Chemical Bonding I: Basic Concepts Chapter 9. Intermolecular Forces 11.2 Intermolecular forces are attractive forces between molecules. Intramolecular

Chemical Bonding I:Basic Concepts

Chapter 9

Intermolecular Forces

11.2

Intermolecular forces are attractive forces between molecules.

Intramolecular forces hold atoms together in a molecule.

Intermolecular vs Intramolecular

• 41 kJ to vaporize 1 mole of water (inter)

• 930 kJ to break all O-H bonds in 1 mole of water (intra)

Generally, intermolecular forces are much weaker than intramolecular forces.

“Measure” of intermolecular force

boiling point

melting point

Hvap

Hfus

Hsub

9.1

Valence electrons are the outer shell electrons of an atom. The valence electrons are the electrons thatparticpate in chemical bonding.

1A 1ns1

2A 2ns2

3A 3ns2np1

4A 4ns2np2

5A 5ns2np3

6A 6ns2np4

7A 7ns2np5

Group # of valence e-e- configuration

9.2

Li + F Li+ F -

The Ionic Bond

1s22s11s22s22p5 1s21s22s22p6[He][Ne]

Li Li+ + e-

e- + F F -

F -Li+ + Li+ F -

9.1

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

9.4

8e-

H HO+ + OH H O HHor

2e- 2e-

Lewis structure of water

Double bond – two atoms share two pairs of electrons

single covalent bonds

O C O or O C O

8e- 8e-8e-double bonds double bonds

Triple bond – two atoms share three pairs of electrons

N N8e-8e-

N N

triple bondtriple bond

or

9.4

Bond Type

Bond Length

(pm)

C-C 154

CC 133

CC 120

C-N 143

CN 138

CN 116

Lengths of Covalent Bonds

Bond Lengths

Triple bond < Double Bond < Single Bond 9.4

9.4

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 an octet for all atoms except hydrogen

4. If structure contains too many electrons, form double and triple bonds on central atom as needed.

Writing Lewis Structures

9.6

Write the Lewis structure of nitrogen trifluoride (NF3).

Step 1 – N is less electronegative than F, put N in center

F N F

F

Step 2 – Count valence electrons N - 5 (2s22p3) and F - 7 (2s22p5)

5 + (3 x 7) = 26 valence electrons

Step 3 – Draw single bonds between N and F atoms and complete octets on N and F atoms.

Step 4 - Check, are # of e- in structure equal to number of valence e- ?

3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons

9.6

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

Step 1 – C is less electronegative than O, put C in center

O C O

O

Step 2 – Count valence electrons C - 4 (2s22p2) and O - 6 (2s22p4) -2 charge – 2e-

4 + (3 x 6) + 2 = 24 valence electrons

Step 3 – Draw single bonds between C and O atoms and complete octet on C and O atoms.

Step 4 - Check, are # of e- in structure equal to number of valence e- ?

3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons

9.6

Step 5 - Too many electrons, form double bond and re-check # of e-

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

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

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

Cl ClBe

2 atoms bonded to central atom0 lone pairs on central atom 10.1

AB2 2 0 linear linear

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR

AB3 3 0trigonal planar

trigonal planar

10.1

10.1


Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


trigonal planar

10.1

AB4 4 0 tetrahedral tetrahedral

10.1


Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


trigonal planar

10.1


AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

10.1


Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


trigonal planar

10.1


AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

AB6 6 0 octahedraloctahedral

10.1

10.1

bonding-pair vs. bonding

pair repulsionlone-pair vs. lone pair

repulsionlone-pair vs. bonding

pair repulsion> >

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


trigonal planar

AB2E 2 1trigonal planar

bent

10.1

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR

AB3E 3 1


tetrahedraltrigonal

pyramidal

10.1

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


10.1

AB3E 3 1 tetrahedraltrigonal

pyramidal

AB2E2 2 2 tetrahedral bent

H

O

H

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR

10.1

AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

AB4E 4 1trigonal

bipyramidaldistorted

tetrahedron

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR

10.1

AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

AB4E 4 1trigonal


tetrahedron

AB3E2 3 2trigonal

bipyramidalT-shaped

ClF

F

F

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR

10.1

AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

AB4E 4 1trigonal


tetrahedron

AB3E2 3 2trigonal

bipyramidalT-shaped

AB2E3 2 3trigonal

bipyramidallinear

I

I

I

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR

10.1


AB5E 5 1 octahedral square pyramidal

Br

F F

FF

F

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR

10.1


AB5E 5 1 octahedral square pyramidal

AB4E2 4 2 octahedral square planar

Xe

F F

FF

10.1

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

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.

