Match List - I with List - II:
List-I                         List-II
(a) NaOH       (i) Acidic
(b) Be(OH)$$_2$$   (ii) Basic
(c) Ca(OH)$$_2$$   (iii) Amphoteric
(d) B(OH)$$_3$$  
(e) Al(OH)$$_3$$
Choose the most appropriate answer from the options given below:

We begin by recalling the usual definitions.

• A hydroxide that produces $$\text{OH}^-$$ ions readily in water and neutralises acids is called basic.
• A hydroxide that donates $$\text{H}^+$$ (or accepts $$\text{OH}^-$$) and neutralises bases is called acidic.
• A hydroxide that can behave both as an acid and as a base is called amphoteric.

Now we examine each hydroxide one by one, using periodic trends and well-known reactions to justify its behaviour.

For sodium hydroxide, $$\text{NaOH}$$:
Sodium belongs to Group 1. Alkali-metal hydroxides are strongly basic because they dissociate completely:
$$\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-$$
The liberation of $$\text{OH}^-$$ makes the solution strongly basic. Hence $$\text{NaOH}$$ → Basic ⇒ (ii).

For beryllium hydroxide, $$\text{Be(OH)}_2$$:
Beryllium sits at the top of Group 2 and possesses a very high charge density. Owing to this high polarising power its hydroxide neither dissociates completely as a base nor remains strictly acidic; instead it can react with both acids and bases:
$$\text{Be(OH)}_2 + 2\,\text{HCl} \rightarrow \text{BeCl}_2 + 2\,\text{H}_2\text{O}$$
$$\text{Be(OH)}_2 + 2\,\text{OH}^- \rightarrow [\text{Be(OH)}_4]^{2-}$$
Because it shows dual behaviour, $$\text{Be(OH)}_2$$ is amphoteric ⇒ (iii).

For calcium hydroxide, $$\text{Ca(OH)}_2$$:
Calcium is lower down the same Group 2. The larger ionic size reduces polarising power, so its hydroxide acts predominantly as a base:
$$\text{Ca(OH)}_2 \rightarrow \text{Ca}^{2+} + 2\,\text{OH}^-$$
Therefore $$\text{Ca(OH)}_2$$ is basic ⇒ (ii).

For boric acid written as $$\text{B(OH)}_3$$:
Boron is a typical non-metal. The molecule accepts $$\text{OH}^-$$ ions (acts as a Lewis acid) forming $$\text{[B(OH)}_4]^-$$. Experimentally its aqueous solution has $$\text{pH}<7$$. So $$\text{B(OH)}_3$$ behaves as an acid ⇒ (i).

For aluminium hydroxide, $$\text{Al(OH)}_3$$:
Like beryllium, aluminium has appreciable charge density and shows dual behaviour:
With acid: $$\text{Al(OH)}_3 + 3\,\text{HCl} \rightarrow \text{AlCl}_3 + 3\,\text{H}_2\text{O}$$
With base: $$\text{Al(OH)}_3 + \text{OH}^- \rightarrow [\text{Al(OH)}_4]^-$$
Thus $$\text{Al(OH)}_3$$ is amphoteric ⇒ (iii).

Collecting all the matches we have
$$(a)\;\text{NaOH}\; \rightarrow\; (ii)\quad\text{Basic}$$ $$(b)\;\text{Be(OH)}_2\; \rightarrow\; (iii)\quad\text{Amphoteric}$$ $$(c)\;\text{Ca(OH)}_2\; \rightarrow\; (ii)\quad\text{Basic}$$ $$(d)\;\text{B(OH)}_3\; \rightarrow\; (i)\quad\text{Acidic}$$ $$(e)\;\text{Al(OH)}_3\; \rightarrow\; (iii)\quad\text{Amphoteric}$$

Comparing with the given options, the sequence (ii, iii, ii, i, iii) corresponds to Option B.

Hence, the correct answer is Option B.