Untitled - Kelly Walsh High School

166 CHEMISTRY FOR THE UTTERLY CONFUSED
know that iron and diamond cannot involve ionic bonding. This is a check on
the validity of our earlier predictions. Several of the remaining substances,
methane, methyl alcohol, methyl fluoride, and krypton difluoride, do not contain
a metal or a polyatomic ion. (Be careful with methyl alcohol. This compound
does not contain the hydroxide ion. The –OH group in this compound is
an alcohol group.) Only two substances remain: ammonium fluoride and sodium
chloride. Ammonium fluoride does not contain a metal, but it does contain
a polyatomic ion, the ammonium ion. The presence of the polyatomic ion
means that the strongest intermolecular force will be ionic bonding. Sodium
chloride consists of a metal and a nonmetal, which means that its strongest
intermolecular force is ionic bonding.
The next strongest type of intermolecular force is hydrogen bonding. This type
of force requires a hydrogen atom bonded to nitrogen, oxygen, or fluorine. We
can eliminate any compound not containing hydrogen (KrF 2) and any compound
not containing nitrogen, oxygen, or fluorine (CH 4). This leaves us with
methyl alcohol, CH 3OH, and methyl fluoride, CH 3F. (You may need to sketch
the atom arrangement for these two compounds to make the prediction.) In
methyl alcohol, there is a bond between one of the hydrogen atoms and the
oxygen atom. This attachment allows the formation of hydrogen bonds. In this
compound, the remaining hydrogen atoms bond to the carbon. In methyl fluoride,
the hydrogen atoms bond to the carbon. Since no hydrogen atoms are
bonded to the fluorine, hydrogen bonding is not possible.
We now have three substances remaining: methane, CH 4, methyl fluoride,
CH 3F, and krypton difluoride, KrF 2. We also have two types of intermolecular
force remaining: dipole-dipole forces and London forces. In order to match
these substances and forces we must know which of the substances are polar
and which are nonpolar. Polar substances utilize dipole-dipole forces, while
nonpolar substances utilize London forces. To determine the polarity of each
substance, we must draw a Lewis structure for the substance (Chapter 9) and
use valence-shell electron pair repulsion (VSEPR) (Chapter 10). The Lewis
structures for these substances are:
H
H C H
H
H
H C F
H
F Kr F
The molecular geometry of methane and of methyl fluoride is tetrahedral. In
the case of methane, this symmetrical arrangement of polar covalent carbonhydrogen
bonds leads to a canceling of the bond polarities resulting in a nonpolar
molecule. As a nonpolar molecule, the strongest intermolecular force in methane
is a London force. In methyl fluoride, a fluorine atom replaces one of the hydrogen