Problem 25 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The pair having the same magnetic moment is: [At. No.: \(\mathrm{Cr}=24, \mathrm{Mn}=25, \mathrm{Fe}=26, \mathrm{Co}=27]\) (a) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) and \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (b) \(\left[\mathrm{CoCl}_{4}\right]^{2-}\) and \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (c) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) and \(\left[\mathrm{CoCl}_{4}\right]^{2-}\) (d) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) and \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\)

Step-by-Step Solution

Answer

The pair with the same magnetic moment is option (d).

Step 1: Determine the oxidation state

For each complex, identify the oxidation state of the central metal ion.- In \([\mathrm{Mn}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}\), the oxidation state of Mn is +2.- In \([\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}\), the oxidation state of Cr is +2.- In \([\mathrm{Fe}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}\), the oxidation state of Fe is +2.- In \([\mathrm{CoCl}_{4}]^{2-}\), the oxidation state of Co is +2.

Step 2: Calculate the number of electron spins

Determine the number of unpaired electrons for each metal in its oxidation state by using its electron configuration. - Mn in +2 oxidation state: [Ar] 3d^5, 5 unpaired electrons. - Cr in +2 oxidation state: [Ar] 3d^4, 4 unpaired electrons. - Fe in +2 oxidation state: [Ar] 3d^6, 4 unpaired electrons. - Co in +2 oxidation state: [Ar] 3d^7, 3 unpaired electrons.

Step 3: Use the magnetic moment formula

For the magnetic moment \(\mu\), use the formula: \(\mu = \sqrt{n(n+2)}\) Bohr Magnetons (B.M.), where \(n\) is the number of unpaired electrons.- For Mn: \(n = 5\), \(\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\) B.M.- For Cr: \(n = 4\), \(\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\) B.M.- For Fe: \(n = 4\), \(\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\) B.M.- For Co: \(n = 3\), \(\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\) B.M.

Step 4: Compare magnetic moments

Compare the magnetic moments calculated for each complex.- \([\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}\) and \([\mathrm{Fe}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}\) both have a magnetic moment of approximately 4.90 B.M.

Key Concepts

Oxidation State

In chemistry, an oxidation state is a measure of the degree of oxidation of an atom in a chemical compound. It's often represented as a positive or negative integer, depending on the nature of the electron transfer. Understanding oxidation states is crucial because they help identify how electrons are distributed in a compound.
An atom's oxidation state can offer insight into its potential chemical reactions.
For transition metals, like the elements considered in this exercise, oxidation states can vary due to their ability to donate or accept various numbers of electrons. In the exercise, the central metal ions in each complex are in the +2 oxidation state. For example:

• Manganese in \([\mathrm{Mn}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}\) has a +2 oxidation state because of the neutral H2O ligands and the overall 2+ charge.
• Similarly, chromium, iron, and cobalt also achieve a +2 oxidation state in their respective complexes.

Identifying the oxidation state helps in further calculations, such as determining electron configuration or the number of unpaired electrons, which are essential for understanding the magnetic properties of compounds.

Unpaired Electrons

Unpaired electrons play a pivotal role in determining the magnetic properties of a molecule. These are the electrons that do not pair up with another electron in an atomic or molecular orbital. They lead to a net magnetic moment due to their spin.

• The more unpaired electrons there are, the stronger the magnetic moment and, consequently, the magnetic property of the atom or molecule.
• In our example, manganese in its +2 state has 5 unpaired electrons, which contributes to its high magnetic moment.
• Chromium and iron both have 4 unpaired electrons in their +2 oxidation states, leading to a similar magnetic moment.

The formula for magnetic moment, \(\mu = \sqrt{n(n+2)}\), where \(n\) is the number of unpaired electrons, highlights why unpaired electron count is crucial for determining magnetic properties. The formula shows us a direct relationship between the number of unpaired electrons and the resulting magnetic moment.

Electron Configuration

Electron configuration is a shorthand notation that shows the distribution of electrons among the orbitals of an atom. It is crucial in determining the chemical and physical properties of elements or ions, including their magnetic properties.
In the context of the transition metals being discussed:

• The electron configuration helps us identify how electrons occupy d-orbitals, which is key for locating unpaired electrons.
• For manganese in the +2 state, we have the configuration \([\mathrm{Ar}] 3d^5\), with 5 unpaired electrons visible.
• Chromium in the +2 state shows \([\mathrm{Ar}] 3d^4\), iron \([\mathrm{Ar}] 3d^6\), and cobalt \([\mathrm{Ar}] 3d^7\), reflecting the number of unpaired electrons and their influence on their respective magnetic moments.

Understanding electron configuration can reveal the expected chemical behavior of elements due to electron arrangement, predictive unpaired electron count, and thus the molecule’s magnetic moment.