Question

Arrange the compounds according to the magnitudes of their lattice energies based on the relative ion charges and sizes.

RbI, LiCl, CaO, Na2O

Answer 1

We know, that the Lattice energy is nothing but the Coulomb's law working between two charged species. Hence, the Lattice energy can be provided by the following equation-

L.E. = k.(Q+.Q_)/r

where, Q+, Q- are the charges of cations and anions respectively and r is the distance between the two. Now, let us analyze the above compounds-

S. No.	Compound	Charges on Cation and Anion
1		+1,-1
2		+1,-1
3		+2,-2
4	ΝαΟ	+1,-2

Now, the following cases emerge-

1. The compounds with charges of '2' in cation or anion (Na2O) or in cation anion both (CaO) will have the highest Lattice energies as the L.E.is directly proportional to the product of charges of cation and anion.

Having said that,
L.E.(CaO) > L.E.(Na2O)
, because in CaO,
L.E. (Cao) « Q.Q_ = 2.2 = 4
and in Na2O,
L.E.(Na20) « Q+Q- = 1.2 = 2
.

2. Now, comes the remaining two compounds which will be compared on the basis of their size(covalent radii) as we know that, increase in the size of the ions will result in an increase of distance between the ions which in turn will decrease the L.E as is evident from the above equation.

As both the cation and anion of RbI is larger in size than that of LiCl, here,

L.E.LICI) > L.E.RbI)

Hence, the order is quite evident now,

.

L.E.(Cao) > L.E.(Na,0) > L.E. Lici) > L.E.RLI)