Question

A metal readily forms its sulphate $\,{\text{MS}}{{\text{O}}_4}\,$, which is water soluble. It forms its oxide $\,MO\,$ which becomes inert on heating. It forms its insoluble hydroxide $\,M{\left( {OH} \right)_2}\,$ which is soluble in $\,NaOH\,$ solution. What would be $M$?
A.$Be$
B.$Ba$
C.$Ca,$
D.$Mg$

Answer 1

Hint: Given in this question that metal is forming its oxides $MO$. All compounds given in the question belong to the same group which is the group $\,2\,$. Moving down a group, the solubility of hydroxides increases.

Complete step by step answer:
Let us analyse each criteria and options one by one;
Beryllium $\,(Be)\,$ lies in the top of the group and hence its hydroxide has the lowest solubility which is actually insoluble. Beryllium oxide becomes inert to heating when heated because of an increase in the degree of polymerization. Beryllium hydroxide $\,Be{(OH)_2}\,$ dissolves in sodium hydroxide $\,(NaOH)\,$ to give sodium beryllate $\,(N{a_2}Be{O_2})\,$ as you can see in the following chemical reaction;
$\,Be{(OH)_2} + 2NaOH \to N{a_2}Be{O_2} + {H_2}O\,$
Hence, this option is correct.
Barium $\,(Ba)\,$ is the lower most element in the group while compared to other elements in the question as a result its hydroxide is also more soluble. However barium hydroxide is soluble is sodium hydroxide. Barium oxide form peroxides on heating. Hence, this option is not correct.
Calcium$\,(Ca)\,$ is below beryllium and above barium. Calcium hydroxide is relatively insoluble in water. When calcium oxide is heated it is not inert, rather it gives a bright light. Calcium hydroxide is soluble in sodium hydroxide. However, this cannot be the right option due to non- inertness showed by its oxide on heating.
Magnesium $\,(Mg)\,$ is below beryllium, but above all the other elements mentioned in the question in their group. Magnesium hydroxide seems to be insoluble in water. On heating magnesium oxide stays inert. However, magnesium hydroxide $\,(Mg{(OH)_2})\,$ is insoluble in sodium hydroxide. Hence, this option is also not correct.
So, the correct answer to this is option A which is $\,Be\,$.

Additional information:
Properties of $Be$
-Beryllium is a silvery-white metal. It is relatively soft and has a low density.
-Beryllium is used in alloys with copper or nickel to make gyroscopes, springs, electrical contacts, spot-welding electrodes and non-sparking tools

Note: $BeO$ does not dissolve in ${H_2}O$ but being a amphoteric dissolve in $NaOH$ and $BeO$ on heating gives inert gas and while reading total question go with the simplest point which saves time in examination When a hydroxide is soluble in $NaOH\,$ it means it is soluble in base so the metal should have amphoteric nature/alkaline nature (which means metal acts as acidic or basic nature) . $Be\,$ has diagonal relation with $Al\,$ . $Al\,$ is amphoteric and so is $\,Be\,$.