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: SF6, ClF5, and XeF4 are three compounds whose central atoms do not follow the octet rule. Draw Lewis structures for these compounds.

Solution: SF6, ClF5, and XeF4 are three compounds whose central atoms do not follow the octet rule. Draw Lewis structures for these compounds.

Problem

SF6, ClF5, and XeF4 are three compounds whose central atoms do not follow the octet rule. Draw Lewis structures for these compounds.

Solution

To draw the Lewis structure of each of the compound, we must first determine how many valence electrons each of them have so we can use those electrons to form bonds or distribute them as lone pairs on the atoms.

SF6 

• S → Group 6A → 6 valence e- 
• F→ Group 7A → 7 valence e- x 6

• Total valence e- = 48 e-

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