ANSWERS TO CUBIC CRYSTAL LATTICES
1. (a)
1
4 circle
(b) 1 circle
(c) i)
1
4
sphere
1
ii)
8
sphere
(d) 4 x
2. (a) 6 [see the 6 shaded atom's surrounding the central shaded atom]
(b) 1 8 [see answer for 1(c) ii.]




(c) 8 x 1 8 = 1 atom
(d) radius =
1
2
1
8
=
1
2
sphere

x (3.345 x 10–8 cm) = 1.673 x 10–8 cm [see diagram for Exercise 1(a)]
(e) 3.345 x 10–8 cm

3. volume of unit cell = V = (5.357 x 10–8 cm)3 = 1.5373 x 10–22 cm3

1mol
174.2 g
mass of unit cell = m =
x
= 2.8925 x 10–22 g
1mol
6.0225 x 1023

2.8925 x 10 22 g
m
=
= 1.882 g/cm3 (or 1.882 g/mL)
-22
3
V
1.5373 x 10 cm


501.8 g
1mol
4. mass of unit cell =
x
= 8.3324 x 10–22 g
23
1mol
6.0225
x
10
 
8.3324 x 10 22 g
m
volume of unit cell = V =
=
= 1.6638 x 10–22 cm3
d
5.008 g / cm3


length of unit cell = (1.6638
x 10–22 cm3)1/3 = 5.500 x 10–8 cm
d=
5. 1 (centre) + 8 x
1
(corners) = 2

+
1
6. 1 Cs (centre) + 8 x 8 Cl– (corners) = CsCl , which agrees
8
1 Cl– (centre)
+8x

1
8
Cs+ (corners) = CsCl , which agrees
– (in centre) has 8 nearest–neighbour Cs+ ions at corners
7. (a) Each Cl
(b) Each Cs+ (in centre) has 8 nearest–neighbour Cl– ions at corners

8. Label the unit cell as follows.
A
C
E
B
X
G
D
F
the Cs+ an d Cl– io ns touch al ong the body d iago nals, su ch as a long AXH
H
A
D
X
E
H
The dia gona l AXH is the d iago nal throu gh the p lane ADHE
. Th e di agon al AXH
+
has a l ength eq ual to 2 tim es the ra dius of Cs
(as sume d to be at the centre)
plu s 2 time s th e rad ius of Cl– (a ssum ed to be at the corn ers).
Therefore: length of AXH = 2 (1.67 x 10–8 + 1.81 x 10–8) = 6.96 x 10–8 cm
The problem requires finding the distance AE, which is the unit cell length. However, since the length of
EH is not known, AE cannot be found yet. Aha! Pythagoras Theorem to the rescue. The bottom plane
of the cube, EFGH, has lengths with the following ratios. Assume the unit cell length, AE, is 1.
E
F
2
1
G
1
Therefore
H
A
The diagona l
pl ane ADHE has
EH = 2
X
E
3
H
an d (AH)2 = (AE)2 + (EH)2
= 1+ 2
so
AH =
3
2
6.96 x 10 8
3
2

1
D

=
AE
, and length AE = 4.02 x 10–8 cm = length of unit cell
1
Hebden : Chemistry AP
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Now to find the density of CsCl. The body centred unit cell contains one Cs+ and one Cl–.
(132.9  35.5) g
1mol
mass =
x
= 2.796 x 10–22 g
1mol
6.0225 x 1023
volume = (4.02 x 10–8 cm)3 = 6.496 x 10–23 cm3

2.796 x 10 22 g
m
density =
= 
= 4.304 g/cm3
-23
3
V
6.496 x 10 cm
9. This exercise uses some of the results from exercise 8. A body–centred unit cell contains TWO titanium
atoms (one at the body centre and one distributed among the 8 corners).
 
