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, )