Chem1000a
Spring 2008
Practice Problems 3: Group 1 Metals and Metal Lattices
Solutions
Additional Practice Problems
1.
Name each of the following compounds:
(a)
CuI
copper(I) oxide
(b)
CuS
copper(II) sulfide
(c)
BaF2
barium fluoride
(d)
NiO
nickel(II) oxide
2.
Give the formula for each of the following compounds:
(a)
iron (III) chloride
FeCl3
(b)
sodium peroxide
Na2O2
(c)
cobalt (III) oxide
Co2O3
(d)
magnesium fluoride
MgF2
3.
Suppose that you want to make 10. g of potassium superoxide. What are the minimum
masses of potassium and oxygen you will need?
K(s)
+
O2(g)
→
KO2(s)
39.0983 g/mol
31.9988 g/mol
71.0971 g/mol
n KO2 = 10.g ×
1mol
= 0.14mol
71.0971g
2 significant figures for all three answers!
n K = n KO2 = 0.14mol
m K = 0.14mol ×
39.0983 g
= 5.5 g
1mol
nO2 = n KO2 = 0.14mol
mO2 = 0.14mol ×
4.
31.9988 g
= 4.5 g
1mol
Write a balanced equation for each of the following reactions and name all products:
(a)
Na and Br2
sodium bromide
2 Na(s) + Br2(l) → 2 NaBr(s)
(b)
Li and N2
lithium nitride
6 Li(s) + N2(g) → 2 Li3N(s)
(c)
Li and O2
4 Li(s) + O2(g) → 2 Li2O(s)
lithium oxide
(d)
K and H2O
1-4
or
5.
2 K(s) + 2 H2O(l) → 2 KOH(aq) + H2(g)
2 K(s) + 2 H2O(l) → 2 K+(aq) + 2 OH-(aq) + H2(g)
potassium hydroxide and hydrogen gas
In the standard notation of closest-packing of spheres, the letters A, B and C refer to
closest-packed layers. Which of the following sequences describe closest-packing in 3
dimensions, and which do not. State your reasoning explicitly:
(a)
ABCABC....
This series describes a cubic closest-packed lattice.
(b)
ABACABA....
This series describes an irregular closest-packed lattice
(combination of hcp and ccp).
(c)
ABBABABBA....
This series does not describe a closest packed lattice.
A “B” layer cannot lie directly on top of another “B” layer.
6.
Sketch a portion of a simple cubic lattice, and use your sketch to show that the
coordination number in this lattice is 6.
see text and/or notes
7.
Calcium metal crystallizes in a face-centered cubic unit cell. The density of the solid is
1.54 g/cm3. Calculate the radius of a calcium atom.
Step 1: Find the mass of one unit cell
***One FCC unit cell contains the equivalent of four atoms***
mcell = 4mCa = 4 ×
40.078 g / mol
= 2.6620 × 10 − 22 g
23
6.02214 × 10 / mol
Step 2: Find the volume of one unit cell
m 2.6620 × 10 −22 g
(10 −2 m) 3
=
= 1.73 × 10 −22 cm 3 ×
g
d
cm 3
1.54 3
cm
− 28
3
= 1.73 × 10 m
V=
x
x
2
Step 3: Find the length of an edge of the cube
x
x = 3 V = 3 1.73 × 10 −28 m 3 = 5.57 × 10 −10 m
Step 4: Find the radius of one atom of calcium
***Use Pythagorean theorem to determine the triangle side shown in the diagram***
2 x = 4rCa
rCa =
2x
2 (5.57 × 10 −10 m)
=
= 1.97 × 10 −10 m = 197 pm
4
4
2-4
8.
Vanadium metal has a density of 6.11 g/cm3. If the atomic radius of vanadium is 1.32 Å,
is the vanadium unit cell simple cubic, body-centered cubic, or face-centered cubic?
Body-Centered Cubic
There are several different approaches you could take to this question.
1. You could calculate the packing fraction and compare it to the values calculated for
each lattice type in question 9.
2. You could calculate the atomic radius of vanadium using the density provided and
molar mass. After doing this for each type of lattice, compare the values obtained to
the value provided and choose the best lattice type.
3. You could calculate the density of vanadium using the atomic radius provided and
molar mass. After doing this for each type of lattice, compare the values obtained to
the value provided and choose the best lattice type.
4. You could calculate the molar mass of vanadium using the density and atomic radius
provided. After doing this for each type of lattice, compare the values obtained to the
value on the periodic table and choose the best lattice type.
5. You could look up the answer in a textbook or other reliable resource; however, it is
strongly recommended that you be capable of using one of the four approaches above
since it is highly unlikely that this option would be available on a test!
9.
Calculate the packing fraction for each of the three cubic lattice types (simple, bcc and
fcc). Recall that the packing fraction is the fraction of space in a unit cell occupied by
the atoms themselves (as opposed to empty space between atoms).
Packing fraction = volume occupied by atoms ÷ total volume of lattice
Strategy: Determine relationship of r to x then calculate both volumes and compare.
(Note that density of a substance refers to the density of the unit cell – not of an
individual atom.)
(a)
Simple cubic lattice
2r = x
Vatoms −in −lattice = Vatom =
4 3
πr
3
Vatoms −in − cell
Vcell
x
Vcell = x 3 = ( 2r ) 3 = 8r 3
4 3
πr
3
=
= 0.5236 = 52.36%
8r 3
3-4
(b)
Body-centered cubic lattice
4r = 3 x
x
2
x
3
4
8
Vatoms −in −lattice = 2 × Vatom = 2 × πr 3 = πr 3
3
3
3
Vcell
⎛ 4r ⎞
64 3
r
⎟⎟ =
= x = ⎜⎜
3 3
⎝ 3⎠
3
(c)
x
8 3
πr
Vatoms −in −cell
3
=
= 0.6802 = 68.02%
Vcell
⎛ 64 3 ⎞
r ⎟⎟
⎜⎜
⎝3 3 ⎠
Face-centered cubic lattice
4
16
Vatoms −in −lattice = 4 × Vatom = 4 × πr 3 = πr 3
3
3
x
x
2
4r = 2 x
3
Vatoms −in −cell
Vcell
x
Vcell
32 3
⎛ 4r ⎞
r
= x =⎜
⎟ =
2
⎝ 2⎠
3
16 3
πr
3
=
= 0.7405 = 74.05%
⎛ 32 3 ⎞
r ⎟
⎜
⎝ 2 ⎠
4-4
Because there is no experimental data for this
question, significant figures are irrelevant.
Anywhere from 2 to 4 digits would be reasonable.