34
1 – Description of ionic crystals
2. There are Z = 4 formula units FeO in the unit cell. If FeO is stoichiometric,
the density ρ th is
N a
ZM
th
Av
3
FeO
ρ =
where M FeO is the molar mass of the oxide.
The calculation gives
6.02 10
4.371 10
4 55.85 16
th
23
8 3
#
#
#
#
ρ =
+
−
^
] ^
h
g
h
ρ th = 5.717 g cm
–3
Because the measured density (ρ = 5.312 g cm
–3
) is less than that of the
stoichiometric monoxide, we conclude that the formula Fe 0.91 O should be
retained with an incompletely filled iron sub-lattice.
3. The equilibrium reaction involving the monoxide and oxygen is
2
1
O 2(g) m O
#
O + 2h
• + V ′′
Fe
4. The presence of trivalent cations Fe
3+
may be explained by hole trapping
by Fe
2+
cations according to the reaction
Fe
#
Fe + h
•
m Fe
•
Fe
Note – The equilibrium reaction with oxygen may also be written as
2Fe
#
Fe + 2
1
O 2(g) m O
#
O + 2Fe
•
Fe + V ′′
Fe
5. Based on one mole of Fe 1−α O, electroneutrality, mass balance, and sitoneutrality lead respectively to
2x + 3y − 2 = 0
x + y = 1 − α
x + y + z = 1
We deduce
x = 1 − 3α
y = 2α
z = α
Finally, the formula for the monoxide is
Fe
Fe V O
1 3
2
2
3
α
α α
−
+
+