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Chapter Thirteen
Thinking It Through
T13.1
First, it is necessary to determine the number of chromium atoms in the unit cell, a body–centered cubic unit
cell, then determine the number of atoms in 26.0 g of chromium using the molecular mass of chromium and
Avogadro's number. Finally, the number of unit cells would be the number of atoms in 26.0 g of chromium
divided by the number of atoms in the unit cell.
T13.2
If the unit cell contains four atoms of A then there must be six atoms of B in the unit cell. The formula for
the compound is A2B3, giving a stoichiometric ratio of 2 A atoms to 3 B atoms. The number of atoms in the
unit cell must have the same ratio.
T13.3
The pentagon (a) cannot be a unit cell in a two-dimensional lattice. When it is moved in one direction
parallel to one edge, this does not align with the unit cell that was move in the direction of the adjacent
edge. The unit cells do not build up a two-dimensional lattice.
T13.4
The structure of (b) is more likely to form a condensation polymer since it has both an acid and a base
functionality. These can react to form a bond an eliminate water. Structure (a) is more likely to form an
addition polymer.
T13.5
The light that has passed through a polarizing filter is less intense than light passed through ordinary glass
because the polarizing filter has removed all of the vibrations except those in one plane.
T13.6
Liquid crystals have a rigid region, are long and thin, and have a strong dipole. The hydrocarbon does not
have the rigid region and it does not have a strong dipole even though it is long and thin.
T13.7
In general, ceramics are made from metals with large positive oxidation states combined with small
nonmetals with large negative oxidation states. For TiBr3, Ti has only a +3 oxidation state and Br– is a
large nonmetal ion with a low negative oxidation state.
T13.8
First, it is necessary to determine the length of the unit cell edge. For NaCl type structures, the cell edge is
always twice the radius of the cation plus twice the radius of the anion. Next, it is necessary to calculate the
volume of the unit cell, which is simply the cube of the cell edge. Finally, the mass of the atoms contained
in one unit cell is divided by the volume of the unit cell in order to calculate density.
T13.9
The unit cell for gold, a face centered cubic structure, using the Pythagorem theorem, a 2 + b2 = c2, the
length of the diagonal across the face to the cube can be determined. This length is 4  radius of the gold
atom.
T13.10 To calculate the percentage of unoccupied space for a simple cubic structure, first determine the volume of
the unit cell. Second, there is the equivalent of one atom in the unit cell, and the radius for the atom is 1/2
the length of the side of the unit cell. Determine the volume of the atom using the equation:
4 3
r
3
Finally subtract the volume of the atom from the volume of the cell; that will be the volume of empty space.
The percentage of empty space is:
volume of empty space
Percentage of empty space = 
 100%
volume of the unit cell
Volume of the atom =
T13.11 Convert 200 m to nm
1000 nm
= 2  105 nm
1 m
Then, assume that if the nanotubes will bundle together such that they will line up across the diameter of the
human hair, and the number of nanotubes is:
200 10 5 nm/ hair
Number of
 nanotubes per hair =
1.4 nm/nanotube
200
m
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
Chapter Thirteen
T13.12 To determine the largest value that the wavelength of the X rays, use equation 13.1
n = 2 d sin
2 d sin

n
 = 18.5°
d = distance between the gold layers = 4.07  105 m
n = integer = 1

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