Ch.9 - Bonding & Molecular StructureWorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Bond Energy

See all sections
Sections
Chemical Bonds
Lattice Energy
Lattice Energy Application
Born Haber Cycle
Dipole Moment
Lewis Dot Structure
Octet Rule
Formal Charge
Resonance Structures
Additional Practice
Bond Energy

Solution: “Inert” xenon actually forms several compounds, especially with the highly electronegative elements oxygen and fluorine. The simple fluorides XeF2, XeF4, and XeF6 are all formed by direct reaction of

Problem

“Inert” xenon actually forms several compounds, especially with the highly electronegative elements oxygen and fluorine. The simple fluorides XeF2, XeF4, and XeF6 are all formed by direct reaction of the elements. As you might expect from the size of the xenon atom, the Xe—F bond is not a strong one. Calculate the Xe—F bond energy in XeF6, given that the enthalpy of formation is −402 kJ/mol.