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
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: Use the formal charge arguments to rationalize why BF3 would not follow the octet rule.

Solution: Use the formal charge arguments to rationalize why BF3 would not follow the octet rule.

Problem

Use the formal charge arguments to rationalize why BF3 would not follow the octet rule.

Solution
  • Analyze the Lewis structure first then deduce how BF3 exists as neutral molecule 
  • Then compare this to the Lewis structures with formal charges but appears as neutral molecule
  • B will appear as the central atom since it is less electronegative than F
  • Boron has valence electron of 3 while F has 7. Total valence electron is 24. 3(1) + 7(3) = 24
  • F atoms will appear as having 3 lone pairs while being single bonded to B
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