Chapter 10. CHEMICAL BONDING

Li NBe O NeC FB

Na PMg S ArSi ClAl

Lewis Electron-Dot Symbols

The Ionic Bonding Model

• electrons are transferred from one element to another and bonds are formed between the two oppositely charged ions

Na Cl+

Cl-

Na+ +

H He

Cl+Mg

Cl

Cl-

Mg2+ + 2

Energy Considerations in Ionic Bonding

Consider the following reactions,

Now we can determine that,

+ -r 1

- -r 2

Li (g) + Li(g) H = -IE

F(g) + F (g) H = -EA

e

e

+ -r r 2 r 1Li(g) + F(g) Li (g) + F (g) H = H - H

= IE - EA

-1 -1

-1

= 520 kJ mol - 328 kJ mol

= 192 kJ mol

• the reaction in the gas phase is endothermic.

• the reaction of the gas phase ions,

The lattice energy is the energy required to break a solid ionic compound into a mole of gaseous ions. (See Chapter 12, Section 12.7 and 12.9)

+ -r latticeLi (g) + F (g) LiF(s) H = H

to form the solid is quite exothermic and is called the lattice energy

• difficult to determine but is important because it gives us a way to compare or quantify the strength of the ionic bond.

• the lattice energy indicates the strength of the ionic interaction which influences the melting point, hardness and solubility of ionic compounds.

• the lattice energy cannot be measured directly but we can use Hess’ Law to determine it.

• lattice energies are determined by means of a Born-Haber cycle which is a series of chosen steps, from elements to ionic compounds, for which all the enthalpies are known except the lattice energy.

• because the enthalpy is a state function we choose hypothetical steps whose enthalpy changes we can measure which take us from reactants to LiF formation. These steps are not the actual steps that occur when lithium reacts with fluorine.

Lattice Energy (Chapters 12.7 and 12.9, Do problems 58, 58, 73-76 from Ch. 12)

olatticeH = 617 kJ 161 kJ 79.5 kJ 520 kJ 328 kJ

= 1050 kJ

We begin with the elements in their standard states Li(s) and F2(g).

Step 1. Vaporize lithium metal.o o1 subLi(s) Li(g) H = H = 161 kJ

Step 2. Break F2 up into atoms,o1 1

2 2 22 2

12

F (g) F(g) H = (bond energy of F )

= 159 kJ = 79.5 kJ

Step 3. Ionize Li.+ o

3Li(g) Li (g) + - H = IE = 520 kJe

Step 4. Ionize F.- o

4F(g) + - F (g) H = -EA = -328 kJe

Step 5. Form crystalline solid from the gaseous reactant ions.+ - o o

5 latticeLi (g) + F (g) LiF(s) H = H

The reaction of solid lithium and fluorine gas to form the salt, LiF is,o1

2 f2Li(s) + F (g) LiF(s) H = 617 kJ

BDE

o1Li(s) Li(g) H = 161 kJ

o12 22 F (g) F(g) H = 79.5 kJ

+ o3Li(g) Li (g) + - H = 520 kJe

- o4F(g) + - F (g) H = -328 kJe

+ - o o5 latticeLi (g) + F (g) LiF(s) H = H

o12 f2Li(s) + F (g) LiF(s) H = 617 kJ

o o o o o of 1 2 3 4 5H = H + H + H + H + H

o o o1f sub 2 lattice2H = H (Li) + BDE(F ) + IE(Li) EA(F) H

or

olatticeH = 617 kJ 161 kJ 79.5 kJ 520 kJ 328 kJ

= 1050 kJ

eg. Determine lattice energy for Na2S(s) given:1st electron affinity of S is 203 kJ mol-1 2nd electron affinity of S is -694 kJ mol-1 ionization energy of Na is 496 kJ mol-1 the heat of sublimation of solid S is 277 kJ mol-1 the heat of sublimation of solid Na is 107 kJ mol-1 and the heat of formation of Na2S is -366 kJ mol-1

Periodic Trends in Lattice Energy

2

charge A charge Belectrostatic force

distance

E F d

charge A charge Belectrostatic energy

distance

olattace

cation charge anion chargeelectrostatic energy H

cation radius anion radius

From Coulomb’s Law,

and since

then

cations and anions lie as close to each other as possible so that the distance between them is the sum of their radii

olattace

cation charge anion chargeH

cation radius anion radius

• as we move down a group of metals or non-metals the ionic size increases so the lattice energy decreases

• as the charge on the ions increases, the lattice energy increases

F-Li+ O2-Mg2+

r 76 pm 133 pm 140 pm72 pm

DlatticeHo 1050 kJ mol-1 3923 kJ mol-1

Using the Model to Explain the Properties of Ionic Solids

• ionic solids are hard, brittle, and rigid

• ionic solids don’t conduct electricity but the liquids do as well as solutions of the ionic compound

• the melting and boiling points of ionic compounds are extremely high because it requires enormous amounts of energy to disrupt the forces between the ions and form ion pairs.

• it should be noted that the vapor exists as ion pairs not individual ions and the condensed phases do not exist as separate molecules.

