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Bonding and Structure

Solubility and conductivity of simple molecular compounds

  • 1 thing to know
  • Components 01, 02 and 03

Overview

Recall questions on why these compounds dissolve in non-polar solvents but usually not in water, and why they conduct in no state.

What you need to know

Answer each one before you reveal it.

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Process

Why are simple molecular compounds generally insoluble in water, why do stronger intermolecular forces give higher boiling points, and what does the higher boiling point of two compounds indicate?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Why does ice have a relatively low melting point compared to giant covalent or ionic structures?

    Model answer

    Because only intermolecular forces between H₂O molecules need to be overcome, rather than strong covalent or ionic bonds throughout a giant structure.

  2. 2

    Under what conditions might a simple molecular compound dissolve in water?

    Model answer

    If it can form intermolecular interactions with water molecules.

  3. 3

    Why are simple molecular compounds often soluble in non-polar solvents?

    Model answer

    Because they have similar intermolecular forces to the non-polar solvent molecules.

  4. 4

    Why are simple molecular compounds generally insoluble in water?

    Model answer

    Unless they can form intermolecular interactions with water molecules, they cannot overcome the strong hydrogen bonding between water molecules.

  5. 5

    Why are molecules in simple molecular compounds electrically neutral overall?

    Model answer

    Because the electrons are shared in covalent bonds within each molecule, giving no overall charge.

  6. 6

    Why are there no ions present in simple molecular compounds?

    Model answer

    Because the atoms are bonded covalently by sharing electrons, not by electron transfer, so no charged particles are formed.

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Foundations in Chemistry
Bonding and Structure
Periodic Table and Energy
Enthalpy Changes

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AThe first step is Cl• + O₃ → ClO• + O₂.
BThe second step is ClO• + O → Cl• + O₂.
CThe overall equation is O₃ + O → 2O₂.

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In ClO⁻, chlorine has oxidation number +1.
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