Modern Chemistry Chapter 6 Chemical Bonding Part V Molecular Geometry and Intermolecular Forces

Modern ChemistryChapter 6

Chemical Bonding

Part VMolecular Geometry and Intermolecular

Forces

Chapter 6 Section 5 Molecular Geometry pages 197-207

2

Chapte

r V

oca

bula

ryVSEPR theoryHybridizationHybrid orbitalsDipoleHydrogen bonding London dispersion

forces

VSEPR Theory

• Valence-Shell Electron-Pair Repulsion

• Predicts the shapes of molecules and ions in which valence shell electron pairs are arranged about each atom so that the electron pairs are kept as far apart as possible, thus minimizing repulsions.

VSEPR Theory

• Molecular geometries are predicted by considering the number of regions of electron density.

• Single, double, and triple bonds, as well as lone pairs are all considered one region of electron density.

• Lone pairs require more space than bonding pairs

VSEPR Theory

• To predict the relative positions of atoms around a given central element using VSEPR theory, first note the arrangement of valence electron pairs around the central atom.

• See VSEPR summary sheet on p. 3 of Ch. 6 note packet pt. 2

6

VSE

PR

& M

ole

cula

r G

eom

etr

yp.

xx

Chapter 6 Section 5 Molecular Geometry pages 197-207

7

Molecular GeometryLINEAR

Example formula: BeF2

Type of molecule: AB2 Bond angle: 180° Shared pairs on the central atom: 2Unshared pairs on the central atom:

0

- Be -

F::F: :

: :

Chapter 6 Section 5 Molecular Geometry pages 197-207

8

TRIGONAL PLANAR

Example formula: BF3

Type of molecule: AB3 Bond angle: 120° Shared pairs on the central atom: 3Unshared pairs on the central atom:

0

Molecular Geometry

F:::

:F::

:F::

B

Chapter 6 Section 5 Molecular Geometry pages 197-207

9

Molecular GeometryTETRAHEDRAL

Example formula: CH4

Type of molecule: AB4 Bond angle: 109.5° Shared pairs on the central atom: 4Unshared pairs on the central atom:

0

CHH

H

H

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Molecular GeometryANGULAR

Example formula: H2O

Type of molecule: AB2E2

Bond angle: 104.5° Shared pairs on the central atom: 2Unshared pairs on the central atom:

2

HO

H

::

Chapter 6 Section 5 Molecular Geometry pages 197-207

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Molecular GeometryTRIGONAL PYRAMIDAL

Example formula: NH3

Type of molecule: AB3EBond angle: 107.5° Shared pairs on the central atom: 3Unshared pairs on the central atom:

1

H H

N

:

H

Chapter 6 Section 5 Molecular Geometry pages 197-207

12

Molecular GeometryTrigonal Bipyramidal

Example formula: PCl5Type of molecule: AB5

Bond angle: 90°, 120° Shared pairs on the central atom: 5Unshared pairs on the central atom: 0

H

:

Chapter 6 Section 5 Molecular Geometry pages 197-207

13

Molecular GeometryOctahedral

Example formula: SF6

Type of molecule: AB6

Bond angle: 90° Shared pairs on the central atom: 6Unshared pairs on the central atom: 0

H

:

Chapter 6 Section 5 Molecular Geometry pages 197-207

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Molecular Geometry• Unshared pairs occupies more

space around the central atom than shared pairs

• Unshared pairs repel other electrons more strongly than shared pairs

• Multiple bonds are treated the same as single bonds

• Polyatomic ions are treated like molecules.

Chapter 6 Section 5 Molecular Geometry pages 197-207

15

Hybridization• The mixing of two or more atomic

orbitals of similar energies on the same atom to produce new hybrid atomic orbitals of equal energy

• Example CH4

C = _ _ __ 1s 2s 2p _ _ _ _ 1s sp3

Chapter 6 Section 5 Molecular Geometry pages 197-207

16

Hybridization• s and p orbitals have different

shapes• The 2s & 2p hybridize to make

four identical orbitals– named sp3

– The 3 is from the three p orbitals used

– But the 1 is not written for the s

Chapter 6 Section 5 Molecular Geometry pages 197-207

17

Hybridization• All sp3 orbitals have the same

energy– Higher than 2s but– Lower than 2p

• Hybrid orbitals – orbitals of equal energy produced by the combination of two or more orbitals.

