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Introductory Chemistry - 1st Canadian Edition: Other Aspects of Covalent Bonds

Introductory Chemistry - 1st Canadian Edition
Other Aspects of Covalent Bonds
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table of contents
  1. Cover
  2. Title Page
  3. Copyright
  4. Table Of Contents
  5. Acknowledgments
  6. Dedication
  7. About BCcampus Open Education
  8. Chapter 1. What is Chemistry
    1. Some Basic Definitions
    2. Chemistry as a Science
  9. Chapter 2. Measurements
    1. Expressing Numbers
    2. Significant Figures
    3. Converting Units
    4. Other Units: Temperature and Density
    5. Expressing Units
    6. End-of-Chapter Material
  10. Chapter 3. Atoms, Molecules, and Ions
    1. Acids
    2. Ions and Ionic Compounds
    3. Masses of Atoms and Molecules
    4. Molecules and Chemical Nomenclature
    5. Atomic Theory
    6. End-of-Chapter Material
  11. Chapter 4. Chemical Reactions and Equations
    1. The Chemical Equation
    2. Types of Chemical Reactions: Single- and Double-Displacement Reactions
    3. Ionic Equations: A Closer Look
    4. Composition, Decomposition, and Combustion Reactions
    5. Oxidation-Reduction Reactions
    6. Neutralization Reactions
    7. End-of-Chapter Material
  12. Chapter 5. Stoichiometry and the Mole
    1. Stoichiometry
    2. The Mole
    3. Mole-Mass and Mass-Mass Calculations
    4. Limiting Reagents
    5. The Mole in Chemical Reactions
    6. Yields
    7. End-of-Chapter Material
  13. Chapter 6. Gases
    1. Pressure
    2. Gas Laws
    3. Other Gas Laws
    4. The Ideal Gas Law and Some Applications
    5. Gas Mixtures
    6. Kinetic Molecular Theory of Gases
    7. Molecular Effusion and Diffusion
    8. Real Gases
    9. End-of-Chapter Material
  14. Chapter 7. Energy and Chemistry
    1. Formation Reactions
    2. Energy
    3. Stoichiometry Calculations Using Enthalpy
    4. Enthalpy and Chemical Reactions
    5. Work and Heat
    6. Hess’s Law
    7. End-of-Chapter Material
  15. Chapter 8. Electronic Structure
    1. Light
    2. Quantum Numbers for Electrons
    3. Organization of Electrons in Atoms
    4. Electronic Structure and the Periodic Table
    5. Periodic Trends
    6. End-of-Chapter Material
  16. Chapter 9. Chemical Bonds
    1. Lewis Electron Dot Diagrams
    2. Electron Transfer: Ionic Bonds
    3. Covalent Bonds
    4. Other Aspects of Covalent Bonds
    5. Violations of the Octet Rule
    6. Molecular Shapes and Polarity
    7. Valence Bond Theory and Hybrid Orbitals
    8. Molecular Orbitals
    9. End-of-Chapter Material
  17. Chapter 10. Solids and Liquids
    1. Properties of Liquids
    2. Solids
    3. Phase Transitions: Melting, Boiling, and Subliming
    4. Intermolecular Forces
    5. End-of-Chapter Material
  18. Chapter 11. Solutions
    1. Colligative Properties of Solutions
    2. Concentrations as Conversion Factors
    3. Quantitative Units of Concentration
    4. Colligative Properties of Ionic Solutes
    5. Some Definitions
    6. Dilutions and Concentrations
    7. End-of-Chapter Material
  19. Chapter 12. Acids and Bases
    1. Acid-Base Titrations
    2. Strong and Weak Acids and Bases and Their Salts
    3. Brønsted-Lowry Acids and Bases
    4. Arrhenius Acids and Bases
    5. Autoionization of Water
    6. Buffers
    7. The pH Scale
    8. End-of-Chapter Material
  20. Chapter 13. Chemical Equilibrium
    1. Chemical Equilibrium
    2. The Equilibrium Constant
    3. Shifting Equilibria: Le Chatelier’s Principle
    4. Calculating Equilibrium Constant Values
    5. Some Special Types of Equilibria
    6. End-of-Chapter Material
  21. Chapter 14. Oxidation and Reduction
    1. Oxidation-Reduction Reactions
    2. Balancing Redox Reactions
    3. Applications of Redox Reactions: Voltaic Cells
    4. Electrolysis
    5. End-of-Chapter Material
  22. Chapter 15. Nuclear Chemistry
    1. Units of Radioactivity
    2. Uses of Radioactive Isotopes
    3. Half-Life
    4. Radioactivity
    5. Nuclear Energy
    6. End-of-Chapter Material
  23. Chapter 16. Organic Chemistry
    1. Hydrocarbons
    2. Branched Hydrocarbons
    3. Alkyl Halides and Alcohols
    4. Other Oxygen-Containing Functional Groups
    5. Other Functional Groups
    6. Polymers
    7. End-of-Chapter Material
  24. Chapter 17. Kinetics
    1. Factors that Affect the Rate of Reactions
    2. Reaction Rates
    3. Rate Laws
    4. Concentration–Time Relationships: Integrated Rate Laws
    5. Activation Energy and the Arrhenius Equation
    6. Reaction Mechanisms
    7. Catalysis
    8. End-of-Chapter Material
  25. Chapter 18. Chemical Thermodynamics
    1. Spontaneous Change
    2. Entropy and the Second Law of Thermodynamics
    3. Measuring Entropy and Entropy Changes
    4. Gibbs Free Energy
    5. Spontaneity: Free Energy and Temperature
    6. Free Energy under Nonstandard Conditions
    7. End-of-Chapter Material
  26. Appendix A: Periodic Table of the Elements
  27. Appendix B: Selected Acid Dissociation Constants at 25°C
  28. Appendix C: Solubility Constants for Compounds at 25°C
  29. Appendix D: Standard Thermodynamic Quantities for Chemical Substances at 25°C
  30. Appendix E: Standard Reduction Potentials by Value
  31. Glossary
  32. About the Authors
  33. Versioning History

