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.
Write Lewis structure of $\mathrm{O}_2{ }^{-}$ion and find out oxidation state of each oxygen atom? What is the average oxidation state of oxygen in this ion?
The Lewis structure of $\mathrm{O}_2{ }^{-}$is
Oxygen atom carrying no charge has six electrons, so its oxidation number is zero. But oxygen atom carrying -1 charge has 7 electrons, so its oxidation number is -1.
Average oxidation number of each oxygen atom $=\frac{1}{2}$
$$\begin{aligned} \mathrm{O}_2^{-} & =2 x=-1 \\ x & =-\frac{1}{2} \end{aligned}$$
Why do beryllium and magnesium not impart colour to the flame in the flame test?
All alkaline earth metals (except Be and Mg ) impart a characteristic colour to the Bunsen flame. The different colours arise due to different energies required for electronic excitation and de-excitation.
Be and Mg atoms due to their small size, bind their electrons more strongly (because of higher effective nuclear charge). Hence, require high excitation energy and are not excited by the energy of the flame with the result that no flame colour is shown by them.