2. a) Born Haber Cycle Examples
The heat of formation of sodium chloride (ΔHf0) from the sodium metal and chlorine gas can be experimentally measured.
Na (s) + 12 Cl2 (g) → NaCl(s) ΔHf0 = -411kJ/mol
The formation of ionic solid sodium chloride form solid sodium metal and gaseous chlorine is not a single step process but goes through several processes. Heat changes of all the processes except the lattice energy can be experimentally measured.
The processes or steps in the formation of sodium chloride are-
1. Solid sodium atom sublimes to gaseous atom by absorbing heat energy (∆Hsub).
Na (s) → Na (g),
Sublimation energy ΔHsub = + 107kJ/mol
2. Gaseous sodium atom absorbs the ionization energy to release one electron and forms gaseous sodium ion.
Na(g) → Na+(g) + 1e– ,
Ionization energy ∆HIE = +502kJ/mol
3. Diatomic gaseous chlorine breaks into two individual atoms by absorbing bond energy, such that each chlorine atom absorbs half of the bond energy of chlorine molecule.
Cl2(g) → 2Cl(g) 1/2
Bond dissociation energy of chlorine = 1/2* ∆Hdiss= 1/2* 242 = +121kJ/mol
4. Chlorine atom accepts an electron to form chloride ion and releases energy equivalent to electron affinity.
Cl(g) + 1e– → Cl– (g)
Electron affinity = ∆HEA = -355kJ/mol
5. Gaseous sodium ion and gaseous chloride ion combine to form solid sodium chloride molecule and releases energy equivalent to lattice energy.
Na+(g) + Cl–(g) → Na+ Cl– (s)
Lattice energy = ∆HLE = U = ?
Summation of enthalpy of all the processes from step 1 to step 5) give the net enthalpy of formation of solid crystalline sodium chloride from sodium and chlorine in their standard conditions of solid and gas respectively. This should be equal to the experimentally measured enthalpy of formation of solid sodium chloride.
The enthalpies are represented as a cycle in the figure.

b. I)
b. i) Relationship between dipole moment & bond angle is :-
μ2=m12+m22+2m1m2Cos
θ 
µ = resultant dipole moment
m1 & m2 are dipole moments
µ2 = (1.2)2 + (1.8)2 + 2 x 1.2 x 1.8 x Cos 1300
µ2 = 1.44 + 3.24 + 4.32 x – 0.6427
µ2 = 4.68 – 2.7764
µ2 = 1.9035
µ = 1.3796 Ans.
ii)

Interaction energy between two rotating dipole is :-
E2=
2P2/4πε°r3

E2 = 9 x 109x 2x 1.5/(10-10 )3
E2 = 27 x 109/10-30
E2 = 27 x 1039
E2 = 2.7 x 1037 KJ/mol
2. a) Use one example to describe Born-Haber cycle (Not the example on your slides). b)...