Chapter 9: The Basics
of Chemical Bonding
Chemistry: The Molecular Nature
of Matter, 6E
Jespersen/Brady/Hyslop
Chemical Bonds
 Attractive forces that hold atoms together
in complex substances
 Molecules and ionic compounds
Why study?
 Changes in these bonding forces are the
underlying basis of chemical reactivity
 During reaction:
 Break old bonds
 Form new bonds
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Chemistry: The Molecular Nature of Matter, 6E
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Two Classes of Bonds
 Covalent bonding
 Occurs in molecules
 Sharing of e’s
 Ionic Bonding




Occurs in ionic solid
e’s transferred from 1 atom to another
Simpler
We will look at this first
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Chemistry: The Molecular Nature of Matter, 6E
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Ionic Bonds
 Result from attractive forces between
oppositely charged particles
Na+
Cl –
 Metal - nonmetal bonds are ionic because:
 Metals have
 Low ionization energies
 Easily lose e– to be stable
 Non-metals have
 Very exothermic electron affinities
 Formation of lattice stabilizes ions
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Chemistry: The Molecular Nature of Matter, 6E
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Ionic Compounds
 Formed from metal and nonmetal
Na + Cl
Na+ + Cl
NaCl(s)
e
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 Ionic Bond
 Attraction between + and – ions in ionic
compound.
 Why does this occur? Why is e
transferred?
 Why Na+ and not Na2+ or Na?
 Why Cl and not Cl2 or Cl+?
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Chemistry: The Molecular Nature of Matter, 6E
5
Ionic Compounds
Ionic crystals:
 Exist in 3-dimensional array of
cations and anions = lattice
structure
Ionic chemical formulas:
 Always written as empirical
formula
 Smallest whole number ratio
of cation to anion
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Chemistry: The Molecular Nature of Matter, 6E
6
Energetics
 Must look at energy of system to answer
these questions
 For any stable compound to form from its
elements
 Potential Energy of system must be lowered.
 Net  in energy Hf° < 0 (negative)
 What are factors contributing to energy
lowering for ionic compound?
 Use Hess’s Law to determine
 Conservation of energy
 Envision two paths
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Chemistry: The Molecular Nature of Matter, 6E
7
Two Paths to Evaluate Energy
1. Na(s) + ½Cl2(g)  NaCl(s)
Hf° = – 411.1 kJ/mol
2. Stepwise path
Na(s)  Na(g)
Hf°(Na, g) = 107.8 kJ/mol
½Cl2(g)  Cl(g)
Hf°(Cl, g) = 121.3 kJ/mol
Na(g)  Na+(g) + e–
IE(Na) =495.4 kJ.mol
Cl(g) + e–  Cl–(g)
EA(Cl) = – 348.8 kJ/mol
Na+(g) + Cl–(g)  NaCl(s)
Hlattice = – 787 kJ/mol

Na(s) + ½Cl2 (g)  NaCl(s) Hf° = – 411 kJ/mol
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Chemistry: The Molecular Nature of Matter, 6E
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Lattice Energy
 Amount that PE of system decreases when 1
mole of solid salt is formed from its gas
phase ions.
 Energy released when ionic lattice forms.
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Lattice Energy
 Always HLattice= – = exothermic
 HLattice gets more exothermic (larger negative
value) as ions of opposite charge in crystal lattice
are brought closer together as they wish to be.
 Ions tightly packed with opposite charged ions next
to each other.
q q 
H Lattice 
d
 Any  in PE due to ionizing atoms is more than met
by  in PE from formation of crystal lattice. Even
for +2 and –2 ions
 overall exothermic to form ionic solids and they
are stable compounds.
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Chemistry: The Molecular Nature of Matter, 6E
10
Your Turn!
Assuming that the separation between cations
and anions in the lattice is nearly identical,
which species would have the greatest lattice
energy?
A. Sodium chloride
B. Calcium chloride
C. Calcium nitride
D. Sodium oxide
E. Calcium oxide
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Why do Metals form Cations and
Nonmetals form Anions?
Nonmetal
Metal
 Left hand side of Periodic  Right hand side of
Periodic Table
Table
 IE large and positive
 IE small and positive
 Little energy required to
remove e to give cations
 EA small and negative
or positive
 Not favorable to attract an
e to it.
 Difficult to remove e
 EA large and negative
 But easy to add e
 Exothermic—large amount
of E given off
 PE of system 
 Least expensive, energy-  Least expensive, energywise, to form cation
wise, to form anion
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Electron Configurations of Ions
 How e'ic structure affects types of ions formed
Ex.
Na 1s2 2s2 2p6 3s1 = [Ne] 3s1
Na+ 1s2 2s2 2p6 = [Ne]
IE1 = 496 kJ/mol
small
not too difficult
IE2 = 4563 kJ/mol
large ~10 x larger very difficult
 Can remove 1st e, as doesn't cost too much
 Can’t remove 2nd e, as can't regain lost energy
from lattice

