Answer

1. For CaF₂, calculate the lattice energy using the Born-Landé equation, the Born-Mayer equation (using d* = 34.5 pm), the Kapustinskii equation, and using a Born-Haber cycle. Cite your source for each datum used in these calculations.

Born-Landé equation: E_lat = –N_oMZ⁺Z^–e²(1 – 1/n) /4πε_or_o

N_o = 6.022×10²³ mol^–1

M = 2.519 (Weller, Overton, Rourke, Armstrong, 7^th ed., Table 4.8)

Z⁺ = +2

Z^– = –1

e = 1.602×10^–10 C

4πε_o = 1.113×10^–10 J^–1C²m^–1

r_o = r₊ + r_– = 112 + 131 = 243 pm

r₊ = 112 pm (Ca²⁺ is eight coordinate in the fluorite structure) (Weller, Overton, Rourke, Armstrong, 7^th ed., Table 1.4)

r_– = 131 pm (F^– is four coordinate in the fluorite structure) (Weller, Overton, Rourke, Armstrong, 7^th ed., Table 1.4)

n = 9.8

κ = 1.23×10^–11 N^–1m² (class notes)

n = (4πε_o)18r_o⁴/Me²κ + 1 = (1.113×10^–1018(2.43×10^–10)⁴/(2.519) (1.602×10^–19)²(1.23×10^–11) + 1 = 9.8

E_lat(B-L) = –(6.022×10²³)(2.519)(+2)(–1) (1.602×10^–19)²(1 – 1/9.8)/(1.113×10¹⁰)(2.43×10^–10) = 2585 kJ/mol

 

Born-Mayer; equation: E_lat = –N_oMZ⁺Z^–e²(1 – 34.5/r_o) /4πε_or_o

N_o = 6.022×10²³ mol^–1

M = 2.519 (Weller, Overton, Rourke, Armstrong, 7^th ed., Table 4.8)

Z⁺ = +2

Z^– = –1

e = 1.602×10^–10 C

4πε_o = 1.113×10^–10 J^–1C²m^–1

r_o = r₊ + r_– = 112 + 131 = 243 pm

r₊ = 112 pm (Ca²⁺ is eight coordinate in the fluorite structure) (Weller, Overton, Rourke, Armstrong, 7^th ed., Table 1.4)

r_– = 131 pm (F^– is four coordinate in the fluorite structure) (Weller, Overton, Rourke, Armstrong, 7^th ed., Table 1.4)

E_lat(B-M) = –(6.022×10²³)(2.519)(+2)(–1) (1.602×10^–19)²(1 – 34.5/243)/(1.113×10¹⁰)(2.43×10^–10) = 2470 kJ/mol

 

Kapustinskii; equation: E_lat = –1202Z⁺Z^–ν(1 – 0.345/r_o) /r_o

N_o = 6.022×10²³ mol^–1

M = 2.519 (Weller, Overton, Rourke, Armstrong, 7^th ed., Table 4.8)

Z⁺ = +2

Z^– = –1

ν = 3

r_o = r₊ + r_– = 112 + 131 = 243 pm = 2.43Å

r₊ = 112 pm (Ca²⁺ is eight coordinate in the fluorite structure) (Weller, Overton, Rourke, Armstrong, 7^th ed., Table 1.4)

r_– = 131 pm (F^– is four coordinate in the fluorite structure) (Weller, Overton, Rourke, Armstrong, 7^th ed., Table 1.4)

E_lat(Kap) = –(1202)(+2)(–1)(3)(1.602×10^–19)² (1 – 0.345/2.43)/(2.43) = 2547 kJ/mol

 

Born-Haber cycle

Ca(s) + F₂(g) → CaF₂(s) ΔH_f° = –1214 kJ/mol (CRC Handbook)

Ca(s → Ca(g) S = 192 kJ/mol (Wikipedia)

Ca(g) → Ca⁺(g) + e^– IE₁ = 589 kJ/mol (Weller, Overton, Rourke, Armstrong, 7^th ed., Table 1.5)

Ca⁺(g) → Ca²⁺(g) + e^– IE₂ = 1145 kJ/mol (Weller, Overton, Rourke, Armstrong, 7^th ed., Table 1.5)

F₂(g) → 2 F(g) BDE = 155 kJ/mol (Weller, Overton, Rourke, Armstrong, 7^th ed., Table 2.7)

2 F(g) + 2 e^– → 2 F^–(g) 2 EA = 2(328) = 656 kJ/mol (Weller, Overton, Rourke, Armstrong, 7^th ed., Table 1.6)

E_lat = S + IE₁ + IE₂ + BDE – 2EA – ΔH_f°

E_lat = (192) + (589) + (1145) + (155) – (656) – (–1214) = 2638 kJ/mol

The textbook (Table 4.7) reports the lattice enthalpy to be 2597 kJ/mol. The theoretical values are all within 5 % of this value.

2. For each of the following transition metal complexes, find the LFSE in units of Dq, indicate if the complex is Jahn-Teller active, and give the point group of the complex.

a. [Fe(CN)₆]^4–

This is Fe²⁺, d⁶ in a strong field, i.e. t_2g⁶ so LFSE = 24Dq – 2P

t_2g⁶ is not Jahn-Teller active so there is no distortion

Since there is no distortion the point group is O_h

b. [Co(CN)₆]^3–

This is Co³⁺, d⁶ in a strong field, i.e. t_2g⁶ so LFSE = 24Dq – 2P

t_2g⁶ is not Jahn-Teller active so there is no distortion

Since there is no distortion the point group is O_h

c. [Ni(CN)₆]^4–

This is Ni²⁺, d⁸ in a strong field, i.e. t_2g⁶e_g² so LFSE = 12Dq

t_2g⁶e_g² is not Jahn-Teller active so there is no distortion

Since there is no distortion the point group is O_h