement of atoms, the “building” plans, not shapeVSEPR: 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 atomsurrounding atomm indicates # of
nonbonding e-
n indicates # of
Possible Bonding SitesPossible Bonding Sites BondsBonds Lone Pair Geometry Formula Bond Angle
2 0 Linear AX2 1800
3 0 Trigonal Planar AX3 1200
2 1 Angular (Bent or V) AX2E < 1200
4 0 Tetrahedral AX4 109.50
3 1 Pyramidal AX3E 1070
2 2 Bent or V AX2E2 1050
5 0 Trigonal Bipyramidal AX5 900, 1200
4 1 Seesaw AX4E 1730, 1020
3 2 T-shaped AX3E2 87.50
2 3 Linear AX2E3 1800
6 0 Octahedral AX6 900
5 1 Square Pyramidal AX5E 850
4 2 Square Planar AX4E2 900
GEOMETRY AROUND CENTRAL ATOM pg 337
Linear
Trigonal Planar (Planar Triangular)
Tetrahedral
Trigonal Bipyramidal
Octahedral
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 AX3E Pyramidal 4 possible bonding sites, tetrahedralin this case; 3 bonds, 1 lone pair
3-D DRAWING
P. .
Cl
Cl
Cl
..:
..:
.. :
..
....
Use VSEPR to Predict
LewisStructure
Molecular Shape
assigne- group
BondAngle
Polarity: take shape into acct; polar bonds present but counterbalanced results in NP (no dipole moment, )
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