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 \(\mathrm{V}^{2+}\) ? (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 Electronic Configuration

To calculate the spin-only magnetic moment, we need to know the electronic configuration of the ions. For \(\mathrm{Mn}^{2+}\), the configuration is \([\mathrm{Ar}] 3d^5\). For \(\mathrm{Cr}^{2+}\), it is \([\mathrm{Ar}] 3d^4\). For \(\mathrm{V}^{2+}\), the configuration is \([\mathrm{Ar}] 3d^3\).

Step 2: Calculate Number of Unpaired Electrons

The number of unpaired electrons is equal to the number of electrons in the unfilled \(d\)-orbitals. \(\mathrm{Mn}^{2+}\) has 5 unpaired electrons, \(\mathrm{Cr}^{2+}\) has 4 unpaired electrons, and \(\mathrm{V}^{2+}\) has 3 unpaired electrons.

Step 3: Use the Spin Only Magnetic Moment Formula

The formula for spin-only magnetic moment is \(\mu = \sqrt{n(n+2)}\), where \(n\) is the number of unpaired electrons. For \(\mathrm{Mn}^{2+}\), \(\mu = \sqrt{5(5+2)} = \sqrt{35}\), for \(\mathrm{Cr}^{2+}\), \(\mu = \sqrt{4(4+2)} = \sqrt{24}\), and for \(\mathrm{V}^{2+}\), \(\mu = \sqrt{3(3+2)} = \sqrt{15}\).

Step 4: Compare Spin Only Magnetic Moments

Calculate the numerical values: \(\sqrt{35} \approx 5.92\), \(\sqrt{24} \approx 4.90\), \(\sqrt{15} \approx 3.87\). Thus, the order by magnetic moment is \(\mathrm{Mn}^{2+} > \mathrm{Cr}^{2+} > \mathrm{V}^{2+}\).

Key Concepts

Electronic Configuration

Electronic configuration is fundamental in understanding the properties of an atom or ion, especially its magnetic behavior. Each ion, including transition metals like \(\mathrm{Mn}^{2+}, \mathrm{Cr}^{2+},\) and \(\mathrm{V}^{2+}\), has its own unique configuration of electrons. An electron configuration describes the distribution of electrons in an atom's atomic orbitals. - For \(\mathrm{Mn}^{2+}\), converting from \(\mathrm{Mn}\)'s neutral state, we remove two electrons, resulting in \([\mathrm{Ar}] 3d^5\).
- \(\mathrm{Cr}^{2+}\) follows similarly, leading to \([\mathrm{Ar}] 3d^4\).
- For \(\mathrm{V}^{2+}\), it is \([\mathrm{Ar}] 3d^3\).Understanding the electronic configuration is crucial because it tells us about the arrangement of electrons, specifically how many unpaired electrons are present, which is key to determining magnetic properties.

Unpaired Electrons

Unpaired electrons are electrons that are alone in an orbital without a partner of opposite spin. These play a critical role in the magnetic properties of an ion. Let's explore the number of unpaired electrons:- In \(\mathrm{Mn}^{2+}\), which has an electronic configuration of \([\mathrm{Ar}] 3d^5\), you have 5 unpaired electrons, one in each of the five 3d orbitals.
- \(\mathrm{Cr}^{2+}\) has \([\mathrm{Ar}] 3d^4\), resulting in 4 unpaired electrons distributed among the orbitals.
- In \(\mathrm{V}^{2+}\), with \([\mathrm{Ar}] 3d^3\), there are 3 unpaired electrons.These unpaired electrons are significant because they contribute directly to the magnetic moment of the ion, as only unpaired electrons generate a magnetic field.

Magnetic Moment Formula

The magnetic moment is a measure of the strength and orientation of a magnet's magnetic field. For calculating the spin-only magnetic moment, a simplified formula is commonly used, especially for transition metals:\[\mu = \sqrt{n(n+2)}\]where \( n \) is the number of unpaired electrons.Using this formula:- For \(\mathrm{Mn}^{2+}\), with 5 unpaired electrons, the calculation is \(\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\) Bohr Magnetons (B.M.).
- For \(\mathrm{Cr}^{2+}\), with 4 unpaired electrons, \(\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\) B.M.
- Lastly, for \(\mathrm{V}^{2+}\), with 3 unpaired electrons: \(\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\) B.M.By using these calculated values, we can compare the magnetic moments and rank them accordingly, revealing the order: \(\mathrm{Mn}^{2+} > \mathrm{Cr}^{2+} > \mathrm{V}^{2+}\). This highlight states that more unpaired electrons contribute to a higher magnetic moment.