Other Aspects of Covalent Bonds

Learning Objectives

  1. Describe a nonpolar bond and a polar bond.
  2. Use electronegativity to determine whether a bond between two elements will be nonpolar covalent, polar covalent, or ionic.
  3. Describe the bond energy of a covalent bond.

Consider the H2 molecule:

\chemfig{\Lewis{0:,H}H}

Because the nuclei of each H atom contain protons, the electrons in the bond are attracted to the nuclei (opposite charges attract). But because the two atoms involved in the covalent bond are both H atoms, each nucleus attracts the electrons by the same amount. Thus the electron pair is equally shared by the two atoms. The equal sharing of electrons in a covalent bond is called a nonpolar covalent bond.

Now consider the HF molecule:

\chemfig{H-\Lewis{0:2:6:,F}}

There are two different atoms involved in the covalent bond. The H atom has one proton in its nucleus that is attracting the bonding pair of electrons. However, the F atom has nine protons in its nucleus, with nine times the attraction of the H atom. The F atom attracts the electrons so much more strongly that the electrons remain closer to the F atom than to the H atom; the electrons are no longer equally balanced between the two nuclei.

Because the electrons in the bond are nearer to the F atom, this side of the molecule takes on a partial negative charge, which is represented by δ− (δ is the lowercase Greek letter delta). The other side of the molecule, the H atom, adopts a partial positive charge, which is represented by δ+:

\Large \chemfig{\overset{\delta +}{H}-\overset{\delta -}{\Lewis{0:2:6:,F}}}

A covalent bond between different atoms that attract the shared electrons by different amounts and cause an imbalance of electron distribution is called a polar covalent bond.

Technically, any covalent bond between two different elements is polar. However, the degree of polarity is important. A covalent bond between two different elements may be so slightly imbalanced that the bond is, essentially, nonpolar. A bond may be so polar that an electron actually transfers from one atom to another, forming a true ionic bond. How do we judge the degree of polarity?

Scientists have devised a scale called electronegativity, a scale for judging how much atoms of any element attract electrons. Electronegativity is a unitless number; the higher the number, the more an atom attracts electrons. A common scale for electronegativity is shown in Figure 9.3 “Electronegativities of the Elements.”

Periodic table of electronegativities of elements.
Figure 9.3 “Electronegativities of the Elements.” Electronegativities are used to determine the polarity of covalent bonds. (Source: from Joanjoc at Wikimedia Commons, public domain.)

The polarity of a covalent bond can be judged by determining the difference between the electronegativities of the two atoms involved in the covalent bond, as summarized in Table 9.2 “Electronegativities of Bond Types.”

Table 9.2 Electronegativities of Bond Types
Electronegativity DifferenceBond Type
0nonpolar covalent
0–0.4slightly polar covalent
0.4–1.9definitely polar covalent
>1.9likely ionic

The unequal sharing of electrons in a covalent bond is usually indicated by partial charge notation as seen earlier, or by a dipole arrow. Dipole arrows depict the unequal sharing by showing the flow of electron density. Dipole arrows have an end with a “+ sign” depicting an electropositive area from which electron density is being pulled, and an end with an arrowhead pointing to the more electronegative atom toward which electron density is being pulled.

