Chemical Bonding: Basic Concepts

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Chemical Bonding: Basic Concepts Chapter 6

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Chemical Bonding: Basic Concepts. Chapter 6. Why do atoms form bonds?. To become more stable. What is a bond?. A mutual electrical attraction between the nuclei and valence electrons of different atoms that binds the atoms together. Bonding Forces. Electron – electron - PowerPoint PPT Presentation

Transcript of Chemical Bonding: Basic Concepts

Page 1: Chemical Bonding: Basic Concepts

Chemical Bonding:Basic Concepts

Chapter 6

Page 2: Chemical Bonding: Basic Concepts

Why do atoms form bonds?

To become more stable

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What is a bond?

• A mutual electrical attraction between the nuclei and valence electrons of different atoms that binds the atoms together.

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Bonding Forces

Electron – electron repulsive forces

Proton – proton repulsive forces

Electron – proton attractive forces

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Three types of bonds

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What type of bond will form?

• Type of bond is dependent on differences in electronegativity

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A measure of the

ability of an atom in a chemical compound to attract electrons

Remember- Electronegativity is …

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Rule of thumb

• Metal and non-metal usually form an ionic bond

• Non-metal and non-metal from a covalent bond

• Metals come together to form metallic bonds

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What type of bond?

ElementsElectro-

negativity difference

Bond Type

More-negative

atom

Chlorine & Calcium

3.5 – 3.0 = 0.5

Covalent

Chlorine

Chlorine & Oxygen

3.0 – 1.0 = 2.0

Ionic Chlorine

Bromine &

Chlorine

3.0 – 2.8 = 0.2

Covalent

Oxygen

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Pause for a CauseCreate a “PFC” page in your notebook to use for this unit.

Pause for a CauseCreate a “PFC” page in your notebook to use for this unit.

PFC #1Determine what type of

bond is present in each of the following

1. Magnesium chloride2. Water (H2O)

3. Chlorine gas (Cl2)

4. Tungsten shavings

Check your answers:

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Properties of each type of bond

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A chemical bond that results in the sharing of electrons between two atoms

Covalent Bond

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Characteristics of Covalent Bonds a) Usually occur between atoms of nonmetals

b) Result in a particle called a molecule

c) Have low melting points (therefore are not solids at room temperature)

d) Do not usually conduct electricity.

Characteristics of Covalent Bonds a) Usually occur between atoms of nonmetals

b) Result in a particle called a molecule

c) Have low melting points (therefore are not solids at room temperature)

d) Do not usually conduct electricity.

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Ionic Bond

• The chemical bond resulting from the electrostatic attraction between

positive and negative ions.

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Electrons are transferred

• Opposite charges attract (electrostatic force)

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Ionic substances have predictable characteristics. a) Have extremely high melting points b) Tend to be soluble in water. c) Usually form a crystalline lattice structure as a solid. d) Are good conductors of electricity in the molten state.

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Metallic BondingBonding that occurs between atoms of a metal due to

delocalized electrons forming an electron sea

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Characteristics of Metallic Bonding

• The highly mobile “sea of electrons” in metals account for its high conductivity.

• Metals are better conductors of heat and electricity than ionic and molecular compounds.

• Metals are also malleable, ductile and shiny.

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Pause for a Cause #2

Tell what kind of bonds I have…

1. Heat me up and I can conduct electricity.

2. My electrons flow like fish in the sea.

3. My bonding partner and I are very dull (not shiny).

4. Write your own riddle about one type of bond.

Check your answers1. Ionic2. Metallic3. Covalent

4. Let’s hear some of yours!

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How strong are bonds?

Ionic (Lattice energy)Dependent on two factors

1. Charge

2. Size

Covalent (bond strength)Dependent on two factors

1. Bond type (single/double/triple)

2. Size (Must remember your size trends)

Shorter bonds are strongerLonger bonds are weaker

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cmpd lattice energy

MgF2

MgO

LiF

LiCl

2957

3938

1036

853

Q= +2,-1

Q= +2,-2

r F < r Cl

Electrostatic (Lattice) Energy

Lattice energy (E) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions.

