Top Banner
28

Chapter 9 notes

Jan 14, 2017

Download

Education

Wong Hsiung
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Page 1: Chapter 9 notes
Page 2: Chapter 9 notes

2

9.1 LEWIS SYMBOLS AND THE OCTET RULE

Page 3: Chapter 9 notes

3

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

1A 1ns1

2A 2ns2

3A 3ns2np1

4A 4ns2np2

5A 5ns2np3

6A 6ns2np4

7A 7ns2np5

Group # of valence e-e- configuration

Page 4: Chapter 9 notes

4

Lewis Dot Symbols for the Representative Elements &Noble Gases

Octet rule states that an atom tends to gain, lose or share e- until it has eight e- in the valence shell

Page 5: Chapter 9 notes

5

Page 6: Chapter 9 notes

6

Page 7: Chapter 9 notes

7

Page 8: Chapter 9 notes

8

9.2 IONIC BONDING AND THE LATTICE ENERGY

Page 9: Chapter 9 notes

9

Li + F Li+ F -

The Ionic Bond

1s22s1

1s22s22p5

1s2

1s22s22p6

[He]

[Ne]

Li Li+ + e-

e- + F F -

F -Li+ + Li+ F -

LiF

Ionic bond: the electrostatic force that holds ions together in an ionic compound.

Page 10: Chapter 9 notes

10

Lattice energy can be calculated using Hess’s Law

Electrostatic (Lattice) Energy

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

Page 11: Chapter 9 notes

11

9.3 COVALENT BONDING AND THE LEWIS STRUCTURES

Page 12: Chapter 9 notes

12

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

Page 13: Chapter 9 notes

13

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

Page 14: Chapter 9 notes

14

9.4 ELECTRONEGATIVITY

Page 15: Chapter 9 notes

15

The Electronegativities of Common Elements

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

Non metal metal

Page 16: Chapter 9 notes

16

• The electronegativity of H is 2.1; Cl is 3.0.

• Since there is a difference in electronegativity between the two elements (3.0 – 2.1 = 0.9), the bond in H – Cl is polar.

• Cl is more electronegative, the bonding electrons are attracted toward the Cl atom and away from the H atom.

• Cl atom = slightly negative charge

• H atom = slightly positive charge.

+ H – Cl –

Page 17: Chapter 9 notes

17

H F

FH

Polar covalent bond or polar bond is a covalent bond with greater electron density around one of the two atoms (unequal sharing of e- )

electron rich

region

electron poor

region

e- riche- poor

+ -

Page 18: Chapter 9 notes

18

Nonpolar covalent bond results when two identical non-metals equally share electrons between them (equal sharing of e-). Eg. H2, N2, Cl2

Page 19: Chapter 9 notes

19

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 Type0 Nonpolar Covalent

2 Ionic0 < and <2 Polar Covalent

•No electronegativity difference between two atoms leads to a pure non-polar covalent bond.•A small electronegativity difference leads to a polar covalent bond.•A large electronegativity difference leads to an ionic bond.

Page 20: Chapter 9 notes

20

H-H = 0.0 H-Cl = 0.9 Na-Cl = 2.1

Page 21: Chapter 9 notes

21

Classify the following bonds as ionic, polar covalent, or covalent: The bond in CsCl; the bond in H2S; and the 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 Non polar covalent

Page 22: Chapter 9 notes

The formal charge is the charge that an atom seems to have in a Lewis structure. formal charge on an atom in a Lewis structure

=12

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.

9.5 LEWIS STRUCTURE & FORMAL CHARGELewis structures are diagrams that show the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule.

Why formal charge ?A Lewis structure in which there are no formal charges is preferred.

Page 23: Chapter 9 notes

23

H C O HC – 4 e-

O – 6 e-

2H – 2x1 e-

12 e-

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

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

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

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

formal charge on an atom in a Lewis structure

=12

total number of bonding electrons( )

total number of valence electrons in the free atom

-total number of nonbonding electrons

-

-1 +1

Page 24: Chapter 9 notes

24

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

=12

total number of bonding electrons( )

total number of valence electrons in the free atom

-total number of nonbonding electrons

-

0 0

Page 25: Chapter 9 notes

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

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 Enthalpy

Bond Enthalpies

Single bond < Double bond < Triple bond

9.6 BOND ENERGIES

Page 26: Chapter 9 notes

26

BE = measure the strength of a bond

the larger the BE, the stronger the bond

Page 27: Chapter 9 notes

27

Use bond enthalpies 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 enthalpy (kJ/mol)

Enthalpy change (kJ/mol)

H H 1 436.4 436.4F F 1 156.9 156.9

Type of bonds formed

Number of bonds formed

Bond enthalpy (kJ/mol)

Enthalpy change (kJ/mol)

H F 2 568.2 1136.4

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

Page 28: Chapter 9 notes