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11/9/2011
Chapter 9
Bonding
Ionic Compounds
• Formed between metal and nonmetal:
– Example: Mg (s) + Cl2 (g) ------> MgCl2 (s)
• Ionic solids: ions are
arranged in a regular lattice
• Strong forces due to attraction of ions for each other
– For example, melting points generally are high
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Lattice Energy
• A measure of the stability of an ionic solid
• Based on Coulomb’s Law
–
–
–
–
q q 
F  C 1 22 
 r 
F = force
q1 and q2 = charges
r = distance between charges
C = constant
• Forces between ions are stronger when there are
– Higher charges (most important)
– Smaller ions
• We will not cover the Born-Haber Cycle
– pp 369 - 371
Ionic Compounds
• Stronger forces between ions for
– Higher charges (most important)
– Smaller ions
• Strong forces => high MP
• Example: Three compounds and three melting points
are shown below.
• Which MP goes with which compound?
– Compounds:
– Melting Points (oC):
LiF,
610,
MgO, LiCl
845, 2800
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Covalent Compound
• Formed from two nonmetals
• Bonding is due to shared electrons
• Covalent bond: a bond in which two electrons are
shared by two atoms
– A covalent compound contains only covalent bonds
• We represent covalent compounds with Lewis
structures
Lewis Structures
• Look at HF
• # valence electrons = Group number
– H: 1 valence electron
– F: 7 valence electrons
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Lewis Structures
• Represent valence electrons as dots
– H: 1 valence electron
– F: 7 valence electrons
• Share electrons to give both elements full shell
configuration (H = 2; F = 8)
– Two shared electrons are a covalent bond
– Bond represented by a line
Lewis Structures
• Need orderly process for complex molecules
• Follow the rules below:
1. Draw a skeleton structure joining atoms by single bond.
2. Count the number of valence electrons, including the charge.
3. Deduct 2 electrons for each bond from step 1.
4. Distribute remaining electrons to give all atoms an octet
of electrons (Octet rule)
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Skeleton Structures
• You should be able to write skeleton structures
– From organic condensed structural formulas
• CH2ClCH2Cl
• CH3COOH
– One central atom
• NH3
Lewis Structures
1.
2.
3.
4.
Draw a skeleton structure joining atoms by single bond
Count the number of valence electrons, including the charge.
Deduct 2 electrons for each bond from step 1.
Distribute remaining electrons to give all atoms an octet of electrons
• Examples:
– OH– CO
– CH2CH2
– CH3CHO
– NH4+
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Bond Lengths
•
•
•
•
•
•
•
Distance between two bonded atoms
Bond lengths vary from compound to compound
Average values below for carbon-carbon bonds
C-C 154 pm
C=C 134 pm
C≡C 120 pm
In general,
– Single bond > double bond > triple bond
Polar and Nonpolar Bonds
• Electrons in bonds not necessarily shared equally
• The F atom attracts
electrons more than
H atom
• Get a polar bond
– e- shared unequally
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Polar and Nonpolar Bonds
• We can picture the formation of a ions as a situation
in which the bond is so polar that electrons have
been completely transferred
• Nonpolar
• Polar
• Ionic
Electronegativity
• Electronegativity: the ability of an atom to attract
electrons when in chemical bond
– High EN = strong attraction for e– Low EN = weak attraction for e– Useful for making predictions of bond polarity
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Electronegativity Difference
• Nature of a bond is related to the electronegativity
difference between elements involved
– O–H
– H-F
DEN = 3.5 – 2.1 = 1.4
DEN = 4.0 – 2.1 = 1.9
Electronegativity Difference
• Rules of Thumb
–
–
–
–
DEN = 0
DEN < 0.5
DEN >= 0.5
DEN >= 2.0
Totally covalent bond
Essentially nonpolar bond (text doesn’t do this)
Polar covalent bond
Ionic bond
• Questions
– Which is more polar: HF or HCl?
– Which bond has greater ionic character: BO or CO?
– Is the C-Se bond
polar?
