1. Identify the relevant group-IV cations
   In classical salt analysis, the common group-IV cations are: $Zn^{2+},\ Mn^{2+},\ Ni^{2+},\ Co^{2+}$ These are precipitated as sulphides by passing $H_2S$ in ammoniacal/basic medium.

2. Given: the cation gives a precipitate with $H_2S$
   So the cation forms its sulphide: $MS$ This is consistent with all four options.

3. The precipitate dissolves in dilute $HCl$
   Now test the resulting solution.

4. Test with $NaOH$ gives a white precipitate
   Let us compare the hydroxides formed:

   • $Co(OH)_2$ : bluish / dirty precipitate
   • $Ni(OH)_2$ : green precipitate
   • $Mn(OH)_2$ : white precipitate, but quickly turns brown on oxidation
   • $Zn(OH)_2$ : white gelatinous precipitate

   So at this stage, the likely ions are: $Mn^{2+} \text{ or } Zn^{2+}$

5. Test with basic potassium ferrocyanide gives a bluish-white precipitate
   Characteristic observations:

   • $Zn^{2+}$ gives a bluish-white precipitate with basic potassium ferrocyanide.
   • $Mn^{2+}$ does not give this characteristic bluish-white precipitate.
   • $Ni^{2+}$ and $Co^{2+}$ give different colored precipitates.

   Therefore, the cation must be: $Zn^{2+}$

6. Evaluate options

   • A: $Co^{2+}$ — incorrect
   • B: $Ni^{2+}$ — incorrect
   • C: $Mn^{2+}$ — incorrect
   • D: $Zn^{2+}$ — correct

Final Answer

$\boxed{Zn^{2+}}$ So, the correct option is D.