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Bond Type

Bond Length(pm)

C-C 154

CC 133

CC 120

C-N 143

CN 138

CN 116

Lengths of Covalent Bonds

Bond LengthsTriple bond < Double Bond < Single Bond

9.4

Shorter bonds are stronger bonds

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Pause for a Cause #3Label each set as covelent or ionic and circle the substance with the greatest lattice or bond energy

Label each set as covelent or ionic and circle the substance with the greatest lattice or bond energy

1. MgCl2 or MgO

2. H2O or H2S (both have single bonds)

3. N2 (triple bond) or H2 (single bond)

4. NaCl or KCl

Check your answers

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H F FH

Polar covalent bond or polar bond is a covalent bond with greater electron density around one of the two atoms

electron richregionelectron poor

region e- riche- poor

d+ d-

9.5

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Lewis Structures

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The Octet RuleChemical compounds tend to form so that each atom, by

gaining, losing, or sharing electrons,

has an octet of electrons in its

highest occupied energy level.

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Li + F Li+ F -

The Ionic Bond

1s22s1 1s22s22p5 1s21s22s22p6[He] [Ne]

Li Li+ + e-

e- +F F -

F -Li+ + Li+ F -

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A covalent bond is a chemical bond in which two or more electrons are shared by two atoms.

Why should two atoms share electrons?

F F+

7e- 7e-

F F

8e- 8e-

F F

F F

Lewis structure of F2

lone pairslone pairs

lone pairslone pairs

single covalent bond

single covalent bond

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1. Draw skeletal structure of compound showing what atoms are bonded to each other. Put least electronegative element in the center.

2. Count total number of valence e-. Add 1 for each negative charge. Subtract 1 for each positive charge.

3. Complete octets for as many atoms as possible, starting with the outer atoms.

4. If all octets are not filled, create double or triple bonds within the structure (with C, S, P, O or N)

Writing Lewis Structures

9.6

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Write the Lewis structure of nitrogen trifluoride (NF3).

F N F

F

9.6

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Write the Lewis structure of the carbonate ion (CO32-).

9.6

2 single bonds (2x2) = 41 double bond = 4

8 lone pairs (8x2) = 16Total = 24

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A resonance structure is one of two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure.

O O O+ -

OOO+-

O C O

O

- -O C O

O

-

-

OCO

O

-

-9.8

What are the resonance structures of the carbonate (CO3

2-) ion?

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Pause for a Cause #5

Draw Lewis structures for the following molecules

1. HBr 2. CF4

3. NH2Cl

4. SiCl3Br

5. NH3

6. H2S

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Violations of the Octet RuleViolations of the Octet Rule

BF3BF3

SF4SF4

B and Sn tends to form bonds with 6 electrons.

3rd row and heavier elements CAN exceed the octet rule using empty valence d orbitals.

Be tends to form bonds with 4 electrons.

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9.7

Two possible skeletal structures of formaldehyde (CH2O)

H C O HH

C OH

An atom’s formal charge is the difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure.

formal charge on an atom in a Lewis structure

=1

2

total number of bonding electrons( )

total number of valence electrons in the free atom

-total number of nonbonding electrons

-

The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion.

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H C O H

formal charge on C

=

formal charge on O

=

formal charge on an atom in a Lewis structure

=total number of bonding pairs

+total number of nonbonding electrons

-1 +1

9.7

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C – 4 e-

O – 6 e-

2H – 2x1 e-

12 e-

2 single bonds (2x2) = 41 double bond = 4

2 lone pairs (2x2) = 4Total = 12

HC O

H

formal charge on C

= 4 - 0 - ½ x 8 = 0

formal charge on O

= 6 - 4 - ½ x 4 = 0

formal charge on an atom in a Lewis structure

=1

2

total number of bonding electrons( )

total number of -

total number of nonbonding electrons

-

0 0

9.7

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Formal Charge and Lewis Structures

9.7

1. For neutral molecules, a Lewis structure in which there are no formal charges is preferable to one in which formal charges are present.

2. Lewis structures with large formal charges are less plausible than those with small formal charges.

3. Among Lewis structures having similar distributions of formal charges, the most plausible structure is the one in which negative formal charges are placed on the more electronegative atoms.

Which is the most likely Lewis structure for CH2O?

H C O H

-1 +1H

C OH

0 0

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What shape do molecules have?

Determined by Valence Shell Electron Pair Repulsion Theory

VSEPR Theory- electrostatic repulsion between the valence-level pairs surrounding an atom causes these pairs to be oriented as far

apart as possible.

What does it mean?

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Valence shell electron pair repulsion (VSEPR) model:

Predict the geometry of the molecule from the electrostatic repulsions between the electron (bonding and nonbonding) pairs.

