Question

Select the correct order of the given properties:
This question has multiple correct options.
A. Mg${ C }_{ 2 }{ O }_{ 4 }$ > Ca${ C }_{ 2 }{ O }_{ 4 }$ > Sr${ C }_{ 2 }{ O }_{ 4 }$ > Ba${ C }_{ 2 }{ O }_{ 4 }$ : Solubility order
B. Be${ S }_{ 2 }{ O }_{ 3 }$ > Mg${ S }_{ 2 }{ O }_{ 3 }$ > Ca${ S }_{ 2 }{ O }_{ 3 }$ > Sr${ S }_{ 2 }{ O }_{ 3 }$ : Solubility order
C. K${ O }_{ 3 }$ > Rb${ O }_{ 3 }$ > Cs${ O }_{ 3 }$ : Thermal stability order
D. LiN${ O }_{ 3 }$ > NaN${ O }_{ 3 }$ > KN${ O }_{ 3 }$ > CsN${ O }_{ 3 }$ : Thermal stability order

Answer 1

Hint: For solubility, you need to compare the size of anion and cation and for thermal stability look at the polarizing power of the cation. Now try to answer the question accordingly.

Complete step by step answer:
We should know that solubility indicates the maximum amount of a substance that can be dissolved in a solvent at a given temperature. Such a solution is called saturated.
Let’s discuss its trends -

Case 1 - When the size of anion and cation is small we can say their size is comparable. Then on moving top to bottom in a group, there is an apparent decrease in hydration energy and lattice but the decrease in lattice energy is found to be more than that of hydration energy and hence solubility increases moving down the group.
Case 2 - When the size of the anion is large its hydration does not occur properly. Thus on moving down the group both lattice energy and hydration energy shows decrease but the decrease in hydration energy is found to be more than that of lattice energy. Hence solubility decreases.

In option A, this order is correct because the size of cations and ${ C }_{ 2 }{ O }_{ 4 }^{ 2- }$ is not comparable (anion is larger). Hence, as we discussed in case 2, solubility must decrease down the group.
Solubility order - Mg${ C }_{ 2 }{ O }_{ 4 }$ >Ca${ C }_{ 2 }{ O }_{ 4 }$ >Sr${ C }_{ 2 }{ O }_{ 4 }$ >Ba${ C }_{ 2 }{ O }_{ 4 }$
In option B, this order is correct because the size of cations and ${ S }_{ 2 }{ O }_{ 3 }^{ 2- }$ is not comparable (anion is larger). Hence, as we discussed in case 2, solubility must decrease down the group.
Solubility order - Be${ S }_{ 2 }{ O }_{ 3 }$ >Mg${ S }_{ 2 }{ O }_{ 3 }$ >Ca${ S }_{ 2 }{ O }_{ 3 }$ >Sr${ S }_{ 2 }{ O }_{ 3 }$
We should know that Thermal stability is defined as the ability of a fluid to resist breaking down under heat stress.
In all cases, for a particular set of Group 1 compounds, the thermal stability increases down the group as the ionic radius of the cation increases, and its polarising power decreases. Less thermal stability means it requires less heat to decompose a compound.

Therefore options C and D are incorrect.
The correct order will be,
K${ O }_{ 3 }$ < Rb${ O }_{ 3 }$ < Cs${ O }_{ 3 }$ : Thermal stability order
LiN${ O }_{ 3 }$ < NaN${ O }_{ 3 }$ < KN${ O }_{ 3 }$ < CsN${ O }_{ 3 }$ : Thermal stability order

Therefore, we can conclude that the correct answers to this question are options A and B.

Note: We should also know that Group 1 compounds tend to be more thermally stable than group 2 compounds because the cation has a smaller charge and a larger ionic radius, and so a lower polarising power. particularly when adjacent metals in the same period are compared.