Lattice energy
- 1. Lattice energy JNC 310 : Structure & Chemical Crystallography Abhishek Rawat (S0750) 20 January 2020 N.C.U. - JNCASR
- 2. Outline Lattice energy & its properties Determination of lattice energy Lattice energy calculations Comparison b/w both approaches Q & A Conclusion
- 3. Where can lattice energy be valid? Covalent Compounds Ionic Compounds Lattice energies are associated with many interactions, as cations and anions pack together in an extended lattice. For covalent bonds, the bond dissociation energy is associated with the interaction of just two atoms. Metallic Compounds??
- 4. • Electrostatic attraction is nondirectional! – No direct anion–cation pair • Therefore, there is no ionic molecule – The chemical formula is an empirical formula, simply giving the ratio of ions based on charge balance Crystal lattice of ionic compounds © 2014 Pearson Education, Inc.
- 5. • I.E. of the metal is endothermic Na(s) → Na+(g) + 1 e ─ DH° = +496 kJ/mol • E.A. of the nonmetal is exothermic ½Cl2(g) + 1 e ─ → Cl─(g) DH° = −349 kJ/mol • I.E. (metal) > E.A. (nonmetal) • Formation of the ionic compound should be endothermic • But the heat of formation of most ionic compounds is exothermic and generally large. Why? Na(s) + ½Cl2(g) → NaCl(s) DH°f = −787 kJ/mol Energetics of ionic bond formation
- 6. • The extra stability that accompanies the formation of the crystal lattice is measured as the lattice energy • The lattice energy is the energy released when the solid crystal forms from separate ions in the gas state • Always exothermic • Hard to measure directly, but can be calculated from knowledge of other processes • Lattice energy ∝ charge ∝ 1/ distance b/w ions Lattice Energy Courses/lumenlearning.com
- 7. • Melting/boiling point (stronger bonds = higher melting point/boiling point) • Hardness (stronger bonds = harder crystals) • Odor (stronger bonds = weaker odor) • State at room temperature (stronger bonds are more likely to be solids) • Solubility?? Properties of Lattice Energy
- 8. Determining Lattice Energy Theoretical approach Experimental approach Born landé equation Born Haber cycle Calculation for Lattice energy
- 9. Born-Landé equation In 1918, Max Born & Alfred Landé proposed the formula for Lattice energy calculation on the basis of electrostatic potential of ionic lattice and repulsive potential energy terms 𝑼 = − 𝑵 𝑨 𝑨 𝒛 + 𝒛 − 𝒆 𝟐 𝟒𝝅𝝐 𝟎 𝒓 𝟎 (𝟏 − 𝟏 𝒏 ) • NA = Avogadro no. = 𝟔. 𝟎𝟐𝟑 × 𝟏𝟎 𝟐𝟑 • A = Madelung const. • z = Numeric charge of respective ion • e = electronic charge = 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 C • 𝝐 𝟎 = permittivity of free space (𝟒𝝅𝝐 𝟎 = 𝟏. 𝟏𝟏𝟐 × 𝟏𝟎−𝟏𝟎 C2/J.m) • r0 = distance to closest ion • n= Born exponent (approx. 5−12) Courses/lumenlearning.com
- 10. Considering NaCl structure For a Na+ ion Madelung constant & Born exponent Courses/lumenlearning.com
- 11. Electrostatic potential Epair = − 𝑧2 𝑒2 4𝜋𝜖0 𝑟 • z = magnitude of charge on each ion • e = electronic charge = 1.6 × 10−19 C • 𝜖0 = permittivity of free space (4𝜋𝜖0 = 1.112 × 10−10 C2/J.m) • r= distance between two oppositely charged ions EA = − 𝑧2 𝑒2 4𝜋𝜖0 𝑟 A • A = Madelung constant, related to geometry of the crystals. Repulsive potential ER = 𝐵 𝑟 𝑛 • B = constant, scaling the strength of the repulsive interaction • r= distance between two oppositely charged ions • n= Born exponent (approx. 5−12), expresses the steepness of the repulsive barrier Derivation of Born-Landé equation Courses/lumenlearning.com
