In the leaching of alumina from bauxite, the ore expected to leach out in the process by reacting with NaOH is:

In the Baeyer’s process, bauxite is digested with concentrated aqueous sodium hydroxide at about 150 °C under pressure. The basic idea is that those oxides which react with NaOH will pass into the solution, while the oxides which do not react will remain as an insoluble residue.

First, alumina itself behaves amphoterically and dissolves:

$$$\mathrm{Al_2O_3 + 2\,NaOH + 3\,H_2O \;\longrightarrow\; 2\,Na[Al(OH)_4]}$$$

Now we examine each oxide given in the options to see whether it will similarly react (leach out) or stay undissolved.

1. For $$\mathrm{Fe_2O_3}$$ and 2. for $$\mathrm{TiO_2}$$ Both of these oxides are essentially basic (or at best very weakly amphoteric) and, under the conditions of the Baeyer’s process, they do not react with concentrated NaOH. Hence they stay behind in the red mud residue and are not leached out.

3. For $$\mathrm{ZnO}$$ While zinc oxide is indeed amphoteric and can dissolve in strong alkali, ZnO is not a common impurity present in bauxite. The question speaks of “the ore expected to leach out in the process”, i.e., the impurity that really occurs in bauxite and actually goes into solution in the industrial step. Therefore, even though ZnO is amphoteric, it is not the correct choice in this context.

4. For $$\mathrm{SiO_2}$$ Silica is an acidic oxide. An acidic oxide readily reacts with a strong base such as NaOH, giving a soluble silicate:

$$$\mathrm{2\,NaOH + SiO_2 \;\longrightarrow\; Na_2SiO_3 + H_2O}$$$

Because $$\mathrm{Na_2SiO_3}$$ remains in the aqueous phase, silica gets leached out along with sodium aluminate. This behaviour exactly matches what is observed during the purification of bauxite.

Comparing all the possibilities, we see that only $$\mathrm{SiO_2}$$ is the impurity that actually dissolves (is leached) in the alkaline solution.

Hence, the correct answer is Option D.