Chemical Bonding
Objectives: (Some of these are covered in detail in Lab.)
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Understand and be able to describe the difference between ionic and covalent bonding.
Be able to apply the Octet Rule as a basis for bonding.
Know the definition of Lattice Energy and be able to calculate Lattice Energy from appropriate
thermodynamic quantities (Born-Haber Cycle: Hess’s Law).
Understand and be able to predict trends in Lattice Energies based on ion size and charge;
know how these trends are related to melting points of ionic compounds.
Draw Lewis Electron-Dot Structures for small molecules and polyatomic ions when the Octet
Rule is satisfied.
Know the three common exceptions to the Octet Rule and be able to draw Lewis Structures for
molecule and polyatomic ions that violate the rule.
Understand covalent bond formation and properties; bond enthalpy (strength) and bond length.
Be able to predict relative bond enthalpies and lengths based on atomic size and bond order.
Be able to draw resonance structures; understand the concept of resonance and “resonance
stabilization”.
Be able to calculate Formal Charge and use Formal Charge as a basis for predicting relative
stability of different resonance forms and different orders of attachment of atoms.
Understand and be able to describe electronegativity; use electronegativity to predict polarity of
bonds and diatomic molecules.
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Chemical Bonding
Octet Rule: Transfer/Sharing of electrons to obtain a complete valence electron shell of 8
electrons (except H has a valence shell of 2 electrons).
Ionic Bonding: Transfer of electrons through redox to obtain 8 valance shell electrons.
Covalent Bonding: Sharing of electrons to obtain 8 valance shell electrons.
(Exceptions where > 8 electrons are found.)
Metallic Bonding: “Electron-Sea Model” (Covered in detail in Chapter 12.)
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Chemical Bond Formation: Covalent - Sharing of Electrons.
Covalent bonding: sharing of electrons by orbital
overlap. Each atom (except H) obtains an octet of
electrons.
Consider the formation of methane:
C(s) + 2 H2(g) —> CH4(g)
(8 electrons now surround C)
Covalent bonding gives us “molecules”, individual
units of CH4. Covalent compounds exist as discrete
molecules.
We will show the bonding in covalent molecules
using Lewis Structures. Lewis Structures are
covered in detail in lab.
a) The attractions and repulsions among
electrons and nuclei in the hydrogen
molecule. (b) Electron distribution in the
H2 molecule. The concentration of
electron density between the nuclei leads
to a net attractive force that constitutes
the covalent bond holding the molecule
together.
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Covalent Bonds: Bond Order, Bond Length and Bond Strength
Bond order: A single bond is described as having a bond order of 1, a double bond as
having a bond order of 2 and a triple bond as having a bond order of three.
Single bond: A covalent bond involving one shared electron pair.
Triple bond: A covalent bond involving three shared electron pairs.
Double bond: A covalent bond involving two shared electron pairs.
Bond length: The distance between the nuclei of the atoms involved in a bond. Bond
length depends on the size of the bonded atoms and on the bond order. Generally, bond
length decreases with a decrease in atomic radii of the bonded atoms and with an increase
in bond order.
Bond Order:!
1!
C–C!
1.54Å!
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2!
C=C!
1.34Å!
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C≡C
1.20Å
Except for N≡N,
these are average
bond lengths.
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Covalent Bonds: Bond Order, Bond Length and Bond Strength
Bond enthalpy (energy, strength): The bond enthalpy, D, is the enthalpy change (∆H) per
mole required to break a bond in the gas phase. Bond enthalpy depends on the two atoms
bonding. Bond enthalpy generally increases with decreasing bond length, shorter bonds
tend to be stronger. A double bond is stronger than a single bond (between the same two
atoms) and a triple bond is stronger than a double.
H2(g) —> 2 H (g)
∆H = bond enthalpy = 436 kJ/mol
Bond breaking is always an endothermic
process, so bond enthalpies are (+) values.
D (kJ/mol): 163!
!
(Average)
418!
