chemical bonding pdf

Larson-Foothill College 1

Chemical BondingObjectives: (Some of these are covered in detail in Lab.)• 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.


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.)


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.

Chemical Bond Formation: Covalent - Sharing of Electrons.

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.


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.

C–C! C=C! C≡C1.54Å! 1.34Å! 1.20Å

Bond Order:! 1! 2! 3

Except for N≡N, these are average bond lengths.

Larson-Foothill College

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.

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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! 418! 941! (Average) (Average)


Average Bond Strengths


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)

= [(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)

1 2


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.


Using Bond Enthalpies to Calculate Heat of Reaction

Text Problem 8.96An 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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• 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.


Polar Bonds in Covalent Molecules

most electronegative atom becomes

partially (-)

least electronegative atom becomes

partially (+)


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

General rules for polar bonds: (You do not need to memorize these.)(∆EN = difference in electronegativity)


Representative Polar Bonds

Bond Type |∆EN|

Nonpolar < 0.3

Weakly Polar Covalent Between 0.3 and 0.5

Polar Covalent 0.5 to 2

Ionic > 2

∆EN = 3.4-2.2 = 1.2

∆EN = 4.0-2.2 = 1.8

∆EN = 3.0-2.2 = 0.8

This computer-generated 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.


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 = QrWe 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 atomcharge of one electron

x 100%

Calculate the percent ionic character for HCl.


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.


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.

Chemical Bond Formation: Ionic


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 chargeand distance between ions in the solid phase;charge having the greater affect.

k is a constant, Q1 and Q2 are the ion charges, and d is the distance between ions in the crystal.

What are the trends in lattice energy shown above in terms of ion size and ion charge?

∆HLattice ∝kQ1Q2

d

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.


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) ∆H°f Na(g) = 108 kJ (sublimation)2. 1/2 Cl2(g) —> Cl(g) ∆H°f Cl(g) = 122 kJ (1/2 bond energy)3. Na(g) —> Na+(g) + e– IE1(Na) = 496 kJ4. Cl(g) + e– —> Cl–(g) EA(Cl) = -349 kJ5. Na+(g) + Cl–(g) —> NaCl(s) –∆Hlattice = ?

Add these 5 steps together: to get step 6, ∆H°f NaCl(s):

6. Na(s)+ 1/2 Cl2(g) —> NaCl(s) ∆H°f NaCl(s) = -411 kJ

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

1.

2.

3.

4.

5.

6.

Start at standard states.

End at standard state.

NaCl(s) —> Na+(g) + Cl–(g)


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/molIE1 Ca(g) = 590 kJ/molIE2 Ca+(g) = 1145 kJ/molCl2(g) bond energy = 242 kJ/molEA Cl(g) = -349 kJ/mol∆H°f CaCl2(s) = -796 kJ/mol

Answer 2253 kJ


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?

a) What is the element?

b) What is the electron configuration of an atom of this element?

Text Problem 8.4:The orbital diagram below shows the highest energy electrons for a 2+ ion of an element.


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.


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.


Molecules and Ions With the Same Lewis StructuresIsoelectronic = same electron arrangement

Note: This table is NOT from our text book.

Text

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.


Oxoacids and their Lewis Structures(Acidic H always bonded to an O atom!)


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.


Resonance StructuresResonance structures have the same connectivity of the atoms but a different arrangement of the bonding electrons.

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”.

Ozone, O3

What are the bond orders in ozone and in the nitrate ion?

• Benzene has two resonance structures; thus, it is resonance stabilized.


Examples of Resonance Structures

• 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–?


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.


Octet “Rule” Exceptions: Greater then Eight Outer ElectronsAll 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!!)


chemical bonding pdf

Documents

bond length

double bond

triple bond

single bond

covalent bond formation

bond strengthbond order

chemical bond formation

relative bond enthalpies