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Chapter 9
Lewis Theory of Chemical Bonding
Lewis Bonding Theory
Emphasizes valence electrons to explain bonding
Lewis structures - Electron Dot Structures
Lewis structures allow us to predict many properties of molecules - molecular stability, shape, size, polarity
Why Do Atoms Bond?
A chemical bond forms when the potential energy of the bonded atoms is less than the potential energy of
the separate atoms.
To calculate this potential energy, you need to consider the following interactions:
nucleus–to–nucleus repulsions
electron–to–electron repulsions
nucleus–to–electron attractions
Types of Bonds
Types of Atoms Type of Bond Bond Characteristic
metals to nonmetals Ionic electrons
transferrednonmetals to
nonmetals Covalent electrons shared
metals to metals Metallic electrons pooled
Types of Bonding
Ionic Bonds
Metal atoms lose an electrons and become
cations.
Nonmetal atoms gain electrons and become
anions.
The oppositely charged ions are then attracted to each other, resulting in an
ionic bond.
Covalent BondsNonmetal atoms have relatively high
ionization energies, so it is difficult to remove electrons from them.
When nonmetals bond together, it is better in terms of potential energy for the atoms to share valence electrons.
The potential energy is lowest when the
electrons are between the nuclei.
Atoms held together because shared electrons are attracted to both nuclei.
Metallic BondsThe relatively low
ionization energy of metals allows them to lose electrons
easily.
Metal atoms release their valence electrons to be
shared as a pool by all the atoms/ions in the metal.
An organization of metal cation islands in a sea of
electrons.
Bonding results from attraction of cation for
the delocalized electrons.
Valence Electrons & Bonding
Because valence electrons are held most loosely, and
Because chemical bonding involves the transfer or sharing of electrons between two or more atoms,
Valence electrons are the most important in
bonding.
Determining the Number of Valence Electrons in an Atom
The column number on the Periodic Table will tell you how many valence electrons a main group atom has.
IA IIA IIIA IVA VA VIA VIIA VIIIA
Li Be B C N O F Ne
1e- 2e- 3e- 4e- 5e- 6e- 7e- 8e-
Lewis Structures of AtomsWe represent the valence electrons of main-group elements
as dots surrounding the symbol for the element.
H
Li Be B C N O F Ne
He
Na Mg Al Si P S Cl Ar
B C N
P
IA
IIA IIIA
IVA VA VIA VIIA
VIIIA
Practice – Write the Lewis structure for
Arsenic
A lithium ion A fluoride ion
•••
••
As
Stable Electron Arrangementsand Ion Charge
Metals form cations by losing enough electrons to get the same electron configuration as the previous noble gas.
Nonmetals form anions by gaining enough electrons to get the same electron configuration as the next noble gas.
The noble gas electron configuration must be very stable.
Main-group ions and the noble gas configurations.
Lewis Bonding Theory ⇒ Octet Rule
When atoms bond, they tend to gain, lose, or share electrons to result in eight valence electrons
ns2np6 (noble gas configuration)
ExceptionsH, Li, Be, B attain an electron configuration like He
He = two valence electrons (a duet)Li loses its one valence electron H may share or gain one electron
It commonly loses its one electron to become H+ Be loses two electrons to become Be2+
It commonly shares its two electrons in covalent bonds, resulting in four valence electrons
B loses three electrons to become B3+
It commonly shares its three electrons in covalent bonds, resulting in six valence electrons
Expanded octets for elements in Period 3 or below
Lewis Theory and Ionic Bonding
Lewis symbols can be used to represent the transfer of electrons from metal atom to nonmetal atom, resulting in ions that are attracted to each other and therefore bond.
Sodium Chloride Formation
Lewis Theory Predictions for Ionic Bonding
Lewis theory predicts the number of electrons a metal atom should lose or a nonmetal atom should gain.
This allows us to predict the formulas of ionic compounds that result.
It also allows us to predict the relative strengths of the resulting ionic bonds from Coulomb’s Law.
Predicting Ionic FormulasUsing Lewis Symbols
Electrons are transferred until the metal loses all its valence electrons and the nonmetal has an octet.
Numbers of atoms are adjusted so the electron transfer comes out even.
Li2O
Ca · · Cl · · ·
· · · ·
Use Lewis theory to predict the chemical formula of calcium chloride
Cl · · ·
· · · · CaCl2
Ca2+
Use Lewis symbols to predict the formula of an ionic compound made from reacting a metal, M, that has two valence
electrons with a nonmetal, X, that has five valence electrons
M3X23M2+
2X3-
Sr3N2
Energetics of Ionic Bond FormationThe ionization energy of the metal is endothermic:
Na(s) → Na+(g) + 1 e ─ " ΔH° = +496 kJ/mol
The electron affinity of the nonmetal is exothermic:
½Cl2(g) + 1 e ─ → Cl─(g)" ΔH° = −244 kJ/mol
Therefore the formation of the ionic compound should be endothermic.