H C O HC – 4 e-

O – 6 e-

2H – 2x1 e-

12 e-



formal charge on C = 4 -2 - ½ x 6 = -1

formal charge on O = 6 -2 - ½ x 6 = +1


=1

2




-

-1 +1

9.7

C – 4 e-

O – 6 e-

2H – 2x1 e-

12 e-



HC O

H

formal charge on C = 4 -0 - ½ x 8 = 0

formal charge on O = 6 -4 - ½ x 4 = 0


=1

2




-

0 0

9.7

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 +1 HC O

H

0 0

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?

Exceptions to the Octet Rule

The Incomplete Octet

H HBeBe – 2e-

2H – 2x1e-

4e-

BeH2

BF3

B – 3e-

3F – 3x7e-

24e-

F B F

F

3 single bonds (3x2) = 69 lone pairs (9x2) = 18

Total = 24

9.9

Exceptions to the Octet Rule

Odd-Electron Molecules

N – 5e-

O – 6e-

11e-

NO N O

The Expanded Octet (central atom with principal quantum number n > 2)

SF6

S – 6e-

6F – 42e-

48e-

S

F

F

F

FF

F

6 single bonds (6x2) = 1218 lone pairs (18x2) = 36

Total = 48

9.9

Covalent

share e-

Polar Covalent

partial transfer of e-

Ionic

transfer e-

Increasing difference in electronegativity

Classification of bonds by difference in electronegativity

Difference Bond Type

0 Covalent

2 Ionic

0 < and <2 Polar Covalent

9.5

Classify the following bonds as ionic, polar covalent, or covalent: The bond in CsCl; the bond in H2S; andthe NN bond in H2NNH2.

Cs – 0.7 Cl – 3.0 3.0 – 0.7 = 2.3 Ionic

H – 2.1 S – 2.5 2.5 – 2.1 = 0.4 Polar Covalent

N – 3.0 N – 3.0 3.0 – 3.0 = 0 Covalent

9.5

H F FH

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

electron richregion

electron poorregion e- riche- poor

+ -

9.5

Electronegativity is the ability of an atom to attract toward itself the electrons in a chemical bond.

Electron Affinity - measurable, Cl is highest

Electronegativity - relative, F is highest

X (g) + e- X-(g)

9.5

9.5

Dipole Moments and Polar Molecules

10.2

H F

electron richregion

electron poorregion

= Q x rQ is the charge

r is the distance between charges

1 D = 3.36 x 10-30 C m

10.2

10.2

10.2

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

O HH

dipole momentpolar molecule

SO

O

CO O

no dipole momentnonpolar molecule

dipole momentpolar molecule

C

H

H

HH

no dipole momentnonpolar molecule

Does CH2Cl2 have a dipole moment?

10.2


Dipole-Dipole Forces

Attractive forces between polar molecules

Orientation of Polar Molecules in a Solid

11.2


Ion-Dipole Forces

Attractive forces between an ion and a polar molecule

11.2

Ion-Dipole Interaction

11.2

Intermolecular ForcesDispersion Forces

Attractive forces that arise as a result of temporary dipoles induced in atoms or molecules

11.2

ion-induced dipole interaction

dipole-induced dipole interaction

Intermolecular ForcesDispersion Forces Continued

11.2

Polarizability is the ease with which the electron distribution in the atom or molecule can be distorted.

Polarizability increases with:

• greater number of electrons

• more diffuse electron cloud

Dispersion forces usually increase with molar mass.

SO

O

What type(s) of intermolecular forces exist between each of the following molecules?

HBrHBr is a polar molecule: dipole-dipole forces. There are also dispersion forces between HBr molecules.