doesn’t form.
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Electron Configurations of Ions
Ex.
Ca
[Ar] 4s2
Ca2+ [Ar]
IE1 small = 590 kJ/mol
not too difficult
IE2 small = 1140 kJ/mol not too difficult
IE3 large = 4940 kJ/mol too difficult
 Can regain by lattice energy ~2000 kJ/mole if
+2, –2 charges.
 But 3rd e too hard to remove
 Can't recoup required energy through lattice
formation.
 doesn't happen
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Chemistry: The Molecular Nature of Matter, 6E
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Electron Configurations of Ions
 Stability of Noble Gas Core below
valence e's effectively limits # e's that
metals lose.
 Ions formed have Noble gas e
configuration
 True for anions and cations
Ex. Cl 1s2 2s2 2p6 3s2 3p5 = [Ne]3s2 3p5
Cl 1s2 2s2 2p6 3s2 3p6 = [Ar]
 Adding another e
 requires putting it into next higher n shell
 Energy cost too high
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Electron Configurations of Ions
Ex.
O
1s2 2s2 2p4
O 1s2 2s2 2p5
EA1 = 141 kJ/mol
O2 1s2 2s2 2p6 = [Ne] EA2 = +844 kJ/mol
EAnet = +703 kJ/mol
endothermic
 Energy required more than made up for by 
in HLattice caused by higher 2 charge
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Chemistry: The Molecular Nature of Matter, 6E
16
Electron Configurations of Ions
Generalization:
 When ions form
 Atoms of most representative elements (s and p
block)
 Tend to gain or lose e's to obtain nearest Noble
gas e configuration
 Except He (2 e's), all noble gases have 8 e's in
highest n shell
Octet Rule
 Atoms tend to gain or lose e's until they
have achieved outer (valence) shell
containing octet (8 e's)
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Octet Rule
 Works well with
 Group IA and IIA
metals
 Al
 Non-metals
 H and He can't obey
 Limited to 2 e's in n
= 1 shell
 Doesn't work with
 Transition metals
 Post transition metals
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Chemistry: The Molecular Nature of Matter, 6E
18
Your Turn!
What is the correct electron configuration for Cs
and Cs+ ?
A. [Xe] 6s1 , [Xe]
B. [Xe] 6s2 , [Xe] 6s1
C. [Xe] 5s1 , [Xe]
D. [Xe] 6s1 , [Xe] 6s2
E. [Xe] 6p2 , [Xe] 6p1
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Chemistry: The Molecular Nature of Matter, 6E
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Post Transition Metals
Ex.
Sn
[Kr] 4d10 5s2 5p2
Sn2+ [Kr] 4d10 5s2
 Neither has Noble Gas e configuration
 Have emptied 5p subshell
Sn4+ [Kr] 4d10
 Does have empty n = 5
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Chemistry: The Molecular Nature of Matter, 6E
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Transition Metals
 1st e's lost from ns orbital of outer shell
 Lose e's from highest n first, then highest ℓ
Ex. Fe
[Ar] 3d6 4s2
Fe2+ [Ar] 3d6
loses 4s e's 1st
Fe3+ [Ar] 3d5
then loses 3d e
 Extra stability due to exactly half-filled d
subshell.
 Consequences
 TM2+ common oxidation state as remove 2 e's
from outer ns shell
 Ions of larger charge result from loss of d e's.
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Chemistry: The Molecular Nature of Matter, 6E
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Transition Metals
 Not easy to predict which ions form and
which are stable
 But ions with exactly filled or half-filled d
subshells are extra stable and tend to form.
 Mn2+
[Ar]3d5
 Fe3+
[Ar]3d5
 Zn2+
[Ar]3d10
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Chemistry: The Molecular Nature of Matter, 6E
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Your Turn!
What is the correct electronic configuration for
Cu and Cu2+ ?
A. [Ar] 3d 9 4s2, [Ar] 3d 9
B. [Ar] 3d 10 4s1, [Ar] 3d 8 4s1
C. [Ar] 3d 10 4s1, [Ar] 3d 9
D. [Ar] 3d 9 4s2, [Ar] 3d 10 4s1
E. [K] 3d 9 4s2, [Ar] 3d 9
 Filled and half-filled orbitals are particularly
stable.
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Chemistry: The Molecular Nature of Matter, 6E
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Predicting Cation Configurations
Consider Bi, whose configuration is
[Xe]6s2 4f14 5d10 6p3. What ions are expected?
Rewrite config’n: [Xe]4f14 5d10 6s2 6p3
Bi3+ and Bi5+
Consider Fe, whose configuration is: [Ar]4s2 3d6
What ions are expected?
Rewrite config’n: [Ar]3d6 4s2
Fe2+ and Fe3+
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Chemistry: The Molecular Nature of Matter, 6E
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Predicting Anion Configurations
Non-metals gain electrons to become
isoelectronic with next larger noble gas
O: [He]2s22p4 + ? 2e– → ?
O2– : [He]2s2 2p6
N: [He]2s22p3 + ? 3e– → ? N3– : [He]2s2 2p6
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Lewis Symbols
 Electron bookkeeping method
 Way to keep track of e–’s
 Write chemical symbol surrounded by dots for each
e–
Group #
Valence e–'s
e– conf'n
IA
1
ns1
IIA
2
ns2
H
Li
He
Na
Jespersen/Brady/Hyslop
Be
Mg
IIIA
3
ns2np1
IVA
4
ns2np2
B
C
Al
Si
Chemistry: The Molecular Nature of Matter, 6E
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Lewis Symbols
Group #
Valence e-'s
e- conf'n
VA
VIA
VIIA
VIIIA
5
6
7
8
ns2np3
ns2np4
ns2np5
ns2np6
He
N
O
F
Ne
P
S
Cl
Ar
For the representative elements
Group # = # valence e–’s
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Chemistry: The Molecular Nature of Matter, 6E
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Lewis Symbols
 Can use to diagram e– transfer in ionic
bonding
Na
Mg
+ Cl
+
O
Jespersen/Brady/Hyslop
Na+
+ Cl
2+
Mg
+ O
Chemistry: The Molecular Nature of Matter, 6E

