Identify Metal Ion Precipitated by Hydrogen Sulfide in Basic Medium

To determine the correct option, let us analyze the cations in qualitative inorganic analysis based on their separation into groups:

1. Group II Cations ($\text{Pb}^{2+}$, $\text{Cu}^{2+}$): These ions are precipitated out as sulfides in an acidic medium using $\text{H}_2\text{S}$ gas in the presence of dilute $\text{HCl}$. The low sulfide ion concentration (controlled by the common ion effect of $\text{H}^+$) is sufficient to exceed the solubility product ($K_{sp}$) of Group II sulfides.

2. Group III Cations ($\text{Fe}^{3+}$): These ions are precipitated out as hydroxides (e.g., $\text{Fe(OH)}_3$) using ammonium hydroxide ($\text{NH}_4\text{OH}$) in the presence of ammonium chloride ($\text{NH}_4\text{Cl}$).

3. Group IV Cations ($\text{Mn}^{2+}$): Group IV cations ($\text{Mn}^{2+}, \text{Zn}^{2+}, \text{Ni}^{2+}, \text{Co}^{2+}$) have higher solubility products for their sulfides. Hence, they require a higher concentration of $\text{S}^{2-}$ ions for precipitation, which is achieved in a basic medium by passing $\text{H}_2\text{S}$ gas in the presence of $\text{NH}_4\text{OH}$ and $\text{NH}_4\text{Cl}$. The reaction for $\text{Mn}^{2+}$ is: $\text{Mn}^{2+} + \text{S}^{2-} \xrightarrow{\text{NH}_4\text{OH}} \text{MnS} \downarrow \text{ (buff-colored precipitate)}$

Thus, the metal ion that gets precipitated out in the presence of $\text{H}_2\text{S}$ and $\text{NH}_4\text{OH}$ is $\text{Mn}^{2+}$.

The corresponding metal is Manganese ($\text{Mn}$), with atomic number $Z = 25$. Its outer electronic configuration is: $\text{Mn}: [\text{Ar}] 3d^5 4s^2$

The highest possible oxidation state of manganese corresponds to the removal or involvement of all seven valence electrons ($5$ from $3d$ and $2$ from $4s$): $\text{Highest Oxidation State} = +7$ (As seen in compounds like $\text{KMnO}_4$ or $\text{Mn}_2\text{O}_7$).

Therefore, the correct option is D.