Step-by-Step Solution:

1. Analyze the properties of metal ion M2. The problem states that an aqueous solution of metal ion M2 always forms tetrahedral complexes with reagents Q and R. This is a characteristic feature of metal ions with a completely filled d-orbital ($d^{10}$ configuration) or an empty d-orbital ($d^0$ configuration). A very common example of a $d^{10}$ ion in qualitative analysis is $Zn^{2+}$. The electronic configuration of $Zn$ is $[Ar] 3d^{10} 4s^2$, so $Zn^{2+}$ is $[Ar] 3d^{10}$. Due to the stable $d^{10}$ configuration, it does not form square planar or octahedral complexes easily and predominantly forms tetrahedral complexes using its $4s$ and $4p$ orbitals for hybridization ($sp^3$), for example, $[Zn(NH_3)_4]^{2+}$ and $[Zn(CN)_4]^{2-}$. Therefore, we can strongly infer that M2 is $Zn^{2+}$.

2. Analyze the reaction of M2 with reagent S. The problem states that an aqueous solution of M2 ($Zn^{2+}$) on reaction with reagent S gives a white precipitate, which then dissolves in an excess of S. This behavior is characteristic of an amphoteric hydroxide. Amphoteric hydroxides are insoluble in water but dissolve in both acids and strong alkalis.

3. Evaluate the options for reagent S based on its reaction with $Zn^{2+}$.

   • Option A: $K_4[Fe(CN)_6]$ (Potassium ferrocyanide) Reaction: $2Zn^{2+} + [Fe(CN)_6]^{4-} \longrightarrow Zn_2[Fe(CN)_6](s)$ This reaction forms a white or bluish-white precipitate of zinc ferrocyanide. This precipitate is insoluble in excess potassium ferrocyanide. So, this option is incorrect.

   • Option B: $Na_2HPO_4$ (Disodium hydrogen phosphate) Reaction in neutral solution: $3Zn^{2+} + 2HPO_4^{2-} \longrightarrow Zn_3(PO_4)_2(s) + 2H^+$ This forms a white precipitate of zinc phosphate. This precipitate does not dissolve in excess $Na_2HPO_4$. So, this option is incorrect.

   • Option C: $K_2CrO_4$ (Potassium chromate) Reaction: $Zn^{2+} + CrO_4^{2-} \longrightarrow ZnCrO_4(s)$ This forms a yellow precipitate of zinc chromate. The problem specifies a white precipitate. So, this option is incorrect.

   • Option D: KOH (Potassium hydroxide) This is a strong alkali. It reacts with $Zn^{2+}$ in two steps: i. Formation of a white precipitate with a limited amount of KOH: $Zn^{2+}(aq) + 2OH^-(aq) \longrightarrow Zn(OH)_2(s)$ Zinc hydroxide, $Zn(OH)_2$, is a white, gelatinous precipitate. ii. Dissolution of the precipitate in excess KOH: $Zn(OH)_2(s) + 2OH^-(aq) \text{ (excess)} \longrightarrow [Zn(OH)_4]^{2-}(aq)$ The precipitate dissolves to form the soluble, colorless tetrahydroxozincate(II) complex ion. This matches the description given in the problem perfectly. The white precipitate ($Zn(OH)_2$) dissolves in excess of the reagent S (KOH).

4. Conclusion Based on the analysis, reagent S must be KOH. For completeness, let's consider M1. The fact that M1 forms both tetrahedral and square planar complexes is characteristic of $Ni^{2+}$. For example, $Ni^{2+}$ forms the tetrahedral complex $[NiCl_4]^{2-}$ with excess $Cl^-$ (reagent Q) and the square planar complex $[Ni(CN)_4]^{2-}$ with excess $CN^-$ (reagent R). The entire scheme is consistent with M1 being $Ni^{2+}$, M2 being $Zn^{2+}$, and S being KOH.