2
28
Covalent Compounds
 Form individual separate molecules
 Atoms bound by sharing e–’s
 Do not conduct electricity
 Often low melting point
Covalent Bonds
 Shared pairs of e–’s between 2 atoms
 2 H atoms come together, Why?
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Chemistry: The Molecular Nature of Matter, 6E
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Covalent Bond
 Attraction of valence e– of 1 atom by nucleus
of other atom
 Shifting of e– density
 As distance between nuclei ,  probability
of finding either e– near either nucleus
 Pulls nuclei closer together
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Covalent Bond
 As nuclei get
close
 Begin to repel
each other
 Both have
high positive
charge
 Final internuclear distance between 2 atoms in bond
 Balance of attractive and repulsive forces
 Net attraction since bond forms
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Electron Pair Bond
 e–’s come together in molecular bond
 In sense 1s orbital of each H is filled
 e–’s share same region of space  their
spins must be paired (1  and 1 )
 Refer to as e– pair bond
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Chemistry: The Molecular Nature of Matter, 6E
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Bond Dissociation Energy, D
 Energy needed to break one mole of bond
in covalent molecule in gas phase
 Symbolized D
 D varies with bond type
 Single, double, triple
 D varies with atoms involved
 D is always positive
 Formation of bonds stabilizes structure
energetically
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Chemical Potential Energy
ΔH°rxn = {D (Bonds broken) }  {D (Bonds formed) }
Bond
Bond Energy
(kJ mol−1)
Bon
d
Bond Energy
(kJ mol−1)
C–C
348
C–Br
276
C═C
612
C–I
238
C≡C
960
H–H
436
C–H
412
H–F
565
C–N
305
H–Cl
431
C═N
613
H–Br
366
C≡N
890
H–I
299
C–O
360
H–N
388
C═O
743
H–O
463
C–F
484
H–S
338
C–Cl
338
H–Si
376
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Covalent Bond
 2 quantities characterize this bond
Bond Length (bond distance)
 Distance between 2 nuclei = rA + rB
Bond Energy
 Also bond strength
 Amount of energy released when bond formed (
PE) or
 Amount of energy must put in to “break” bond
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Chemistry: The Molecular Nature of Matter, 6E
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Your Turn!
Which species is most likely covalently bonded?
A. CsCl
B. NaF
C. CaF2
D. CO
E. MgBr2
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Lewis Structures
 Molecular formula drawn with Lewis Symbols
 Method for diagramming electronic structure
of covalent bonds
 Uses dots to represent e–s
 Covalent bond
 Shared pair of e–s
 Each atom shares e–s so has complete octet
ns2np6
 Noble Gas e– configuration
 Except H which has complete shell with 2 e–s
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Octet Rule:
When atoms form covalent bonds, tend to
share sufficient e–’s so as to achieve outer
shell having 8 e–’s
 Indicates how all atoms in molecule are attached
to one another
 Accounts for ALL valence e–’s in ALL atoms in
molecule
Let’s look at some examples
Noble Gases: 8 valence e–’s
 Full octet ns2 np6
 Stable monatomic gases
 Don’t form compounds
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Chemistry: The Molecular Nature of Matter, 6E
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Lewis Structures
Diatomic Gases:
 H and Halogens
H2