941
(Average)
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Average Bond Strengths
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Using Bond Enthalpies (Energies) to Estimate ∆Hrxn
The heat of reaction between covalent molecules can be estimated from bond energies, D.
∆Hrxn = ∑ (bond energies broken) - ∑ (bond energies formed)
∆Hrxn= ∑ (Dreactants) - ∑ (Dproducts)
= [D(C-H) + D(Cl-Cl)]-[D(C-Cl)+D(H-Cl)
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2
= [(413+242)-[328+431] kJ = –104 kJ
The actual value is -99.8 kJ, so our
estimate is close.
CH4(g) + Cl2 (g) —> CH3Cl (g) + HCl (g)
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Using Bond Energies to Estimate ∆Hrxn
Text Problem 8.73: Ammonia is produced directly from nitrogen and hydrogen by using the
Haber process. The chemical reaction is
3 H2(g) + N2(g) —> 2 NH3(g)
(a) Use bond enthalpies (Table 8.4) to estimate the enthalpy change for the reaction. Is
it exothermic or endothermic.
(b) Compare the enthalpy change you calculated in (a) to the true enthalpy change as
obtained using ∆H°f values.
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Using Bond Enthalpies to Calculate Heat of Reaction
Text Problem 8.96
An important reaction for the conversion of natural gas to other useful hydrocarbons is
the conversion of methane to ethane.
In practice, this reaction is carried out in the presence of oxygen, which converts the
hydrogen produced to water.
Use bond enthalpies (Table 8.4) to estimate ΔH for these two reactions. Why is the
conversion of methane to ethane more favorable when oxygen is used?
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Polar Bonds in Covalent Molecules
• A polar covalent bond is
formed when two bonding
atoms have a large
difference in
electronegativity.
• Electronegativity is a
measure of the ability of an
atom, when bonded to
another atom, to attract
electrons to itself.
• The greater the difference
in electronegativity
between the atoms the
greater the bond polarity.
most
electronegative
atom becomes
partially (-)
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least
electronegative
atom becomes
partially (+)
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Dipole Moments: µ = Qr
The dipole moment of a bond can be calculated by estimating the charge in coulombs, Q,
on each atom and the distance between the charge, r in meters.
The units of the dipole moment are debyes, D: 1D = 3.34x10–30 C•m
A relative charge of ±1 (proton and electron) has a value of 1.60x10-19 Coulombs, C
Trend: Dipole moments increase with larger charges, i.e., larger difference in
electronegativity. (Changes in r are less important.)
Arrange the following bonds in order of increasing polarity: N–F, Be–F and O–F
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Representative Polar Bonds
General rules for polar bonds: (You do not need to memorize these.)
(∆EN = difference in electronegativity)
Bond Type
|∆EN|
Nonpolar
< 0.3
Weakly Polar Covalent
Between 0.3 and 0.5
Polar Covalent
0.5 to 2
Ionic
>2
∆EN = 4.0-2.2 = 1.8
This computergenerated rendering
shows the calculated
electron-density
distribution on the
surface of the F2, HF,
and LiF molecules. The
regions of relatively low
electron density (net
positive charge) appear
blue, those of relatively
high electron density
(net negative charge)
appear red, and
regions that are close
to electrically neutral
appear green.
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∆EN = 3.4-2.2 = 1.2
∆EN = 3.0-2.2 = 0.8
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Dipole Moments and Percent Ionic Character
Problem: Calculate the effective charge on the hydrogen and chlorine atoms in HCl.
Hint: We will need to use the formula:
µ = Qr
We also need the following information:
charge of one electron: 1.60x10–19 C
H–Cl bond length: 1.27Å
HCl Dipole Moment: 1.08 D
1D = 3.34x10–30 C•m
The more polar a bond is, the greater its “percent ionic character”. The percent ionic
character of a bond can be defined as:
charge on bonded atom
x 100%
charge of one electron
Calculate the percent ionic character for HCl.
