s-Block Elements

Solubility of fluorides of alkaline earth metals decreases down the group due to more increase in value of lattice enthalpies as compared to hydration-enthalpy.

Thus, the order of solubility of given compounds is as follows :

$\mathrm{BeF}_2>\mathrm{MgF}_2>\mathrm{CaF}_2>\mathrm{SrF}_2$

Hence, $\mathrm{BeF}_2$ has maximum solubility.

Both assertion and reason are correct and reason is the correct explanation of the assertion. $\mathrm{BeO}$ is more stable than $\mathrm{BeCO}_3$ due to small size and high polarising power of $\mathrm{Be}^{2+}$.

As $\mathrm{BeCO}_3$ is unstable and $\mathrm{BeO}$ is more stable thus, when $\mathrm{BeCO}_3$ is kept in an atmosphere of $\mathrm{CO}_2$, a reversible process takes place and stability of $\mathrm{BeCO}_3$ increases. The reaction is shown below

$\mathrm{BeCO}_3 \rightleftharpoons \mathrm{BeO}+\mathrm{CO}_2$