AB2 2 0

Class

# of atomsbonded to

central atom

# lonepairs on

central atomArrangement of electron pairs

MolecularGeometry

10.1

linear linear

B B

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Cl ClBe

2 atoms bonded to central atom0 lone pairs on central atom10.1

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AB2 2 0 linear linear

Class

# of atomsbonded to

central atom

# lonepairs on

central atomArrangement of electron pairs

MolecularGeometry

VSEPR

AB3 3 0trigonal planar

trigonal planar

10.1

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10.1

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AB2 2 0 linear linear

Class

# of atomsbonded to

central atom

# lonepairs on

central atomArrangement of electron pairs

MolecularGeometry

VSEPR

AB3 3 0trigonal planar

trigonal planar

10.1

AB4 4 0 tetrahedral tetrahedral

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10.1

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AB2 2 0 linear linear

Class

# of atomsbonded to

central atom

# lonepairs on

central atomArrangement of electron pairs

MolecularGeometry

VSEPR

AB3 3 0trigonal planar

trigonal planar

10.1

AB4 4 0 tetrahedral tetrahedral

AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

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10.1

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AB2 2 0 linear linear

Class

# of atomsbonded to

central atom

# lonepairs on

central atomArrangement of electron pairs

MolecularGeometry

VSEPR

AB3 3 0trigonal planar

trigonal planar

10.1

AB4 4 0 tetrahedral tetrahedral

AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

AB6 6 0 octahedraloctahedral

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10.1

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10.1

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bonding-pair vs. bondingpair repulsion

lone-pair vs. lone pairrepulsion

lone-pair vs. bondingpair repulsion

> >

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Class

# of atomsbonded to

central atom

# lonepairs on

central atomArrangement of electron pairs

MolecularGeometry

VSEPR

AB3 3 0trigonal planar

trigonal planar

AB2E 2 1trigonal planar bent

10.1

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Class

# of atomsbonded to

central atom

# lonepairs on

central atomArrangement of electron pairs

MolecularGeometry

VSEPR

AB3E 3 1

AB4 4 0 tetrahedral tetrahedral

tetrahedraltrigonal

pyramidal

10.1

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Class

# of atomsbonded to

central atom

# lonepairs on

central atomArrangement of electron pairs

MolecularGeometry

VSEPR

AB4 4 0 tetrahedral tetrahedral

10.1

AB3E 3 1 tetrahedraltrigonal

pyramidal

AB2E2 2 2 tetrahedral bent

H

O

H

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Class

# of atomsbonded to

central atom

# lonepairs on

central atomArrangement of electron pairs

MolecularGeometry

VSEPR

10.1

AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

AB4E 4 1trigonal

bipyramidaldistorted

tetrahedron

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Class

# of atomsbonded to

central atom

# lonepairs on

central atomArrangement of electron pairs

MolecularGeometry

VSEPR

10.1

AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

AB4E 4 1trigonal

bipyramidaldistorted

tetrahedron

AB3E2 3 2trigonal

bipyramidal T-shaped

ClF

F

F

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Class

# of atomsbonded to

central atom

# lonepairs on

central atomArrangement of electron pairs

MolecularGeometry

VSEPR

10.1

AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

AB4E 4 1trigonal

bipyramidaldistorted

tetrahedron

AB3E2 3 2trigonal

bipyramidal T-shaped

AB2E3 2 3trigonal

bipyramidal linear

I

I

I

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Class

# of atomsbonded to

central atom

# lonepairs on

central atomArrangement of electron pairs

MolecularGeometry

VSEPR

10.1

AB6 6 0 octahedraloctahedral

AB5E 5 1 octahedralsquare

pyramidal

Br

F F

FF

F

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Class

# of atomsbonded to

central atom

# lonepairs on

central atomArrangement of electron pairs

MolecularGeometry

VSEPR

10.1

AB6 6 0 octahedraloctahedral

AB5E 5 1 octahedralsquare

pyramidal

AB4E2 4 2 octahedralsquare planar

Xe

F F

FF

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10.1

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Predicting Molecular Geometry1. Draw Lewis structure for molecule.

2. Count number of lone pairs on the central atom and number of atoms bonded to the central atom.

3. Use VSEPR to predict the geometry of the molecule.

What are the molecular geometries of SO2 and SF4?

SO O

AB2E

bent

S

F

F

F F

AB4E

distortedtetrahedron

10.1

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Pause for a CausePg. 211 #48

48. Draw the Lewis Structure and predict the molecular geometry for each of the following:

a. SCl2

b. PI3

c. Cl2O

d. NH2Cl

e. SiCl3Br

f. ONCl

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Dipole Moments and Polar Molecules

10.2

H F

electron richregion

electron poorregion

+d -d

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10.2

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10.2

Which of the following molecules have a dipole moment?H2O, CO2, SO2, and CH4

OH

H

dipole momentpolar molecule

S

OO

CO O

no dipole momentnonpolar molecule

dipole momentpolar molecule

C

H

H

HH

no dipole momentnonpolar molecule

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Polar or non-polar?Let’s draw the Lewis

StructuresAnd find out!a. HF

b. H2O

c. NH3

d. CO2

e. SO3

f. H2S

g. CCl4

Page 67: Chemical Bonding: Basic Concepts

Valence Bond Theory and NH3

N – 1s22s22p3

3 H – 1s1

If the bonds form from overlap of 3 2p orbitals on nitrogenwith the 1s orbital on each hydrogen atom, what would the molecular geometry of NH3 be?

