Discussion Forum

When metal 'M' is treated with NaOH , a while 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

Answer:

Explanation:

Among the given metals Al forms white gelatinous ppt. with NaOH . Hence , the probable metal can be Al. This ppt. is dissolved in excess of NaOH due to the formation of sodium metal Aluminate . Both the reactions are shown below,

$Al^{3+}$          $\underrightarrow{NaOH}$      $Al(OH)_{3+}\underrightarrow{ excess -of-NaOH}$         $NaAlO_{2}$

Aluminum hydroxide on strong heating gives alumina (Al₂O₃) which is used as an adsorbent in chromatography. This reaction can be seen as :

 $2Al(OH)_{3}$        $\underrightarrow{\triangle}$          $Al_{2}O_{3}+3H_{2}O $

Thus, metal M is Al

Ca  being below sodium in electrochemical reactivity series, cannot displace Na from its aqueous solution.

Zn reacts with NaOH to form sodium zincate which is a soluble compound.

Fe  reacts with sodium hydroxide to form tetrahydroferrate (II) sodium which is again a soluble complex.