H· + ·H  H:H
or
HH
Each H has 2 e–’s through sharing
Can write shared pair of e–’s as line ()
:=
covalent bond
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Chemistry: The Molecular Nature of Matter, 6E
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Lewis Structures
Diatomic Gases:
F2
F
+
F
FF
F
F
 Each F has complete octet
 Only need to form one bond to complete
octet
 Pairs of e–’s not included in covalent bond
are called Lone Pairs
 Same for rest of Halogens: Cl2, Br2, I2
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Lewis Structures
Diatomic Gases:
HF
H + F
HF
H F
 Same for HCl, HBr, HI.
 Molecules are diatomics as need only 1e– to
complete octet
 Separate molecules
 Gas in most cases because very weak
intermolecular forces
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Your Turn!
How many electrons are required to complete
the octet around nitrogen, when it forms N2 ?
A. 2
B. 3
C. 1
D. 4
E. 6
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Lewis Structures
 Many nonmetals form more than 1 covalent bond
C
O
N
Needs 4 e-’s
Forms 4 bonds
H
Needs 3 e-’s
Forms 3 bonds
H N H
H C H
O H
H
H
H
H
H C H
H N H
H
H
methane
ammonia
Jespersen/Brady/Hyslop
Needs 2 e-’s
Forms 2 bonds
O H
H
water
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Multiple Bonds
Single Bond
 Bond produced by sharing one pair of
e–’s between 2 atoms
 Many molecules share more than one pair
of e–’s between 2 atoms
 Multiple bonds
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Double Bonds
 2 pairs of e–’s shared between 2 atoms
Ex. CO2
O
C
O
O C O
O C O
Triple bond
 3 pairs of e–’s shared between 2 atoms
Ex. N2
N
N
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N
N
Chemistry: The Molecular Nature of Matter, 6E
N
N
45
Your Turn!
Which species is most likely to have multiple
bonds ?
A. CO
B. H2O
C. PH3
D. BF3
E. CH4
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Carbon Compounds
 Carbon-containing compounds
 Exist in large variety
 Mostly due to multiple ways in which C can form
bonds
 Functional groups
 Groups of atoms with similar bonding
 Commonly seen in C compounds
 Molecules may contain more than 1
functional group
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Important Compounds of Carbon
Alkanes
 Hydrocarbons
 Only single bonds
Isomers
Ex.
CH4
CH3CH3
CH3CH2CH3
 Same molecular formula
 Different physical properties
 Different connectivity
(structure)
H H H H
H C C C C H
H
H C H
H
H
H C C C H
H H H
butane
iso-butane
H H H H
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methane
ethane
propane
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Hydrocarbons
H C C H Ethylene
 Alkenes
(ethene)
H H
 Contain at least one
H H
double bond
H C C C C H
butene
H H H H
H C C H
 Alkynes
 Contain at least one
triple bond
H
Jespersen/Brady/Hyslop
H H
acetylene
(ethyne)
C C C C H butyne
H H
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Oxygen Containing Organics
 Alcohols
 Replace H with OH
H
H H
H C O
H C C O
H H
H H H
methanol
 Ketones
 Replace CH2 with
C=O
 Carbonyl group
ethanol
H O H
H C C C H
H
H
acetone
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Carbonyl Group
 Carbon in hydrocarbon with double
bond to oxygen




Aldehydes
Ketones
Carboxylic acids
Amides
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Oxygen Containing Organics
H O
 Aldehydes
 At least 1 atom
attached to C=O is H
H
acetaldehyde
 Organic Acids
 Contains carboxyl
group
 COOH
 Esters
H C C H
H O
H C C O
H
H
acetic acid
H O
methyl acetate
(methylethanoate)
 Alkyl group replaces H C C O
H on carboxylic acid
H H C H
H
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Nitrogen Containing Organics
 Amines
 derivatives of NH3 with H’s replaced by alkyl groups
H H
H H H
H C N H
H C N C H
H
methylamine
H
H
dimethylamine
 Amides
 Contain -NH2 on carbon chain with carbonyl
H O
H C C N H
H
H
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acetamide (ethanamide)
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Your Turn!
How many isomers are there of butanol?
A. none
B. 2
C. 3
D. 6
E. 4
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Electronegativity and Bond
Polarity
 2 atoms of same
element form bond
 Equal sharing of e’s
 2 atoms of different
elements form bond
 Unequal sharing of
e’s
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Why?
 1 atom usually attracts es
more strongly than other
Result:
 Unbalanced distribution
of e density within bond




e cloud tighter around Cl in HCl
Slight + charge around H
Slight – charge around Cl
Not complete transfer of e from 1 atom to
another.
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Electronegativity and Bond
Polarity
 Leads to concept of
Partial charges
+

