The correct match of the group reagents in List-I for precipitating the metal ion given in List-II from solutions, is

In qualitative inorganic analysis, cations are separated into groups by adding specific group reagents. Each reagent precipitates only those ions whose analytical product (sulfide, hydroxide, carbonate, etc.) has a very low solubility under the given conditions. We first recall the standard group scheme:

• Group II reagent  : pass $$H_2S$$ in dilute $$HCl$$ (acidic medium) ⇒ precipitates insoluble sulfides such as $$CuS, CdS, PbS$$ …
• Group III reagent : add $$NH_4OH$$ in the presence of excess $$NH_4Cl$$ (buffered weakly basic) ⇒ precipitates gelatinous hydroxides $$Fe(OH)_3, Al(OH)_3, Cr(OH)_3$$ …
• Group IV reagent : pass $$H_2S$$ in basic medium (presence of $$NH_4OH$$) ⇒ precipitates the less‐insoluble sulfides $$ZnS, NiS, CoS, MnS$$ …
• Group V reagent  : add $$(NH_4)_2CO_3$$ in presence of $$NH_4OH$$ ⇒ precipitates carbonates of alkaline-earth metals $$CaCO_3, SrCO_3, BaCO_3$$ …

Now match each List-I reagent with the metal ion of List-II whose salt is precipitated by that reagent.

Reagent : pass $$H_2S$$ in presence of $$NH_4OH$$ (alkaline) ⇒ Group IV.
Among the given ions, $$Mn^{2+}$$ forms the insoluble sulfide $$MnS$$ under these conditions.
Hence P → 3 ($$Mn^{2+}$$).

Reagent : $$(NH_4)_2CO_3$$ in presence of $$NH_4OH$$ ⇒ Group V.
Of the listed ions, $$Ba^{2+}$$ is an alkaline-earth metal and is precipitated as $$BaCO_3$$.
Hence Q → 4 ($$Ba^{2+}$$).

Reagent : $$NH_4OH$$ with $$NH_4Cl$$ buffer ⇒ Group III.
This reagent precipitates trivalent hydroxides; $$Al^{3+}$$ gives $$Al(OH)_3$$.
Hence R → 2 ($$Al^{3+}$$).

Reagent : pass $$H_2S$$ in dilute $$HCl$$ (acidic) ⇒ Group II.
$$Cu^{2+}$$ is a typical group-II cation, precipitated as black $$CuS$$.
Hence S → 1 ($$Cu^{2+}$$).

Collecting the matches: P → 3, Q → 4, R → 2, S → 1.

Therefore the correct option is:
Option A which is: P → 3; Q → 4; R → 2; S → 1.