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Hint:In this question, we have to identify the chemical formula of Beryllium sulphate when it is in aqueous solution. Beryllium sulphate is prepared by dissolving beryllium oxide in hot concentrated sulphuric acid $\,\left( {{H_2}S{O_4}} \right)$.
Complete step by step answer:
$BeS{O_4}$ (Beryllium sulphate) crystallizes from an aqueous solution as the tetrahydrate $BeS{O_4}.4{H_2}O$. Now we will discuss it in detail.
In $BeS{O_4}$ (Beryllium sulphate) , the $B{e^{2 + }}$ is a small cation with more hydration energy due to which it will be hydrated by four water molecules and in aqueous solution it forms a tetrahydrate ($BeS{O_4}.4{H_2}O$) which is highly soluble in water.
All the four water molecules in $BeS{O_4}.4{H_2}O$ are coordinated to $B{e^{2 + }}$ ion and the actual structure is $\left[ {Be{{\left( {{H_2}O} \right)}_4}} \right]S{O_4}$.
The $B{e^{2 + }}$ ion utilizes four $s{p^3}$ hybrid orbitals to accommodate 8 electrons from the four water molecules and the structure of the ion ${\left[ {Be{{\left( {{H_2}O} \right)}_4}} \right]^{2 + }}$ is tetrahedral.
Thus, the $SO_4^{2 - }$ anion combines with ${\left[ {Be{{\left( {{H_2}O} \right)}_4}} \right]^{2 + }}$ cation and forms $\left[ {Be{{\left( {{H_2}O} \right)}_4}} \right]S{O_4}$ in aqueous solution.
Hence, the correct option is (A).
Additional Information: Beryllium due to its small size can be hydrated more than the other alkaline metals due to which it does not liberate water molecules. It is attained as a product of the large atomic nuclei having collision with the cosmic rays. Beryllium is used to make cell phones, aircrafts and missiles.
Note:
Hydration energy is the amount of energy released when one mole of ion undergoes hydration.
Smaller the size of an atom more will be the attraction, more hydration energy will be released.
Bigger the size of an atom less will be the attraction, less hydration energy will be released.
Complete step by step answer:
$BeS{O_4}$ (Beryllium sulphate) crystallizes from an aqueous solution as the tetrahydrate $BeS{O_4}.4{H_2}O$. Now we will discuss it in detail.
In $BeS{O_4}$ (Beryllium sulphate) , the $B{e^{2 + }}$ is a small cation with more hydration energy due to which it will be hydrated by four water molecules and in aqueous solution it forms a tetrahydrate ($BeS{O_4}.4{H_2}O$) which is highly soluble in water.
All the four water molecules in $BeS{O_4}.4{H_2}O$ are coordinated to $B{e^{2 + }}$ ion and the actual structure is $\left[ {Be{{\left( {{H_2}O} \right)}_4}} \right]S{O_4}$.
The $B{e^{2 + }}$ ion utilizes four $s{p^3}$ hybrid orbitals to accommodate 8 electrons from the four water molecules and the structure of the ion ${\left[ {Be{{\left( {{H_2}O} \right)}_4}} \right]^{2 + }}$ is tetrahedral.
Thus, the $SO_4^{2 - }$ anion combines with ${\left[ {Be{{\left( {{H_2}O} \right)}_4}} \right]^{2 + }}$ cation and forms $\left[ {Be{{\left( {{H_2}O} \right)}_4}} \right]S{O_4}$ in aqueous solution.
Hence, the correct option is (A).
Additional Information: Beryllium due to its small size can be hydrated more than the other alkaline metals due to which it does not liberate water molecules. It is attained as a product of the large atomic nuclei having collision with the cosmic rays. Beryllium is used to make cell phones, aircrafts and missiles.
Note:
Hydration energy is the amount of energy released when one mole of ion undergoes hydration.
Smaller the size of an atom more will be the attraction, more hydration energy will be released.
Bigger the size of an atom less will be the attraction, less hydration energy will be released.
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