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Quiz 8. Ionic Compounds

Skills to develop

Ionic Compounds

The common table salt, NaCl, is a representative of an important class of compounds called salts. A salt consists of positive and negative ions, Na+ and Cl- in NaCl. Salts such as KCl, CsCl, CaCl2, CsF, KClO4 NaNO3, and CaSO4 are generally considered ionic compounds, which are composed of (positive) cations and (negative) anions. When they dissolve in water, the solutions contain hydrated positive and negative ions.

A salt is produced in a neutralization reaction or in a direct reaction. For example, NaCl solid can be obtained by

NaOH + HCl = H2O + NaCl (solid after evaporation of solvent)
2 Na + Cl2(g) = NaCl (solid) (burning metallic sodium in a chlorine gas.)

Monoatomic Ions

Halide ions are formed when a group 7A element acquires an electron to attain the electronic configuration of a noble gas. Group 6A elements need to acquire two, instead of one, electron to achieve the same, whereas a nitride needs to gain three electrons. Similarly, alkali metals, alkali earth metals, and group 3A elements lose 1, 2, and 3 electrons respectively to attain the same stablilty. The following groups of ions have the same electronic configurations.

[He] = 1s2 H- He Li+ Be2+ B3+
[Ne] = 2s22p6 N3- O2- F- Ne Na+ Mg2+ Al3+
[Ar] S2- Cl- Ar K+ Ca2+
[Kr] Se2- Br- Kr Rb+ Sr2+
Cu+ Zn3+ Ga3+ Ge4+
Ag+ Cd2+ In3+ Sn4+
[Xe] I- Xe Cs+ Ba2+

This list gives you positive and negative ions. Some of them have electronic configurations of some noble gases. Thus, the formation of these ions can be attributed to the tendency of an atom to acquire the electronic configuration of its nearest noble gas on the periodic table of elements.

Ions of Transition Metals

Transition metal ions also appear frequently, some of the common mono-atomic ions of transition metals are given below with their electronic configurations indicated. The stable cores of noble gases are indicated by [], whereas the number of d electrons are indicated by a superscript number after d.

Ions Electronic
Configuration
Remark
Sc [Ar]4s2 3d1Metal
Sc2+ [Ar] 3d1Color ion
Sc3+ [Ar] Colorless
Ti [Ar]4s2 3d2
Ti4+ [Ar] Colorless
V [Ar]4s2 3d3
V2+ [Ar] 3d3 V2+ (aq)
V3+ [Ar] 3d2 V3+ (aq)
Cr [Ar]4s1 3d5
Cr3+ [Ar] 3d3 Cr3+ (aq)
Mn [Ar]4s2 3d5
Mn2+ [Ar] 3d5 Mn2+ (aq)
Fe [Ar]4s2 3d6
Fe2+ [Ar] 3d6 Fe2+ (aq)
Fe3+ [Ar] 3d5 Fe3+ (aq)
Co [Ar]4s2 3d7
Co2+ [Ar] 3d7 Co2+ (aq)
Ni [Ar]4s2 3d8
Ni2+ [Ar] 3d8 Ni2+ (aq)
Cu [Ar]4s1 3d10
Cu1+ [Ar] 3d10 Colorless
Cu2+ [Ar] 3d9 Cu2+ (aq)
Zn [Ar]4s2 3d10
Zn2+ [Ar] 3d10 Colorless

Ionic Radii

Polyatomic Ions

The ammonium ion and some anions from oxy-acids are typical poly-atomic ions. Some of these are given so that you will be familar with them.

NH4+ ammonium ion
N(CH3)H3+ Methyl ammonium ion,
any number of H can be replaced by a methy group
N(CH3)4+ Tetramethyl ammonium
SO42- Sulfate
SO32- Sulfite
PO43- Phosphate
PO33- Phosphite
ClO4- Perchlorate
ClO3-Chlorate
ClO2-Chlorite
NO3- Nitrate
NO2- Nitrite

A salt consists of positive and negative ions, and the stoichiometry is determined by balancing the charges so that the salt as a whole is neutral.

Lattice Energy and Energy of Crystallization

The Lattice energy, U, is the amount of energy requried to separate a mole of the solid (s) into a gas (g) of its ions. NaCl(s) ® Na+(g) + Cl-(g)       U kJ mol-1 Wishful thinking would be to separate the ions in a salt so that there is no interaction between the positive and negative ions. After the separation, the ions must be in a gaseous sate. However, the process requires energy, and the lattice energies are positive.

The same amount of energy will be released when the ions are condensed from a gaseous state to a solid state. In this view, the released energy is called energy of crystallization, Ecryst. Since energy is released, the sign is negative for such a quantity. Thus, we have

a Mb+(g) + b Xa- (g) ® MaXb(s)       Ecryst kJ mol-1
Therefore, we have this relationship: Ecryst = - U To carry out either process experimentally is impossible, let alone measuring the energy involved. One way to evaluate lattice energy is called the Bohn-Haber cycle.

Born-Haber Cycle for the Calculation of Lattice Energy

Since the concepts of lattice energy and energy of crystallization are sound, and lattice energy is a useful quantity to judge the chemical properties of the salt, we can and usually calculate it by applying the law of conservation of energy, in a manner similar to the Hess's law. For the calculation of lattice energy, this process is called the Born-Haber cycle.

