chapter 8 chemical bonding i: basic conceptsinfo.psu.edu.sa/psu/maths/chapter08_lecture.pdf · 8.6...
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Chapter 8
Chemical Bonding I:
Basic Concepts
Copyright McGraw-Hill 2009 1
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Copyright McGraw-Hill 2009 2 Copyright McGraw-Hill 2009
8.1 Lewis Dot Symbols
• Valence electrons determine an element’s
chemistry.
• Lewis dot symbols represent the valence electrons
of an atom as dots arranged around the atomic
symbol.
• Most useful for main-group elements
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Copyright McGraw-Hill 2009 3 Copyright McGraw-Hill 2009
Lewis Dot Symbols of the Main Group Elements
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Copyright McGraw-Hill 2009 4 Copyright McGraw-Hill 2009
Write Lewis dot symbols for the following:
(a) N
(b) S2
(c) K+
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Copyright McGraw-Hill 2009 5 Copyright McGraw-Hill 2009
Write Lewis dot symbols for the following:
(a) N
(b) S2
(c) K+ K+
N• ••
• •
••
•• S
••
•• 2
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Copyright McGraw-Hill 2009 6 Copyright McGraw-Hill 2009
8.2 Ionic Bonding
Na• Cl• ••
••
•• + Na+ Cl ••
••
••
••
+
IE1 + EA1 = 496 kJ/mol 349 kJ/mol = 147 kJ/mol
fH = – 410.9 kJ/mol m.p. = 801oC
• Ionic bond: electrostatic force that holds
oppositely charge particles together
• Formed between cations and anions
• Example
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Copyright McGraw-Hill 2009 7
Microscopic View of NaCl Formation
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Copyright McGraw-Hill 2009 8 Copyright McGraw-Hill 2009
NaCl(s) Na+(g) + Cl(g) Hlattice = +788 kJ/mol
Because they are defined as an amount of energy,
lattice energies are always positive.
+
-
-
-
-
-
-
-
- +
+
+ +
+
+
+
• Lattice energy = the energy required to completely
separate one mole of a solid ionic
compound into gaseous ions
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Q = amount of charge
d = distance of separation
d
Q1 Q2
• Coulombic attraction:
2
21
d
QQF
• Lattice energy (like a coulombic force) depends on
• Magnitude of charges
• Distance between the charges
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Copyright McGraw-Hill 2009 10
Lattice energies of alkali metal iodides
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Copyright McGraw-Hill 2009 11 Copyright McGraw-Hill 2009
The ionic radii sums for LiF and MgO are 2.01 and 2.06 Å,
respectively, yet their lattice energies are 1030 and 3795 kJ/mol. Why
is the lattice energy of MgO nearly four times that of LiF?
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• Born-Haber cycle: A method to determine
lattice energies
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From Gas
to Solid
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Lattice Energy
e.g., formation of LiF(s):
Heat of formation: Li (s) + ½ F2 (g) LiF (s)
This equation is the result of the following:
1- ∆Hsub = 161 Sublimation: Li(s) Li(g)
2- ∆HIE = 520 Ionization: Li(g) Li+(g) + e
-
3- ∆HDis = 77 Dissociation: ½ F2(g) F(g)
4- ∆HEA = -328 Anion formation F(g) + e- F
-
5- ∆HLE = -1047 Solid formation Li+(g) + ½F
-(g) LiF(s)
∆HTotal = -617 kJ/mol Li (s) + ½ F2(g) LiF (s)
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• Born-Haber cycle for CaO
Ca(s) + (1/2) O2(g) CaO(s)
Ca(g)
#1
#1 Heat of sublimation = Hf[Ca(g)] = +178 kJ/mol
Ca2+(g)
#2
#2 1st & 2nd ionization energies = I1(Ca) + I2(Ca) = +1734.5 kJ/mol
O(g)
#3
#3 (1/2) Bond enthalpy = (1/2) D(O=O) = Hf[O(g)] = +247.5 kJ/mol
O2(g)
#4
#4 1st & 2nd electron affinities = EA1(O) + EA2(O) = +603 kJ/mol
+ #5
#5 (Lattice Energy) = Hlattice[CaO(s)] = (the unknown)
#6
#6 Standard enthalpy of formation = Hf[CaO(s)] = 635 kJ/mol
+178 +1734.5 +247.5 +603 Hlatt = 635 Hlattice = +3398 kJ/mol
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8.3 Covalent Bonding
• Atoms share electrons to form covalent
bonds.
