Problem 49 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Amongst the following, the lowest degree of paramagnetism per mole of the compound at \(298 \mathrm{~K}\) will be shown by (a) \(\mathrm{MnSO}_{4} \cdot 4 \mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{FeSO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{NiSO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}\)

Step-by-Step Solution

Answer

(b) CuSO₄ · 5H₂O has the lowest paramagnetism per mole.

Step 1: Identify the Metal Ions

Examine each compound to determine the metal ion present. (a) \(\mathrm{MnSO}_{4} \cdot 4 \mathrm{H}_{2} \mathrm{O}\) - metal ion Mn (manganese).(b) \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) - metal ion Cu (copper).(c) \(\mathrm{FeSO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) - metal ion Fe (iron).(d) \(\mathrm{NiSO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) - metal ion Ni (nickel).

Step 2: Determine the Oxidation States

Find the oxidation state of each metal ion in the compounds:(a) Mn in \(\mathrm{MnSO}_{4}\) is in the +2 oxidation state.(b) Cu in \(\mathrm{CuSO}_{4}\) is in the +2 oxidation state.(c) Fe in \(\mathrm{FeSO}_{4}\) is in the +2 oxidation state.(d) Ni in \(\mathrm{NiSO}_{4}\) is in the +2 oxidation state.

Step 3: Determine the Electron Configuration

Identify the electron configuration of each metal ion:(a) Mn\(^{2+}\): \([\mathrm{Ar}] \, 3d^5\)(b) Cu\(^{2+}\): \([\mathrm{Ar}] \, 3d^9\)(c) Fe\(^{2+}\): \([\mathrm{Ar}] \, 3d^6\)(d) Ni\(^{2+}\): \([\mathrm{Ar}] \, 3d^8\)

Step 4: Identify Unpaired Electrons

Count the number of unpaired electrons in each ion based on their electron configurations:(a) Mn\(^{2+}\): 5 unpaired electrons.(b) Cu\(^{2+}\): 1 unpaired electron.(c) Fe\(^{2+}\): 4 unpaired electrons.(d) Ni\(^{2+}\): 2 unpaired electrons.

Step 5: Compare Paramagnetism

Recall that the degree of paramagnetism is directly related to the number of unpaired electrons. Less unpaired electrons indicate lower paramagnetism.The compounds have the following number of unpaired electrons:- Mn\(^{2+}\): 5- Cu\(^{2+}\): 1- Fe\(^{2+}\): 4- Ni\(^{2+}\): 2Cu\(^{2+}\) has the fewest unpaired electrons.

Key Concepts

Unpaired Electrons

Unpaired electrons are key players in determining the magnetic properties of atoms and molecules. In the presence of unpaired electrons, compounds typically exhibit paramagnetism. This occurs because unpaired electrons have magnetic moments that can align with an external magnetic field. Hence, more unpaired electrons lead to stronger magnetic properties.
To understand if a compound is paramagnetic, we often refer to its electron configuration. For example, here are the electron configurations for the metal ions given in the compounds:

• Mn\(^{2+}\) has 5 unpaired electrons with the configuration \([\text{Ar}] \ 3d^5\).
• Cu\(^{2+}\) has only 1 unpaired electron: \([\text{Ar}] \ 3d^9\).
• Fe\(^{2+}\) comes with 4 unpaired electrons in \([\text{Ar}] \ 3d^6\).
• Ni\(^{2+}\) possesses 2 unpaired electrons as \([\text{Ar}] \ 3d^8\).

As you see, compounds with fewer unpaired electrons show less paramagnetism, like Cu\(^{2+}\), which has only one unpaired electron and thus is the least paramagnetic of the group.

Metal Ions

Metal ions are atoms of metallic elements that have lost one or more electrons and carry a net positive charge. They play an important role in many chemical reactions and determine a compound's properties. In the context of the given problem, the metal ions Mn\(^{2+}\), Cu\(^{2+}\), Fe\(^{2+}\), and Ni\(^{2+}\) are the central atoms in their respective compounds.

The charges on these ions arise from the loss of electrons. For instance, Mn\(^{2+}\), Fe\(^{2+}\), Ni\(^{2+}\), and Cu\(^{2+}\) have each lost two electrons. This is indicative of their +2 oxidation state. The behavior of metal ions under magnetic influences largely depends on their electronic configuration, impacting properties like paramagnetism. Understanding the characteristics of these ions helps us predict how the substance as a whole will interact with a magnetic field.

Oxidation States

Oxidation states indicate the degree of oxidation or loss of electrons of an atom. They are essential for understanding many properties of compounds, including magnetic properties and chemical reactivity. Each of the metals in the provided problem is in a +2 oxidation state. This means each metal ion has lost two electrons compared to its neutral state.

• Manganese in Mn\(^{2+}\) is in the +2 state, meaning it generally started as a manganese atom and lost two electrons.
• Copper, iron, and nickel in Cu\(^{2+}\), Fe\(^{2+}\), and Ni\(^{2+}\), respectively, also share this +2 state.

The oxidation state not only tells us about electron loss but also helps determine the electron configuration of the ion. As a result, we can find how many unpaired electrons remain. Since we know that fewer unpaired electrons lead to lower paramagnetism, the oxidation state helps us quickly determine a compound's magnetic characteristics. For instance, Cu\(^{2+}\) having the lowest number of unpaired electrons aligns with its lowest degree of paramagnetism among the given options.