Chem 101 week 10 ch9

Chapter 9Chemical Bonding I: Basic

Concepts

Chemical Bonds• Three basic types of

bonds:– Ionic

• Electrostatic attraction between ions

– Covalent• Sharing of electrons

– Metallic• Metal atoms bonded to

several other atoms

Ionic Bonding

Energetics of Ionic Bonding

As we saw in the last chapter, it takes 495 kJ/mol to remove electrons from sodium.


We get 349 kJ/mol back by giving electrons to chlorine.


• But these numbers don’t explain why the reaction of sodium metal and chlorine gas to form sodium chloride is so exothermic!

Energetics of Ionic Bonding• There must be a third

piece to the puzzle.• What is as yet

unaccounted for is the electrostatic attraction between the newly formed sodium cation and chloride anion.

Lattice Energy

• This third piece of the puzzle is the lattice energy:The energy required to completely separate a mole of a

solid ionic compound into its gaseous ions.

• The energy associated with electrostatic interactions is governed by Coulomb’s law:

Eel = Q1Q2

d

Lattice Energy

• Lattice energy, then, increases with the charge on the ions.

• It also increases with decreasing size of ions.


By accounting for all three energies (ionization energy, electron affinity, and lattice energy), we can get a good idea of the energetics involved in such a process.


• These phenomena also helps explain the “octet rule.”

• Metals, for instance, tend to stop losing electrons once they attain a noble gas configuration because energy would be expended that cannot be overcome by lattice energies.

Covalent Bonding• In these bonds atoms share

electrons.• There are several electrostatic

interactions in these bonds:– Attractions between electrons

and nuclei– Repulsions between electrons– Repulsions between nuclei

Polar Covalent Bonds

• Although atoms often form compounds by sharing electrons, the electrons are not always shared equally.

• Fluorine pulls harder on the electrons it shares with hydrogen than hydrogen does.

• Therefore, the fluorine end of the molecule has more electron density than the hydrogen end.

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

Electronegativity:

• The ability of atoms in a molecule to attract electrons to itself.

• On the periodic chart, electronegativity increases as you go…– …from left to right across

a row.– …from the bottom to the

top of a column.


• When two atoms share electrons unequally, a bond dipole results.

• The dipole moment, , produced by two equal but opposite charges separated by a distance, r, is calculated:

= Qr• It is measured in debyes (D).


The greater the difference in electronegativity, the more polar is the bond.

9.5

The Electronegativities of Common Elements

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

Lewis Structures

Lewis structures are representations of molecules showing all electrons, bonding and nonbonding.

Writing Lewis Structures

1. Find the sum of valence electrons of all atoms in the polyatomic ion or molecule.– If it is an anion, add one

electron for each negative charge.

– If it is a cation, subtract one electron for each positive charge.

PCl35 + 3(7) = 26


2. The central atom is the least electronegative element that isn’t hydrogen. Connect the outer atoms to it by single bonds.

Keep track of the electrons:

26 6 = 20


3. Fill the octets of the outer atoms.


26 6 = 20 18 = 2


4. Fill the octet of the central atom.


26 6 = 20 18 = 2 2 = 0


5. If you run out of electrons before the central atom has an octet…

…form multiple bonds until it does.

Writing Lewis Structures• Then assign formal charges.

– For each atom, count the electrons in lone pairs and half the electrons it shares with other atoms.

– Subtract that from the number of valence electrons for that atom: The difference is its formal charge.


• The best Lewis structure…– …is the one with the fewest charges.– …puts a negative charge on the most

electronegative atom.

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

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

Technology

Chem 101 week 10 ch9