Te

mpe

ratu

re /

oC

400

600

800

1000

1200

1400

1600

1800

2000

2200

boiling point

melting point

NaF NaCl NaBr NaI

Covalent Bonding

• sharing electrons is the principle way that atoms interact chemically, based on the number of known covalent compounds compared to the number of known ionic compounds

H H

H H

or

H-H

Basic Lewis Diagrams

H F H F or H F

Systematic method for drawing Lewis diagrams

eg. Draw the Lewis diagram for CH4

1. Count up the number of valence electrons C - 4e4H - 4e

8e

2. Decide which atoms are bonded together and draw a bond, which is equivalent to two electrons between the bonded atoms.

C

H

H

HH

3. Determine how many electrons are left and add them in as lone pairs about the appropriate nucleus trying to obey the “octet rule” or ensuring a full outer shell. For the present example we are finished.

eg. Draw the Lewis structure for NH3

eg. Draw the Lewis structure for N2

eg. Draw the Lewis structure for ethyne, C2H2.

eg. Draw the Lewis structure for ethanol, C2H5OH.

Bond Strengths and Bond Lengths

orA B(g) A(g) + B(g) H = BE(AB)

C N O F

Bo

nd

En

erg

y /

kJ m

ol-1

280

300

320

340

360

380

400

420

440

460

480

Bo

nd

Len

gth

/ p

m

130

135

140

145

150

155

160

F Cl Br I

Bo

nd

En

erg

y /

kJ m

ol-1

200

250

300

350

400

450

500

Bo

nd

Len

gth

/ p

m

120

140

160

180

200

220

C-X bond lengths and bond strengths, X=C, N, O, F

C-X bond lengths and bond strengths, X=F, Cl, Br, I

single double triple

Bo

nd

En

erg

y /

kJ m

ol-1

200

400

600

800

1000

1200

Bo

nd

Len

gth

/ p

m

110

115

120

125

130

135

140

145

Bond lengths and bond strengths for different CC bonds

eg. Using the periodic table to rank the bonds in each set in order of decreasing bond length and strength,

a) S-F, S-Br, S-Cl

b) C O C O C O, ,

Using the Model to Explain the Properties of Covalent Solids

• the model proposes that electron sharing between pairs of atoms leads to strong bonds between those atoms within a molecule

• there are two types of compounds with covalent bonds that have very different properties

Molecular solids

• low mpt (large range)

• soft

Network Covalent Solids

SiO2

Diamond

• very high mpt

1550 oC

3550 oC

• hard

• compounds held together by covalent bonds are poor electrical conductors in all phases or when dissolved in water

Infrared (IR) Spectroscopy

• characterization of substances

wavenumber / cm-1

500 1000 1500 2000 2500 3000 3500 4000

inte

nsity

/ ar

bitr

ary

1033 cm-1

1746 cm-1

820 cm-1

2917 cm-1

1049 cm-1

732 cm-1

611 cm-1

533 cm-1

O

O

S

F

Cl

Br

I

+

Polar Covalent Bonds

Electronegativity: the relative ability of a bonded atom to attract shared electrons.

• the bond energies of H-H and F-F are 432 and 159 kJ mol-1, respectively.

• we might consider, then, that the bond strength of H-F is intermediate between the two, ~296 kJ mol-1.

• the bond strength of H-F is, however, 565 kJ mol-1. Linus Pauling reasoned that the difference is due to an electrostatic (charge) contribution to the HF bond energy.

H F+

fluorine holds the shared electrons closer to it than does H, thereby giving the bond some “ionic character”

Pauling Electronegativity Scale.

in general the smaller the atom the higher the electronegativity

Bond Polarity

H F+

• the HF bond is polar. F has a higher electronegativity so the electrons in the bond are “attracted” more to F than H. the arrow points to the negative end of the polar bond or to the more electronegative bond.

Bond Character

• it was remarked earlier that HF had “partial ionic character”

• due to differences in electronegativities, covalent compounds, other than homonuclear diatomics, have some partial ionic character

“0 % ionic character”

“non-zero covalent

character”

: always some sharing of electrons

: H2, N2, Cl2, etc.

The difference in electronegativity, then, can give a measure of the polarity of the bond.

eg. Use the polar arrow to indicate the polarity of each bond, N-H, F-N, I-Cl.

eg. Rank the following bonds in order if increasing polarity; H-N, H-O, H-C

Properties of period 3 chlorides

Metallic Bonding: in a piece of metal what holds the atoms together?

• the “electron-sea model” proposes that all the metal atoms in the sample contribute their valence electrons to form an electron sea which is delocalized throughout the metal. The positive cores are submerged in this electron sea in an orderly array.

• the valence electrons are “shared” among all atoms in the substance

• the solid is held together by the mutual attraction of the metal cations for the mobile, highly delocalized electrons

Using the Model to Explain the Properties of Metallic Solids

Mechanical Properties

• the picture of the metallic solid is a regular array of ionic cores in a mobile sea of electrons

Conductivity

• the sea of electrons is mobile so electrons are easily transferred through the solid making metals excellent conductors of electricity

• metals are also good conductors of heat because the delocalized electrons can transfer heat better than localized electrons in a covalent bond or in ionic solids

Melting Point

• typically melting points (and boiling points) are quite high since the cationic core and its electron(s) must break away from the others

• the alkali earth metals boil at a higher temperature because they form 2+ ions and twice as many electrons meaning stronger bonding