Question

The correct order of thermal stability of hydroxide is:
(a)$Mg{\left(OH\right)}_2<Sr{\left(OH\right)}_2<Ca{\left(OH\right)}_2<Ba{\left(OH\right)}_2$
(b)$Mg{\left(OH\right)}_2<Ca{\left(OH\right)}_2<Sr{\left(OH\right)}_2<Ba{\left(OH\right)}_2$
(c)$Ba{\left(OH\right)}_2<Sr{\left(OH\right)}_2<Ca{\left(OH\right)}_2<Mg{\left(OH\right)}_2$
(d)$Ba{\left(OH\right)}_2<Ca{\left(OH\right)}_2<Sr{\left(OH\right)}_2<Mg{\left(OH\right)}_2$

Answer 1

Hint: An attempt to this question can be made by determining the factors affecting the thermal stability of a compound. In the periodic table, as we go down in the group the lattice enthalpy increases, which leads to an increase in thermal stability of the hydroxide. Now determine the position of the elements as they are present in group number 2.

Complete answer:
In the given question, the hydroxides of group 2 are under consideration.

	Group 2
I
II	Be
III	Mg
IV	Ca
V	Sr
VI	Ba

As we go down in the group, the size of the element increases due to increase in the shells. As the size increases, ionization energy of the element decreases, i.e. it’s easy for one element to give electrons to another, which leads to increase in lattice enthalpy. And lattice enthalpy value is directly proportional to the thermal stability of hydroxide.
In the given case of alkaline earth metal, the thermal stability of hydroxide goes on increasing down the group due to the above stated reason.
From the above statements we can conclude that the correct order of increasing lattice enthalpy and thermal stability is: $Mg{\left(OH\right)}_2<Ca{\left(OH\right)}_2<Sr{\left(OH\right)}_2<Ba{\left(OH\right)}_2$

So, the correct answer is option (b) and the respective order is, $Mg{\left(OH\right)}_2<Ca{\left(OH\right)}_2<Sr{\left(OH\right)}_2<Ba{\left(OH\right)}_2$

Note:
Lattice enthalpy is the change in enthalpy occurs when an ionic compound forms from gaseous ions. Greater the lattice enthalpy of a compound, greater will be the stability of the compound.