Calculate the lattice energy of TiO2, Al2O3, and SiO2 using the Born-Haber cycle and Born-Lande relation. Comparing the lattic energy values of these two methods, comment on the type of bonding of these materials with the help of fraction of ionic character expression: 1-exp(-0.25(Xm-Xx)^2). Use the data given below for the calculations. Assume reasonable values for any missing data.
Material: Titanium/TiO2 Heat of sublimation or dissociation of pure element, kJ/mol: 421
Heat of oxide formation, kJ/mol: -944.7 Madelung constant of oxide structures: 2.408 (rutile)
Material: Aluminum/Al2O3 Heat of sublimation or dissociation of pure element, kJ/mol: 293.4
Heat of oxide formation, kJ/mol: -1669.8 Madelung constant of oxide structures: 4.1719 (corundum)
Material: Silicon/SiO2 Heat of sublimation or dissociation of pure element, kJ/mol: 384.22
Heat of oxide formation, kJ/mol: -910.86 Madelung constant of oxide structures: 2.2 (beta-quartz)
Material: Oxygen Heat of sublimation or dissociation of pure element, kJ/mol: 498
Born Exponent Principal Quantum Number of Outermost Electrons of Ion Examples
5 1 H-, Li+
7 3 F-, Na+
9 3 Cl-, K+, Zn2+, Ga3+
10 4 Br-, Rb+, Cd2+, In3+
12 5 I-, Cs+, Au+, Tl3+
The Correct Answer and Explanation is :
Let’s break down the calculation and approach for the lattice energy of TiO2, Al2O3, and SiO2 using both the Born-Haber cycle and Born-Lande relation. We will also determine the ionic character of each compound using the given formula.
Step 1: Using the Born-Lande Equation
The Born-Lande equation is:
[
U = \frac{N_A Z_+ Z_- e^2}{4 \pi \varepsilon_0 r_0} \left( 1 – \frac{1}{n} \right)
]
Where:
- ( U ) is the lattice energy.
- ( N_A ) is Avogadro’s number (6.022 × 10²³ mol⁻¹).
- ( Z_+ ) and ( Z_- ) are the charges on the cation and anion.
- ( e ) is the charge of the electron (1.602 × 10⁻¹⁹ C).
- ( \varepsilon_0 ) is the permittivity of free space (8.854 × 10⁻¹² C²/N·m²).
- ( r_0 ) is the distance between ions in the lattice.
- ( n ) is the Born exponent, which corresponds to the ion type (given in the table).
For TiO2, Al2O3, and SiO2:
- Ti²⁺, Al³⁺, and Si⁴⁺ are the cations.
- O²⁻ is the common anion.
- ( n ) values are taken from the given table based on ion pairs.
Step 2: Born-Haber Cycle for Each Material
The Born-Haber cycle for calculating lattice energy is:
[
U = \text{Sublimation energy} + \text{Ionization energy} + \text{Electron affinity} + \text{Heat of formation}
]
For each material, we sum the given enthalpy values:
- Sublimation energies for the metals are given.
- Heat of formation for each metal oxide is given.
- Ionization energies for Ti, Al, and Si are assumed to follow their general trends, based on periodicity.
- Electron affinity of oxygen is also assumed, typically around -1.46 eV (but needs to be converted).
Step 3: Fraction of Ionic Character Calculation
The ionic character is calculated using:
[
f_{\text{ionic}} = 1 – \exp \left( -0.25 \left( X_m – X_x \right)^2 \right)
]
Where:
- ( X_m ) is the electronegativity of the metal.
- ( X_x ) is the electronegativity of the anion (oxygen in this case).
The electronegativity values for Ti, Al, Si, and O are:
- Ti: 1.54
- Al: 1.61
- Si: 1.90
- O: 3.44
Now let’s calculate the lattice energies for these compounds, calculate the ionic character, and compare their bonding nature.
I’ll generate the calculations and comparisons.
The table shows the lattice energies for TiO2, Al2O3, and SiO2 calculated using the Born-Lande equation, along with their ionic character values based on the electronegativity difference between the metal cation and oxygen.
Explanation:
- Lattice Energy: The lattice energy represents the strength of the ionic bonds in the crystal. As expected, SiO2 has the highest lattice energy due to the large charge on Si⁴⁺ and the close packing of oxygen ions. Al2O3 has a relatively higher lattice energy compared to TiO2, reflecting the stronger bonding due to the higher charge on Al³⁺.
- Ionic Character: The ionic character quantifies how ionic or covalent the bonding is. A higher ionic character indicates more ionic bonding. TiO2 has the highest ionic character (about 59%), indicating that it has relatively stronger ionic bonding compared to Al2O3 and SiO2, which are more covalent.
Conclusion:
- TiO2 has the most ionic bonding, reflected in its high ionic character (59%). The lattice energy is also moderate among the three materials, suggesting strong ionic interactions in the structure.
- Al2O3 has slightly lower ionic character and higher lattice energy than TiO2, indicating that its structure involves both ionic and some covalent bonding.
- SiO2, with the lowest ionic character and highest lattice energy, suggests a more covalent character in bonding, although it still retains some ionic interaction due to the Si⁴⁺ and O²⁻ pairing.
These results help us understand the bonding nature in these materials, where TiO2 is mostly ionic, Al2O3 shows a mixed ionic-covalent nature, and SiO2 is more covalent.