Chapter 6 Section 5 Molecular Geometry pages 197-207

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HybridizationN = _ _ _ 1s 2s 2p _ _ _ 1s sp3

O = _ _ 1s 2s 2p _ _ 1s sp3

Chapter 6 Section 5 Molecular Geometry pages 197-207

19

HybridizationBe = 1s 2s _ _ __ 1s spB = _ __ __ 1s 2s 2p _ _ _ __ 1s sp2

Uses one p orbital

Uses two p orbitals

Chapter 6 Section 5 Molecular Geometry pages 197-207

20

H

yb

rid

izati

on

p.

xx

Chapter 6 Section 5 Molecular Geometry pages 197-207

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Comparing Molecular & Ionic Compoundsp

. xx

Chapter 6 Section 5 Molecular Geometry pages 197-207

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Molecular Polarity

• Dipole: created by equal but opposite charges that are separated by a short distance

H - ClLower EN

Higher ENpolar bond =

dipole

2.1

3.0

δ+ δ-

Chapter 6 Section 5 Molecular Geometry pages 197-207

23

Molecule Polarity• Molecule polarity for

compounds with more than one bond depends on …

bond polarity and

molecule geometry.

Chapter 6 Section 5 Molecular Geometry pages 197-207

24

Molecule Polarity1. Draw the Lewis Structure true

to shape. Example NH3

NH

HH

:

Chapter 6 Section 5 Molecular Geometry pages 197-207

25

Molecule Polarity2. Find all the partial positive and

negatives for each atom in the molecule

HH

H

Look at each bond.High EN = δ- Low EN = δ+

δ-

δ+

δ+

δ+

2.1

3.0N

:

Chapter 6 Section 5 Molecular Geometry pages 197-207

26

Molecule Polarity3. Look at around the “outside” of

the molecule.

NH

HH

:

All the same δ = NP; Different δ = P

δ-

δ+

δ+

δ+

Chapter 6 Section 5 Molecular Geometry pages 197-207

27

Molecule Polarity1. Draw the Lewis Structure true

to shape. Example CH4

CH

HH

H

Chapter 6 Section 5 Molecular Geometry pages 197-207

28

Molecule Polarity2. Find all the partial positive and

negatives for each atom in the molecule

Look at each bond.High EN = δ- Low EN = δ+

δ+

δ+

δ+

δ+

2.1

2.5C

HH

H

H

2.12.

1

2.1

δ-

Chapter 6 Section 5 Molecular Geometry pages 197-207

29

Molecule Polarity3. Look at around the “outside” of

the molecule.

δ+

δ+

δ+

δ+

CH

HH

H

δ-

All the same δ = NP; Different δ = PCarbon is not on the “outside”.

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Intramolecular Forces• In covalent bonding, atoms form

stable units called molecules by sharing electrons.

- termed intramolecular (within the

molecule) bonding

- covalent bonding is an

intramolecular force

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Intermolecular Forces• Forces between neutral molecules

also exist which are termed intermolecular forces.

- London Dispersion Forces (LDF)

- Dipole-Dipole Interactions (DD)

- Hydrogen Bonding (HB)

• DD and LDF together are often termed Van der Waals forces.

32

Intermolecular Forces

• Intermolecular forces (IMF) are responsible for the ability of substances to exist in condensed states (i.e. liquids and solids)

- all substances have intermolecular forces (or interatomic as in the cases of monatomic gases - i.e. noble gases)

- when intermolecular forces are broken (e.g. substances melt or vaporize), the molecules remain intact - covalent bonds are not broken.

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London Dispersion Forces• London Dispersion Forces (LDF) –

intermolecular forces resulting from the instantaneous dipoles created by the constant motion of electrons.

- present between all atoms and molecules.

- increase with increasing numbers of

electrons 

- LDF are the weakest IMF• Consider Helium:

Chapter 6 Section 5 Molecular Geometry pages 197-207

34

London Dispersion Forces• Nonpolar molecules don’t have

dipoles• However at any instance the

electron distribution may be uneven.

