Question

A metal M readily forms water-soluble sulphate $\text{MS}{{\text{O}}_{\text{4}}}$, water insoluble hydroxide $\text{M(OH}{{\text{)}}_{\text{2}}}$ and oxide $\text{MO}$which becomes inert on heating.
The hydroxide is soluble in NaOH. The metal M is:
(A) Be
(B) Mg
(C) Ca
(D) Sr

Answer 1

Hint: The metal undergoing the reaction is dependent on the size of its cationic form which will help to determine its chemical nature as its sulphate is water-soluble and $\text{NaOH}$ soluble hydroxide, indicating its acidic nature.

Complete step by step solution:
In the group, 2 elements of the alkaline earth metals, the cations $({{M}^{+2}})$ on moving down the group increases in size. Due to which their lattice enthalpy and the hydration enthalpy decreases down the group. As the bond between the cations and anions becomes longer with a decrease in the attractive force and also, they get less hydrated moving down the group.
But, in case of sulphates of alkaline earth metals, the decrease in the hydration enthalpy is more compared to the lattice enthalpy because of the large size of sulphate ion, causing a decrease in their solubility down the group. Thus, only hydrated beryllium sulphate and magnesium sulphate are soluble.
$Be+{{H}_{2}}S{{O}_{4}}\to BeS{{O}_{4}}+{{H}_{2}}$
Whereas, in case of the hydroxides of alkaline earth metals, due to the small size of the $O{{H}^{-}}$ ions, the cations make up for the distance between them. Thus, the decrease in lattice enthalpy is more prominent than the hydration enthalpy leading to more solubility increase down the group. Only magnesium hydroxide and beryllium hydroxide are insoluble in water.
As given, the hydroxide of the metals is soluble in $\text{NaOH}$, it is possible only in case of beryllium which is amphoteric in nature, that it acts as both an acid or base. Thus, they dissolve in sodium hydroxide by acting as a strong acid. The oxide of beryllium is also inert on heating.
\[Be{{\left( OH \right)}_{2}}+2NaOH\to N{{a}_{2}}Be{{O}_{2}}+{{H}_{2}}O\]

Hence, the answer is option (A)- $Be$.

Note: The hydration enthalpy is solely dependent on the size of the ions, whereas the lattice enthalpy governs both the size and the distance between the ions. The hydroxides of alkaline earth metals are generally basic in nature, but due to the smaller size of beryllium, it has a peculiar nature, acting as amphoteric hydroxide.