 Calcium carbonate is
one of the most
interesting and
versatile compounds
on the planet.
 Roughly 4% of earth’s
crust
 The major component
of rocks such as
limestone and marble
 The white cliffs of
Dover are one of the
most famous natural
formations of CaCO3.
 THE PROPERTIES OF SUBSTANCES &
CHEMICAL BONDS
 The properties of substances are determined in large
part by the chemical bonds (O2 vs N2)
 What determines the type of bonding in each substance?
 The electronic structures of the atoms are the key to
answering the question
 We will examine the relationship
between the electronic structures
of atoms and the chemical bonds
they form
8.1 CHEMICAL BONDS, LEWIS SYMBOLS,
AND THE OCTET RULE
 Three general types of chemical
bonds
• Ionic bond: electrostatic forces
• Covalent bond: the sharing of
electrons
• Metallic bond: free electrons
8.1 CHEMICAL BONDS, LEWIS SYMBOLS, AND THE OCTET RULE
 LEWIS SYMBOLS
 The valence electrons
• The electrons involved in chemical bonding
• In most atoms, they reside in the outermost occupied shell
 The Lewis symbol
• Diagrams that show the bonding between atoms of a
molecule and the lone pairs of electrons that may exist in the
molecule
8.1 CHEMICAL BONDS, LEWIS SYMBOLS, AND THE OCTET RULE
 THE OCTET RULE
 The noble gases have very stable electron arrangements
(high ionization energies)
 Atoms often gain, lose, or share electrons to achieve the
same number of electrons as the noble gas
 The octet rule
• Atoms tend to gain, lose or share electrons until they are
surrounded by eight valence electrons
• The rule provides a useful framework to introduce concepts of
bonding although there are many exceptions
8.2 IONIC BONDING
 Consider a reaction of Na(s) and Cl2(g)
 The reaction can be explained by
the low ionization energy of Na and
the high electron affinity of Cl
 We can also explain the reaction by
the octet rule
8.2 IONIC BONDING
 FORMATION OF SODIUM CHLORIDE
8.2 IONIC BONDING
 ENERGETICS OF IONIC BOND FORMATION
 The formation of sodium chloride is very exothermic
 The heat of formation of other ionic substances is also quite
negative
 The first ionization energy of Na(g) is 496 kJ/mol
 The electron affinity of Cl(g) is −349 kJ/mol
 If we consider only the electron transfer for the formation
reaction, the energy change would be:
496 − 349 = 147 kJ/mol
How can we explain this?
8.2 IONIC BONDING
 THE LATTICE ENERGY
 The lattice energy is the energy required to completely
separate a mole of a solid ionic compound into its gaseous
ions
 The potential energy of two interacting charged particles
 The magnitude of lattice energies depends predominantly on
the ionic charges
8.2 IONIC BONDING
 THE LATTICE ENERGY
Sample Exercise 8.1 Magnitudes of Lattice Energies
Without consulting Table 8.2, arrange the following ionic compounds in order
of increasing lattice energy: NaF, CsI, and CaO.
Answer: CsI < NaF < CaO.
Practice Exercise
Which substance would you expect to have the greatest lattice energy, MgF2,
CaF2, or ZrO2?
