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: Draw Lewis structures of all the important resonance forms of (b) HBrO4 (HOBrO3)

Solution: Draw Lewis structures of all the important resonance forms of (b) HBrO4 (HOBrO3)

Problem

Draw Lewis structures of all the important resonance forms of 

(b) HBrO4 (HOBrO3)


Solution
  • For the Lewis structure of HBrO4, Br appears to have 7 valence electron, H as 1 and O has 6. Total valence electrons is 32. ((1(1) + 6(4) + 7(1)  = 32)
  • Br will appear as central atom since its less electronegative than O (H can't be central atom)
  • Draw the initial structure of HBrOwhere all O atoms is attached to Br and H on one of the O atom. This structure will try follow the octet rule to complete 8 electrons around each atom except for Br which has an expanded octet
  • Os around Br (except the OH) will try to have 1 electron (3 lone pairs in total) to complete the octet while Br will also take another electron forming 2 lone pairs
  • Calculating for the formal charges:

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