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Practice Question
Text Problem 8.7:
The partial Lewis structure below is for a hydrocarbon molecule. In the full Lewis structure,
each carbon atom satisfies the octet rule, and there are no unshared electron pairs in the
molecule. The carbon–carbon bonds are labeled 1, 2, and 3.
a)
Determine where the hydrogen atoms are in the molecule. Draw them in.
b)
Rank the carbon–carbon bonds in order of increasing bond length.
c)
Rank the carbon–carbon bonds in order of increasing bond enthalpy.
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Chemical Bond Formation: Ionic
Ionic bonding: transfer of electrons (through redox) to form ions. For main group elements,
the ions form noble gas configurations. The metal is oxidized while a nonmetal is reduced.
Na(s) + (1/2)Cl2(g) —> [Na+ + Cl–] —> NaCl(s);
∆H°f = –410.9 kJ
All ionic compounds are solids at room temperature!
The compounds exist as crystal lattices. Uniform solids with a predictable structure.
There are no “individual” NaCl units, each Na+ ion is attracted to 6 nearest neighbor Cl– ions, and
each Cl– ion to 6 nearest neighbor Na+ ions.
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Energetics of Ionic Bond Formation – Lattice Energy of Ionic Crystals
The lattice energy is the energy needed to break
apart the solid crystal lattice and form ions in the
gas phase. This is always ENDOTHERMIC!
NaCl(s) —> Na+(g) + Cl–(g)
∆Hlattice = +788 kJ/mol
The lattice energy IS NOT the opposite of ∆H°f.
For NaCl(s) ∆H°f is –410.9 kJ/mol
The lattice energy depends on the charge
and distance between ions in the solid phase;
charge having the greater affect.
∆ H Lattice ∝
kQ1Q2
d
k is a constant, Q1 and Q2 are the ion
charges, and d is the distance
between ions in the crystal.
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What are the trends in lattice energy shown
above in terms of ion size and ion charge?
Lattice energies can be used to predict
relative melting point for ionic substances. In
general, the greater the lattice energy, the
higher the melting point.
Predict the order of melting points for:
KCl, CaO, CsCl, CaCl2 and BaO.
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Lattice Energy cannot be easily measured directly. It is calculated
using the Born-Haber cycle and Hess’ Law.
Steps in the cycle:
1. Na(s) —> Na(g)
2. 1/2 Cl2(g) —> Cl(g) 3. Na(g) —> Na+(g) + e– 4. Cl(g) + e– —> Cl–(g) 5. Na+(g) + Cl–(g) —> NaCl(s)
4.
3.
∆H°f Na(g) ∆H°f Cl(g)
IE1(Na)
EA(Cl)
–∆Hlattice
= 108 kJ (sublimation)
= 122 kJ (1/2 bond energy)
= 496 kJ
= -349 kJ
=?
Add these 5 steps together: to get step 6, ∆H°f NaCl(s):
5.
6. Na(s)+ 1/2 Cl2(g) —> NaCl(s)
∆H°f NaCl(s)= -411 kJ
2.
1.
Start at standard states.
6.
Using Hess’ Law and solving for the energy of step 5:
5 = 6 - {1 + 2 + 3 + 4}
So we have:
–∆Hlattice = 5 = -411 kJ - {108 + 122 +496 + -349} kJ = –788 kJ
∆Hlattice = +788 kJ
NaCl(s) —> Na+(g) + Cl–(g)
End at standard state.
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Lattice Energy Calculations Using the Born-Haber Cycle
Write the appropriate equations and perform the calculations to determine the lattice energy
for the crystal CaCl2. Data are as follows:
Ca(s) to Ca(g) ∆Hsublimation = 178 kJ/mol
IE1 Ca(g) = 590 kJ/mol
IE2 Ca+(g) = 1145 kJ/mol
Cl2(g) bond energy = 242 kJ/mol
EA Cl(g) = -349 kJ/mol
∆H°f CaCl2(s) = -796 kJ/mol
Answer 2253 kJ
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Conceptual Questions
Text Problem 8.2: Illustrated below are four ions (A1, A2, Z1 and Z2), showing their relative
ionic radii. The ions shown in red carry a 1+ charge, and those shown in blue carry a 1–
charge.