• In forming the bond the atoms achieve a
more stable electron configuration.
• Often found between two nonmetals
•H H• +
•• H H H–H or
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• Octet: Eight is a “magic” number of electrons.
• Octet Rule: Atoms will gain, lose, or share
electrons to acquire eight valence electrons
Na• Cl• ••
••
•• + Na+ Cl ••
••
••
••
+
Examples:
H• •O• ••
•• H• O
••
••
•• •• H H + +
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•Lewis Structures
•H H• +
•• H H H–H
Cl• ••
••
•• + Cl• ••
••
•• Cl ••
••
••
•• Cl ••
•• •• Cl
••
••
•• Cl ••
••
•• –
Shared electrons Bonds
Non-bonding valence electrons Lone pairs
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Copyright McGraw-Hill 2009 19 Copyright McGraw-Hill 2009
• Multiple Bonds
- The number of shared electron pairs is the number
of bonds.
Cl ••
••
••
•• Cl ••
••
•• Cl ••
••
•• Cl ••
••
•• – Single Bond
O ••
•• C
•• ••
•• •• O
••
•• O ••
•• O ••
•• =C= Double Bond
•• •• N
••
••
•• N N
••
•• N Triple Bond
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• Bond strength and bond length
bond strength single < double < triple
bond length single > double > triple
N–N N=N NN
Bond Strength 163 kJ/mol 418 kJ/mol 941 kJ/mol
Bond Length 1.47 Å 1.24 Å 1.10 Å
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Covalent Bond
• Often found between two nonmetals
• Covalent bond: electrons are shared equally
between two identical atoms.
• Ionic Bond: electrons transfer completely to form
oppositely charged ions
• In between are polar covalent bonds.
• The electrons are not shared equally
• One end is slightly positive, the other is slightly
negative.
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Copyright McGraw-Hill 2009 22
8.4 Electronegativity and Polarity
• Nonpolar covalent bond = electrons are shared
equally by two bonded atoms
• Polar covalent bond = electrons are shared
unequally by two bonded atoms
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red high electron density
green intermediate electron density
blue low electron density
• Electron density distributions
+ -
H – F
alternate
representations
H – F
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Polar covalent bond
The density of electron cloud is shifted towards one of the two
bonded atoms
- Dipole Moment
- Polarity
H - F + -
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• Electronegativity: ability of an atom to draw shared
electrons to itself.
- More electronegative elements attract electrons more
strongly.
• relative scale
• related to IE and EA
• unitless
• smallest electronegativity: Cs 0.7
• largest electronegativity: F 4.0
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Copyright McGraw-Hill 2009 26 Copyright McGraw-Hill 2009
Electronegativity: The Pauling Scale
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Variation in Electronegativity with Atomic Number
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Copyright McGraw-Hill 2009 28 Copyright McGraw-Hill 2009
• Polar and nonpolar bonds
2.1 - 2.1 = 0.0 4.0 - 2.1 = 1.9 4.0 - 0.9 = 3.1
nonpolar
covalent
polar
covalent ionic
> 2.0 is ionic
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Bond Polarity and Dipole Moment
The greater the difference in electronegativity, the
more polar is the bond.
H - F
+ -
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Copyright McGraw-Hill 2009 30 Copyright McGraw-Hill 2009
• Dipole moments and partial charges
- Polar bonds often (not always) result in polar molecules.
- A polar molecule possesses a dipole.
- dipole moment () = the quantitative measure of a
dipole
= Q r
r
+Q – Q
+ -
H – F
SI unit: coulomb•meter (C • m)
common unit: debye (D)
1 D = 3.34 1030 C • m
HF 1.82 D
HCl 1.08 D
HBr 0.82 D
HI 0.44 D
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8.5 Drawing Lewis Structures
1) Draw skeletal structure with the central atom being the
least electronegative element.