But the heat of formation of most ionic compounds is exothermic and generally large.
Na(s) + ½Cl2(g) → NaCl(s)" ΔH°f = −411 kJ/mol Why?
Na(s) + ½Cl2(g) → NaCl(s)" ΔH°f = + “something”
Ionic Bonding & the Crystal LatticeThe extra energy that is released comes
from the formation of a structure in which every cation is surrounded by anions.
This structure is called a crystal lattice.
The crystal lattice is held together by electrostatic attractions.
The crystal lattice maximizes these attractions between cations and anions, leading to the most stable arrangement.
Crystal Lattice
Electrostatic attraction is nondirectional!!
There is no direct anion–cation pair “bond”
Therefore, there is no ionic molecule.
The chemical formula for an ionic compound is an empirical formula, simply giving the ratio of ions based on charge balance.
Lattice EnergyThe extra stability that accompanies the formation of
the crystal lattice is measured as the lattice energy.
The lattice energy is the energy released when the solid crystal forms from separate ions in the gas state
1) Always exothermic 2) Can be calculated from knowledge of other processes
Lattice energy depends directly on size of charges and inversely on distance between ions.
Practice – Given the information below, determine the lattice energy of NaCl
Na(s) → Na(g) +108 kJ½ Cl2(g) → Cl(g) +½(244 kJ)Na(g) → Na+(g) +496 kJCl (g) → Cl−(g) −349 kJNa(s) + ½ Cl2(g) → NaCl(s) −411 kJ
Na+ (g) + Cl−(g) → NaCl(s) ΔH (NaCl lattice)?
Determining Lattice EnergyThe Born–Haber Cycle
The Born–Haber Cycle is a hypothetical series of reactions that represents the formation of an ionic compound from its constituent elements.
The reactions are chosen so that the change in
enthalpy of each reaction is known except for the last one, which is the lattice energy.
Naº (s) + ½ Cl2 (g) NaCl (s)
Born–Haber Cyclefor NaCl
ΔH°f (metal atoms, g)
ΔH°f (nonmetal atoms, g)
ΔH°f (cations, g)
ΔH°f (anions, g)
ΔH°(crystal lattice)
ΔH°f (salt)
separating atoms
forming ions
forming lattice
Born–Haber Cycle
Use Hess’s Law to add up enthalpy changes of other reactions to determine the lattice energy.
ΔH°f(salt) = ΔH°f(metal atoms, g) + ΔH°f(nonmetal atoms, g)
+ ΔH°f(cations, g) + ΔH°f(anions, g)
+ ΔH°(crystal lattice)
ΔH°f(NaCl, s) = ΔH°f [Na(s)--->Na(g)] (Heat of vaporization) + ΔH°f (Cl–Cl bond energy) + Na 1st Ionization Energy + Cl Electron Affinity
+ NaCl Lattice Energy
ΔH°f(NaCl, s) =
ΔH°f(Na atoms,g) + ΔH°f(Cl atoms,g)
+ ΔH°f(Na+,g) + ΔH°f(Cl−,g)
+ ΔH°(NaCl lattice)
Na(s) → Na(g) +108 kJ½ Cl2(g) → Cl(g) +½(244 kJ)Na(g) → Na+(g) +496 kJCl (g) → Cl−(g) −349 kJNa+ (g) + Cl−(g) → NaCl(s) ΔH (NaCl lattice)Na(s) + ½ Cl2(g) → NaCl(s) −411 kJ (measured in an experiment)
NaCl Lattice Energy = (−411 kJ) − [(+108 kJ) + (+122 kJ) + (+496 kJ) + (−349 kJ) ]
= −788 kJ
NaCl Lattice Energy = ΔH°f(NaCl, s) − [ΔH°f(Na atoms,g) + ΔH°f(Cl–Cl bond energy) + Na 1st Ionization Energy + Cl Electron Affinity]
Practice – Given the information below, determine the lattice energy of MgCl2
Mg(s) ➔ Mg(g) ΔH1°f = +147.1 kJ/mol½ Cl2(g) ➔ Cl(g) ΔH2°f = +122 kJ/molMg(g) ➔ Mg+(g) ΔH3°f = +738 kJ/molMg+(g) ➔ Mg2+(g) ΔH4°f = +1450 kJ/molCl(g) ➔ Cl−(g) ΔH5°f = −349 kJ/molMg(s) + Cl2(g) ➔ MgCl2(s) ΔH6°f = −641 kJ/mol
Practice – Given the information below, determine the lattice energy of MgCl2
Mg(s) ➔ Mg(g) ΔH1°f = +147.1 kJ/mol2{½ Cl2(g) ➔ Cl(g)} 2ΔH2°f = 2(+122 kJ/mol)Mg(g) ➔ Mg+(g) ΔH3°f = +738 kJ/molMg+(g) ➔ Mg2+(g) ΔH4°f = +1450 kJ/mol2{Cl(g) ➔ Cl−(g)} 2ΔH5°f = 2(−349 kJ/mol)Mg2+(g) + 2 Cl−(g) ➔ MgCl2(s) ΔH° lattice energy = ? kJ/mol
Mg(s) + Cl2(g) ➔ MgCl2(s) ΔH6°f = −641 kJ/mol
Trends in Lattice EnergyIon Size
The force of attraction between charged particles is inversely proportional to the
distance between them.
