Problem 33 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The correct option(s) to distinguish nitrate salts of \(\mathrm{Mn}^{2+}\) and \(\mathrm{Cu}^{2+}\) taken separately is (are) (a) \(\mathrm{Mn}^{2+}\) shows the characteristic green colour in the flame test (b) Only \(\mathrm{Cu}^{2+}\) shows the formation of precipitate by passing \(\mathrm{H}_{2} \mathrm{~S}\) in acidic medium (c) Only \(\mathrm{Mn}^{2+}\) shows the formation of precipitate by passing \(\mathrm{H}_{2} \mathrm{~S}\) in faintly basic medium (d) \(\mathrm{Cu}^{2+} / \mathrm{Cu}\) has higher reduction potential than \(\mathrm{Mn}^{2+} / \mathrm{Mn}\) (measured under similar conditions)

Step-by-Step Solution

Answer

Options (b), (c), and (d) are correct for distinction.

Step 1: Analyze Flame Test for Mn²⁺

In a flame test, manganese (II) ions, \( \mathrm{Mn}^{2+} \), generally do not show a distinct green color. Manganese does not produce a notable color in the flame test, so this option (a) is likely incorrect for distinguishing \( \mathrm{Mn}^{2+} \) nitrate.

Step 2: Precipitation with H₂S in Acidic Medium

Copper (II) ions, \( \mathrm{Cu}^{2+} \), form a black precipitate of copper (II) sulfide, \( \mathrm{CuS} \), when \( \mathrm{H}_2\mathrm{S} \) is passed in an acidic medium. Manganese (II) does not form a sulfide precipitate under acidic conditions. Thus, option (b) is correct for distinguishing \( \mathrm{Cu}^{2+} \) nitrate.

Step 3: Precipitation with H₂S in Faintly Basic Medium

In a faintly basic medium (around neutral pH), \( \mathrm{Mn}^{2+} \) can form a precipitate of manganese (II) sulfide, \( \mathrm{MnS} \). Copper does not precipitate under these conditions. Therefore, option (c) is correct for distinguishing \( \mathrm{Mn}^{2+} \) nitrate.

Step 4: Compare Reduction Potentials

The standard reduction potential for \( \mathrm{Cu}^{2+}/\mathrm{Cu} \) is higher than for \( \mathrm{Mn}^{2+}/\mathrm{Mn} \), meaning copper is more easily reduced than manganese under similar conditions. This makes option (d) correct, but it serves better for theoretical comparisons rather than a practical test of distinction in the lab.

Key Concepts

Nitrate Salts

Nitrate salts are composed of nitrate ions, \(\mathrm{NO}_3^-\), combined with metal cations like \(\mathrm{Mn}^{2+}\) or \(\mathrm{Cu}^{2+}\). These salts are important in inorganic chemistry for several reasons. Firstly, nitrate salts are highly soluble in water, which makes them suitable for various chemical reactions in aqueous solutions.
Nitrate ions are stable and act as a good spectator ion during reactions, not typically engaging in redox or precipitation reactions themselves.
Their presence in a salt influences the solubility and reactivity of the compound without directly interfering with the substance's main reaction pathways.
Nitrate salts are prevalent in fertilizer production, pyrotechnics, and some explosives due to their oxidizing properties. Examples include potassium nitrate \(\mathrm{KNO}_3\) and ammonium nitrate \(\mathrm{NH}_4\mathrm{NO}_3\). Understanding nitrate salts help students identify the role of each component in a compound's reactivity.

Flame Test

The flame test is a qualitative analysis technique used to detect the presence of certain metal ions based on the color they emit when heated in a flame. Each metal ion produces a characteristic color due to the excitation of electrons when exposed to the high heat of a flame.
However, not all metal ions yield distinct or vibrant colors, making the flame test more useful for some metals than others. For instance, manganese (II) ions \(\mathrm{Mn}^{2+}\) do not result in a notable green color, making it an ineffective test for identifying manganese nitrate salts. Meanwhile, copper ions may show a faint bluish-green color in a flame test, but it is more effectively distinguished by other chemical techniques.

• Convenient for identifying alkali metals and certain alkaline earth metals.
• Not suitable for metals without a clear bright flame color.

This test is especially useful for educational purposes and preliminary identification in laboratory settings.

Precipitation Reactions

Precipitation reactions involve the formation of an insoluble solid, called a precipitate, when two aqueous solutions react. This is a type of double displacement reaction where one of the resulting products is not soluble in water.
In distinguishing manganese and copper ions, \(\mathrm{H}_2\mathrm{S}\) can be added to the solution. Different conditions influence precipitation:

• In an acidic medium, copper \(\mathrm{Cu}^{2+}\) ions precipitate as copper (II) sulfide \(\mathrm{CuS}\), a black solid.
• In a faintly basic medium, manganese \(\mathrm{Mn}^{2+}\) ions precipitate as manganese (II) sulfide \(\mathrm{MnS}\).

Understanding precipitation reactions is crucial because it allows chemists to remove specific ions from solution or separate components in a mixture. The solubility rules, which tell us which salts are soluble or insoluble in water, are fundamental to predicting when a precipitation reaction will occur.

Reduction Potential

Reduction potential measures a substance's tendency to gain electrons and thus be reduced. It is quantified in volts and is a crucial concept in electrochemistry. A higher reduction potential indicates a greater likelihood that a substance will be reduced under given conditions.
In comparing \(\mathrm{Cu}^{2+}/\mathrm{Cu}\) and \(\mathrm{Mn}^{2+}/\mathrm{Mn}\), copper has a higher reduction potential, meaning it will reduce more readily than manganese. This characteristic isn't just a theoretical concept; it affects how these metals behave in redox reactions.
Reduction potentials are used to predict the direction of electron flow in electrochemical cells, influencing battery design and other applications where controlled redox reactions are required. By understanding reduction potentials, students can anticipate which metals will react in particular conditions and plan for procedures such as galvanic cell construction or metal extraction.