Problem 155 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following ions has the maximum magnetic moment? (a) \(\mathrm{Mn}^{2+}\) (b) \(\mathrm{Fe}^{2+}\) (c) \(\mathrm{Ti}^{2+}\) (d) \(\mathrm{Cr}^{2+}\)

Step-by-Step Solution

Answer

The ion with the maximum magnetic moment is \( \mathrm{Mn}^{2+} \).

Step 1: Determine the Electron Configuration

We need to identify the electronic configuration for each ion. For this, we consider the atomic number and the loss of electrons typical for a +2 ion.- \( \mathrm{Mn} \ : 3d^5 4s^2 \rightarrow \mathrm{Mn^{2+}} \ : 3d^5 \)- \( \mathrm{Fe} \ : 3d^6 4s^2 \rightarrow \mathrm{Fe^{2+}} \ : 3d^6 \)- \( \mathrm{Ti} \ : 3d^2 4s^2 \rightarrow \mathrm{Ti^{2+}} \ : 3d^2 \)- \( \mathrm{Cr} \ : 3d^5 4s^1 \rightarrow \mathrm{Cr^{2+}} \ : 3d^4 \)

Step 2: Calculate the Number of Unpaired Electrons

The magnetic moment arises from unpaired electrons. For each ion, count the unpaired electrons in the 3d orbitals:- \( \mathrm{Mn^{2+}} \ : 3d^5 \), with 5 unpaired electrons.- \( \mathrm{Fe^{2+}} \ : 3d^6 \), with 4 unpaired electrons.- \( \mathrm{Ti^{2+}} \ : 3d^2 \), with 2 unpaired electrons.- \( \mathrm{Cr^{2+}} \ : 3d^4 \), with 4 unpaired electrons.

Step 3: Use the Magnetic Moment Formula

The formula for the magnetic moment \( \mu \) is given by \( \mu = \sqrt{n(n+2)} \) where \( n \) is the number of unpaired electrons.- \( \mathrm{Mn^{2+}} \ : \mu = \sqrt{5(5+2)} = \sqrt{35} \)- \( \mathrm{Fe^{2+}} \ : \mu = \sqrt{4(4+2)} = \sqrt{24} \)- \( \mathrm{Ti^{2+}} \ : \mu = \sqrt{2(2+2)} = \sqrt{8} \)- \( \mathrm{Cr^{2+}} \ : \mu = \sqrt{4(4+2)} = \sqrt{24} \)

Step 4: Compare Magnetic Moments

Upon comparing the calculated magnetic moments, it is clear that \( \mathrm{Mn^{2+}} \) has the highest magnetic moment of \( \sqrt{35} \), followed by \( \mathrm{Fe^{2+}} \) and \( \mathrm{Cr^{2+}} \) which have equal moments, and \( \mathrm{Ti^{2+}} \) with the lowest moment.

Key Concepts

Electron Configuration

Electron configuration is a method used to describe the arrangement of electrons within an atom or ion. This is crucial for understanding the chemical behavior of elements. In a neutral atom, electrons fill orbitals starting from the lowest energy level to higher ones, following the Aufbau principle. The specific order typically follows the sequence of 1s, 2s, 2p, 3s, 3p, 4s, and 3d. For transition metals, the 3d orbitals, which form part of the valence shell, are often occupied after the 4s orbitals.

When an atom becomes a positive ion, it loses electrons from its outermost shell. For transition metal ions, that usually means electrons are lost first from the 4s orbital before the 3d orbital. For instance, consider the neutral manganese atom, which has the electron configuration of \( ext{Mn: 3d}^5 ext{4s}^2 \). When it becomes \( ext{Mn}^{2+} \), it loses two electrons from the 4s orbital leaving it as \( ext{Mn}^{2+}: 3d^5 \).

Understanding electron configurations helps predict the number of unpaired electrons in an ion, crucial for calculating magnetic moments.

Unpaired Electrons

Unpaired electrons play a significant role in determining the magnetic properties of an atom or ion. Electrons in an atom or ion occupy orbitals according to the Pauli exclusion principle and Hund’s rule, which dictate that electrons fill degenerate orbitals singly before pairing up.

Unpaired electrons in the d-orbitals of transition metals are particularly important. These unpaired electrons contribute to the ion's overall magnetic moment because each unpaired electron generates a magnetic field. An ion's magnetic properties can be described as paramagnetic if there are unpaired electrons, contributing to a stronger magnetic moment.

Taking \( ext{Mn}^{2+}: 3d^5 \) as an example, it has 5 unpaired electrons. This large number of unpaired electrons results in the highest magnetic moment among other transition metal ions like \( ext{Fe}^{2+} \) and \( ext{Cr}^{2+} \), which have 4 unpaired electrons each.

Magnetic Moment Formula

The magnetic moment of an ion is calculated using a simple formula related to the number of unpaired electrons. The magnetic moment \( \mu \) in units of Bohr magneton (\( \mu_B \)) can be computed using the formula:\[ \mu = \sqrt{n(n+2)} \]where \( n \) is the number of unpaired electrons.

This formula gives a direct way to compare the magnetic properties of different ions based solely on their electronic configuration and unpaired electrons. For example, for \( \text{Mn}^{2+} \) with 5 unpaired electrons, the calculation becomes \( \mu = \sqrt{5(5+2)} = \sqrt{35} \). This value is larger compared to that for ions like \( \text{Ti}^{2+} \) with 2 unpaired electrons, resulting in \( \mu = \sqrt{8} \), showing how \( \text{Mn}^{2+} \) has a significantly higher magnetic moment.

Transition Metal Ions

Transition metal ions are unique in their electronic structure, especially regarding their d-orbitals. These elements, found in the d-block of the periodic table, can exhibit variable oxidation states. This results from the fact that both their s and d electrons can be involved in bonding or ionization.

As ions, transition metals often lose s-electrons first, as seen with ions like \( \text{Fe}^{2+} \) or \( \text{Mn}^{2+} \), which impacts their electron configuration. This electron loss alters magnetic properties, making the ions more likely to interact magnetically due to unpaired electrons in the d-orbitals. Students of chemistry find these behaviors especially interesting because they lead to unique properties such as high magnetic moments, complex coordination chemistry, and vibrant coloration in compounds.

For instance, the \( \text{Cr}^{2+} \) ion with a configuration of \( 3d^4 \) displays distinct magnetic properties and reactivity, revealing the versatility and importance of transition metals in both chemistry and practical applications.