Qualitative Analysis - Result Question 2
2. 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
(2018 Main)
(a) $Zn$
(b) $Ca$
(c) $Al$
(d) $Fe$
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Answer:
Correct Answer: 2. (c)
Solution:
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.
Aluminium hydroxide on strong heating gives alumina $\left(Al _2 O _3\right)$ which is used as an adsorbent in chromatography. This reaction can be seen as :
$2 Al(OH) _3 \stackrel{\Delta}{\longrightarrow} Al _2 O _3+3 H _2 O$
Thus, metal $M$ is $Al$.
Ca, being below sodium in electrochemical reactivity series, cannot displaces $Na$ 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.
BETA
