4. Depending on temperature, RbCl can exist in either the rock-salt or caesium c

Question

4. Depending on temperature, RbCl can exist in either the rock-salt or caesium chloride structure. a) What is the coordination number of the cation and anion in each of these structures? b) In which of these structures will Rb have the larger apparent radius?

5. Calculate the lattice enthalpy for magnesium bromide using the Born-Haber cycle method, using the provided table. Process Enthalpies, H /(kJ mol-1 ) Sublimation of Mg(s) +148 Ionization of Mg(g) +2187 to Mg2+(g) Vaporization of Br2(l) +31 Dissociation of Br2(g) +193 Electron gain by Br(g) -331 Formation of MgBr2(s) -524

Explanation / Answer

4) At high temperature RbCl exists in Cesium chloride structure and aat low temperature it has Na CL structure.

a) The coordination number of cation and anion are 8 in CsCl and 6 in NaCl.

b) Rb does have large apparent radius in CsCl structure ,as it is surrounded by 8 chloride ions and th eradius ratio is high in CsCl structure (> 0.732)

5) Using Born-Haber cycle we can calulate the lattice enthalpy or electron affinities of elements indirectly.

For MgBr2

Mg (s) + Br2(l) --------> MgBr2(s) delta H = -524 kJ

The same reaction can be made in several steps and principle of Born-Haber cycle is the sum of enthalpies of these reactons is equal to enthalpy of the above reaction (Hess law)

The steps are

i) Mg (s) -----> Mg(g) enthlapy = +148 (sublimation)

ii) Mg(g) -------> Mg+2(g) enthalpy = +2187 (ionizaion)

iii) Br2(l) ----------> Br2 (g) enthalpy = +31 (vaporization)

iv) Br2(g) --------> 2Br (g) enthalpy = +193 (dissociation)

v) 2Br(g) ----------> 2 Br-(g) enthalpy = 2 x(-331) (electron affinity)

vi) Mg+2(g) + 2Br- (g) -----> MgBr2(s) enthalpy = U ( lattice energy )

Sum of all these = +148+ 2187 +31 +193 +2(-331) + U

This equal to enthalpy of formationof MgBr2 = -524

Equating and solving for U ,we get lattice energy = -2421 kJ/mol

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