- 12. Total energy, U (r) = EA + ER U (r) = − 𝑧2 𝑒2 4𝜋𝜖0 𝑟 𝐴 + 𝐵 𝑟 𝑛 Applying the minimization condition, 𝑑𝑈(𝑟) 𝑑𝑟 = 0, then r = r0 𝑑𝑈(𝑟) 𝑑𝑟 = 𝑧2 𝑒2 4𝜋𝜖0 𝑟0 2 𝐴 − 𝑛𝐵 𝑟0 𝑛+1 = 0 r0= ( 4𝜋𝜖0 𝑛𝐵 𝑧2 𝑒2 𝐴 ) 1 𝑛−1 𝐵 = 𝑧2 𝑒2 𝐴 4𝜋𝜖0 𝑛 𝑟0 𝑛−1 𝑈(𝑟0) = − 𝐴𝑧2 𝑒2 4𝜋𝜖0 𝑟0 (1 − 1 𝑛 ) Derivation of Born-Landé equation Courses/lumenlearning.com
- 13. Drawbacks of Born-Landé equation Calculations for lattice energy was based on Coulomb’s law which considers ions as point charges In deriving lattice energy the reduction of charges due to their interactions are not considered Courses/lumenlearning.com
- 14. Modification of Born-Landé equation Born-Mayer equation Mayer showed that e–r/ρ , where ρ is a constant dependent on the compressibility of the crystal, gives a better repulsion term than 1 𝑟 𝑛 𝑈 𝐵𝑀 = − 𝑁 𝐴 𝐴 𝑧 + 𝑧 − 𝑒2 4𝜋𝜖0 𝑟0 (1 − 𝜌 𝑟0 ) ρ = 30 pm works well for all alkali metal halides and other simple cases when r0 values are in pm. Courses/lumenlearning.com
- 15. Modification of Born-Landé equation In the absence of detailed structural data, Kapustinskii's equation can be used to estimate U 𝑈 𝐾 = 𝐾 𝑣𝑧 + 𝑧 − 𝑟++ 𝑟− (1− 𝑑 𝑟++ 𝑟−) K = 1.202 × 10−4 𝐽. 𝑚/𝑚𝑜𝑙 𝑑 = 3.45 × 10−11 𝑚 𝑣 = no. of ions in the emperical formula r = ionic radii of ions, in meter Kapustinskii’s equation Courses/lumenlearning.com
- 16. • Gives a good estimation. Real value differs in most cases by less than 5% • Used with ionic compounds containing polyatomic ions as a means of calculating their thermochemical radii, in which the ions are treated as spheres • Useful for complex ions like sulfate (SO4 2−) or phosphate (PO4 3−) Advantages of Kapustinskii’s equation Courses/lumenlearning.com
- 17. Experimentally determining Lattice Energy • The Born–Haber cycle is a hypothetical series of reactions that represents the formation of an ionic compound from its constituent elements • The reactions are chosen so that the change in enthalpy of each reaction is known except for the last one, which is the lattice energy NCERT – Chemistry Textbook (Part 1), Ch-6
- 18. Born–Haber Cycle • Use Hess’s law to add up enthalpy changes of other reactions to determine the lattice energy DH°f(salt) = DH°f(metal atoms, g) + DH°f(nonmetal atoms, g) + DH°f(cations, g) + DH°f(anions, g) + DH°(crystal lattice) DH°(crystal lattice) = lattice energy For metal atom(g) cation(g), DH°f = first ionization energy Add together all the ionization energies to get to the desired cation M2+ = 1st IE + 2nd IE For nonmetal atoms (g) anions (g), DH°f = electron affinity NCERT – Chemistry Textbook (Part 1), Ch-6
- 19. © 2014 Pearson Education, Inc. Born–Haber Cycle
- 20. According to Hess’s law, total energy change during the complete course of a chemical reaction is the same whether the reaction is made in one step or multi steps. ΔHf = Δ Hsub + 1 2 𝐷 + IE + EA + Δ H(crystal) Na (s) + 𝟏 𝟐 Cl2 (g) → NaCl (s) ΔHf = - 410.9 kJ Na (s) → Na (g) ΔHsub = 107.7 kJ Cl2 (g) → 2Cl (g) D = 243.4 kJ Na (g) → Na+ (g) IE = 496 kJ Cl (g) → Clˉ (g) EA = - 349 kJ Lattice Energy calculation for NaCl chem.libretexts.org
