9-1 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 9 Models of Chemical Bonding

9-1

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Chapter 9

Models of Chemical Bonding

9-2


Models of Chemical Bonding

9.1 Atomic Properties and Chemical Bonds

9.2 The Ionic Bonding Model

9.3 The Covalent Bonding Model

9.4 Bond Energy and Chemical Change

9.5 Between the Extremes: Electronegativity and Bond Polarity

9-3


A general comparison of metals and nonmetals.Figure 9.1

9-4


Types of Chemical Bonding

1. Metal with nonmetal:

electron transfer and ionic bonding

2. Nonmetal with nonmetal:

electron sharing and covalent bonding

3. Metal with metal:

electron pooling and metallic bonding

9-5


Figure 9.2 The three models of chemical bonding.

9-6


Lewis Electron-Dot Symbols

For main group elements -

Example:

Nitrogen, N, is in Group 5A and therefore has 5 valence electrons.

N:.

..

:

N .. ..N :.

. :N ...

The A group number gives the number of valence electrons.

Place one dot per valence electron on each of the four sides of the element symbol.

Pair the dots (electrons) until all of the valence electrons are used.

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Figure 9.4

Lewis electron-dot symbols for elements in Periods 2 and 3.

9-8


SAMPLE PROBLEM 9.1 Depicting Ion Formation

PLAN:

SOLUTION:

PROBLEM: Use partial orbital diagrams and Lewis symbols to depict the formation of Na+ and O2- ions from the atoms, and determine the formula of the compound.

Draw orbital diagrams for the atoms and then move electrons to make filled outer levels. It can be seen that 2 sodiums are needed for each oxygen.

3s 3p

Na

3s 3p

Na2s 2p

O2s 2p

O2-

2 Na+

:Na

Na+ O

.:

..

.2Na+ + O 2-

:: ::

9-9


Electron configurations

Li 1s22s1

Orbital diagrams

Lewis electron-dot symbols

+ F 1s22s22p5 Li+ 1s2 + F- 1s22s22p6

Three ways to represent the formation of Li+ and F- through electron transfer.

Figure 9.5

Li

1s 2s 2p

F

1s 2s 2p

+

Li+

1s 2s 2p

F-

1s 2s 2p+

.+ F: ::Li . Li+ + F -::

::

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Periodic Trends in Lattice Energy

Coulomb’s Law

charge A X charge B

electrostatic force distance2

energy = force X distance therefore

charge A X charge B

electrostatic energy distance

cation charge X anion charge

electrostatic energy cation radius + anion radius

H0lattice

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Figure 9.7 Trends in lattice energy.

9-12


Figure 9.8

Electrostatic forces and the reason ionic compounds crack.

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Figure 9.9 Electrical conductance and ion mobility.

Solid ionic compound

Molten ionic compound

Ionic compound dissolved in water

9-14


Table 9.1 Melting and Boiling Points of Some Ionic Compounds

Compound mp (0C) bp (0C)

CsBr

661

1300

NaI

MgCl2

KBr

CaCl2

NaCl

LiF

KF

MgO

636

714

734

782

801

845

858

2852

1304

1412

1435

>1600

1413

1676

1505

3600

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Figure 9.10 Covalent bond formation in H2.

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Figure 9.11 Distribution of electron density of H2.

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back to previous slide

Table 9.2

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Figure 9.12

Internuclear distance(bond length)




Bond length and covalent radius.

Covalent radius

72 pm

Covalent radius

114 pm

Covalent radius

133 pm

Covalent radius

100 pm

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Table 9.3

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SAMPLE PROBLEM 9.2 Comparing Bond Length and Bond Strength

SOLUTION:

PROBLEM: Using the periodic table, but not Tables 9.2 and 9.3, rank the bonds in each set in order of decreasing bond length and bond strength:

(a) S - F, S - Br, S - Cl (b) C = O, C - O, C O

PLAN: (a) The bond order is one for all and sulfur is bonded to halogens; bond length should increase and bond strength should decrease with increasing atomic radius. (b) The same two atoms are bonded but the bond order changes; bond length decreases as bond order increases while bond strength increases as bond order increases.