Answer: ZrO2
8.2 IONIC BONDING
 The lattice energy can not be
determined directly by
experiment
8.2 IONIC BONDING
 ELECTRON CONFIGURATIONS OF IONS
 The formation of Na2+ and Cl2− are energetically very unfavorable
 Groups 1A, 2A, and 3A atoms form 1+, 2+, and 3+ ions,
respectively
 Groups 5A, 6A, and 7A atoms form 1−, 2−, and 3− ions,
respectively
 Transition metals do not observe the octet rule: Fe2+ and Fe3+
8.3 COVALENT BONDING
 A chemical bond formed by
sharing a pair of electrons
 The attractions and repulsions
among electrons and nuclei in
the H2 molecule
 Quantum mechanical
calculation tells us that the
concentration of electron
density between the nuclei
leads to a net attractive force
that constitutes the covalent
bond holding the molecule
together
8.3 COVALENT BONDING
 LEWIS STRUCTURES
 The formation of covalent bonds can be represented using
Lewis symbols
 For the nonmetals, the number of valence electrons in a
neutral atom is the same as the group number
the number of
covalent bonds
1
2
3
4
8.3 COVALENT BONDING
 MULTIPLE BONDS
 The Lewis structures of CO2 and N2
 Bond length: the distance between the nuclei of atoms
involved in a bond
8.4 BOND POLARITY AND ELECTRONEGATIVITY
 Compare the electron-sharing of Cl2 and H2 with the
sharing of H2O and NaCl
 Polar covalent bond and nonpolar covalent bond
 If the difference of the covalent bond-forming atoms in
relative ability to attract electrons is large enough,
an ionic bond is formed
 Electronegativity is defined as the ability of an atom in a
molecule to attract electrons to itself
8.4 BOND POLARITY AND ELECTRONEGATIVITY
 ELECTRONEGATIVITY
 Linus Pauling (1901-1994) developed the first and most
widely used electronegativity scale
 Electronegativity can be used to estimate whether a given
bond will be nonpolar or polar covalent, or ionic
 The electronegativity of an atom in a molecule is related to its
ionization energy and electron affinity
 Fluorine, the most electronegative element, has an
electronegativity of 4.0
 The least electronegative element, cesium, has an
electronegativity of 0.7
8.4 BOND POLARITY AND ELECTRONEGATIVITY
 ELECTRONEGATIVITY
8.4 BOND POLARITY AND ELECTRONEGATIVITY
 ELECTRONEGATIVITY AND BOND POLARITY
 The greater the difference in electronegativity between two
atoms, the more polar their bond
8.4 BOND POLARITY AND ELECTRONEGATIVITY
 ELECTRONEGATIVITY AND BOND POLARITY
8.4 BOND POLARITY AND ELECTRONEGATIVITY
 DIPOLE MOMENTS
 We can indicate the polarity of the HF molecule in two ways:
 Polarity helps determine many of the properties of substances
such as hydrogen bonding and solvation
 How can we quantify the polarity of molecule? For the two
equal and opposite charges, the dipole moment is:
• , dipole moment (C·m
or debye, D)
• Q, charge (C)
• r, distance (m)
1 D = 3.34 × 10-30 C·m
Sample Exercise 8.5 Dipole Moments of Diatomic Molecules
The bond length in the HCl molecule is 1.27 Å. (a) Calculate the dipole
moment, in debyes, that would result if the charges on the H and Cl atoms
were 1+ and 1–, respectively. (b) The experimentally measured dipole
moment of HCl(g) is 1.08 D. What magnitude of charge, in units of e, on the H
and Cl atoms would lead to this dipole moment?
Solution
8.4 BOND POLARITY AND ELECTRONEGATIVITY
 DIPOLE MOMENTS
 The actual charges on the atoms decrease from 0.41 in HF to 0.057 in HI
8.4 BOND POLARITY AND ELECTRONEGATIVITY
 DIFFERENTIATING IONIC AND COVALENT BONDING
 The ability to quickly categorize the predominant bonding
interactions in a substance as covalent or ionic imparts
considerable insight into the properties of that substance
 By considering the interaction between a metal and a
nonmetal
• SnCl4: colorless liquid, mp -33 ˚C, bp 114 ˚C, polar covalent
 To use the difference in electronegativity
• SnCl4: 1.2;
NaCl: 2.1
• MnO: 2.0, ionic; Mn2O7: 2.0, polar covalent
• The increase in the oxidation state of a metal leads to an
increase in the degree of covalent character in the bonding
8.5 DRAWING LEWIS STRUCTURES
 Lewis structures can help us understand the bonding in many
compounds and are frequently used when discussing the
properties of molecules
PCl3
5
+ 3(7) = 26
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.
8.5 DRAWING LEWIS STRUCTURES
2. Arrange atoms.
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
8.5 DRAWING LEWIS STRUCTURES
3. Complete the octets
around all the atoms
bonded to the central
atom. Hydrogen is an
exception
Keep track of the electrons:
26 - 6 = 20; 20 - 18 = 2
8.5 DRAWING LEWIS STRUCTURES
4. Place any leftover
electrons on the
central atom
Keep track of the electrons:
26 - 6 = 20; 20 - 18 = 2; 2 - 2 = 0
8.5 DRAWING LEWIS STRUCTURES
5. If there are not enough
electrons to give the central
atom an octet, try multiple
bonds
Sample Exercise 8.8 Lewis Structure for a Polyatomic Ion
Draw the Lewis structure for the BrO3– ion.
Solution
The total number of valence electrons is, therefore, 7 + (3  6) + 1 = 26.