H——Cl
+ on H = +0.17
 on Cl = 0.17
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Polar Covalent Bond
 Aka Polar bond
 Bond that carries partial + and – charges at
opposite ends
 Bond is dipole
 2 poles or 2 charges involved
Polar Molecule
 Molecule has partial + and – charges at
opposite ends due to a polar bond
http://web.visionlearning.com/custom/chemistry/animations/CHE1.7-an-H2Obond.shtml
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
58
Dipole Moment
 Quantitative measure of extent to which bond is
polarized.
 Dipole moment = Charge on either end  distance
between them




μ=q×r
Units = debye (D)
1D = 3.34 × 10–30 C·m (Coulombs·meter)
The size of the dipole moment or the
degree of polarity in the bond depends
on the differences in abilities of bonded
atoms to attract e’s to themselves
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
59
Table 9.3 Dipole Moments and Bond
Lengths for Some Diatomic
Molecules
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
60
Electronegativity ()
 Relative attraction of atom for es in bond
 Ability of bonded atom to attract es to itself
 Quantitative basis
 Table of electronegativities - fig 8.5
 Difference in electronegativity
 = estimate of bond polarity
  = |1  2|
 Ex. N—H
Si—F

+
Jespersen/Brady/Hyslop
+

Chemistry: The Molecular Nature of Matter, 6E
61
Electronegativity Table
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
62




Electronegativities
Must know for 2nd row and H.
(H) ~ 2.0
(F) ~ 4.0
 by 0.5 for each element as go to left
H
2

Li Be B
C
N
O
F
1
1.5
2
2.5
3
3.5
4

P
S
Cl
2
2.5 3.0
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
63
Learning Check
 Calculate the bond polarity of the following
 HCl
 CO
 CH
 LiF
 BeO
(H)~2 (Cl)~3
 = 3 – 2 = 1
(C)~2.5 (O)~3.5
(C)~2.5 (H)~2
 = 3.5 – 2.5 = 1
 = 2.5 – 2 = 0.5
(Li)~1 (F)~4
 = 4 – 1 = 3
(Be)~1.5 (O)~3.5  = 1.5 – 3.5 = 2
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
64
Your Turn!
Which of the following species has the least
polar bond?
A. HCl
B. HF
C. HI
D. HBr
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
65
Trends in Electronegativity
  from left to right
 across period as Zeff 
  from top to bottom
 down group as n 
Ionic and Covalent Bonding
 2 extremes of bonding
 Actual is usually somewhere in between.
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
66
Using Electronegativities
   A  B
 Difference in electronegativity
 Measure of ionic character of
bond
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
67
Using Electronegativities
 Nonpolar Covalent Bond
 No difference in electronegativity
 Ionic Character of bond
 Degree to which bond is polar
 > 1.7 means mostly ionic
 >50% ionic
 More electronegative element almost completely
controls e
  < 0.5
 Means almost purely covalent
 Nonpolar; < 5% ionic
 0.5 <  < 1.7 polar covalent
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
68
Result
 Elements in same region of Periodic Table
 i.e., 2 nonmetals
 Have similar ’s
 Bonding more covalent
 Elements in different regions of Periodic
Table
 i.e., metal and nonmetal
 Have different ’s
 Bonding predominantly ionic
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
69
Reactivities of Elements Related
to Electronegativities
 Parallels between  and its reactivity
 Tendency to undergo redox reactions
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
70
Reactivities of Elements Related
to Electronegativities
Metals
 Low , means easy to oxidize (groups I and IIA)
 High , means hard to oxidize (Pt, Ir, Rh, Au, Pd)
 Reactivity  across row as  
Nonmetals
 Oxidizing power of nonmetal (how easily reduced)
parallels .
 Oxidizing power  across row as  
 Oxidizing power  down column as  
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
71
Drawing Lewis Structures




Very useful
Way of diagramming structure
Used to describe structure of molecules
Can be used to make reasonable accurate
predictions of shapes of molecules
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
72
Drawing Lewis Structures
 Not all molecules obey the Octet Rule.
 Holds rigorously for 2nd row elements like C,
N, O, and F
 B and Be sometimes have less than octet
BeCl2, BCl3
 2nd row can never have more than 8 e’s
 3rd row and below, atoms often exceed octet
 Why?
 n = 3 shell can have up to 18 e’s as now
have d orbitals in valence shell
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
73
Method for Drawing Lewis Structure
1. Decide how atoms are bonded
Skeletal structure = arrangement of atoms.
Central atom