If use the3 2p orbitalspredict 900

Actual H-N-Hbond angle is

107.30

10.4

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Hybridization – mixing of two or more atomic orbitals to form a new set of hybrid orbitals.

1. Mix at least 2 nonequivalent atomic orbitals (e.g. s and p). Hybrid orbitals have very different shape from original atomic orbitals.

2. Number of hybrid orbitals is equal to number of pure atomic orbitals used in the hybridization process.

3. Covalent bonds are formed by:

a. Overlap of hybrid orbitals with atomic orbitals

b. Overlap of hybrid orbitals with other hybrid orbitals

10.4

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10.4

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10.4

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10.4

Predict correctbond angle

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Formation of sp Hybrid Orbitals

10.4

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Formation of sp2 Hybrid Orbitals

10.4

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# of Lone Pairs+

# of Bonded AtomsHybridization Examples

2

3

4

5

6

sp

sp2

sp3

sp3d

sp3d2

BeCl2

BF3

CH4, NH3, H2O

PCl5

SF6

How do I predict the hybridization of the central atom?

Count the number of lone pairs AND the numberof atoms bonded to the central atom

10.4

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10.4

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10.5

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10.5

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10.5

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10.5

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10.5

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Sigma (s) and Pi Bonds (p)

Single bond 1 sigma bond

Double bond 1 sigma bond and 1 pi bond

Triple bond 1 sigma bond and 2 pi bonds

How many s and p bonds are in the acetic acid(vinegar) molecule CH3COOH?

C

H

H

CH

O

O Hs bonds = 6 + 1 = 7

p bonds = 1

10.5

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Dipole Moments and Polar Molecules

10.2

H F

electron richregion

electron poorregion

+d -d

m = Q x rQ is the charge

r is the distance between charges

1 D = 3.36 x 10-30 C m

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10.2

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10.2

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10.2

Which of the following molecules have a dipole moment?H2O, CO2, SO2, and CH4

OH

H

dipole momentpolar molecule

S

OO

CO O

no dipole momentnonpolar molecule

dipole momentpolar molecule

C

H

H

HH

no dipole momentnonpolar molecule

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Molecular Forces

• Intramolecular Forces– The forces of attraction between atoms in

a molecule.– Covalent, Ionic, Metallic bonds

• Intermolecular Forces– The forces of attraction between

molecules.– Dipole-dipole, Hydrogen bonding, and

London dispersion forces.

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Dipole-Dipole Forces- force of attraction found when the positive end of one polar molecule is attracted to the negative end of another polar molecule.

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Dipole- equal but opposite charge separated by a short distance.Polarity is dependent upon differences in electronegativity and the symmetry of the molecule

- NH3, HF, H2O, CH3Cl are all polar - CO2, CCl4 are nonpolar

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. Hydrogen Bonding- a particularly strong dipole-

dipole attractive force formed when

hydrogen bonds with a strongly electronegative

nonmetal (F, O, or N) 1. Polar Bonds cause

hydrogen to leave its electrons almost exposed.

ie. water

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A nonpolar molecule is transformed into a dipole due to a distortion of its’ electron cloud in response to an approaching dipole. -Negative end of the dipole molecule causes an imbalance of the electron cloud in the nonpolar molecule and therefore a bulging away. * ie. water solutions of iodine. Water is polar but iodine molecules are nonpolar

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London Dispersion Forces

Intermolecular attractive force that exist from the creation of instananeous or temporary dipoles .

-They are the only force of attraction between nonpolar molecules.

- They increase in significance as molecularsize increases.

Fritz London 1900-1954

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London Forces in Hydrocarbons

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EXAPLE 1

EX2

EX3

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Relative magnitudes of forces

The types of bonding forces vary in their strength as measured by average

bond energy.

Covalent bonds (400 kcal)

Hydrogen bonding (12-16 kcal )

Dipole-dipole interactions (2-0.5 kcal)

London forces (less than 1 kcal)

Strongest

Weakest

Ionic bonds