H-F_dipole_arrow

Example 9.7

What is the polarity of each of the following bonds?

  1. C–H
  2. O–H

Solution

Using Figure 9.3 “Electronegativities of the Elements,” we can calculate the electronegativity differences of the atoms involved in the bond.

  1. For the C–H bond, the difference in electronegativities is 2.5 − 2.1 = 0.4. Thus we predict that this bond will be slightly polar covalent.
  2. For the O–H bond, the difference in electronegativities is 3.5 − 2.1 = 1.4, so we predict that this bond will be definitely polar covalent.

Test Yourself

What is the polarity of each of these bonds?

  1. Rb–F
  2. P–Cl

Answers

  1. likely ionic
  2. polar covalent

The polarity of a covalent bond can have significant influence on the properties of the substance. If the overall molecule is polar, the substance may have a higher melting point and boiling point than expected; also, it may or may not be soluble in various other substances, such as water or hexane.

It should be obvious that covalent bonds are stable because molecules exist. However, the bonds can be broken if enough energy is supplied to a molecule. To break most covalent bonds between any two given atoms, a certain amount of energy must be supplied. Although the exact amount of energy depends on the molecule, the approximate amount of energy to be supplied is similar if the atoms in the bond are the same. The approximate amount of energy needed to break a covalent bond is called the bond energy of the covalent bond. Table 9.3 “Bond Energies of Covalent Bonds” lists the bond energies of some covalent bonds.

Table 9.3 Bond Energies of Covalent Bonds
BondEnergy (kJ/mol)
C–C348
C=C611
C≡C837
C–O351
C=O799
C–Cl328
C–H414
F–F159
H–Cl431
H–F569
H–H436
N–N163
N=N418
N≡N946
N–H389
O–O146
O=O498
O–H463
S–H339
S=O523
Si–H293
Si–O368

A few trends are obvious from Table 9.3. For bonds that involve the same two elements, a double bond is stronger than a single bond, and a triple bond is stronger than a double bond. The energies of multiple bonds are not exact multiples of the single-bond energy; for carbon-carbon bonds, the energy increases somewhat less than double or triple the C–C bond energy, while for nitrogen-nitrogen bonds the bond energy increases at a rate greater than the multiple of the N–N single bond energy. The bond energies in Table 9.3 are average values; the exact value of the covalent bond energy will vary slightly among molecules with these bonds but should be close to these values.

To be broken, covalent bonds always require energy; that is, covalent-bond breaking is always an endothermic process. Thus the ΔH for this process is positive:

\text{molecule }\ce{OH}\rightarrow \text{molecule O}+\ce{H}\hspace{5mm}\Delta H\approx +463\text{ kJ/mol}

However, when making a covalent bond, energy is always given off; covalent-bond making is always an exothermic process. Thus ΔH for this process is negative:

\text{molecule S}+\ce{H}\rightarrow \text{molecule }\ce{SH}\hspace{5mm}\Delta H\approx -339\text{ kJ/mol}

Bond energies can be used to estimate the energy change of a chemical reaction. When bonds are broken in the reactants, the energy change for this process is endothermic. When bonds are formed in the products, the energy change for this process is exothermic. We combine the positive energy change with the negative energy change to estimate the overall energy change of the reaction. For example, in

2H2 + O2 → 2H2O

We can draw Lewis electron dot diagrams for each substance to see what bonds are broken and what bonds are formed:

\begin{matrix} \chemfig{H-H}&&&&\chemfig{H-\Lewis{2:6:,O}-H} \\ &+&\chemfig{\Lewis{2:6:,O}=\Lewis{2:6:,O}}&\rightarrow& \\ \chemfig{H-H}&&&&\chemfig{H-\Lewis{2:6:,O}-H} \\ \end{matrix}

(The lone electron pairs on the O atoms are omitted for clarity.) We are breaking two H–H bonds and one O–O double bond and forming four O–H single bonds. The energy required for breaking the bonds is as follows:

\begin{array}{rl} 2 \ \chemfig{H-H}\text{ bonds:}&2\times(+436\text{ kJ/mol}) \\ 1 \ \chemfig{O=O}\text{ bond:}&\phantom{2\times(}+498\text{ kJ/mol} \\ \hline \text{Total:}&\phantom{2\times(}+1,370\text{ kJ/mol} \end{array}

The energy given off during the formation of the four O–H bonds is as follows:

\begin{array}{rl} 4 \ \chemfig{O-H}\text{ bonds:}&4\times (-463\text{ kJ/mol}) \\ \hline \text{Total:}&-1,852\text{ kJ/mol} \end{array}

Combining these two numbers:

\begin{array}{rl} &+1,370\text{ kJ/mol} \\ +&(-1,852\text{ kJ/mol}) \\ \hline \text{Net Change:}&-482\text{ kJ/ml}\approx \Delta H \end{array}

The actual ΔH is −572 kJ/mol; we are off by about 16% — although not ideal, a 16% difference is reasonable because we used estimated, not exact, bond energies.