Usually given first
Usually least electronegative
2. Count all valence es. (All atoms)
3. Place 2 es between each pair of atoms

Draw in single bonds
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
74
Method for Drawing Lewis Structure
4. Complete octets of terminal atoms
(atoms attached to central atom) by
adding es in pairs
5. Place any remaining e’s on central atom
in pairs
6. If central atom does not have octet


Form double bonds
If necessary, form triple bonds
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
75
Ex. N2F2
2 N = 2  5e = 10 e
2F = 2  7e = 14 e
Total = 24 e
single bonds  6 e
18 e
F lone pairs  12 e
6 e
N electrons
 6 e
0 e
Skeletal Structure
F
N
F
Complete terminal
atom octets
F
N
N
F
Put remaining es on
central atom
F
Jespersen/Brady/Hyslop
N
N
N
Chemistry: The Molecular Nature of Matter, 6E
F
76
Ex. N2F2
 Not enough electrons to complete N octets
 Must form double bond between N’s to
satisfy both octets.
F
N
N
F
Jespersen/Brady/Hyslop
F
N
Chemistry: The Molecular Nature of Matter, 6E
N
F
77
Ex. SiF4
1 Si = 1 
= 4
4F = 4  7e = 28 e
Total = 32 e
single bonds  8 e
24 e
F lone pairs  24 e
0 e
4e
Skeletal Structure
F
e
F
Si
F
F
Complete terminal
atom octets
F
F
Si
F
F
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
78
Ex. H2CO3
 CO32 oxoanion, so C central, and O’s
around, H+ attached to two O’s
O
1 C = 1  4e = 4 e
3 O = 3  6e = 18 e
2 H = 2  1e = 2 e
Total = 24 e
single bonds  10 e
14 e
O lone pairs  14 e
0 e
Jespersen/Brady/Hyslop
H
O
C
O
H
O
H
O
H
O
C
 But C only has 6 e
Chemistry: The Molecular Nature of Matter, 6E
79
Ex. H2CO3 (cont.)
 Too few e
 Must convert 1 of
lone pairs on O to
2nd bond to C
 Form double bond
between C and O
O
H
C
O
H
O
H
O
H
Jespersen/Brady/Hyslop
O
O
C
Chemistry: The Molecular Nature of Matter, 6E
80
Ex. PCl5
1 P = 1  5 e = 5 e 
5 Cl = 5  7e = 35 e
Total = 40 e
single bonds  10 e
30 e
Cl lone pairs
 30 e
0 e
 P has 10 e
 OK as 3rd row element
 Can expand its shell
Jespersen/Brady/Hyslop
Cl
Cl
P
Cl
Cl
Cl
Cl
Cl
P
Cl
Chemistry: The Molecular Nature of Matter, 6E
Cl
Cl
81
Ex. BBr3
1 B = 1  3 e = 3 e 
3 Br = 3  7e = 21 e
Total = 24 e
single bonds
 6 e
18 e
Br lone pairs  18 e
0 e
Br
B
Br
Br
B
e
 B has only 6
 Does not form double bond
 Has incomplete octet
Jespersen/Brady/Hyslop
Br
Br
Chemistry: The Molecular Nature of Matter, 6E
Br
82
Ex. CO2
O C O
1 C = 1  4e = 4 e
2 O = 2  6e = 12 e
O C O

Total = 16 e
single bonds  4 e
O C O
O C O
12 e
O lone pairs  12 e
O C O
O O O

0e
 C only has 4 e
O C O
O C O
 Must form 2 double
bonds to O to complete  Which of these is correct?
C’s octet
 Need another criteria
 3 ways you can do this  Come back to this
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
83
Lewis Structures—Theory
 Lewis Structures meant to describe how
atoms share es in chemical bonds
 Theory
What experimental observations back up
this theory?
 Look at properties related to number of e
pairs shared between 2 atoms
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
84
Your Turn!
What is wrong with the following structure?
A. Too few total electrons
B. Too many total electrons
C. Lack of octet around nitrogen
D. Too many electrons around the N atom
E. Structure is correct as written
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
85
Review of Covalent Bond
Properties
Bond Length
 Distance between nuclei of bonded atoms
Bond Energy
 Energy required to separate bonded atoms
into neutral particles
Bond Order
 Number of pairs of es shared between 2
atoms
 Measure of amount of e density in bond
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
86
Bond Order, Bond Length and Bond
Energy
Greater Bond Order = Greater e density
 Nuclei held together more tightly
 Larger bond energy, D
 Larger D means
 Nuclei drawn closer together
 Shorter bond length
 As bond order , bond length , and
bond energy 
 Assumes comparing bonds between same 2
elements
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
87
Ex. H2SO4
H O
1 S = 1  6e = 6 e
4 O = 4  6e = 24 e
2 H = 2  1e = 2 e
Total = 32 e
H O
single bonds  12 e
20 e
O lone pairs  20 e
0 e
H O
• n=3, has empty d orbitals
• Could expand it's octet
• Could write structure with double bonds.
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
O
S
O
H
O
H
O
H
O
O
S
O
O
S
O
88
How Do We Know Which is Accurate?
Experimental evidence
In this case bond lengths from X-ray data
S—O bonds (no H attached) are shorter 142 pm
S—O—H, S—O longer 157 pm
Indicates that 2 bonds are shorter than other 2
Structure with S=O for 2 O’s without H’s is more
accurate
 Preferred Lewis Structure
 Even though it seems to violate octet rule
unnecessarily





Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
89
Need Criteria to Predict When This
Occurs
 Covalent bond = shared e pair
 If es shared equally
 Each atom “owns” ½ of e pair or 1 e
 When “break” bond
 ½ of these es going to each atom in bond
 So for S
 have 4 bonds
 Have 4 es
 2 es less than 6 es, S has as atom
 In bookkeeping sense
 S has 2+ charge if it obeys octet rule
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
90
Formal Charge (FC)
 Apparent charge on atom
 Bookkeeping method
 Does not represent real charges
FC = # valence e  # lone pair e  ½ (#
bonding e)
FC = # valence e  [# bonds to an atom + #
unshared e ]
 Indicate Formal charges by placing them in
circles around atoms
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
91
FC = #valence e  [#bonds to atom
+ # unshared e ]
O
H
O
-1
+2
S
O
O
H
-1
O
H
O
S
O
O
Jespersen/Brady/Hyslop
H
Structure 1
FCS = 6  (4 + 0) = 2
FCH = 1  (1 + 0) = 0
FCO(s) = 6  (1 + 6) =  1
FCO(d) = 6  (2 + 4) = 0
Structure 2
FCS = 6  (6 + 0) = 0
FCH = 1  (1 + 0) = 0
FCO(s) = 6  (2 + 4) = 0
FCO(d) = 6  (2 + 4) = 0
Chemistry: The Molecular Nature of Matter, 6E
92
H2SO4
 No formal charges on any atom in Structure 2
Conclusion:
 When several Lewis Structures are possible
 Those with smallest formal charges
 Most stable
 Preferred
Most Stable Lewis Structure
1. Least number of FC's best
2. All FC  1
3. Any negative FC on most electronegative element
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
93
Can Use FC to Explain B Chemistry





BCl3
Why doesn’t a double bond form here?
FCB = 3 – 0 – 3 = 0
FCCl = 7 – 6 – 1 = 0
All FC's = 0 so stable, doesn't need to form
double bond
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
94
CO2 Which Structure is Best
 Use FC to determine
which structure is best
1
+1
FCC = 4  (4 + 0) = 0
FCO(s) = 6  (1 + 6) = 1
FCO(d) = 6  (2 + 4) = 0
FCO(t) = 6 – (3 + 2) = +1
O
C
O
O
C
O
-1
 Central structure Best
 All FC’s = 0
Jespersen/Brady/Hyslop
+1
O
C
Chemistry: The Molecular Nature of Matter, 6E
O
95
Your Turn!
What is the formal charge on Xe for the
following ?
A. +2, +4
B. +2, +3
C. +4, 0
D. +4, +2
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
96
Resonance: When Single Lewis
Structure Fails

O
N = 3
O = 3.5
1N=1
= 5
3 O = 3  6e = 18 e
1 charge
= 1 e
Total = 24 e
single bonds  6 e
18 e
O lone pairs 18 e
0 e
5e
Jespersen/Brady/Hyslop
e
N
O
O

O
N
O
O
Chemistry: The Molecular Nature of Matter, 6E
97
Ex. NO3
 Lewis structure predicts
1 bond shorter than
other 2

O
N
O
Experimental observation:
O
 All 3 N—O bond lengths are same
 All shorter than N—O single bonds
 Have to modify Lewis Structure
 e can't distinguish O atoms
 Can write 2 or more possible structures
simply by moving where e are
 Changing placement of e
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
98
What are Resonance Structures?
 Multiple Lewis Structures for single molecule
 No single Lewis structure is correct
 Structure not accurately represented by any 1 Lewis
Structure
 Actual structure = "average" of all possible structures
 Double headed arrow between resonance structures
used to denote resonance
1