2 x 47.9 g
1mol
mass of unit cell =
x
= 1.591 x 10–22 g
1mol
6.0225 x 1023
1.591 x 10 22 g
m
=
= 3.535 x 10–23 cm3
3
d
4.50 g / cm


length of unit cell = (3.535 x 10–23 cm3)1/3 = 3.282 x 10–8 cm
volume =
The body
has a distance equal to four times the radius of a titanium atom. Using the diagrams
 diagonal

from exercise 8:
length of body diagonal
length of unit cell
=
1
3
length of body diagonal =

3 x 3.282 x 10–8 cm = 5.684 x 10–8 cm

length of body diagonal = 4 x radius ,

so
radius =
5.684 x 10 8 cm
= 1.42 x 10–8 cm
4
10. number of corner atoms = 8 x 1 8 = 1 atom , number of face–centred atoms = 6 x
total number of atoms = 4 atoms

number of nearest neighbours = 12. To see this, look at the diagrams below.

1
2
= 3 atoms

Since the corne r and face-cente red
ato ms a re consi dered to be
interch ange able, th e corner atom s
als o ha ve 12 neare st n eigh bours.
4 n eigh bours
in plan e
4 face -cen tered
nei ghbo urs i n
cu be a bove
11. # of X ions at corners = 8 x
1
8
=1,
4 face -cen tered
nei ghbo urs i n
cu be b elow
# of Y ions at faces = 6 x
1
2
=3 ,
formula of compound = XY3
12. number of Ni atoms = 8 x 1 8 (corners) + 4 x 1 4 (edges) = 2
number of As atoms = 2 (completely inside unit cell)
formula of nickel arsenide = NiAs


13. number of Ca atoms at corners = 8 x 1 8 = 1
number of O atoms at faces = 6 x 1 2 = 3


number of Ti atoms at body centre = 1
formula of perovskite = CaTiO3


AP : CUBIC CRYSTAL LATTICES ANSWERS
3
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14. Since there are 4 particles per unit cell then the molar mass of the unit cell = 4 x 1.0087 g = 4.0348 g
The diagonal on each face is 4 neutron radii or 2 neutron diameters = 2 x 10 –13 cm
length of diagonal
2 x 10 13 cm
length of unit cell edge
=
=
= 1.41 x 10–13 cm
1
2
2
volume of unit cell = (1.41 x 10–13 cm)3 = 2.83 x 10–39 cm3
volume of 1 mol of unit cells = 2.83 x 10–39 cm3 x 6.0225 x 1023 = 1.70 x 10–15 cm3


4.0348 g
m
density =
=
= 2 x 1015 g/cm3
V
1.70 x 10 -15 cm3

15. Let N = Avogadro's Number. The unit cell contains 1 (corners) + 3 (faces) = 4 atoms


mass of unit cell =
4 x 192.2 g
1mol
768.8 g
x
=
1mol
N
N
volume of unit cell = (3.833 x 10–8 cm)3 = 5.6314 x 10–23 cm3
768.8
m
 g / N  = 22.61 g/cm3
density =  =
V
5.6314 x 10 -23 cm3
768.8 g
Solving for N:
= 22.61 g/cm3 x 5.6314 x 10–23 cm3 = 1.273 x 10–21 g
N

768.8 g
N = 
= 6.038 x 1023
1.273 x 10 -21 g

16. mass of unit cell = 21.45 g/cm3 x (3.92 x 10–8 cm)3 = 1.292 x 10–21 g
6.0225 x 1023
778.1g
molar massof unit cell = 1.292 x 10–21 g x
=
1mol
mol
A face–centred unit cell has 4 atoms, so that
778.1g
= 
194.5 g (which suggests
Pt: 195.1 g)

4
The face of the unit cell has atoms touching across the face diagonal. Therefore
2
diagonal length
=
1
unit cell
edge length
molar mass of atoms =
diagonal length =
2 x unit cell edge length = 5.544 x 10–8 cm