- 21. Mg (s) + F2 (g) → MgF2 (s) ΔHf = - 1096.5 kJ Mg (s) → Mg (g) ΔHsub = 146.4 kJ F2 (g) → 2F (g) D = 158.8 kJ Mg (g) → Mg+ (g) IE (1) = 737 kJ Mg+ (g) → Mg2+ (g) IE (2) = 1149 kJ F (g) → Fˉ (g) EA = - 328 kJ –1096.5 = 146.4 + 158.8 + 737 + 1149 + 2 × (–328) + Δ H(crystal) Δ H(crystal) = – 2631.7 kJ/mol Lattice Energy calculation for MgF2 chem.libretexts.org ΔHf = Δ Hsub + D + IE (1) + IE (2) + 2EA + Δ H(crystal)
- 22. Trends in Lattice Energy: Ion Charge • The force of attraction between oppositely charged particles is directly proportional to the product of the charges • Larger charge means the ions are more strongly attracted – Larger charge = stronger attraction – Stronger attraction = larger lattice energy • Of the two factors, ion charge is generally more important NCERT – Chemistry Textbook (Part 1), Ch-6
- 23. Lattice Energy versus Ion Size NCERT – Chemistry Textbook (Part 1), Ch-6
- 24. © 2014 Pearson Education, Inc. Comparison b/w experimental & theoretical calculated Lattice Energy values
- 25. Which compound will have the greatest lattice energy, MgS or LiF? • Magnitude of charge is the first thing we should look at – magnesium (2+ charge), sulfide (2- charge) – lithium (1+ charge), fluoride (1- charge) – MgS has the greater individual charges, so MgS has the greater lattice energy. Q & A
- 26. Which compound will have lower melting point, Na2S or BeO? • A lower melting point means we need to select the compound with the lower lattice energy. We check the charges first. – sodium (1+), sulfide (2-) – beryllium (2+), oxide (2-) – sodium sulfide has smaller individual charges, so it has the lower melting point. Q & A
- 27. Which compound has harder crystals, CaCl2 or MgCl2? • Harder crystals require a higher lattice energy. First, check the charges. – calcium (2+), chloride (1-) – magnesium (2+), chloride (1-) – the charges are the same, so we need to see which ions have fewer energy levels. Q & A
- 28. Which compound has harder crystals, CaCl2 or MgCl2? – the charges are the same, so we need to see which ions have fewer energy levels. – calcium (4), chloride (3) – magnesium (3), chloride (3) – fewer energy levels give MgCl2 the higher lattice energy and therefore, the harder crystals. Q & A
- 29. Which compound has the lower boiling point, AgNO3 or K2SO4? • Lower boiling point means a lower lattice energy. – silver (1+), nitrate (1-) – potassium (1+), sulfate (2-) – Charges indicate a lower lattice energy for AgNO3. Lower lattice energy means a lower boiling point. Q & A
- 30. Conclusion We have learnt • Lattice energy & its significance • Approaches to determine it • Modification of the theoretical approaches • Experimental calculation • Trends in lattice energy determination
- 31. Thank you
- 32. Assignment Q1. Will lattice energy concept be valid for metallic compounds? Why or Why not? Q2. Assume the interionic distance for NaCl2 to be the same as those of NaCl (r = 282 pm), and assume the structure to be of the fluorite type (M = 2.512). Evaluate the energy of crystallization, Ecryst .