Problem 56 - A Complete Resource Book in Chemistry for JEE Main

Question

What is the current order of spin only magnetic moment (in B.M.) of \(\mathrm{Mn}^{2+}, \mathrm{Cr}^{2+}\) and \(\mathrm{V}^{2+\text { ? }}\) (a) \(\mathrm{Mn}^{2+}>\mathrm{V}^{2+}>\mathrm{Cr}^{2+}\) (b) \(\mathrm{V}^{2+}>\mathrm{Cr}^{2+}>\mathrm{Mn}^{2+}\) (c) \(\mathrm{Mn}^{2+}>\mathrm{Cr}^{2+}>\mathrm{V}^{2+}\) (d) \(\mathrm{Cr}^{2+}>\mathrm{V}^{2+}>\mathrm{Mn}^{2+}\)

Step-by-Step Solution

Answer

(c) \( \mathrm{Mn^{2+}} > \mathrm{Cr^{2+}} > \mathrm{V^{2+}} \)

Step 1: Determine the Electron Configuration

Identify the electron configurations for each transition metal ion. For \( \mathrm{Mn^{2+}} \), the configuration is \( [\mathrm{Ar}] \, 3d^5 \).For \( \mathrm{Cr^{2+}} \), the configuration is \( [\mathrm{Ar}] \, 3d^4 \).For \( \mathrm{V^{2+}} \), the configuration is \( [\mathrm{Ar}] \, 3d^3 \).

Step 2: Count the Unpaired Electrons

Calculate the number of unpaired electrons based on the electron configurations:- \( \mathrm{Mn^{2+}} \) has 5 unpaired electrons.- \( \mathrm{Cr^{2+}} \) has 4 unpaired electrons.- \( \mathrm{V^{2+}} \) has 3 unpaired electrons.

Step 3: Calculate Spin-Only Magnetic Moment

Use the formula for spin-only magnetic moment: \[ \mu = \sqrt{n(n+2)} \] where \( n \) is the number of unpaired electrons.Calculate for each ion:- For \( \mathrm{Mn^{2+}} \): \( \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, \text{B.M.} \)- For \( \mathrm{Cr^{2+}} \): \( \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \, \text{B.M.} \)- For \( \mathrm{V^{2+}} \): \( \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, \text{B.M.} \)

Step 4: Order the Magnetic Moments

Compare the calculated magnetic moments:- \( \mathrm{Mn^{2+}} \) has the highest magnetic moment (5.92 B.M.).- \( \mathrm{Cr^{2+}} \) has the second-highest (4.90 B.M.).- \( \mathrm{V^{2+}} \) has the lowest (3.87 B.M.).Therefore, the order is \( \mathrm{Mn^{2+}} > \mathrm{Cr^{2+}} > \mathrm{V^{2+}} \).

Key Concepts

Transition Metal Ions

Transition metal ions are particularly interesting due to their unique electron configurations and magnetic properties. These ions are derived from transition metals, which are elements found in the center of the periodic table, specifically in the d-block. They are known for having partially filled d-orbitals. This characteristic gives rise to a variety of oxidation states. Transition metal ions, such as - \( \mathrm{Mn^{2+}} \)- \( \mathrm{Cr^{2+}} \)- \( \mathrm{V^{2+}} \)are created by the loss of electrons, typically from the outermost s and d sub-shells. This loss of electrons changes the electron configuration and hence alters the chemical and physical properties of the metal. Understanding these ions is crucial for predicting behavior in chemical reactions and properties such as magnetism.

Electron Configurations

Electron configurations are a method to represent the arrangement of electrons in an atom or ion. Electrons are placed in different energy levels, sublevels, and orbitals following specific rules and principles like the Aufbau principle, Hund's rule, and Pauli exclusion principle. For transition metal ions, the electron configuration becomes especially interesting once they lose electrons.
For example, in the transition to a 2+ charge, manganese \( \mathrm{Mn^{2+}} \) has an electron configuration of - \( [\mathrm{Ar}] \, 3d^5 \), indicating it has lost two electrons from the 4s and 3d orbitals. Similarly, chromium \( \mathrm{Cr^{2+}} \) and vanadium \( \mathrm{V^{2+}} \) have electron configurations of - \( [\mathrm{Ar}] \, 3d^4 \)- \( [\mathrm{Ar}] \, 3d^3 \), respectively. An accurate understanding of these configurations is necessary to determine properties such as the number of unpaired electrons.

Unpaired Electrons

The number of unpaired electrons in an ion significantly affects its magnetic properties. Electrons in atoms and ions can be paired or unpaired within their orbitals. Unpaired electrons, with their spin, give rise to magnetism. Transition metals, by nature of having partially filled d orbitals, often exhibit unpaired electrons, making them paramagnetic.
For the ions in this exercise:- \( \mathrm{Mn^{2+}} \) has 5 unpaired electrons.- \( \mathrm{Cr^{2+}} \) has 4 unpaired electrons.- \( \mathrm{V^{2+}} \) has 3 unpaired electrons.The number of unpaired electrons will directly influence the spin-only magnetic moment, which is an intrinsic property related to the ion's magnetism. More unpaired electrons generally mean a higher magnetic moment.

Magnetic Moment Formula

The spin-only magnetic moment is a measure of the magnetism of a system arising from its unpaired electrons. Calculated using the formula:\[ \mu = \sqrt{n(n+2)} \]where \( \mu \) is the magnetic moment expressed in Bohr Magnetons (B.M.) and \( n \) is the number of unpaired electrons. This formula helps predict the magnetic behavior of transition metal ions.
For instance:- For \( \mathrm{Mn^{2+}} \), \( n = 5 \), resulting in \( \mu \approx 5.92 \, \text{B.M.} \).- For \( \mathrm{Cr^{2+}} \), \( n = 4 \), resulting in \( \mu \approx 4.90 \, \text{B.M.} \).- For \( \mathrm{V^{2+}} \), \( n = 3 \), resulting in \( \mu \approx 3.87 \, \text{B.M.} \).These calculations reveal that the order of magnetic moment is \( \mathrm{Mn^{2+}} > \mathrm{Cr^{2+}} > \mathrm{V^{2+}} \), providing a quantifiable means to compare the magnetism of different ions. Understanding how to apply and calculate this formula is important for students learning about the magnetic properties of atoms and ions.