Question

A metal $\mathrm{M}$readily forms its sulfate ${\mathrm{MSO}}_4$ which is water-soluble. It forms its oxide $\mathrm{MO}$ which becomes inert on heating. It forms an insoluble hydroxide $\mathrm{M}{\left(\mathrm{OH}\right)}_2$ which is soluble in $\mathrm{NaOH}$ solution. Then $\mathrm{M}$ is

• A. $\mathrm{Ba}$
• B. $\mathrm{Be}$
• C. $\mathrm{Ca}$
• D. $\mathrm{Mg}$

Answer 1

The correct option is B

$\mathrm{Be}$

Explanation:

Beryllium:

• Beryllium is a chemical element having an atomic number $4$ and an atomic symbol $\mathrm{Be}$.
• The oxides and hydroxides of alkaline earth metal are basic but there is an exception in the case of Beryllium $\left(\mathrm{Be}\right)$ which is amphoteric.
• Only beryllium and magnesium sulfates are soluble in water.
• The reaction of Beryllium with sulfuric acid:

$$\begin{gathered} \mathrm{Be}\left(\mathrm{s}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\to {\mathrm{BeSO}}_4\left(\mathrm{aq}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right) \\ \left(\mathrm{Beryllium}\right)(\mathrm{Sulfuric}\mathrm{acid})(\mathrm{Beryllium}\mathrm{sulfate})\left(\mathrm{Hydrogen}\right) \end{gathered}$$

• Beryllium hydroxide$\left(\mathrm{Be}{\left(\mathrm{OH}\right)}_2\right)$ reacts with base i.e. sodium hydroxide$\left(\mathrm{NaOH}\right)$ to produce sodium beryllate$\left({\mathrm{Na}}_2{\mathrm{BeO}}_2\right)$ along with water.
• The chemical reaction can be depicted as:

$$\begin{gathered} \mathrm{Be}{\left(\mathrm{OH}\right)}_2\left(\mathrm{aq}\right)+2\mathrm{NaOH}\left(\mathrm{aq}\right)\to {\mathrm{Na}}_2{\mathrm{BeO}}_2\left(\mathrm{aq}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ (\mathrm{Beryllium}\mathrm{hydroxide})(\mathrm{Sodium}\mathrm{hydroxide})(\mathrm{Sodium}\mathrm{beryllate})\left(\mathrm{Water}\right) \end{gathered}$$

• The oxide of Beryllium $\left(\mathrm{BeO}\right)$ is insoluble in pure water but reacts with either an acid (like a base would) or a base. Thus, it is amphoteric in nature.

Hence, the correct option is B i.e. $\mathrm{Be}$.