Ch.11 - Liquids, Solids & Intermolecular ForcesWorksheetSee 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

Solution: Rank the following molecules from the least polar to the most polar:CH2Br2, CF2Cl2, CH2F2, CH2Cl2, CBr4, CF2Br2.

Solution: Rank the following molecules from the least polar to the most polar:CH2Br2, CF2Cl2, CH2F2, CH2Cl2, CBr4, CF2Br2.

Problem

Rank the following molecules from the least polar to the most polar:

CH2Br2, CF2Cl2, CH2F2, CH2Cl2, CBr4, CF2Br2.

Solution
  • The Lewis structures for this set of compounds are similar since we are using C as central atom which will follow a tetrahedral structure

  • We have to analyze each of the atoms attached to C to determine which will contribute more to the polarity of the molecule

  • Trend for electronegativity is increasing from bottom to top. Therefore F > Cl > Br will be our guide

  • For the polarity of the molecules, due to a big difference in polarity with H and halogens, this combination will appear as our most polar combination

  • In this case, we have CH2Br2, CH2Fand CH2Cl2

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