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. Then metal $M$is:
A. $Al$
B. $Fe$
C. $Zn$
D. $Ca$

Answer 1

Precipitation occurs due to the production of metal hydroxide.
Metal hydroxide upon heating results in metal oxide with the evaporation of water molecules.

Complete step by step answer:
In the question, it is given that when metal $M$ is treated with$NaOH$, a white gelatinous precipitate $X$ is obtained. Except Fe other three metals are forming white precipitate with sodium hydroxide and upon heating form metal oxide.
But, among these three metals $\left( {Al,{\text{ }}Zn,{\text{ }}Ca} \right)$aluminium oxide is used as an adsorbent in chromatography. Chromatography is the method of separating or purifying solids and liquids. Adsorbent is the stationary phase used in chromatography which helps the molecules adhering to one another without making chemical bonds $\left( {the{\text{ }}process{\text{ }}is{\text{ }}called{\text{ }}as{\text{ }}adsorption} \right).$
When aluminium is treated with$NaOH$, it will give aluminium hydroxide and the reaction is given by, $Al + 3NaOH \to 3N{a^ + } + Al{(OH)_3}$, which is white gelatinous precipitate.
When it is treated with excess $NaOH$, sodium salt of aluminium hydroxide is formed.
$Al{(OH)_3} + NaOH \to NaAl{(OH)_4}$
When aluminium hydroxide is strongly heated, aluminium oxide is formed with the evaporation of water molecules.
$2Al{(OH)_3}\xrightarrow{\Delta }A{l_2}{O_3} + 3{H_2}O$

So, option$- \left( A \right)$ is the correct answer.

Note:
The common adsorbents used in chromatography are alumina ($A{l_2}{O_3}$) and silica gel ($Si{O_2}$).