Name an element from group 2 which forms an amphoteric oxide and a water soluble sulphate.
An element from group 2 which forms an amphoteric oxide and a water soluble sulphate is beryllium.
Beryllium forms oxides of formula BeO . All other alkaline earth metal oxides are basic in nature. BeO is amphoteric in nature i.e., it reacts with acids and bases both.
$$\begin{gathered} \mathrm{Al}_2 \mathrm{O}_3+2 \mathrm{NaOH} \longrightarrow 2 \mathrm{NaAlO}_2+\mathrm{H}_2 \mathrm{O} \\ \mathrm{Al}_2 \mathrm{O}_3+6 \mathrm{HCl} \longrightarrow 2 \mathrm{AlCl}_3+3 \mathrm{H}_2 \mathrm{O} \end{gathered}$$
Sulphate of beryllium is a white solid which crystallises as hydrated salts ( $\left.\mathrm{BeSO}_4 \cdot 4 \mathrm{H}_2 \mathrm{O}\right)$. $\mathrm{BeSO}_4$ is fairly soluble in water due to highest hydration energy in the group (small size). For $\mathrm{BeSO}_4$, hydration energy is more than lattice energy and so, they are readily soluble.
Discuss the trend of the following
(i) Thermal stability of carbonates of Group 2 elements.
(ii) The solubility and the nature of oxides of Group 2 elements.
(i) All the alkaline earth melals form carbonates $\left(\mathrm{MCO}_3\right)$. All these carbonates decompose on heating to give $\mathrm{CO}_2$ and metal oxide. The thermal stability of these carbonates increases down the group i.e., from Be to Ba .
$$\mathrm{BeCO}_3<\mathrm{MgCO}_3<\mathrm{CaCO}_3<\mathrm{SCO}_3<\mathrm{BaCO}_3$$
$\mathrm{BeCO}_3$ is unstable to the extent that it is stable only in atmosphere of $\mathrm{CO}_2$. These carbonates however show reversible decomposition in closed container.
$$\mathrm{BeCO}_3 \rightleftharpoons \mathrm{BeO}+\mathrm{CO}_2$$
Hence, more is the stability of oxide formed, less will be stability of carbonates. Stability of oxides decreases down the group is beryllium oxide i.e., high stable making $\mathrm{BeCO}_3$ unstable.
(ii) All the alkaline earth metals form oxides of formula $M O$. The oxides are very stable due to high lattice energy and are used as refractory material.
Except BeO (predominantly covalent) all other oxides are ionic and their lattice energy decreases as the size of cation increases.
The oxides are basic and basic nature increases from BeO to BaO (due to increasing ionic nature).
$\underbrace{\mathrm{BeO}}_{\text {Amphoteric }}<\underbrace{\mathrm{MgO}<}_{\text {Weak basic }} \underbrace{\mathrm{CaO}<\mathrm{SrO}<\mathrm{BaO}}_{\text {Strong basic }}$
BeO dissolves both in acid and alkalies to give salts and is amphoteric The oxides of the alkaline earth metals (except BeO and MgO ) dissolve in water to form basic hydroxides and evolve a large amount of heat. BeO and MgO possess high lattice energy and thus insoluble in water.
Why are $\mathrm{BeSO}_4$ and $\mathrm{MgSO}_4$ readily soluble in water while $\mathrm{CaSO}_4$, $\mathrm{SrSO}_4$ and $\mathrm{BaSO}_4$ are insoluble?
The lattice energy of alkaline earth metal sulphates is almost constant due to large size of sulphate ion. Thus, their solubility is decided by hydration energy which decreases on moving down the group.
The greater hydration enthalpies of $\mathrm{Be}^{2+}$ and $\mathrm{Mg}^{2+}$ ions overcome the lattice enthalpy factor and therefore, their sulphates are soluble in water.
However, hydration enthalpy is low for $\mathrm{Ca}^{2+}, \mathrm{Sr}^{2+}$ ions and cannot overcome the lattice energy factor. Hence, these are insoluble.
All compounds of alkali metals are easily soluble in water but lithium compounds are more soluble in organic solvents. Explain.
Smallest size of $\mathrm{Li}^{+}$ion among all alkali metals and its high polarising power are the two factors which develop covalent character in the lithium compounds (Fajan's rule). Compounds of other alkali metals are ionic in nature. So, they are soluble in water. Since lithium compounds being relatively covalent are soluble in alcohol and other organic solvents in accordance with "like dissolve like".
In the Solvay process, can we obtain sodium carbonate directly by treating the solution containing $\left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3$ with sodium chloride? Explain.
$\mathrm{No},\left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3$ reacts with NaCl as
$$\left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3+2 \mathrm{NaCl} \rightleftharpoons \mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{NH}_4 \mathrm{Cl}$$
Because the products obtained $\mathrm{Na}_2 \mathrm{CO}_3$ and $\mathrm{NH}_4 \mathrm{Cl}$ are highly soluble and the equilibrium will not shift in forward direction.
That's why in the Solvay process, we cannot obtain sodium carbonate directly by treating the solution containing $\left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3$ with sodium chloride.