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An in-depth exploration of ionic bonding, a type of chemical bonding where electrons are transferred from one atom to another. Learn about the differences between ionic and covalent bonding, the structures of common ionic compounds, and how to predict their lattice energies using the Born-Mayer and Kapustinskii equations.
Typology: Exercises
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Na: [Ne]3s
1
Na
+^ : [Ne]
Ca: [Ar]4s
2
Ca
: [Ar]
Cl: [Ne]3s
2
3p
5
Cl
-^ : [Ar]
O: [He]2s
2
2p
4
: [Ne]
Chem 59-
Because electrostatic attraction is not directional in the same way as is covalentbonding, there are many more possible structural types. However, in the solid state, allionic structures are based on
infinite lattices of cations and anions
. There are some
important classes that are common and that you should be able to identify, including:
CsCl
NaCl
Zinc Blende
Fluorite
Wurtzite
And others…Fortunately, we can use the size of the ions to find out what kind ofstructure an ionic solid should adopt and we will use the structural arrangement todetermine the energy that holds the solid together - the crystal lattice energy,
F^ Ca
F
Chem 59-
Some common arrangements for simple ionic salts:
Cesium chloride structure 8:8 coordinationPrimitive Cubic (52% filled)e.g. CsCl, CsBr, CsI, CaS
Rock Salt structure 6:6 coordinationFace-centered cubic (fcc)e.g. NaCl, LiCl, MgO, AgCl
Zinc Blende structure 4:4 coordinationfcce.g. ZnS, CuCl, GaP, InAs
Wurtzite structure 4:4 coordinationhcpe.g. ZnS, AlN, SiC, BeO
Chem 59-
Fluorite structure 8:4 coordinationfcce.g. CaF
, BaCl 2
, UO 2
, SrF 2
2
Anti-fluorite structure 4:8 coordinatione.g. Li
O, Na 2
Se, K 2
S, Na 2
S 2
Rutile structure 6:3 coordinationBody-centered cubic (bcc)(68% filled)e.g. TiO
, GeO 2
, SnO 2
, NiF 2
2
Nickel arsenide structure 6:6 coordinationhcpe.g. NiAs, NiS, FeS, PtSn There are many other common forms of ionic structures but it is more important to beable to understand the reason that a salt adopts the particular structure that it doesand to be able to predict the type structure a salt might have.
You can determine empiricalformula for a structure by countingthe atoms and partial atoms withinthe boundary of the unit cell (thebox). E.g. in the rutile structure,two of the O ions (green) are fullywithin the box and there are fourhalf atoms on the faces for a totalof 4 O ions. Ti (orange) one ion iscompletely in the box and thereare 8 eighth ions at the corners;this gives a total of 2 Ti ions in thecell. This means the empiricalformula is TiO
; the 6:3 ratio is 2
determined by looking at thenumber of closest neighboursaround each cation and anion.
F^ Ca
F
Chem 59-
The energy that holds the arrangement of ions together is called the
lattice energy
, o
and this may be determined experimentally or calculated.^ U
o^
is a measure of the energy released as the
gas phase ions
are assembled into a
crystalline lattice. A lattice energy must always be exothermic.E.g.: Na
(g)
Cl
NaCl
(s)
o^
= -788 kJ/mol
(s)
(s)
(g)
+(g)
2(g)
(g)
ea
d
ie
sub
f
o
Chem 59-
(s)
(s)
(g)
+(g)
2(g)
(g)
ea
d
ie
sub
f
o
f^
sub
ie
d^
ea
o
o
o^
You must use the correct stoichiometry and signs to obtain the correct lattice energy. Practice Born-Haber cycle analyses at: http://chemistry2.csudh.edu/lecture_help/bornhaber.html
Chem 59-
0
2
A
B
*^
0
A
B
*****^
in kJ/mol
0
A
B
*****^
Where:e is the charge of the electron,
ε
0
is the permittivity of a vacuum
is Avogadro’s number z
A^
is the charge on ion “A”, z
B^
is the charge on ion “B”
d
0
is the distance between the cations and anions (in Å) = r
+^
is a Madelung constant d
*^
= exponential scaling factor for the repulsive term = 0.345 Å n
= the number of ions in the formula unit
The equations that we will use to predict lattice energies for crystalline solids arethe Born-Mayer equation and the Kapustinskii equation, which are very similar toone another. These equations are simple models that calculate the attraction andrepulsion for a given arrangement of ions.
Chem 59-
In this case the energy of coulombic forces (electrostatic attraction and repulsion) are:E
coul
= (e
2
π ε
) * (z 0
A^
z
B^
/ d) * [+2(1/1) - 2(1/2) + 2(1/3) - 2(1/4) + ....]
because for any given ion, the two adjacent ions are each a distance of d away, the next two ions are 2
d, then 3
d, then 4
d etc. The series in the square brackets can
be summarized to give the expression:E
coul
= (e
2
π ε
) * (z 0
A^
z
B^
/ d) * (2 ln 2)
where (2 ln 2) is a
geometric factor
that is adeqate for describing the 1-D nature of the
For an Infinite Chain of Alternating Cations and Anions:infinite alternating chain of cations and anions.
The origin of the equations for lattice energies.
The lattice energy
0
is composed of both coulombic (electrostatic) energies and an
additional close-range repulsion term - there is some repulsion even between cationsand anions because of the electrons on these ions. Let us first consider the coulombicenergy term:
0
coul
rep
Chem 59-
The numerical values of Madelung constants for a variety of different structures arelisted in the following table. CN is the coordination number (cation,anion) and
n
is the
total number of ions in the empirical formula e.g. in fluorite (CaF
) there is one cation 2
and two anions so
n
= 1 + 2 = 3. lattice
A
CN
stoich
A
/^
n
CsCl
(8,8)
AB
NaCl
(6,6)
AB
Zinc blende
(4,4)
AB
wurtzite
(4,4)
AB
fluorite
(8,4)
AB
2
rutile
(6,3)
AB
2
CdI
2
(6,3)
AB
2
Al
O 2
3
(6,4)
A
B 2
3
Notice that the value of A is fairly constant for each given stoichiometry and that thevalue of A/
n
is very similar regardless of the type of lattice.
Chem 59-
This is the Born-Mayer equation, when the constants are evaluatedwe get the form of the equation that we will use:
0
= 1390 (z
A^
z
B
/ d
*** (1 - (d**
*****^
/ d
in kJ/mol
Note: d
*^
is the exponential scaling factor for the repulsive term and a
value that we will use for this is
If only the point charge model for coulombic energy is used to estimate the lattice energy (i.e. if U
0
= E
coul
) the calculated values are much higher than the experimentally measured lattice
energies.E.g. for NaCl (r
Na+
= 0.97Å, r
Cl-
0
= 1390 (z
A^
z
B^
/ d
= 1390 ((1)(-1)/2.78) * (1.748) kJ/mol = - 874 kJ/mol
But the experimental energy is -788 kJ/mol. The difference in energy is caused by the repulsionbetween the electron clouds on each ion as they are forced close together. A correction factor,E
rep
, was derived to account for this.
rep
= - (e
2
π ε
) * (z 0
A^
z
B^
d*/ d
and since
coul
= (e
2
π ε
) * (z 0
A^
z
B^
/ d) *
the total is given by
0
= (e
2
π ε
) * (z 0
A^
z
B^
/ d