Answer

a. Estimate the lattice energy using the Born-Landé equation.

Both ions in CsCl are 8-coordinate. Since an anionic radius for Cl^– is not give, the 6-coordinate value must be used. Thus, r_o = r₊ + r_– = 174 + 181 = 355 pm

The Born exponent, n, is found from the compressibility: n = 1 + (4πε_o)18r_o⁴/Me²κ = 1 + (1.113×10^–10 J^–1C²m^–1)18(3.55×10^–10 m)/ (1.763)(1.602×10^–19 C)²(5.55×10^–11 m²N^–1) = 13.7

E_lat = –(6.022×10²³)(1.763)(+1)(–1)(1.602×10^–19)²/ (1.113×10^–10)(3.55×10^–10)[1 – 1/13.7] = 639 kJ/mole

b. Estimate the lattice energy using the Born-Mayer equation using d* = 34.5 pm.

E_lat = –(6.022×10²³)(1.763)(+1)(–1)(1.602×10^–19)²/ (1.113×10^–10)(3.55×10^–10)[1 – 34.5/355] = 623 kJ/mole

c. Estimate the lattice energy using the Kapustinskii equation.

E_lat = –1202(2)(+1)(–1)(1 – 0.345/3.55)/3.55 = 611 kJ/mole

d. Estimate the lattice energy using a Born-Haber cycle.

Cs(s) + ½ Cl₂(g) → CsCl(s) ΔH_f° = –433 kJ/mole

Cs(s) → Cs(g) S = 79 kJ/mole

Cs(g) → Cs⁺(g) + e^– IE = 375 kJ/mole

½ Cl₂(g) → Cl(g) ½BDE = ½×242 = 121 kJ/mole

Cl(g) + e^– → Cl^–(g) EA = 349 kJ/mole

Cs⁺(g) + Cl^–(g) → CsCl(s) E_lat

E_lat = S + IE + ½BDE – EA – ΔH_f° = 79 + 375 + 121 – 349 –(–433) = 659 kJ/mole

e. Compare the different determinations of the lattice energy. Which is most reliable? Why?

The Born-Haber cycle should give the most reliable value for the lattice energy since it the experimental measurement. The theoretical models are all lower than the experimental value (by 3 - 7.3 %) because they do not include any contribution accounting for covalency or van der Waal's interactions.

f. Do the ionic radii given in the table below accurately account for the lattice parameter? Why or why not?

The ionic radii give a Cs-Cl distance of 355 pm. The body diagonal of the cubic structure is twice this value, 710 pm. Using the geometry formula given 710 = √3a so that a = 409.9 pm. This is reasonably close to the experimental value of 412.3 pm, only 0.6 % error. This implies that the anionic radius for Cl^– for CN = 8 is only slightly larger than the CN = 6 value, consistent with the change observed for Cs⁺. If r_– (CN = 8) = 183 pm, the lattice parameter is matched exactly.

g. Is this a closest packed structure? Why or why not?

In a closest packed structure the holes are either T_d (CN = 4) or O_h (CN = 6). Since the CN = 8 for both ions, this cannot be a closest packed structure.

Answer

a. [Cr(NH₃)₆]³⁺

Point Group: O_h

LFSE: this is a d³ system so LFSE = 12Dq

Unpaired spins: 3

b. [Fe(H₂O)₄en]³⁺

Point Group: en is a chelating ligand so it must bond to adjacent sites giving a ring that is nonplanar, giving C₂

LFSE: this is a d⁵ system that has 4 weak ligands and one strong ligand that chelates, netting a weak field so LFSE = 0Dq

Unpaired spins: 5 (high spin)

c. [Fe(CN)₄en]^–

Point Group: C₂

LFSE: this is also d⁵ but all of the ligands are strong so LFSE = 20Dq – 2P

Unpaired spins: 1

d. fac-[CoF₃Cl₃]^4– (fac indicates that three ligands occupy the face of the octahedron)

Point Group: C_3v

LFSE: this is a d⁷ system in a weak field so LFSE = 8Dq

Unpaired spins: 3 (high spin)

e. fac-[Co(NH₃)₃(CN)₃]^–

Point Group: C_3v

LFSE: this is a d⁷ system in a strong field so LFSE = 18Dq – P

Unpaired spins: 1