2) Sum the valence electrons. Add 1 electron for each
negative charge and subtract 1 electron for each positive
charge.
3) Subtract 2 electrons for each bond in the skeletal
structure.
4) Complete electron octets for atoms bonded to the central
atom except for hydrogen.
5) Place extra electrons on the central atom.
6) Add multiple bonds if atoms lack an octet.
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Copyright McGraw-Hill 2009 32 Copyright McGraw-Hill 2009
What is the Lewis structure of NO3 ?
1) Draw skeletal structure with central
atom being the least
electronegative.
O
O – N – O
–
4) Complete electron octets for atoms
bonded to the central atom except
for hydrogen.
:O:
:O – N –O:
–
: :
: :
:
18 e
– 6) Add multiple bonds if atoms
lack an octet.
:O:
:O – N = O:
– :
: :
:
24 e
5) Place extra electrons on the
central atom.
2) Sum valence electrons. Add 1 for each negative
charge and subtract 1 for each positive charge.
NO3 (1 5) + (3 6) + 1 = 24 valence e 24 e
3) Subtract 2 for each bond in the skeletal structure. 6 e
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Example
Write Lewis Structure for HCN
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8.6 Lewis Structures and Formal Charge
• The electron surplus or deficit, relative to the free atom, that is assigned to an atom in a Lewis structure.
Formal charges are not “real” charges.
H: orig. valence e = 1
non-bonding e = 0
1/2 bonding e = 1
formal charge = 0
O: orig. valence e = 6
non-bonding e = 4
1/2 bonding e = 2
formal charge = 0
Example: H2O = H : O : H : :
Total
valence
electrons
Formal
Charge =
Total non-
bonding
electrons
Total
bonding
electrons
11 2
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Copyright McGraw-Hill 2009 35 Copyright McGraw-Hill 2009
Example: Formal charges on the atoms in ozone
6 4 12 4
0
6 2 12 6
1
6 6 12 2
1
O
O
O
O OO
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Formal charge guidelines
− A Lewis structure with no formal charges is generally better than one with formal charges.
− Small formal charges are generally better than large formal charges.
− Negative formal charges should be on the more electronegative atom(s).
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Copyright McGraw-Hill 2009 37 Copyright McGraw-Hill 2009
Identify the best structure for the isocyanate ion below:
(a) :C = N = O: –
:C N – O: –
(c)
(b)
:C – N O: –
2 +1
3
+1
+1 +1
1 1
0
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Copyright McGraw-Hill 2009 38 Copyright McGraw-Hill 2009 38 Copyright McGraw-Hill 2009
Identify the best structure for the isocyanate ion below:
(a) :C = N = O: –
:C N – O: –
(c)
(b)
:C – N O: –
2 +1
3
+1
+1 +1
1 1
0
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8.7 Resonance
Two resonance structures, their average or the resonance
hybrid, best describes the nitrite ion.
:O – N = O:
:
: : : –
:O = N – O:
: :
: : – Solution:
The double-headed arrow indicates resonance.
:O – N = O:
:
: : : –
These two bonds are known to be identical.
• Resonance structures are used when two or more equally valid
Lewis structures can be written.
Example: NO2
The electron pairs are not localized, but they are delocalized
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Benzene: C6H6
Additional Examples
Carbonate: CO32
or
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Copyright McGraw-Hill 2009
8.8 Exceptions to the Octet Rule
• Exceptions to the octet rule fall into three categories:
− Molecules with an incomplete octet
− Molecules with an odd number of electrons
− Molecules with an expanded octet
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Copyright McGraw-Hill 2009 42 Copyright McGraw-Hill 2009
• Incomplete Octets
Example: BF3 (boron trifluoride)
BF3 (1 3) + (3 7) = 24 val. e
:F:
:F – B = F:
–
:
:
: :
+1
-1
− Common with Be, B and Al compounds, but they
often dimerize or polymerize.