1
O
N
O
+1
O
O
1
1
Jespersen/Brady/Hyslop
N+1
O
O
1
O
N
O

+1
O
1
Chemistry: The Molecular Nature of Matter, 6E
99
Resonance Structures
 Lewis structures assume e is localized
between 2 atoms
 In cases where need resonance structures,
e are delocalized
 Smeared out over all atoms
 Can move around entire molecule to give
equivalent bond distances
Resonance Hybrid
 Way to depict
resonance delocalization
Jespersen/Brady/Hyslop

O
N
O
Chemistry: The Molecular Nature of Matter, 6E
O
100
C = 2.5
Ex. CO32
O
O = 3.5
1 N = 1  5e = 5 e
3 O = 3  6e = 18 e
1 charge
= 1 e
Total = 24 e
single bonds  6 e
18 e
O lone pairs 18 e
0 e
C only has 6 e so form
double bond
Jespersen/Brady/Hyslop

C
O
O

O
C
O
O

O
C
O
Chemistry: The Molecular Nature of Matter, 6E
O
101
Three Equivalent Resonance
Structures

O
O
C
O
1
1

O
C
O
O
1
1

C
O
1
O 1
O
 All have same net formal charges on C and O’s
 FC = 1 on singly bonded O’s
 FC = O on doubly bond O and C
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
102
Resonance Structures Not Always
Equivalent
 2 or more Lewis Structures for same compound
may or may not represent e distributions of equal
energy
How Do We Determine Which are Good
Contributors?
1. All Octets are satisfied
2. All atoms have as many bonds as possible
3a. FC  1
3b. Any negative charges are on electronegative
atoms.
4. As little charge separation as possible.
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
103
Drawing Good Resonance Structures
1. All must be valid Lewis Structures
2. Only e are shifted



Usually double or triple bond and lone pair
Nuclei can't be moved
Bond angles must remain the same
3. Number of unpaired electrons, if any, must remain
the same
4. Major contributors are the ones with lowest E (see
above)
5. Resonance stabilization is most important when
serves to delocalize charge onto 2 or more atoms
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
104
Ex. NCO
N = 3 C = 2.5 O = 3.5
1 C = 1  4e = 4 e
1 N = 1  5e = 5 e
1 O = 1  6e = 6 e
1 charge
= 1 e
Total = 16 e
single bonds  4 e
12 e
O lone pairs 12 e
0 e
Jespersen/Brady/Hyslop
N
C
O


N
N
N
N
C
C
C
C
Chemistry: The Molecular Nature of Matter, 6E
O
O
O
O



105
Ex. NCO
FCN = 5 – 2 – 3 = 0
FCC = 4 – 0 – 4 = 0
FCO = 6 – 6 – 1 = –1
FCN = 5 – 4 – 2 = –1
FCC = 4 – 0 – 4 = 0
FCO = 6 – 4 – 2 = 0
FCN = 5 – 6 – 1 = –2
FCC = 4 – 0 – 4 = 0
FCO = 6 – 2 – 3 =
+1
Jespersen/Brady/Hyslop
N
1
C
O

Best
1
N
C
2
N
O
+1
C
Chemistry: The Molecular Nature of Matter, 6E
O

OK
Not

Acceptable
106
Resonance Stabilization
 Actual structure is more stable than either
single resonance structure
 For Benzene
 The extra stability is ~146 kJ/mol
 Resonance Energy
 Extra stabilization energy from resonance
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
107
Your Turn!
Which of the structures below exhibit
resonance?
A. NO2
B. H2O
C. N3D. N2O (nitrogen is central atom)
E. CH3CH2CH
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
108
Coordinate Covalent Bonds
Ammonia
 Normal covalent bonds
 One electron from each atom shared
between the two
H
3H
+ N
N
H
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
H
109
Coordinate Covalent Bond
Ammonium Ion
 H+ has no electrons
 N has lone pair
 Can still get 2es shared between them
H
+ H+
N
H
+
H
H
Jespersen/Brady/Hyslop
H
N
H
H
Chemistry: The Molecular Nature of Matter, 6E
110
Coordinate Covalent Bond
 Both electrons of shared pair come from just
one of two atoms
 Once bond formed, acts like any other
covalent bond
 Can't tell where electrons came from after
bond is formed
 Useful in understanding chemical reactions
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
111
Coordinate Covalent Bond
 Especially boron (electron deficient
molecule) reacts with nitrogen compounds
that contain lone pair of electrons
H
Cl
+
N
H
H
H
B
Cl
Jespersen/Brady/Hyslop
Cl
H
Cl
N
B
H
Cl
Chemistry: The Molecular Nature of Matter, 6E
Cl
112
Learning Check
Ex. What is the best Lewis Structure for
HClO4?
XeF4
I3–
BrF5
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
113