(a) Atomic size increases going down a group.

Bond length: S - Br > S - Cl > S - F

Bond strength: S - F > S - Cl > S - Br

(b) Using bond orders we get

Bond length: C - O > C = O > C O

Bond strength: C O > C = O > C - O

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Figure 9.13

Strong covalent bonding forces within molecules

Weak intermolecular forces between molecules

Strong forces within molecules and weak forces between them.

9-22


Figure 9.14 Covalent bonds of network covalent solids.

9-23


Figure 9.15 The infrared (IR) spectra of diethyl ether and 2-butanol.

9-24


Figure 9.16 Using bond energies to calculate H0rxn.

H0rxn = H0

reactant bonds broken + H0product bonds formed

H01 = + sum of BE H0

2 = - sum of BE

H0rxn

Ent

halp

y, H

BOND BREAKING

BOND FORMATION

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Figure 9.17 Using bond energies to calculate H0rxn of methane.

Ent

halp

y,H

BOND BREAKING

4BE(C-H)= +1652kJ

2BE(O2)= + 996kJ

H0(bond breaking) = +2648kJBOND FORMATION

4[-BE(O-H)]= -1868kJ

H0(bond forming) = -3466kJ

H0rxn= -818kJ

2[-BE(C O)]= -1598kJ

9-26


SAMPLE PROBLEM 9.3 Calculating Enthalpy Changes from Bond Energies

SOLUTION:

PROBLEM: Use Table 9.2 (button at right) to calculate H0rxn for the

following reaction:

CH4(g) + 3Cl2(g) CHCl3(g) + 3HCl(g)

PLAN: Write the Lewis structures for all reactants and products and calculate the number of bonds broken and formed.

H

C H

H

H + Cl Cl3

Cl

C Cl

Cl

H + H Cl3

bonds broken bonds formed

9-27


SAMPLE PROBLEM 9.3 Calculating Enthalpy Changes from Bond Energies

continued

bonds broken bonds formed

4 C-H = 4 mol(413 kJ/mol) = 1652 kJ

3 Cl-Cl = 3 mol(243 kJ/mol) = 729 kJ

H0bonds broken = 2381 kJ

3 C-Cl = 3 mol(-339 kJ/mol) = -1017 kJ

1 C-H = 1 mol(-413 kJ/mol) = -413 kJ

H0bonds formed = -2711 kJ

3 H-Cl = 3 mol(-427 kJ/mol) = -1281 kJ

H0reaction = H0

bonds broken + H0bonds formed = 2381 kJ + (-2711 kJ) = - 330 kJ

9-28


Table 9.4

9-29


Figure 9.20 The Pauling electronegativity (EN) scale.

9-30


SAMPLE PROBLEM 9.4 Determining Bond Polarity from EN Values

PLAN:

SOLUTION:

PROBLEM: (a) Use a polar arrow to indicate the polarity of each bond: N-H, F-N, I-Cl.

(b) Rank the following bonds in order of increasing polarity: H-N, H-O, H-C.

(a) Use Figure 9.19(button at right) to find EN values; the arrow should point toward the negative end.

(b) Polarity increases across a period.

(a) The EN of N = 3.0, H = 2.1; F = 4.0; I = 2.5, Cl = 3.0

N - H F - N I - Cl

(b) The order of increasing EN is C < N < O; all have an EN larger than that of H.

H-C < H-N < H-O

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Figure 9.21 Electron density distributions in H2, F2, and HF.

9-32


Figure 9.22

EN

3.0

2.0

0.0

Boundary ranges for classifying ionic character of chemical bonds.

9-33


Figure 9.23

Properties of the Period 3 chlorides.

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9-1 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 9 Models of Chemical Bonding