Example:
Cl Cl Cl
Be Be Be Be
Cl Cl Cl
:F:
:F – B – F:
–
:
: :
: :
no octet
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43
Incomplete Octet
H―Be―H
Be – 2e- 2H – 2x1e-
4e-
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44
• Odd Numbers of Electrons
Example: NO (nitrogen monoxide or nitric oxide)
NO (1 5) + (1 6) = 11 valence e
Example: NO2 (nitrogen dioxide)
NO2 (1 5) + (2 6) = 17 val. e
:N = O: .
:N = O: . Are these both
equally good?
:O = N – O: .
:O – N = O: .
:O = N – O: .
:O – N = O: .
Are these all equally good?
better
0 0 1 +1
0 0 0 0 0 0 0 +1 1 1 +1 0
best
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Copyright McGraw-Hill 2009 45 Copyright McGraw-Hill 2009
• Expanded Octet
− Elements of the 3rd period and beyond have
d-orbitals that allow more than 8 valence electrons.
XeF2 = :F – Xe – F: : :
: : :
(Xe has 10 valence
electrons)
22 valence e
(S has 12 valence
electrons )
F F
S
F F
:F:
–
–
:F:
SF6 = 48 valence e
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46
8.9 Bond Enthalpy
• Bond enthalpy is the energy associated with breaking a
particular bond in one mole of gaseous molecules.
HCl(g) H(g) + Cl(g) Ho = 431.9 kJ
Cl2(g) Cl(g) + Cl(g) Ho = 243.4 kJ
O2(g) O(g) + O(g) Ho = 495.0 kJ
N2(g) N(g) + N(g) Ho = 945.4 kJ
single bonds
double bond
triple bond
− For diatomic molecules these are accurately
measured quantities.
− Bond enthalpy is one measure of molecular stability.
− Symbol: Ho
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− Bond enthalpies for polyatomic molecules depend upon
the bond’s environment.
− Average bond enthalpies are used for polyatomic molecules.
• Provide only estimates
H = 435 kJ
H
H – C – H
H
–
–
H
H – C
H
–
–
+ H
H = 410 kJ
6% less
+ H
H
H – C
H
–
–
H
– C – H
H
–
–
H
H – C
H
–
–
H
– C
H
–
–
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48
Covalent bond energies and chemical reactions
• Consider stepwise decomposition of CH4
• Each C-H bond has a different energy.
• CH4 → CH3 + H ∆H = 435 kJ/mol
• CH3 → CH2 + H ∆ H = 453 kJ/mol
• CH2 → CH + H ∆ H = 425 kJ/mol
• CH → C + H ∆ H = 339 kJ/mol
• Each bond is sensitive to its environment but in an
unsymmetrical way
• Average C-H bond energy = 1652/4=413kJ/mol
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Copyright McGraw-Hill 2009 49 Copyright McGraw-Hill 2009
• Prediction of bond enthalpy
reactants
atoms
products BE(r)
BE(p) enth
alp
y
Ho = BE(reactants) BE(products)
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50
Use bond energies to calculate the enthalpy change for:
H2 (g) + F2 (g) 2HF (g)
H = 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
H = 436.4 + 156.9 – 2 x 568.2 = -543.1 kJ
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Copyright McGraw-Hill 2009 51 Copyright McGraw-Hill 2009
Example: Calculate the enthalpy of reaction for
CH4(g) + Br2(g) CH3Br(g) + HBr(g)
Solution: Consider ONLY bonds broken or formed.
H
H – C – H
H
–
–
Br – Br H – Br + +
H
H – C – Br
H
–
–
Hrxn = [ BE(C–H) + BE(Br–Br) ] – [ BE(C–Br) + BE(H–Br) ]
= [ (413) + (193) ] – [ (276) + (366) ]
= – 36 kJ/mol
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Copyright McGraw-Hill 2009 52 Copyright McGraw-Hill 2009
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Copyright McGraw-Hill 2009 53
Key Points
• Lewis dot symbols
• Ionic bonding
• Lattice energy
• Born-Haber cycle
• Covalent bonding
• Octet rule
• Lewis structures
• Bond order
• Bond polarity
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Copyright McGraw-Hill 2009 54
Key Points
• Electronegativity
• Dipole moment
• Drawing lewis structures
• Formal charge
• Resonance structures
• Incomplete octets
• Odd numbers of electrons
• Expanded octets
• Bond enthalpy