Question

When metal ‘M’ is treated with NaOH, a white gelatinous precipitate ‘X’ is obtained, which is soluble in excess of NaOH. Compound ‘X’ when heated strongly gives an oxide which is used in chromatography as an adsorbent. The metal ‘M’ is

• A. $\mathrm{Ca}$
• B. $\mathrm{Al}$
• C. $\mathrm{Fe}$
• D. $\mathrm{Zn}$

Answer 1

The correct option is B

$\mathrm{Al}$

• The above statement can be represented as a reaction and is given as follows: $$\begin{gathered} {\mathrm{Al}}^{3+}\left(\mathrm{s}\right)\stackrel{\mathrm{NaOH}}{\to }\mathrm{Al}{\left(\mathrm{OH}\right)}_3\downarrow \left(\mathrm{s}\right)\stackrel{\mathrm{Excess}\mathrm{NaOH}}{\to }\mathrm{Na}\left[\mathrm{Al}{\left(\mathrm{OH}\right)}_4\right]\left(\mathrm{aq}\right) \\ \left(\mathrm{Aluminium}\right)(\mathrm{White}\mathrm{gelatinous}\mathrm{ppt}(\mathrm{Soluble}\mathrm{Sodium}\mathrm{aluminate}) \\ \mathrm{of}\mathrm{Aluminium}\mathrm{hydroxide}) \end{gathered}$$.
• When compound is strongly heated, we get:$$\begin{gathered} 2\mathrm{Al}{\left(\mathrm{OH}\right)}_3\left(\mathrm{s}\right)\stackrel{\mathrm{Strong}∆}{\to }{\mathrm{Al}}_2{\mathrm{O}}_3\left(\mathrm{g}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ (\mathrm{Aluminium}\mathrm{hydroxide})(\mathrm{Aluminium}\mathrm{trioxide})\left(\mathrm{Water}\right) \end{gathered}$$.
• From the above reaction it is clear that, when a metal ‘M’ i.e., $\mathrm{Al}$ (Aluminium) is treated with $\mathrm{NaOH}$, a white gelatinous precipitate ‘X’ i.e., $\mathrm{Al}{\left(\mathrm{OH}\right)}_3$(Aluminium hydroxide) is obtained, which is soluble in excess of $\mathrm{NaOH}$ and form $\mathrm{Na}\left[\mathrm{Al}{\left(\mathrm{OH}\right)}_4\right]$ (Sodium aluminate).The formed Compound $\mathrm{Al}{\left(\mathrm{OH}\right)}_3$, when heated strongly gives an oxide which is used in chromatography as an adsorbent. Hence, the metal ‘M’ is $\mathrm{Al}$.

Therefore, among the given options, option (b) $\mathrm{Al}$ is correct.