Covalent Bonding Study Notes

COVALENT BONDING

C60 Fullerene Overview

  • Solubility:   - C60 fullerene is capable of dissolving in some solvents unlike diamond and graphite.   - The dissolution occurs due to relatively weak intermolecular forces needing to be broken for solvation.  

  • Hardness:   - Compared to diamond, C60 fullerene is not as hard.   - The energy required to break intermolecular forces in C60 fullerene is significantly lower than the energy needed to break the strong covalent bonds found in diamond.

  • Electrical Conductivity:   - C60 fullerene does not conduct electricity.   - Although each carbon atom forms three bonds, the fourth electron is constrained to movement within its own C60 molecule.   - Electrons are unable to jump between different C60 molecules, preventing conductivity.

Chapter Questions

Skills: Critical Thinking
Skills: Interpretation
Skills: Reasoning

  • It is important to utilize the Periodic Table found in Appendix A on page 320 for answering certain questions.

  1. Determine the Nature of the Compounds: State whether each of the following compounds is ionic or covalent:    - a. MgO
       - b. CH3Br
       - c. H2O2
       - d. FeCl2
       - e. NaF
       - f. HCN

  2. Covalent Bonds:
       a. Define a covalent bond and explain how it binds two atoms together.
       b. Draw dot-and-cross diagrams to illustrate the covalent bonding in:
          - i. Methane (CH4)
          - ii. Hydrogen Sulfide (H2S)
          - iii. Phosphine (PH3)
          - iv. Silicon Tetrachloride (SiCl4)
       c. Draw dot-and-cross diagrams to illustrate the covalent bonding in:
          - i. O2
          - ii. N2

  3. Drawing Diagrams for Organic Compounds:    - Draw dot-and-cross diagrams for the covalent bonding in the following:      - a. Ethane (C2H6)
         - b. Ethene (C2H4)
         - c. Chloroethane (CH3CH2Cl)

  4. Sublimation Temperatures: Explain the differing sublimation temperatures of carbon dioxide at -78.5°C compared to diamond at approximately 4000°C based on their structures and bonding characteristics.

  5. Hexane Properties:
       - Hexane has the formula CH₁₄ and is a liquid at room temperature.
       a. Determine if hexane has a simple molecular or giant structure.
       b. Predict if pentane (C5H12) would have a higher or lower boiling point than hexane.
       c. Discuss the conductivity of hexane.

  6. Explanation of Properties Based on Structure and Bonding:    - Explain the following properties:      a. Why diamond is harder than graphite.
         b. Why C60 fullerene has a lower melting point than graphite.
         c. Why graphite conducts electricity.
         d. Why diamond does not conduct electricity.