(a) Would you expect to find an ionic compound of formula A1A2
Explain.
(b) Which combination of ions leads to the ionic compound having
the largest lattice energy?
(c) Which combination of ions leads to the ionic compound having
the smallest lattice energy?
Text Problem 8.4:
The orbital diagram below shows the highest energy electrons for a 2+ ion of an element.
a) What is the element?
b) What is the electron configuration of an atom of this element?
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Conceptual Questions
Text Problem 8.27: Energy is required to remove two electrons from Ca to form Ca2+ and is
also required to add two electrons to O to form O2−. Why, then, is CaO stable relative to the
free elements?
Question: Given that lattice energy increases as ionic charge increases, explain why
compounds such as NaO, where Na is +2, or BaCl, where Cl is –2, do not
form.
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The following slides are for reference. The details regarding Lewis
Structures are covered in lab. Therefore, little regular lecture time will
be allotted to this topic and these slides will not be discussed during
lecture. However, questions regarding Lewis Structures will be
included on the lecture exam covering Chapter 8 material.
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Molecules and Ions With the Same Lewis Structures
Isoelectronic = same electron arrangement
Text
Note: This table is NOT from our text book.
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Note: These three can be
drawn with an expanded
octet that minimizes formal
charges; thus a more stable
structure. Expanded octet
structures
tend to be
consistent with the known
bond orders.
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Oxoacids and their Lewis Structures
(Acidic H always bonded to an O atom!)
Note: This table is NOT from our text book.
Many of these oxoacids can also be
written with expanded octets that
minimize formal charges; their
expanded octet structures are
generally considered “more stable”.
Expanded octet structures tend to
be consistent with the known bond
orders.
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Resonance Structures
Resonance structures have the same connectivity of the atoms but
a different arrangement of the bonding electrons.
Ozone, O3
In ozone, the double bond can be placed either on the left or right.
Neither is preferred so both resonance structures contribute equally
to the electron arrangement of ozone.
Notice the bond lengths are equivalent in ozone. This indicates only
one type of bond is present in the molecule, not a single and a
double bond. Furthermore, the individual single and double bonds
do not “resonate” back and forth as shown by the two Lewis
Structures. The “true” structure is a “resonance hybrid” where the
double bond is shared between more that two atoms. The resulting
structure is more stable than one where the double bond is
localized between just two atoms. The molecule is said to be
“resonance stabilized”.
What are the bond
orders in ozone and
in the nitrate ion?
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Examples of Resonance Structures
• Benzene has two resonance structures; thus, it is resonance stabilized.
• For inequivalent resonance forms, formal charge can be used to predict the most
stable form:
Question: Based on the formal charges shown above, what do you predict for the bond orders
of the N–C and C–S bonds in NCS–?
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Octet “Rule” Exceptions; Odd Number of Electrons
A few molecules or polyatomic ions have an odd number of valence electrons, thus the octet
rule cannot be satisfied for all atoms in the molecule or ion; pairing all electrons is
impossible. In these cases, consideration of formal charges can help determine the
“preferred” Lewis Structure.
Examples: NO2 and ClO2
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Octet “Rule” Exceptions; Fewer than Eight Valance Electrons
(Boron, B, is one example.)
Boron usually only has six (6) electrons in its outer shell when bonding.
However, boron will accept another pair of electrons to make a fourth bond to complete the
octet.
We could complete the octet around boron by forming a double bond. In so doing, we see that there are three equivalent
resonance structures. These three resonance structures force a fluorine atom to share additional electrons with the
boron atom, which is inconsistent with the high electronegativity of fluorine. In fact, the formal charges tell us that this is
an unfavorable situation.
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Octet “Rule” Exceptions: Greater then Eight Outer Electrons
All these elements have a d sublevel in their valence shell (3d, 4d, etc.) to use in bonding.
Notice row 2 elements, B, C, N, O and F CANNOT do this (no 2d orbitals!!)
Note: This table is NOT from our text book.
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