Lattice Energy and Born-Haber Cycles: Comprehensive Review

Lattice Energy and Enthalpy Change of Atomisation

Enthalpy change ($\Delta H$) is defined as the amount of heat energy transferred during a chemical reaction at a constant pressure.

Lattice energy ($\Delta H_{latt}$) is the enthalpy change when 1 mole of an ionic compound is formed from its gaseous ions under standard conditions. This process is always exothermic as energy is released when ions combine to form an ionic solid lattice, meaning the enthalpy change will always have a large negative value. This large negative value indicates that the ionic compound is significantly more stable than its constituent gaseous ions due to the presence of strong electrostatic forces of attraction between oppositely charged ions in the solid lattice. In the gas phase, however, there are no electrostatic forces between the ions, making them less stable. The more exothermic (more negative) the lattice energy value, the stronger the ionic bonds within the lattice are.

Lattice energy cannot be determined directly by a single experiment. Instead, multiple experimental values and energy cycles are used. Examples of lattice energy equations include:

1. Magnesium oxide: $Mg^{2+}(g) + O^{2-}(g) \rightarrow MgO(s)$

2. Lithium chloride: $Li^{+}(g) + Cl^{-}(g) \rightarrow LiCl(s)$

3. Magnesium chloride: $Mg^{2+}(g) + 2Cl^{-}(g) \rightarrow MgCl_2(s)$

The standard enthalpy change of atomisation ($\Delta H_{at}$) is the enthalpy change when 1 mole of gaseous atoms is formed from its element under standard conditions. This process is always endothermic, requiring energy to break the bonds between atoms in the element to produce gaseous atoms. Consequently, the value of $\Delta H_{at}$ is always positive. Examples include:

1. Sodium: $Na(s) \rightarrow Na(g)$ where $\Delta H_{at} = +107\,kJ\,mol^{-1}$.

2. Potassium: $K(s) \rightarrow K(g)$

3. Mercury: In its elemental form, mercury is a liquid ($Hg(l)$), so the equation is $Hg(l) \rightarrow Hg(g)$.

Standard conditions for these processes are defined as a temperature of $298\,K$ and a pressure of $101\,kPa$. It is crucial to use correct state symbols throughout these calculations.

Electron Affinity and Trends of Group 16 and 17 Elements

The first electron affinity ($EA_1$) is the enthalpy change when 1 mole of electrons is added to 1 mole of gaseous atoms to form 1 mole of gaseous ions, each with a single negative charge, under standard conditions. This is represented as: $X(g) + e^{-} \rightarrow X^{-}(g)$ First electron affinity is usually exothermic (negative value) as energy is released. Elements can also undergo successive electron affinities. The second ($EA_2$) and third ($EA_3$) electron affinities involve adding an electron to an already negative ion, such as: $X^{-}(g) + e^{-} \rightarrow X^{2-}(g)$ $X^{2-}(g) + e^{-} \rightarrow X^{3-}(g)$ These successive affinities are endothermic (positive values) because energy is required to overcome the repulsive forces between the incoming electron and the negatively charged ion.

Electron affinity depends on how strongly the nucleus attracts an incoming electron, which is determined by three factors:

1. Nuclear Charge: A higher nuclear charge exerts a stronger pull, leading to a more exothermic electron affinity.

2. Distance (Atomic Radius): A larger radius reduces the attraction for the incoming electron, making the affinity less exothermic.

3. Shielding: More inner electron shells increase shielding, weakening the attraction and leading to a less exothermic affinity.

In Groups 16 and 17, electron affinities generally become less exothermic as you move down the group because the outermost electrons are farther from the nucleus and shielding increases. However, fluorine is an exception; it has a very small atomic radius creating high electron density, which causes repulsion between the incoming electron and existing electrons. This makes fluorine's first electron affinity ($-328\,kJ\,mol^{-1}$) less exothermic than chlorine's ($-345\,kJ\,mol^{-1}$).

Electron Affinity Data Values ($kJ\,mol^{-1}$):

• Group 16: O ($-141$), S ($-200$), Se ($-195$), Te ($-190$)

• Group 17: F ($-328$), Cl ($-345$), Br ($-325$), I ($-295$) Note that chlorine has a higher nuclear charge than sulfur, resulting in a stronger attraction and a more exothermic $EA_1$.