• An instantaneous dipole can occur and induce dipoles in other molecules

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Dipole-Dipole Interactions• Dipole-Dipole Interactions (DD) –

attractive forces between polar molecules due to the electrostatic attraction of partial positive and negative charges.

- present only between polar molecules

- stronger than LDF, but weaker than

H-bonding

- explained by Coulomb’s Law

Chapter 6 Section 5 Molecular Geometry pages 197-207

36

Molecular Polarity

• Dipole: created by equal but opposite charges that are separated by a short distance

H - ClLower EN

Higher ENpolar bond =

dipole

2.1

3.0

δ+ δ-

37

Dipole-Dipole Interactions

• Dipole-Dipole (DD) Interactions are due to the electrostatic attractions between δ+ and δ- in different polar molecules.

H - Clδ+ δ-

H - Clδ+ δ-||||||||||||||||||||

Dipole-Dipole Interaction

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Comparing Dipole Dipole Forcesp

. xx

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Induced Dipole• Polar molecules cause a dipole

in a nonpolar molecule

O

H

H

δ+

δ- O O::

::

::

δ+ δ

+δ-

|||||||||||||||||||||||||

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Hydrogen Bonding

• Hydrogen Bonding (HB) is an extreme form of dipole-dipole interaction between the hydrogen atom in a polar bond, specificallyO-H, N-H or H-F, and the electronegative O, N or F atom of another molecule.- strongest IMF

41

Hydrogen Bonding

• Due to the combined effect of the large dipole between the hydrogen and the electronegative element (O, N or F) and the presence of the lone pair of electrons on the O, N or F.

42

Hydrogen Bonding• Consider H2O:

43

Hydrogen Bonding• H-F, H-O or H-N bonds have a

large electronegativity difference

• These bonds are very polar.

• Molecules with these bonds have very strong dipole-dipole forces

44

Hyd

rog

en

Bon

din

g

p.

xx

Chapter 6 Section 5 Molecular Geometry pages 197-207

45

Hydrogen Bonding• Compare

PH3 & NH3 H2O & H2S on p. 204

Chapter 6 Section 5 Molecular Geometry pages 197-207

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Hydrogen Bonding• Compare the boiling points of

CH4, NH3, HF and H2O:

47

Hydrogen Bonding• The double helix of DNA is

possible due to hydrogen bonding

48

Hydrogen Bonding• Quick Reference for Predicting

IMF for a Particular Molecule:

Type of Molecule Intermolecular Forces

Non-Polar London Dispersion

Polar without H-O, H-N or H-FLondon Dispersion

Dipole-Dipole

Polar with H-O, H-N or H-FLondon Dispersion

Dipole-DipoleHydrogen Bonding

Chapter 6 Section 5 Molecular Geometry pages 197-207

49

Modern ChemistryChapter 6

Chemical Bonding

Sections 1-5Introduction to Chemical Bonding

Covalent Bonding & Molecular CompoundsIonic Bonding & Ionic Compounds

Metallic BondingMolecular Geometry

Chapter 6 Section 5 Molecular Geometry pages 197-207

50

Do Now

Draw the Structures of AlCl3and CH4.

What would the bond angles need to be between the

valence electron pairs in each molecule to minimize electron

repulsion.

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Do Now

For the following molecules:- Draw the Lewis Structure- Assign the A,B,E form- Assign the molecular geometry- Determine the polarity- Determine the hybridization of the central element.

SiCl4 H2S

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Do Now

For the following molecules/ions:- Draw the Lewis Structure- Assign the A,B,E form- Assign the molecular geometry- Determine the polarity- Determine the hybridization of the central element.

CHCl3 IF4+

53

Do Now

For the following molecules:- Draw the Lewis Structure- Assign the A,B,E form- Assign the molecular geometry- Determine the polarity- Discuss the intermolecular forces between each molecule.

SiCl4 NH2Cl COH2

54

Do Now

For the following molecules:- Draw the Lewis Structure- Assign the A,B,E form- Assign the molecular geometry- Determine the polarity- Discuss the intermolecular forces between each molecule.

SbCl5 SeCl4 N2H4