In 1919, M. Born and F. Haber separately devised a method to calculate the lattice energy from known thermodynamic data such as,

Hsub Enthalpy of sublimation,
Hmelt Enthalpy of melting,
Hvap Enthalpy of vaporization,
Hf Enthalpy of formation,
U Lattice energy,
Ecryst Energy of crystallization,
H Enthalpy of reaction,
D Bond (dissociation) energy
IP Ionization potential, or ionization energy, IE
EA Electron affinity,
For example, to calculate the lattice energy of NaBr, we start with Na metal and Br2 liquid. To make NaBr, each of the reactants need to go through a series of processes in order to calculate the lattice energy:

2 Na (s) --2 Hsub ® 2 Na (g) --2 IP ® 2 Na+(g)

Br2(l) --Hvap ® Br2 (g) --D ® 2 Br(g) --2EA ® 2 Br-(g)

One other reaction is required

2 Na + Br2 ® 2NaBr, 2Hf

The lattice energy corresponds to the reaction

2 NaBr(s) ® 2Na+(g) + 2 Br-(g)

To put all the information together, we may now draw the cycle:

          THE BORN-HABER CYCLE
-----------2Na+ + 2 Br(g)-------- ­ | |2IP + D |2EA | ¯ 2Na(g) + Br2(g) 2Na+(g) + 2Br-(g) ­ | |2Hsub+Hvap | | | 2Na(s) + Br2(l) |2Ecryst | | |2Hf | ¯ ¯ ----------2 NaBr(s)---------------
Thus,

2Hf - (2Hsub + Hvap + 2IP + D) = 2EA + 2Ecryst

Note that we have used two moles of NaBr in the above diagram.

This scheme shows that we can calculate the lattice energy of NaBr from some known thermodynamic data. The same can be calculated from reaction equations and their associated energies. This is illustrated below

2 Na(s) + Br2(l) ® 2 NaBr(s) 2 Hf
2 Na(g) ® 2 Na(s)
or 2 Na(s) ¬ 2 Na(g)
- 2 Hsub
2 Na+(g) + 2e ® 2 Na(g)
or 2 Na(g) ¬ 2 Na+ + 2e
- 2 IP
Br2(g)®Br2(l)
or Br2(l) ¬ Br2(g)
- Hvap
2 Br(g) ® Br2(g)
or Br2(g) ¬ 2 Br(g)
- D
2 Br-(g) ® 2 Br(g) + 2 e
or 2 Br(g) + 2 e ¬ 2 Br-(g)
- 2 EA
Add all the above equations leading to
2 Na+(g) + 2 Br-(g) ® 2NaBr(s) 2 Ecryst
Thus, 2 Hf - 2Hsub - 2 IP - Hvap - D - 2 EA = 2 Ecryst
Ecryst = Hf - Hsub - IP - (Hvap + D) / 2 - EA
The same result as shown in the diagram

Example 1.

Calculate the lattice energy of NaCl.

Solution
To solve a problem like this one, you have to know what data are required, and where to find them. A handbook is a good source, but you will have to look them up in several tables. To solve this problem requires the following data:

Hsub of Na = 108 kJ/mol
D of Cl2 = 244
IP of Na(g) = 496
EA of Cl(g) = -349
Hf of NaCl = -411
The Born-Haber cycle to evaluate Elattice is shown below:

     -----------Na+ + Cl(g)--------
         ­                       |
         |                       |-349
         |496+244/2              ¯
         |                 Na+(g) + Cl-(g)
         |                       |
   Na(g) + 0.5Cl2(g)             |
         ­                       |
         |108                    |
         |                       |Ecryst= -788 
   Na(s) + 0.5Cl2(g)             |
         |                       |
         |-411                   |
         ¯                       ¯
     -------------- NaCl(s) --------------
Ecryst = -411-(108+496+244/2)-(-349) kJ/mol
    = -788 kJ/mol

U = - Ecryst
    = 788 kJ/mol   (lattice energy)

Discussion
The value calculated for U depends on the data used. Data from various sources differ slightly, and so is the result. The lattice energies for NaCl most often quoted in other texts is about 765 kJ/mol.

Compare with the method shown below

Na(s) + 0.5 Cl2(l) ® NaCl(s) - 411 Hf
Na(g) ® Na(s)
- 108 -Hsub
Na+(g) + e ® Na(g) - 496 -IP
Cl(g) ® 0.5 Cl2(g) - 0.5 * 244 -0.5*D
Cl-(g) ® Cl(g) + 2 e 349 -EA
Add all the above equations leading to
Na+(g) + Cl-(g) ® NaCl(s) -788 kJ/mol = Ecryst

There is a another method based on principle of physics to evaluate the lattice energy, and some examples are given in the discussion enthalpy of hydration and lattice energy.

Born-Haber cycle enables us to calculate lattice energies of various compounds. For salts containing polyatomic ions, the Born-Haber cycle is not as useful. Some other means have to be used to evaluate the lattice energy or energy of crystallization.

Comparison of Lattice Energies (U in kJ/mol) of Some Salts
SolidUSolidUSolidU SolidU
LiF 1036LiCl 853LiBr 807LiI757
NaF 923NaCl 786 NaBr747NaI704
KF 821KCl 715KBr682KI649
MgF22957MgCl22526 MgBr2 2440 MgI22327
The lattice energies of some salts are given in the Table on the right here.

Among the mono-valent salts, the lattice energy decrease when the sizes of the ions increase.

Comparing the lattice energy for salts with one divalent ions leads to the same conclusion. The lattice energy decrease when the sizes of the ions increase.

The lattice energy of salts involving a divalent ion are much higher than those of monovalent salts, because much more energy is required to separate these ions.

Confidence Building Questions

Quiz 8.

©cchieh@uwaterloo.ca