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:

We have a metal, denoted as $$M$$, which reacts with aqueous $$\text{NaOH}$$ to give a white gelatinous precipitate $$X$$. The information tells us that $$X$$ dissolves when an excess of $$\text{NaOH}$$ is added. Later, when $$X$$ is heated strongly, it produces an oxide that is employed in chromatography as an adsorbent.

First, let us recall what usually happens when different metal ions are treated with dilute $$\text{NaOH}$$:

1. Iron (Fe) gives green or reddish-brown precipitates (depending on its oxidation state) that are not white and do not dissolve in excess $$\text{NaOH}$$.
2. Zinc (Zn) forms a white precipitate of $$\text{Zn(OH)}_2$$, and this precipitate does dissolve in excess alkali, giving $$\text{[Zn(OH)}_4]^{2-}$$.
3. Calcium (Ca) generally gives a faint, sometimes milky precipitate of $$\text{Ca(OH)}_2$$ that does not dissolve appreciably in excess alkali.
4. Aluminium (Al) produces a white gelatinous precipitate of $$\text{Al(OH)}_3$$, and this precipitate does dissolve in excess alkali, forming the soluble aluminate ion $$\text{[Al(OH)}_4]^-$$.

So, on the basis of mere colour and solubility, both zinc and aluminium appear to fit. We therefore use the additional clue: “the precipitate $$X$$, on strong heating, yields an oxide that is used in chromatography as an adsorbent.”

The oxide obtained from zinc hydroxide is $$\text{ZnO}$$, zinc oxide. Zinc oxide is certainly industrially important, but it is not the standard stationary phase adsorbent used for column chromatography. Instead, the common adsorbent mentioned in basic chemistry laboratories for chromatography is activated alumina, i.e. $$\text{Al}_2\text{O}_3$$.

Let us write the sequence of reactions explicitly for aluminium to check consistency.

When aluminium metal (or an aluminium salt) is treated with aqueous sodium hydroxide, the initial precipitation step is

$$\text{Al}^{3+}_{(aq)} + 3\,\text{OH}^-_{(aq)} \rightarrow \text{Al(OH)}_3\, (s)$$

This $$\text{Al(OH)}_3$$ is a white, gelatinous solid. Now, because aluminium hydroxide is amphoteric, if we add an excess of $$\text{OH}^-$$ ions, it dissolves, giving

$$\text{Al(OH)}_3\, (s) + \text{OH}^-_{(aq)} \rightarrow \text{[Al(OH)}_4]^-_{(aq)}$$

This exactly matches the solubility behaviour described in the question.

Next, we heat the hydroxide strongly. The decomposition (or calcination) reaction is

$$2\,\text{Al(OH)}_3\, (s) \xrightarrow{\;\Delta\;} \text{Al}_2\text{O}_3\, (s) + 3\,\text{H}_2\text{O}\,(g)$$

The product $$\text{Al}_2\text{O}_3$$ is aluminium oxide, commonly called alumina. Activated alumina is widely used as an adsorbent in chromatography columns because of its high surface area and polar character. Therefore the oxide mentioned in the question is clearly alumina.

Since the only metal among the options whose hydroxide behaves in this precise way and whose oxide is the standard chromatographic adsorbent is aluminium, we conclude that

$$M = \text{Al}$$

Option D corresponds to aluminium.

Hence, the correct answer is Option D.