Match List - I with List - II:
List - I (Metal ion)     List - II (Group in Qualitative analysis)
(a) Mn$$^{2+}$$     (i) Group - III
(b) As$$^{3+}$$     (ii) Group - IIA
(c) Cu$$^{2+}$$     (iii) Group - IV
(d) Al$$^{3+}$$     (iv) Group - IIB
Choose the most appropriate answer from the options given below:

In the scheme of qualitative inorganic analysis, metal cations are separated group-wise by taking advantage of differences in the solubility of their salts. The sequence of groups and the reagents employed are fixed, so each radical always appears in the same group. Before matching, let us recall the standard group distribution:

We have

Group I: insoluble chlorides with dilute $$\mathrm{HCl},$$ which contains $$\mathrm{Ag^+,\; Pb^{2+},\; Hg_2^{2+}}.$$

Next,

Group II: sulphides precipitated by passing } \mathrm{H_2S} \text{ in acidic medium. This group is split into two sub-groups:

Group IIA (Cu group): $$\mathrm{Cu^{2+},\; Cd^{2+},\; Pb^{2+},\; Bi^{3+},\; Hg^{2+}},$$

Group IIB (As group): $$\mathrm{As^{3+/5+},\; Sb^{3+/5+},\; Sn^{2+/4+}}.$$

Then, on adding $$\mathrm{NH_4Cl}$$ and excess $$\mathrm{NH_4OH},$$ the cations whose hydroxides are insoluble are brought down as follows:

Group III: $$\mathrm{Fe^{3+},\; Al^{3+},\; Cr^{3+}}.$$

If any cations still remain in solution, bubbling $$\mathrm{H_2S}$$ through the alkaline filtrate (containing $$\mathrm{NH_4OH}$$ and $$\mathrm{NH_4Cl}$$) precipitates sulphides of

Group IV: $$\mathrm{Zn^{2+},\; Mn^{2+},\; Co^{2+},\; Ni^{2+}}.$$

Using this standard information, let us place each ion from List - I into its correct qualitative group from List - II.

(a) $$\mathrm{Mn^{2+}}$$
From the above summary, $$\mathrm{Mn^{2+}}$$ is precipitated as $$\mathrm{MnS}$$ in the alkaline medium of Group IV. So, $$\mathrm{Mn^{2+}} \longrightarrow \text{Group IV} \; (iii).$$

(b) $$\mathrm{As^{3+}}$$
Arsenic belongs to the As sub-group; its sulphide $$\mathrm{As_2S_3}$$ is precipitated by $$\mathrm{H_2S}$$ in acid medium and is soluble in yellow ammonium sulphide, a characteristic of Group IIB. Thus, $$\mathrm{As^{3+}} \longrightarrow \text{Group IIB} \; (iv).$$

(c) $$\mathrm{Cu^{2+}}$$
Copper falls in the Cu sub-group; its black sulphide $$\mathrm{CuS}$$ precipitates directly with $$\mathrm{H_2S}$$ in acid medium and does not dissolve in yellow ammonium sulphide, placing it firmly in Group IIA. Hence, $$\mathrm{Cu^{2+}} \longrightarrow \text{Group IIA} \; (ii).$$

(d) $$\mathrm{Al^{3+}}$$
Aluminium hydroxide $$\mathrm{Al(OH)_3}$$ is gelatinous and precipitates on adding $$\mathrm{NH_4OH}$$ in the presence of $$\mathrm{NH_4Cl},$$ i.e. during Group III testing. Therefore, $$\mathrm{Al^{3+}} \longrightarrow \text{Group III} \; (i).$$

Collecting all the pairings, we obtain

$$(a)-(iii)$$, $$\; (b)-(iv)$$, $$\; (c)-(ii)$$, $$\; (d)-(i).$$

This sequence corresponds exactly to Option A in the given choices.

Hence, the correct answer is Option A.