Quiz 8. Ionic Compounds

Skills to develop

Explain stability of ions in terms of electronic configuration.
Define lattice energy.
Point out the trend of lattice energies in a series of compounds.
Draw the Born Haber Cycle and calculate the energy that can not be measured
for a process.

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	N^3-	O^2-	F^-	Ne	Na⁺	Mg²⁺	Al³⁺
[Ar]	S^2-	Cl^-	Ar	K⁺	Ca²⁺
[Kr]	Se^2-	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.

Transition-metals lose n s electrons first, and then they may lose one or more (n-1) d electreons to form ions.
Transition metal ions are usually colored, due to the d-d transitions. The d-sub shell is split for ions in solutions.
Electronic configurations for transition metal ions.

Ions Electronic
Configuration Remark
Sc [Ar]4s² 3d¹ Metal
Sc²⁺ [Ar] 3d¹ Color 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

Ions	Electronic Configuration	Remark
Sc	[Ar]4s² 3d¹	Metal
Sc²⁺	[Ar] 3d¹	Color 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

Ionic radii of the same charge increase for a group as the atomic numbers increase.
Isoelectronic ions are ions with the same electronic configuration.
Across a period, the cations decrease in radius, even if they have the same charge. The higher the charge, the smaller the ion.
Sizes of anions are much larger than their isoelectronic cations.

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
SO₄^2- Sulfate
SO₃^2- Sulfite
PO₄^3- Phosphate
PO₃^3- 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, E_cryst. Since energy is released, the sign is negative for such a quantity. Thus, we have

a M^b+(g) + b X^a- (g) ® M_aX_b(s) E_cryst kJ mol^-1
Therefore, we have this relationship: E_cryst = - 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,

H_sub	Enthalpy of sublimation,
H_melt	Enthalpy of melting,
H_vap	Enthalpy of vaporization,
H_f	Enthalpy of formation,
U	Lattice energy,
E_cryst	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 H_sub ® 2 Na (g) --2 IP ® 2 Na⁺(g)

Br₂(l) --H_vap ® Br₂ (g) --D ® 2 Br(g) --2EA ® 2 Br^-(g)

One other reaction is required

2 Na + Br₂ ® 2NaBr, 2H_f

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:

2H_f - (2H_sub + H_vap + 2IP + D) = 2EA + 2E_cryst

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 H_f
2 Na(g) ® 2 Na(s) or 2 Na(s) ¬ 2 Na(g)	- 2 H_sub
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)	- H_vap
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 E_cryst

Thus, 2 H_f - 2H_sub - 2 IP - H_vap - D - 2 EA = 2 E_cryst
E_cryst = H_f - H_sub - IP - (H_vap + 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:

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

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

E_cryst = -411-(108+496+244/2)-(-349) kJ/mol
= -788 kJ/mol

U = - E_cryst
= 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	H_f
Na(g) ® Na(s)	- 108	-H_sub
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 = E_cryst

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
Solid U Solid U Solid U Solid U
LiF 1036 LiCl 853 LiBr 807 LiI 757
NaF 923 NaCl 786 NaBr 747 NaI 704
KF 821 KCl 715 KBr 682 KI 649
MgF₂ 2957 MgCl₂ 2526 MgBr₂ 2440 MgI₂ 2327
The lattice energies of some salts are given in the Table on the right here.

Comparison of Lattice Energies (U in kJ/mol) of Some Salts
Solid	U	Solid	U	Solid	U	Solid	U
LiF	1036	LiCl	853	LiBr	807	LiI	757
NaF	923	NaCl	786	NaBr	747	NaI	704
KF	821	KCl	715	KBr	682	KI	649
MgF₂	2957	MgCl₂	2526	MgBr₂	2440	MgI₂	2327

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

Which of the following is not an ionic compound?
NaCN, CaSO₄, LiCl, MgO, HCl, CO, NH₄NO₃

Discussion: Carbon monoxide is a gas, and C-O bond is predominant covalent. Some covalent character remain in the ionic compound
Give the symbol of the noble gas that has the same electronic configuration as the following ions.
S^2-, Cl^-, K⁺, Ca²⁺, Ga³⁺

Skill: Identify the electronic configuration of ions
The noble gas beside Cl is argon.
What ions have the same electronic configuration as Ne?
Choose the type of energy for the process Mg(s) -> Mg(l).
a Enthalpy of sublimation
b Enthalpy of melting
c Enthalpy of vaporization
d Enthalpy of formation
e Enthalpy of reaction

Discussion: Is the process endothermic or exothermic?
Choose the type of energy for the process
Mg(g) -> Mg⁺(g) + e^-.
a Enthalpy of formation
b Enthalpy of reaction
c Bond (dissociation) energy
d Ionization potential
e Electron affinity

Skill: Give appropriate names for various chemical processes and know if the process endothermic or exothermic?
The energy called electron affinity is the energy released for which one of the following steps?
a F₂(g) = 2 F(g)
b Cl₂ = 2 Cl
c Na(s) = Na⁺(s) + e^-
d F₂(g) + e^- = F₂^-
e Br(g) + e^- = Br^-(g)

Discussion: EA is the amount of energy released when a atom acquires an electron. Is the process exothermic?
Which one of the following is more than a single step for making up the Born-Haber cycle?
a Mg(s) -> Mg(l)
b Mg(g) -> Mg⁺(g) + e^-
c Br₂(l) -> 2 Br(g)
d Ca(s) + Cl(g) -> Ca(g) + Cl(g)
e Ca(s) + Cl₂(g) -> CaCl₂

Skill: Identify the chemical reactions involving these types of energy:
H_sub, H_melt, H_vap, H_f, H (enthalpy of reaction), D, IP, and EA
Note: Br₂(l) -> Br₂(g) -> 2Br(g)

Quiz 8.

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
SO₄^2-	Sulfate
SO₃^2-	Sulfite
PO₄^3-	Phosphate
PO₃^3-	Phosphite
ClO₄^-	Perchlorate
ClO₃^-	Chlorate
ClO₂^-	Chlorite
NO₃^-	Nitrate
NO₂^-	Nitrite