Problem 133 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following ions has the highest magnetic moment? (a) \(\mathrm{Fe}^{2+}\) (b) \(\mathrm{Mn}^{2+}\) (c) \(\mathrm{Cr}^{3+}\) (d) \(\mathrm{V}^{3+}\)

Step-by-Step Solution

Answer

\( \mathrm{Mn}^{2+} \) has the highest magnetic moment.

Step 1: Determine Electron Configuration

First, determine the electron configuration of each ion. Fe is in group 8, so - \( \mathrm{Fe}^{2+} \) has the electron configuration \([\mathrm{Ar}] 3d^6\) - \( \mathrm{Mn}^{2+} \) has the electron configuration \([\mathrm{Ar}] 3d^5\) - \( \mathrm{Cr}^{3+} \) has the electron configuration \([\mathrm{Ar}] 3d^3\) - \( \mathrm{V}^{3+} \) has the electron configuration \([\mathrm{Ar}] 3d^2\).

Step 2: Calculate Unpaired Electrons

Determine the number of unpaired electrons in the outermost shell, as these contribute to the magnetic moment. - \( \mathrm{Fe}^{2+} \) has 4 unpaired electrons - \( \mathrm{Mn}^{2+} \) has 5 unpaired electrons - \( \mathrm{Cr}^{3+} \) has 3 unpaired electrons - \( \mathrm{V}^{3+} \) has 2 unpaired electrons.

Step 3: Apply Magnetic Moment Formula

Use the formula for magnetic moment: \[ \mu = \sqrt{n(n+2)} \] where \( n \) is the number of unpaired electrons. - For \( \mathrm{Fe}^{2+} \), \( \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\)- For \( \mathrm{Mn}^{2+} \), \( \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\)- For \( \mathrm{Cr}^{3+} \), \( \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\)- For \( \mathrm{V}^{3+} \), \( \mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83\).

Step 4: Compare Magnetic Moments

Compare the calculated magnetic moments to determine which ion has the highest value. The calculated magnetic moment for \( \mathrm{Mn}^{2+} \) is the highest at approximately 5.92.

Key Concepts

Electron Configuration

In the world of chemistry, electron configuration is a way of describing the arrangement of electrons in an atom or ion. It's akin to having a detailed map that shows where each electron resides. For an ion, the electron configuration may vary from that of its neutral atom because the gain or loss of electrons changes this layout.

The electron configuration fundamentally defines the chemical properties of the element in question. In our exercise, we are handling transition metal ions like \( \mathrm{Fe}^{2+} \) and \( \mathrm{Mn}^{2+} \). These ions stem from iron and manganese, respectively, and have electron configurations as follows:

• \( \mathrm{Fe}^{2+} \) has the electron configuration \([\mathrm{Ar}] 3d^6 \)
• \( \mathrm{Mn}^{2+} \) has the electron configuration \([\mathrm{Ar}] 3d^5 \)
• \( \mathrm{Cr}^{3+} \) corresponds to \([\mathrm{Ar}] 3d^3 \)
• \( \mathrm{V}^{3+} \) corresponds to \([\mathrm{Ar}] 3d^2 \)

The configurations show us which electron orbitals are filled, helping us to glean insights into their chemical behavior.

Unpaired Electrons

Unpaired electrons are electrons that occupy an orbital singly rather than as part of an electron pair. These are crucial in determining the magnetic properties of an element or compound because unpaired electrons create an uneven distribution of electron spins, which contribute to magnetism.

By looking at the electron configuration of each ion, we can swiftly determine the number of unpaired electrons. Transition metals often have interesting magnetic properties because of the various ways their electrons can fill the \(d\) orbitals, often resulting in unpaired electrons:

• \( \mathrm{Fe}^{2+} \) contains 4 unpaired electrons.
• \( \mathrm{Mn}^{2+} \) contains 5 unpaired electrons.
• \( \mathrm{Cr}^{3+} \) contains 3 unpaired electrons.
• \( \mathrm{V}^{3+} \) contains 2 unpaired electrons.

The higher the number of unpaired electrons, the higher the potential magnetic moment for the ion.

Magnetic Moment Formula

The magnetic moment is a commonly used factor to quantify the magnetic strength and orientation of a magnetic entity, like an ion. The measurement of magnetic moment is directly influenced by the number of unpaired electrons.

To calculate the magnetic moment, we use the formula:
\[ \mu = \sqrt{n(n+2)} \]
where \( n \) is the number of unpaired electrons. This relationship suggests that the magnetic moment rises with an increase in the number of unpaired electrons. Let's apply it to our examples:

• \( \mathrm{Fe}^{2+} \) has a magnetic moment \( \mu = \sqrt{4(4+2)} \approx 4.90 \)
• \( \mathrm{Mn}^{2+} \) has a magnetic moment \( \mu = \sqrt{5(5+2)} \approx 5.92 \)
• \( \mathrm{Cr}^{3+} \) has a magnetic moment \( \mu = \sqrt{3(3+2)} \approx 3.87 \)
• \( \mathrm{V}^{3+} \) has a magnetic moment \( \mu = \sqrt{2(2+2)} \approx 2.83 \)

The ion with the highest magnetic moment, therefore, would be \( \mathrm{Mn}^{2+} \), as it has the highest number of unpaired electrons, yielding the strongest magnetic properties.