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Bond Polarity
• Don’t always have electronegativity tables available
• You should recognize that
– F, N, O, and Cl all have high EN. Almost all their bonds are
polar
– C and H have about the same EN. The C-H bond is
essentially nonpolar
Resonance
• Earlier we did Lewis structure for formate ion, HCO2-
• Would you predict that both CO bonds have the
same length?
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Resonance
• The C-O bond lengths in the formate ion are equal
– Bond length: 126 pm
– Midway between C-O (135 pm) and C=O (120 pm)
• Actual structure is what we call a resonance hybrid
of two separate resonance structures
Resonance
• A resonance hybrid basically is a weakness of the
Lewis Structure model
– Cannot use one structure to describe a molecule
– Use more than one
• Actual structure is intermediate between the two
resonance structures
– Mule = horse ↔ donkey
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Resonance
• If you have a choice of ways to distribute multiple
bonds in a structure, then actual structure is
resonance hybrid of the choices
• Example: azide ion N3• Example: carbonate ion CO32-
Benzene
• Important organic chemical, C6H6
• Skeletal structure:
• Frequent representation:
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Formal Charges
• Book-keeping method to help describe distribution
of electrons in a molecule
– Not real charges
– Frequently useful in organic chemistry
– Starting point for predicting chemical and physical
properties
Formal Charges
• Example: Formaldehyde
H2C=O
– Draw circle around each atom, dividing bonds
• Two electrons per bond
• Each bonded atom is assigned one of them
• Both electrons in a lone pair are assigned
to one atom
• Call these the “assigned electrons”
–
Formal charge = - (# assigned e- - # valence e-)
• C:
• O:
• H:
4 assigned, 4 valence
6 assigned, 6 valence
1 assigned, 1 valence
Formal charge = 0
Formal charge = 0
Formal charge = 0
– Sum of formal charges must = charge of species
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Formal Charges
• Consider one of carbonate resonance
structures
• Oxygen has 6 valence e- and carbon 4
• Formal charge of oxygen atoms with 3 lone pairs is -1
• This is as it should be
– Sum of formal charges = -2
– Charge of ion = -2
Formal Charges
• Formal charges help gauge importance of resonance
hybrid structures.
• If possible, don’t want
– Positive charge on highly electronegative element
– Large separation of charge
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Formal Charges
• Look at two important examples from Organic Chem
• What is formal charge of carbon on both structures?
– What do these formal charges tell us?
Formal Charges
• Formal charges on carbon are the below
• This implies that each species must be an ion
– CH3+ (a carbocation)
– CH3- (a carboanion)
• Carbocations and carboanions are very important in
Organic
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Exceptions to Octet Rule
• The octet rule works best for second period elements
(C, N, O)
• There are exceptions
• Compounds of B and Be
– Example: Diborane B2H6
Exceptions to Octet Rule
• Odd-electron molecules
–
–
–
–
Impossible to satisfy octet rule
Example: NO
Can write two Lewis structures with incomplete octets
Use formal charges to choose the best one
5 assigned
5 valence
6 assigned
6 valence
6 assigned 5 assigned
5 valence 6 valence
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Exceptions to the Octet Rule
• Expanded octets
• Examples: SF6
• How is this possible?
– 3rd, 4th, etc. period elements have empty d-orbitals available
– S: 1s22s22p63s23p4
–
n quantum number = 3 for valence electrons
–
If n = 3, l can be 0, 1, or 2.
–
3d orbitals are available for bonding
Exceptions to Octet Rule
• For 3rd period (and 4th, 5th, etc) elements the central
atom can have more than four bonds or pairs of
electrons if necessary
• Usually easy to identify
– Central atom bonded to more than four atoms
• Example: PF5
• Example: XeF4
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Bond Enthalpies
• Enthalpy required to
break a bond
• Apply only to gases
• Average values
Bond Enthalpies
• ΔH = -[bonds formed – bonds broken]
• CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
• ΔH = -[2BE(C=O) + 4BE(OH) -4BE(C-H)-2BE(O=O)]
•
= -[2(803) + 4(467) – 4(416) – 2(498)]
•
= -814 kJ/mole
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