For oxyanions— BrO3–, SO42–, NO3–, CO32–, and so forth—the oxygen atoms surround the central
nonmetal atom.
Practice Exercise
Draw the Lewis structure for (a) ClO2–, (b) PO43–
Answers: (a)
(b)
8.5 DRAWING LEWIS STRUCTURE
 FORMAL CHARGE
 The charge the atom would have if all the atoms in the
molecule had the same electronegativity
 How to 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.
8.5 DRAWING LEWIS STRUCTURE
 FORMAL CHARGE
 The concept of formal charge helps us choose a
preferred Lewis structure
 How to choose the correct structure.
• Choose the Lewis structure in which the atoms bear
formal charges closest to zero
• Choose the Lewis structure in which any negative
charges reside on the more electronegative atoms
preferred
8.5 DRAWING LEWIS STRUCTURE
 FORMAL CHARGE
preferred
8.5 DRAWING LEWIS STRUCTURE
Oxidation number
+1
−1
Formal charge
0
0
Actual partial charge
+0.18
−0.18
+0.178
−0.178
Dipole moment
8.6 RESONANCE STRUCTURES
 Consider the Lewis structure of
ozone, O3
 Both Lewis structures are not
corresponding to the structure of ozone
 Resonance structure: An alternate way
of drawing a Lewis dot structure for a
compound
8.6 RESONANCE STRUCTURES
 How can we understand the resonance structure?
 The rules for drawing Lewis structures do not allow us to
have a single structure that adequately represents the
ozone molecule
8.6 RESONANCE STRUCTURES
Draw the resonance structures for HCO2− and NO3− .
8.6 RESONANCE STRUCTURES
 RESONANCE IN BENZENE
All six C-C bonds are of equal length 1.40 Å
C-C single bond 1.54 Å; C=C double bond 1.34 Å
8.7 EXCEPTIONS TO THE OCTET RULE
 ODD NUMBER OF ELECTIONS
 In a few molecules and polyatomic ions, such as ClO2, NO,
NO2, and O2−, the number of valence electrons is odd
 Complete pairing of these electrons is impossible
 An octet around each atom cannot be achieved
 NO contains 5 + 6 =11 valence electrons
8.7 EXCEPTIONS TO THE OCTET RULE
 LESS THAN AN OCTET OF VALENCE ELECTRONS
 Consider boron trifluoride, BF3
 In the Lewis structure of BF3, there are only six
electrons around the boron atom and the
formal charges are zero
 If you try to complete the octet around boron:
Most important
8.7 EXCEPTIONS TO THE OCTET RULE
 MORE THAN AN OCTET OF VALENCE ELECTRONS
 Consider phosphorus pentachloride, PCl5
 Elements from the third period and
beyond have unfilled nd orbitals
that can be used in bonding
The orbital diagram for
the valence shell of a
phosphorus atom
 The larger the central atom is, the larger the number of
atoms that can surround it
Draw the Lewis structure of SF4, ICl4−, and PO43−
8.8 STRENGTH OF COVALENT BONDS
 The stability of a molecule is related to the strengths of the
covalent bonds it contains
 The strength of a bond is measured by determining how
much energy is required to break the bond
 The bond enthalpy for a Cl-Cl bond, D(Cl-Cl), is measured
to be 242 kJ/mol
8.8 STRENGTH OF COVALENT BONDS
8.8 STRENGTH OF COVALENT BONDS
 The bond enthalpy is always positive as energy is always
required to break chemical bonds
 A molecule with strong chemical bonds generally has less
tendency to undergo chemical change than does with weak
bonds
8.8 STRENGTH OF COVALENT BONDS
 BOND ENTHALPIES AND
THE ENTHALPIES OF REACTION
 We can use average bond enthalpies to estimate the
enthalpies of reactions in which bonds are broken and new
bonds are formed
 Consider a gas-phase reaction:
CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g)
8.8 STRENGTH OF COVALENT BONDS
 BOND ENTHALPIES AND
THE ENTHALPIES OF REACTION
8.8 STRENGTH OF COVALENT BONDS
 BOND ENTHALPY AND BOND LENGTH
 Enormous amounts of energy can be stored in chemical
bonds → can be used as an explosives
 Characteristics of an explosive
• Very exothermic decomposition
• Gaseous products
• Rapid decomposition
• Controllably stable
+
−
2−
+
+
−
+