I. Lithium is the lightest metal and does not exist in pure form in nature due to its high reactivit — Inorganic Chemistry — Solid State Chemistry Question
Calculations of lattice energy of ionic compounds
I. Lithium is the lightest metal and does not exist in pure form in nature due to its high reactivity to water, moisture, oxygen... Lithium readily forms ion with a 1+ charge when reacting with nonmetals. Write down the following chemical reactions at room temperature:
Lithium reacts with water.
Model Answer
2 Li(s) + 2 H2O(l) → 2 Li+(aq) + 2 OH-(aq) + H2(g)
Lithium reacts with halogens, e.g. Cl2.
Model Answer
2 Li(s) + Cl2(g) → 2 LiCl(s)
Lithium reacts with dilute sulfuric acid and concentrated sulfuric acid.
Model Answer
With dilute sulfuric acid: 2 Li(s) + H2SO4(aq) → 2 Li+(aq) + SO4 2-(aq) + H2(g)
With concentrated acid: 2 Li(s) + 3 H2SO4(aq) → 2 LiHSO4(aq) + SO2(g) + 2 H2O(l)
II. The change in enthalpy of a particular reaction is the same whether it takes place in one step or in a series of steps (Hess’s law). Use the following data:
Sublimation enthalpy of Li(s), ∆SH = 159 kJ mol–1.
Ionization energy of Li(g), I = 5.40 eV.
Dissociation enthalpy of Cl2, ∆DH = 242 kJ mol–1.
Electron affinity of Cl(g), E = –3.84 eV.
Formation enthalpy of LiCl(s), ∆fH = – 402.3 kJ mol–1.
rLi+ = 0.62 Å; rCl- = 1.83 Å; NA = 6.02×1023 mol–1.
2.4 Establish the Born-Haber cycle for lithium chloride crystal.
Model Answer
To calculate Uo in accordance with Born-Haber cycle, the following cycle is constructed:
∆fH Li(s) + ½ Cl2(g) → LiCl(s)
∆sH ½ ∆DH U0
I Li(g) Li+(g) + Cl–(g) E
Calculate the lattice energy Uo (kJ mol-1) using the Born-Haber cycle.
Model Answer
Based on this cycle and Hess’s law, we have:
∆fH = ∆sH + 1/2 ∆DH + I + E + U0
or U0 = ∆fH – (∆sH + 1/2 ∆DH + I + E) a
After converting all the numerical data to the same unit, we have:
U0 = – 402.3 – 159 – 121 – (5.40 – 3.84) × 1.6 ⋅10-19 × 1 ⋅ 10-3 × 6.022 ⋅ 1023
U0 = – 832.56 kJ mol-1.
III. In practice, experimental data may be employed to calculate lattice energies in addition to the Born-Haber cycle. One of the semi empirical formulae to calculate the lattice energy Uo for an ionic compound, which was proposed by Kapustinskii, is as follows:
where: ν is the number of ions in the empirical formula of ionic compound,
r+ and r− are the radii of the cation and anion, respectively, in Å,
Z+ and Z− are cation and anion charges, respectively,
U0 is the lattice energy, in kcal mol-1.
2.6 Use the Kapustinskii empirical formula to calculate Uo (in kJ mol–1) of LiCl crystal, given that 1 cal = 4.184 J.
Model Answer
For LiCl crystal, we have:
U0 = – 287.2 (2 × 1 × 1 / (0.62 + 1.83)) (1 - 0.345 / (0.62 + 1.83)) = – 201.43 kcal mol-1
To conveniently compare the results, we convert the obtained result to SI units:
U0 = – 201.43 × 4.184 = – 842.78 kJ/mol
IV.
2.7 Based on the results of two calculation methods in sections II and III, choose the appropriate box:
- According to the Born-Haber cycle and Kapustinskii empirical formula for lithium chloride crystal structure, both methods are close to the experimental value.
- Only the calculated result of the Born-Haber cycle is close to the experimental value.
- Only the calculated result of the Kapustinskii empirical formula is close to the experimental value.
Data: Given the experimental value of lattice energy for LiCl is 849.04 kJ mol-1.
Model Answer
Correct answer:
According to the Born-Haber cycle and Kapustinskii empirical formula for lithium chloride crystal structure, both methods are close to the experimental value.
V. In the formation of LiCl crystal, it is found out that the radius of lithium cation is smaller than that of chloride anion. Thus, the lithium ions will occupy the octahedral holes among six surrounding chloride ions. Additionally, the body edge length of LiCl cubic unit cells is 5.14 Å. Assume that Li+ ions just fit into octahedral holes of the closest packed chloride anions.
2.8 Calculate the ionic radii for the Li+ and Cl- ions.
Model Answer
The geometry diagram for octahedral holes is shown below.
where, R and r are the radii of Cl– and Li+ ions, respectively.
Based on the diagram, we have:
cos 45° = 2R / (2R + 2r) = R / (R + r)
0.707 = R / (R + r) → R = 0.707 (R + r) → r = 0.414 R
The body edge length of the unit cell LiCl = 2 R + 2 r = 5.14 Å
2 R + 2 (0.414 R) = 5.14 Å → R = 1.82 Å (radius of Cl-)
2 (1.82 Å) + 2 r = 5.14 Å → r = 0.75 Å (radius of Li+)
Compare the calculated (theoretical) radii with the experimental radii given below, and choose the appropriate box:
- Both calculated radii of lithium and chloride ions are close to the experimental values.
- Only the calculated radius of lithium ion is close to the experimental value.
- Only the calculated radius of chloride ion is close to the experimental value.
The experimental radii of Li+ and Cl– are 0.62 Å and 1.83 Å, respectively.
Model Answer
Correct answer:
Only the calculated radius of chloride ion is close to the experimental value.