Estimate the sum of the first three electron affinities for nitrogen using data from your textbook and the following: aluminum nitride crystallizes in the wurtzite structure and has a heat of formation of

1. Estimate the sum of the first three electron affinities for nitrogen using data from your textbook and the following: aluminum nitride crystallizes in the wurtzite structure and has a heat of formation of –318.0 kJ/mol; the sublimation energy of aluminum is 330.0 kJ/mol.

Set up a Born-Haber cycle for aluminum nitride:

Al(s) + ½N₂(g)→ AlN(s) ΔH_f^o = –318.0 kJ

Al(s) →Al(g) S = 330.0 kJ

½N₂(g) →N(g) ½BDE = ½(941.66) = 470.8 kJ (Table 2.8)

Al(g) →Al³⁺(g) + 3e^– IP₁ + IP₂ + IP₃ = (5.986 + 18.828 + 28.447)×96.4869
= 5139.0 kJ (Table 1.7)

N(g) ) + 3e^– →N^3–(g) –(EA₁ + EA₂ + EA₃)

Al³⁺(g) + N^3–(g) →AlN(s) –E_lat

–ΔH_f^o + S + ½BDE + (IP₁ + IP₂ + IP₃) – (EA₁ + EA₂ + EA₃) – E_lat = 0

(EA₁ + EA₂ +EA₃) = –ΔH_f^o + S + ½BDE + (IP₁ + IP₂ +IP₃) – E_lat

E_lat must be estimated using the Born-Landé equation. For the wurtzite structure M = 1.64132 (Table 5.12). The Born exponent can be estimated using electron configurations: both Al³⁺ and N^3– have the [Ne] configuration so n = 7. The ions have CN = 4 in the wurtzite structure so the ionic radii are r₊ = 53 pm (Table 5.8) and r_– ~ 132 pm (Table 5.10 for CN = 6, CN = 4 is unavailable), so r_o ~ 53 + 132 = 185 pm.

E_lat = 9506800 J/mol = 9506.8 kJ/mol

Now, substituting all the known values:

(EA₁ + EA₂ +EA₃) = –(–318.0) + (330.0) + (470.8) + (5139.0) – (9506.8) = –3249.0 kJ

This is a reasonable value; EA₁ for N is –7 kJ/mol and adding subsequent electrons to a negative ion should be increasingly disfavored thermodynamically.

Al(s) + ½N₂(g)→ AlN(s)	ΔH_f^o = –318.0 kJ
Al(s) →Al(g)	S = 330.0 kJ
½N₂(g) →N(g)	½BDE = ½(941.66) = 470.8 kJ (Table 2.8)
Al(g) →Al³⁺(g) + 3e^–	IP₁ + IP₂ + IP₃ = (5.986 + 18.828 + 28.447)×96.4869 = 5139.0 kJ (Table 1.7)
N(g) ) + 3e^– →N^3–(g)	–(EA₁ + EA₂ + EA₃)
Al³⁺(g) + N^3–(g) →AlN(s)	–E_lat