Larger ions mean the center of positive charge (nucleus of the cation) is farther away from the negative charge (electrons of the anion).
larger ion = weaker attractionweaker attraction = smaller lattice energy
Lattice Energy vs. Ion Size
Trends in Lattice EnergyIon Charge
Lattice Energy =
−910 kJ/mol
Lattice Energy =
−3414 kJ/mol
The force of attraction between oppositely charged particles is directly proportional to the product of the charges.
Larger charge means the ions are more strongly attracted.
larger charge = stronger attractionstronger attraction = larger lattice energy
Of the two factors, ion charge is generally more important
Lattice Energies of Some Ionic Solids (kJ/mole)(M+ + X- → MX)
Cations F- Cl- Br- I- O2-
Li+ 1036 853 807 757 2,925
Na+ 923 787 747 704 2,695
K+ 821 715 682 649 2,360
Be2+ 3,505 3,020 2,914 2,800 4,443
Mg2+ 2,957 2,524 2,440 2,327 3,791
Ca2+ 2,630 2,258 2,176 2,074 3,401
Al3+ 5,215 5,492 5,361 5,218 15,916
Anions
(2M+ + X2- → M2X)(M2+ + 2X- → MX2)(M2+ + X2- → MX)(M3+ + X- → MX3)(2M3+ + 3X- → M2X3)
Order the following ionic compounds in order of increasing magnitude of lattice energy:
CaO, KBr, KCl, SrO
First examine the ion charges and order by sum of the charges
(KBr, KCl) < (CaO, SrO)
Then examine the ion sizes of each group and order by radius larger < smaller
KBr < KCl < SrO < CaO
Order the following ionic compounds in order of increasing magnitude of lattice energy:
MgS, NaBr, LiBr, SrS
First examine the ion charges and order by sum of the charges
(NaBr, LiBr) < (MgS, SrS)
Then examine the ion sizes of each group and order by radius larger < smaller
NaBr < LiBr < SrS < MgS
Ionic Bonding-Model vs. ObservationsLewis theory implies strong attractions between ions.
Lewis theory predicts high melting points and boiling points for ionic compounds.
The stronger the attraction (larger the lattice energy),
the higher the melting point.
Ionic compounds have high melting points and boiling points (MP generally > 300 °C).
All ionic compounds are solids at room temperature.
Melting an ionic solid
Ionic Compounds Melt
Lewis theory implies that the positions of the ions in the crystal lattice are critical to the stability of the structure
Lewis theory predicts that moving ions out of
position should therefore be difficult, and ionic solids should be hard
Ionic solids are relatively hard(compared to most molecular solids)
Ionic Bonding-Model vs. Observations
Lewis theory implies that if the ions are displaced from their position in the crystal lattice, that repulsive forces should occur
This predicts the crystal will become unstable and break apart. Lewis theory predicts ionic solids will be brittle.
Ionic solids are brittle. When struck they shatter.
Ionic Bonding-Model vs. Observations
To conduct electricity, a material must have charged particles that are able to flow through the material
Lewis theory implies that, in the ionic solid, the ions are locked in position and cannot move around
Lewis theory predicts that ionic solids should not
conduct electricity
Ionic solids do not conduct electricity
Ionic Bonding-Model vs. Observations
Conductivity of NaCl
in NaCl(s), the
ions are stuck in
position and not
allowed to move
to the charged
rods
Lewis theory implies that, in the liquid state or when dissolved in water, the ions will have the ability to move around
Lewis theory predicts that both a liquid ionic compound
and an ionic compound dissolved in water should conduct electricity
Ionic compounds conduct electricity in the liquid state or when dissolved in water
Ionic Bonding-Model vs. Observations
Conductivity of NaCl
in NaCl(aq),
the ions are
separated and
allowed to
move to the
charged rods
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