Question

can you please answer this??

(Explain the following about problem 9.31.

a. Why do you use the electron affinity of F twice, but two separate ionization energies for Mg?

b. Why is the lattice energy of this so much larger than the 788kJ/mol that you calculated for NaCl)

KNOWING THAT PROBLEM 9.31 is :

Use the following information to calculate the lattice energy of MgF2. This is actually Problem 9.31

Mg(s)  Mg(g) ΔH = 148 kJ

F2(g)  2 F(g) ΔH = 159 kJ

Mg(g)  Mg+(g) + e- ΔH = 738 kJ

Mg(g)  Mg2+ (g) + e- ΔH = 1450 kJ

F(g) + e-  F-(g) ΔH = -328 kJ

Mg(s) + F2(g)  MgF2(s) ΔH = -1123kJ

Answer 1

Solution : The overall reaction is a Mg(8) + F78) MgGz (8) DH=-1123 k3 /mol. @ since we are dealing with 2f atoms (f2), and the electron affinity of two 'f'atoms will be equal, as they are some compounds types of atoms. on the other hand, we can't use twice for mg, since, in this case we need to use two ichization energy of the same 'Mg otom, which can not be equal, - Mg(g) Mgt te Ifa mot Mg2+te- Ita I£27 Itz 16 Griven Alf (total) = -1123 kJ | mol To find out lattice energy (U), first we need to make Born haber cycle for this reaction, t Mgtq) + 2fG) The content of tg)+2Fg). Election affinity=-328 $ 1450Kg 7 2%E1° - Tlf = 736 M99) +269) Mghtg2+28 g) Matomizabon ) = 159 Kg Mgca) +f268) Tatoub, = 148 kJ Mg(s) + F2 (g) Alf mg f266)

Thus, we have AH47 148 + 9+138 +140-32880 = -1123 kJ/mole = 1839 tu or o=11839 + 1123) kJ / mole =F-2962 kJ/moll This is much larger then the lattice energy of wall 1-788 kat mol), This is because, Booth-Landeequeting Da section 1 orzechamento (1-) Here both terms (higher charge Mg2+) and smaller radium (Mg-f) favour high lattice energy.
note- in born lande equation the symbols have their usual meanings, and n= born exponent.