Ionic Molecules And The Born Haber Cycle Page 2
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4The role of the lattice energy in stabilizing ions
Consider the formation of CaO:
(g) CaO(s)
Ca(s) + ½ O
2
The following step are required:
Ca(s) Ca(g)
sublimation
+178 kJ/mol
Ca(g) Ca
+
-
(g) + e
first ionication
+589 kJ/mol
(g) Ca
+
2+
-
Ca
(g) + e
second ionization
+1040 kJ/mol
(g) O(g)
½ O
bond energy
½ (+247 kJ/mol)
2
O
-
-
O(g) + e
(g)
electron affinity of O
-141 kJ/mol
O
-
-
2-
-
O
+ e
(g)
electron affinity of O
+878 kJ/mol
2-
2+
(g) CaO(s)
O
(g) + Ca
lattice energy
-3,100 kJ/mol
(g) CaO(s)
Ca(s) + ½ O
enthalpy of reaction
-309 kJ
2
2-
2-
Note that the formation of O
is endothermic (-141 kJ/mol + 878 kJ/mol = 737 kJ/mol) and the gaseous O
ion
2+
is not stable. However, it can be formed, if it is stabilized by association with Ca
. Note the large exothermic
lattice energy.
If the lattice energy increases with charge and the association of oppositely charged ions is a stabilizing force,
+2
-2
can NaF form, with Na
and F
?
(g) NaF(s)
2+
2-
Na
(g) + F
Na(s) Na(g)
sublimation
109 kJ/mol
Na(g) Na
+
-
(g) + e
first ionization
495 kJ/mol
(g) Na
+
2+
-
Na
+ e
second ionization
4560 kJ/mol
(g) F(g)
½ F
bond energy
77 kJ/mol
2
-
F
-
F(g) + e
(g)
electron affinity
-328 kJ/mol
-
-
F
2-
F
(g) + e
(g)
electron affinity
+1250 kJ/mol
(g) NaF
2+
2-
Na
(g) + F
lattice energy
-3692 kJ/mol
(g) NaF(s)
2+
2-
Na
(g) + F
enthalpy of reaction
+2471 kJ
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