Example 9.8

Estimate the energy change of this reaction:

    \begin{equation*} \chemfig{C(-[:-135]H)(-[:135]H)=C(-[:-45]H)(-[:45]H)}+\chemfig{H-H}\longrightarrow \chemfig{H-C(-[:90]H)(-[:-90]H)-C(-[:90]H)(-[:-90]H)-H} \end{equation*}

Solution

Here, we are breaking a C–C double bond and an H–H single bond and making a C–C single bond and two C–H single bonds. Bond breaking is endothermic, while bond making is exothermic. For the bond breaking:

\begin{array}{rl} 1 \ \chemfig{C=C}:&+611\text{ kJ/mol} \\ 1 \ \chemfig{H-H}:&+436\text{ kJ/mol} \\ \hline \text{Total:}&+1,047\text{ kJ/mol} \end{array}

For the bond making:

\begin{array}{rl} 1 \ \chemfig{C-C}:&\phantom{2\times}-348\text{ kJ/mol} \\ 2 \ \chemfig{C-H}:&2\times(-414\text{ kJ/mol}) \\ \hline \text{Total: }&-1,176\text{ kJ/mol} \end{array}

Overall, the energy change is +1,047 + (−1,176) = −129 kJ/mol.

Test Yourself

Estimate the energy change of this reaction:

    \begin{equation*} \chemfig{H-C~C-H}+\chemfig{2H-H}\longrightarrow \chemfig{H-C(-[:90]H)(-[:-90]H)-C(-[:90]H)(-[:-90]H)-H} \end{equation*}

Answer

−295 kJ/mol

Key Takeaways

  • Covalent bonds can be nonpolar or polar, depending on the electronegativities of the atoms involved.
  • Covalent bonds can be broken if energy is added to a molecule.
  • The formation of covalent bonds is accompanied by energy given off.
  • Covalent bond energies can be used to estimate the enthalpy changes of chemical reactions.

Exercises

Questions

  1. Give an example of a nonpolar covalent bond. How do you know it is nonpolar?
  2. Give an example of a polar covalent bond. How do you know it is polar?
  3. How do you know which side of a polar bond has the partial negative charge? Identify the negatively charged side of each polar bond.
    1. H–Cl
    2. H–S
  4. How do you know which side of a polar bond has the partial positive charge? Identify the positively charged side of each polar bond.
    1. H–Cl
    2. N–F
  5. Label the bond between the given atoms as nonpolar covalent, slightly polar covalent, definitely polar covalent, or likely ionic.
    1. H and C
    2. C and F
    3. K and F
  6. Label the bond between the given atoms as nonpolar covalent, slightly polar covalent, definitely polar covalent, or likely ionic.
    1. S and Cl
    2. P and O
    3. Cs and O
  7. Which covalent bond is stronger: a C–C bond or a C–H bond?
  8. Which covalent bond is stronger: an O–O double bond or an N–N double bond?
  9. Estimate the enthalpy change for this reaction. Start by drawing the Lewis electron dot diagrams for each substance.

    N2 + 3H2 → 2NH3

  10. Estimate the enthalpy change for this reaction. Start by drawing the Lewis electron dot diagrams for each substance.

    HN=NH + 2H2 → 2NH3

  11. Estimate the enthalpy change for this reaction. Start by drawing the Lewis electron dot diagrams for each substance.

    CH4 + 2O2 → CO2 + 2H2O

  12. Estimate the enthalpy change for this reaction. Start by drawing the Lewis electron dot diagrams for each substance.

    4NH3 + 3O2 → 2N2 + 6H2O

Answers

  1. H–H; it is nonpolar because the two atoms have the same electronegativities (answers will vary).
    1. Cl side
    2. S side
    1. slightly polar covalent
    2. definitely polar covalent
    3. likely ionic
  1. C–H bond
  1. −80 kJ
  1. −798 kJ

Annotate

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Violations of the Octet Rule
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Copyright © 2014

                                by Jessie A. Key

            Introductory Chemistry - 1st Canadian Edition by Jessie A. Key is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.
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