Distinguishing Nitrate Salts of Manganese and Copper Ions

To determine the correct statements that allow us to distinguish between nitrate salts of $\text{Mn}^{2+}$ and $\text{Cu}^{2+}$, let us analyze each option individually based on chemical principles:

1. Analysis of Option (A):

• In flame test analysis, $\text{Cu}^{2+}$ ions impart a characteristic green (or bluish-green) color to the flame.
• $\text{Mn}^{2+}$ ions do not impart a characteristic green color to the flame.
• Therefore, statement (A) is incorrect.

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2. Analysis of Option (B):

• In qualitative inorganic analysis, cations are classified based on the solubility products ($K_{sp}$) of their sulfides:
  • $\text{Cu}^{2+}$ is a Group II cation. The solubility product of copper sulfide is extremely low ($K_{sp}(\text{CuS}) \approx 6 \times 10^{-36}$). In an acidic medium, the dissociation of $\text{H}_2\text{S}$ is suppressed due to the common ion effect of $\text{H}^+$ ions: $\text{H}_2\text{S} \rightleftharpoons 2\text{H}^+ + \text{S}^{2-}$ Even at this very low concentration of $\text{S}^{2-}$, the ionic product $[\text{Cu}^{2+}][\text{S}^{2-}]$ exceeds $K_{sp}(\text{CuS})$, resulting in the precipitation of black $\text{CuS}$.
  • $\text{Mn}^{2+}$ is a Group IV cation with a relatively higher solubility product ($K_{sp}(\text{MnS}) \approx 2.5 \times 10^{-13}$). The low $\text{S}^{2-}$ concentration in an acidic medium is insufficient to precipitate $\text{MnS}$.
• Hence, passing $\text{H}_2\text{S}$ in an acidic medium produces a precipitate only with $\text{Cu}^{2+}$.
• Therefore, statement (B) is correct.

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3. Analysis of Option (C):

• In a faintly basic medium (such as in the presence of $\text{NH}_4\text{OH}$), $\text{OH}^-$ ions react with $\text{H}^+$ from $\text{H}_2\text{S}$ to form water, removing $\text{H}^+$ and shifting the equilibrium to produce a high concentration of $\text{S}^{2-}$ ions.
• Under these basic conditions, the ionic product exceeds the $K_{sp}$ values of both $\text{CuS}$ and $\text{MnS}$. Thus, both ions form sulfide precipitates ($\text{CuS}$ as a black precipitate and $\text{MnS}$ as a buff/flesh-colored precipitate).
• Since both ions precipitate, it is incorrect to state that only $\text{Mn}^{2+}$ precipitates.
• Therefore, statement (C) is incorrect.

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4. Analysis of Option (D):

• Standard reduction potentials ($E^\circ$) at $298\text{ K}$ for the two redox couples are: $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\text{ V}$ $E^\circ_{\text{Mn}^{2+}/\text{Mn}} = -1.18\text{ V}$
• Since $+0.34\text{ V} > -1.18\text{ V}$, the reduction potential of $\text{Cu}^{2+}/\text{Cu}$ is higher than that of $\text{Mn}^{2+}/\text{Mn}$ under identical conditions.
• Therefore, statement (D) is correct.

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Conclusion:

The correct statements to distinguish between the two salts are (B) and (D).