Question

The correct increasing order of the lattice energy of alkali metal carbonates is:
A.${\text{C}}{{\text{s}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} < {\text{R}}{{\text{b}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} < {{\text{K}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} < {\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} < {\text{L}}{{\text{i}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$
B.\[{\text{R}}{{\text{b}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} < {\text{C}}{{\text{s}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} < {{\text{K}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} < {\text{L}}{{\text{i}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} < {\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}\]
C.\[{\text{L}}{{\text{i}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} < {\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} < {{\text{K}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} < {\text{R}}{{\text{b}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} < {\text{C}}{{\text{s}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}\]
D.None of these

Answer 1

Hint: Lattice energy is the energy which is required to separate one mole of an ionic solid into gaseous ions. The lattice energy can be determined by the Born-Haber cycle. The lattice energy is inversely proportional to the distance between the constituent ions of the ionic compound.

Complete step by step answer:
The two factors which generally affect lattice energy of an ionic compound are the magnitude of the charge associated with the ions and the distance between the ions. In other words, the lattice energy of an ionic compound is directly proportional to the strength of the ionic bond.The individual ions in an ionic lattice are attracted towards each other because of the electrostatic forces of attraction between them. The strength of this electrostatic force of attraction is directly proportional to the magnitude of the charge held by the constituent ions. But this electrostatic force decreases with increase in the distance between the ions in a lattice. More the distance between the ions in the lattice, weaker is the electrostatic force of attraction holding them together and so, lower will be the lattice energy. The carbonate ion is a large sized anion. The sizes of the alkali metal cations with respect to carbonate anion are insignificant. The size of the cations increases from lithium to caesium down the group. Thus, the distance between the carbonate anion and the alkali metal cation decreases from lithium to caesium down the group. Thus, the stability of the ionic compounds will decrease down the group. Hence, the lattice energy of the carbonates of alkali metals also decreases down the group.
Hence, among the alkali metal carbonates, the lattice energy order will be ${\text{C}}{{\text{s}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} < {\text{R}}{{\text{b}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} < {{\text{K}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} < {\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} < {\text{L}}{{\text{i}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ and

so option A is correct.

Note: Some of the important consequences of the lattice energies are:
a)Greater lattice energy means higher stability of the ionic compound.
b)It also affects the solubility of the ionic compounds. Since solubility is inversely proportional to stability which in turn is directly proportional to lattice energy, we can say that the solubility of the ionic compounds is inversely proportional to their lattice energies.
c)Thus, greater the lattice enthalpy, the lesser the solubility.