5 Steps to a 5 AP Chemistry

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Bonding  161


  1. B––The Lewis (electron-dot) structure has five
    bonding pairs around the central Sb and no lone
    pairs. VSEPR predicts this number of pairs to
    give a trigonal bipyramidal structure.

  2. B––All the bonds except in CO are single bonds.
    The CO bond is a triple bond. Triple bonds are
    shorter than double bonds, which are shorter
    than single bonds. Drawing Lewis structures
    might help you answer this question.

  3. A––Answers B–E contain molecules or ions with
    double or triple bonds. Double and triple bonds
    contain πbonds. Water has only single (σ) bonds.
    If any are not obvious, draw a Lewis structure.

  4. A––The VSEPR model predicts all the other
    molecules to be nonpolar.

  5. C––All the other answers involve species contain-
    ing only single bonds. Substances without double
    or triple bonds seldom need resonance structures.

  6. C––Resonance causes bonds to have the same
    average length.

  7. E––Many organic molecules are nonpolar.
    Nonpolar substances are held together by weak
    London dispersion forces.

  8. D––Lewis structures are required. You do not
    need to draw all of them. A and B have one
    unshared pair, while C and E have no unshared
    pairs. D has two unshared pairs of electrons.

  9. E––Tetrahedral molecules are normally sp^3
    hybridized.

  10. D––SiCl 4 , is tetrahedral. BrF 4 −is square planar.
    C 2 H 2 is linear. TeF 6 is octahedral. NO 3 −is trigo-
    nal planar. If you are uncertain about any of
    these, Lewis structures and VSEPR are needed.

  11. A––Only A and E are ionic. The chloride ion has
    no internal bonds, so σand πbonds are not
    possible.


12–14––Sketch a molecular orbital energy-level dia-
gram. Use the same diagram to save time, unless
it becomes too messy.


  1. D

  2. A

  3. D

  4. A––This hybridization requires a geometrical
    shape with three corners.

  5. B––One or more Lewis structures may help you.
    I and III are tetrahedral, and II is an irregular
    tetrahedron (see-saw).

  6. B––A has 0. B has 2. C and D have 0. E has 1. You
    may need to draw one or more Lewis structures.

  7. B––All the molecules are polar except B.

  8. E––The structure is:


All the carbon atoms have one double and two
single bonds. This combination is sp^2.


  1. E—Drawing one or more Lewis structures may
    help you. Only C and E are polar. Only the
    ammonia has hydrogen bonding, which is very,
    very polar.

  2. E—C is metallic bonding. All the others involve
    covalently bonded molecules.


H—C=C—C=C—H


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