Chapter 6 Source Answers and Examples
Chapter problems → · Original reading →
This chapter has no answer key. Its worked examples follow.
Worked examples and reviewed revisions
Example 6.1.1
Worked example
AP edition reviewed replacement.
Using supplied electronegativities H2.1,O3.5,Na0.9,Cl3.0,F4.0, compare ΔEN for H–H,O–H,Na–Cl andF–H. Explain bond polarity and the distinction between molecularHF and solidNaCl. Do not infer bond energy from ΔEN.
New independently computed worked answer.
The differences are0,1.4,2.1and1.9 respectively, so ΔEN increases H–H<O–H<F–H<Na–Cl. H–H is nonpolar covalent; O–H andF–H are polar covalent. HF is molecular with a strongly polar covalent bond; ordinaryNaCl is an extended ionic lattice. A numerical cutoff is a heuristic, not a universal test, and this ΔEN ranking is not a bond-strength or stability ranking.
Example Problem 6.1.1
Worked example
Consider atoms with the following electron configurations:
Which atom has the largest second ionization energy? Explain
The answer lies in the electron configurations. Removing two electrons from magnesium is not as hard since its two valence electrons are shielded from the core electrons and are far away from the nucleus. Neon’s first IE will be very high due to ineffective shielding in the 2p orbitals, but the second one will be a little higher.
Sodium’s first IE is low for the same reasons as magnesium, but its second IE must be high due to highly inefficient shielding in the 2p orbitals plus the electrons are much closer to the nucleus.
At this point then its between Neon and sodium. It’s a little tricky here just because you really have to know your your periodic trends, specifically atomic radii. At this point, since sodium has a larger nucleus with more protons than electrons, there will be more pull for those electrons to prevent them from leaving. Neon has a smaller nucleus so the pull for the electrons will not be as great.
Therefore, sodium is our answer!!
Example Problem 6.3.1
Worked example
AP edition reviewed replacement.
Identify the particles and major attractive interactions in pureNH₃, pureC₆H₁₄ and solidKBr. Explain why the presence ofH alone does not guarantee hydrogen bonding, and whyKBr should not be treated as a molecule.
New independently computed worked answer.
NH₃ molecules are trigonal pyramidal and polar; dispersion and dipole interactions act, with N–H and the nitrogen lone pair permitting intermolecular hydrogen bonding. Hexane molecules are approximately nonpolar; dispersion dominates, and C–H alone does not establish hydrogen bonding. SolidKBr is an extended array ofK⁺ andBr⁻ ions held principally by ionic electrostatic attractions. DissolvedKBr ions interact with polar water through ion–dipole interactions, but that is a different phase/context.