CH4

CH4 is nonpolar: dispersion forces.

SO2

SO2 is a polar molecule: dipole-dipole forces. There are also dispersion forces between SO2 molecules.

11.2

Intermolecular ForcesHydrogen Bond

11.2

The hydrogen bond is a special dipole-dipole interaction between they hydrogen atom in a polar N-H, O-H, or F-H bond and an electronegative O, N, or F atom.

A H…B A H…Aor

A & B are N, O, or F

Hydrogen Bond

11.2

Why is the hydrogen bond considered a “special” dipole-dipole interaction?

Decreasing molar massDecreasing boiling point

11.2

Types of Crystals

Molecular Crystals• Lattice points occupied by molecules• Held together by intermolecular forces• Soft, low melting point• Poor conductor of heat and electricity

11.6

Types of Crystals

Metallic Crystals• Lattice points occupied by metal atoms• Held together by metallic bonds• Soft to hard, low to high melting point• Good conductors of heat and electricity

11.6

Cross Section of a Metallic Crystal

nucleus &inner shell e-

mobile “sea”of e-

Types of Crystals

Ionic Crystals• Lattice points occupied by cations and anions• Held together by electrostatic attraction• Hard, brittle, high melting point• Poor conductor of heat and electricity

CsCl ZnS CaF2

11.6

9.3

Lattice energy (E) increases as Q increases and/or

as r decreases.

cmpd lattice energyMgF2

MgO

LiF

LiCl

2957

3938

1036

853

Q= +2,-1

Q= +2,-2

r F < r Cl

Electrostatic (Lattice) Energy

E = kQ+Q-r

Q+ is the charge on the cation

Q- is the charge on the anionr is the distance between the ions

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

Types of Crystals

Covalent Crystals• Lattice points occupied by atoms• Held together by covalent bonds• Hard, high melting point• Poor conductor of heat and electricity

11.6diamond graphite

carbonatoms

Types of Crystals

11.6

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

10.4

10.4

10.4

Predict correctbond angle

Formation of sp Hybrid Orbitals

10.4

Formation of sp2 Hybrid Orbitals

10.4

# of Lone Pairs+

# of Bonded Atoms Hybridization 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

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

10.5

10.5

Sigma () and Pi Bonds ()

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 and bonds are in the acetic acid(vinegar) molecule CH3COOH?

C

H

H

CH

O

O H bonds = 6 + 1 = 7

bonds = 1

10.5

The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond energy.

H2 (g) H (g) + H (g) H0 = 436.4 kJ

Cl2 (g) Cl (g)+ Cl (g) H0 = 242.7 kJ

HCl (g) H (g) + Cl (g) H0 = 431.9 kJ

O2 (g) O (g) + O (g) H0 = 498.7 kJ O O

N2 (g) N (g) + N (g) H0 = 941.4 kJ N N

Bond Energy

Bond Energies

Single bond < Double bond < Triple bond

9.10

Average bond energy in polyatomic molecules

H2O (g) H (g) + OH (g) H0 = 502 kJ

OH (g) H (g) + O (g) H0 = 427 kJ

Average OH bond energy = 502 + 427

2= 464 kJ

9.10

Bond Energies (BE) and Enthalpy changes in reactions

H0 = total energy input – total energy released= BE(reactants) – BE(products)

Imagine reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products.

9.10

9.10

H2 (g) + Cl2 (g) 2HCl (g) 2H2 (g) + O2 (g) 2H2O (g)

Use bond energies to calculate the enthalpy change for:H2 (g) + F2 (g) 2HF (g)

H0 = BE(reactants) – BE(products)

Type of bonds broken

Number of bonds broken

Bond energy (kJ/mol)

Energy change (kJ)

H H 1 436.4 436.4

F F 1 156.9 156.9

Type of bonds formed

Number of bonds formed

Bond energy (kJ/mol)

Energy change (kJ)

H F 2 568.2 1136.4

H0 = 436.4 + 156.9 – 2 x 568.2 = -543.1 kJ

9.10

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Chemical Bonding I: Basic Concepts Chapter 9. Intermolecular Forces 11.2 Intermolecular forces are attractive forces between molecules. Intramolecular