Constructing Born-Haber Cycles

A Born-Haber cycle is an application of Hess's Law for ionic compounds, allowing the calculation of lattice enthalpy which cannot be measured directly. The diagram represents energy increasing upwards. The cycle includes elements in their standard states as a starting point.

For sodium chloride ($NaCl$), the steps are:

1. Start with elements in standard states: $Na(s) + \frac{1}{2}Cl_2(g)$.

2. Create gaseous atoms (Atomisation):

   • $Na(s) \rightarrow Na(g)$ where $\Delta H_{at} = +108\,kJ\,mol^{-1}$

   • $\frac{1}{2}Cl_2(g) \rightarrow Cl(g)$ where $\Delta H_{at} = +121\,kJ\,mol^{-1}$

3. Create gaseous ions (Ionisation and Electron Affinity):

   • Sodium ionisation: $Na(g) \rightarrow Na^{+}(g) + e^{-}$ ($IE_1 = +500\,kJ\,mol^{-1}$)

   • Chlorine electron affinity: $Cl(g) + e^{-} \rightarrow Cl^{-}(g)$ ($EA_1 = -364\,kJ\,mol^{-1}$)

4. Complete the cycle:

   • Formation of solid: $Na(s) + \frac{1}{2}Cl_2(g) \rightarrow NaCl(s)$ ($\Delta H_{f} = -411\,kJ\,mol^{-1}$)

   • Lattice formation: $Na^{+}(g) + Cl^{-}(g) \rightarrow NaCl(s)$ ($\Delta H_{latt}$)

For magnesium oxide ($MgO$), the cycle is more complex involving two ionisation energies for Mg and two electron affinities for O. The first electron affinity for oxygen is exothermic, but the second is endothermic.

Calculations using Born-Haber Cycles

Using Hess's Law, the lattice energy can be calculated by rearranging the energy cycle. The general equation is: $\Delta H_{latt} = \Delta H_{f} - \Delta H_{1}$ where $\Delta H_{1}$ represents the sum of the enthalpy changes required to convert elements in their standard states to gaseous ions (atomisation, ionisation energy, and electron affinity).

Worked Example: Calculating $\Delta H_{latt}$ for Potassium Chloride ($KCl$):

• $\Delta H_{at}$ (K) = $+90\,kJ\,mol^{-1}$

• $\Delta H_{at}$ (Cl) = $+122\,kJ\,mol^{-1}$

• $IE_1$ (K) = $+418\,kJ\,mol^{-1}$

• $EA_1$ (Cl) = $-349\,kJ\,mol^{-1}$

• $\Delta H_{f}$ ($KCl$) = $-437\,kJ\,mol^{-1}$ $\Delta H_{latt} = (-437) - [(+90) + (+122) + (+418) + (-349)] = -718\,kJ\,mol^{-1}$.

When calculating values for compounds like $MgCl_2$, the electron affinity of chlorine must be doubled because there are two moles of chlorine atoms being converted into two moles of ions. For magnesium oxide, you must include both the first and second ionisation energies for magnesium and both the first and second electron affinities for oxygen.

Factors Affecting Lattice Energy

Two key factors affect lattice energy: ionic charge and ionic radius.

Ionic Radius: Lattice energy becomes less exothermic (less negative) as ionic radius increases. In larger ions, the charge is spread over a greater volume, leading to lower charge density. Furthermore, the distance between the centers of the ions increases. This weakens the electrostatic attraction, resulting in less energy release during lattice formation. For example, since $Cs^{+}$ is larger than $K^{+}$, the lattice energy of $CsF$ is less exothermic than that of $KF$.

Ionic Charge: Lattice energy becomes more exothermic (more negative) as ionic charge increases. Higher charges lead to higher charge density and much stronger electrostatic attractions between oppositely charged ions, releasing more energy upon formation. For example, $CaO$ contains $Ca^{2+}$ and $O^{2-}$ while $KCl$ contains $K^{+}$ and $Cl^{-}$. Because the ions in $CaO$ have higher charges (and are smaller), the lattice energy of $CaO$ is significantly more exothermic than that of $KCl$.