5.544 x 10 8 cm
diagonal length = 4 x radius , so radius =
= 1.39 x 10–8 cm
4

17. Using the geometric arguments from exercise 16, above:

2
4 x radius
diagonal length
4 x radius

=
=
=
1
unit cell edge length
unit cell edge length
3.62 x 10 8 cm
and
radius = 1.28 x 10–8 cm
so

Each unit cell has 4 atoms,
 that

1mol
4 x 63.5 g
mass of unit cell =
x
= 4.218 x 10–22 g
mol
6.0225 x 1023
volume of unit cell = (3.62 x 10–8 cm)3 = 4.744 x 10–23 cm3
4.218 x 10 22 g
m
density =  =
= 8.89 g/cm3

V
4.744 x 10 -23 cm3
4
Hebden : Chemistry AP
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

18. Diagonal length = 4 x radius = 4 x 1.44 x 10–8 cm = 5.76 x 10–8 cm
Using the geometric arguments from exercise 15, above:
1
unit cell edge length
unit cell edge length
unit cell edge length
=
=
=
diagonal length
4 x radius
5.76 x 10 8 cm
2
and
unit cell edge length = 4.07 x 10–8 cm
so
Each unit cell has 4 atoms,
 that


1mol
4 x 107.9 g
mass of unit cell =
x
= 7.166 x 10–22 g
mol
6.0225 x 1023
volume of unit cell = (4.07 x 10–8 cm)3 = 6.742 x 10–23 cm3
density =

7.166 x
10 22 g
m
=
= 10.6 g/cm3
-23
3
V
6.742 x 10 cm
19. Number of negative charges in each molecule = 2; if 1 4 of negative charges are now chlorides, then
1Cl ion
 = 3 x (–2) x 1OH group = 1.5
# of OH
and
# of Cl = 41 x (–2) x
= 0.50
4
1charg e
1charg e
Formula = Cu(OH)1.5Cl0.5
and clearfractions by doubling subscripts to get Cu2(OH)3Cl
20. Assume
 there is 1 Fe atom, altogether, in the formula. The overall molecule must be electrically neutral.
 = ( 1 )(+2) + ( 2 )(+3) = 8

Total positive charge
3
3
3
1oxide ion
Hence, total negative charge = – 8 3
and
number of oxide ions = (
)(– 8 3 ) = 4 3
2 charg e



2
3
Fe
O
Formula = Fe1/
and
multiplying subscripts by 3 to clear fractions: Fe 2 Fe 3
3
2/ 3 4/ 3
2 O4



(Note: magnetite is usually shown in the simpler, but less correctform, as Fe3O4 or even Fe2O3•FeO.)
21. First, account for the substitution of Si4+ by Cu2+.


2
2
Initial formula = Si4
and double the subscripts to clear fractions: Si4 Cu 2 O 2
0.5 Cu 0.5 O 2
4
The overall charge on this molecule is (+4) + (+2) + (–8) = –2
Since the deficit in positive charge is made up by H+ ions, the final formula is H2SiCuO4
(Note: the
 formula for dioptase is sometimes shown as CuSiO2(OH)2 .)
22. Na+ is smaller than Cl– (recall what you learned in Chem 11). Na+ is face–centred cubic; Cl– is face–
centred cubic.
23. Each Cl– is surrounded by 6 Na+ ions arranged in an octahedral shape around the Cl– at the centre.
Each Na+ is surrounded by 6 Cl– ions arranged in an octahedral shape around the Na+ at the centre.
24. number of Na+ = 12 x
number of
ratio of
Cl–
Na+
:
=8x
Cl–
1
1
8
4
(edges) + 1 (body centre) = 4
(corners) + 6 x
1
2
(faces) = 4
= 1:1

25. number of nearest neighbours (see bottom diagram on page 7 of notes):

 layer) + 3 (bottom layer) = 12.
6 (central layer) + 3 (top
This is the same number of neighbours found for face–centred cubic (note the final comment on p.7).
26. Since both hexagonal close–packed and face–centred cubic are similar structures with an identical number
of identical neighbours, the densities are expected to be almost identical.