Problem 59 - A Complete Resource Book in Chemistry for JEE Main

Question

The pair in which both species have the same magnetic moment (spin only value) is (a) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+},\left[\mathrm{CoCl}_{4}\right]^{2-}\) (b) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+},\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (c) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+},\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (d) \(\left[\mathrm{CoCl}_{4}\right]^{2-},\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\)

Step-by-Step Solution

Answer

(b) \([\text{Cr}(\text{H}_2\text{O})_6]^{2+}, [\text{Fe}(\text{H}_2\text{O})_6]^{2+}\)

Step 1: Determine the oxidation states

First, determine the oxidation states of the central metal ions for each complex. In \([\text{Cr}(\text{H}_2\text{O})_6]^{2+}\), chromium is in the +2 oxidation state. In \([\text{CoCl}_4]^{2-}\), cobalt is in the +2 oxidation state. In \([\text{Fe}(\text{H}_2\text{O})_6]^{2+}\), iron is in the +2 oxidation state. In \([\text{Mn}(\text{H}_2\text{O})_6]^{2+}\), manganese is in the +2 oxidation state.

Step 2: Determine the electron configuration

Find the electron configuration for the metal ions. After losing electrons to achieve the +2 oxidation state: - Cr in \([\text{Cr}(\text{H}_2\text{O})_6]^{2+}\) becomes 3d\(^4\).- Co in \([\text{CoCl}_4]^{2-}\) becomes 3d\(^7\).- Fe in \([\text{Fe}(\text{H}_2\text{O})_6]^{2+}\) becomes 3d\(^6\).- Mn in \([\text{Mn}(\text{H}_2\text{O})_6]^{2+}\) becomes 3d\(^5\).

Step 3: Calculate the magnetic moment

Use the formula for magnetic moment, \(\mu = \sqrt{n(n + 2)}\) where \(n\) is the number of unpaired electrons. - \([\text{Cr}(\text{H}_2\text{O})_6]^{2+}\) has 4 unpaired electrons, so \(\mu = \sqrt{4(4 + 2)} = \sqrt{24} \approx 4.9 \mu_B\).- \([\text{CoCl}_4]^{2-}\) has 3 unpaired electrons, so \(\mu = \sqrt{3(3 + 2)} = \sqrt{15} \approx 3.9 \mu_B\).- \([\text{Fe}(\text{H}_2\text{O})_6]^{2+}\) has 4 unpaired electrons, so \(\mu = \sqrt{4(4 + 2)} = \sqrt{24} \approx 4.9 \mu_B\).- \([\text{Mn}(\text{H}_2\text{O})_6]^{2+}\) has 5 unpaired electrons, so \(\mu = \sqrt{5(5 + 2)} = \sqrt{35} \approx 5.9 \mu_B\).

Step 4: Identify matching pairs

Compare the magnetic moments of the complexes. \([\text{Cr}(\text{H}_2\text{O})_6]^{2+}\) and \([\text{Fe}(\text{H}_2\text{O})_6]^{2+}\) both have a magnetic moment of approximately 4.9 \(\mu_B\), indicating they have the same spin-only magnetic moment values.

Key Concepts

Oxidation States

In chemistry, understanding oxidation states is crucial for gaining insights into a compound's reactivity and bonding structure. Each element in a compound is assigned an oxidation state which helps indicate the number of electrons lost or gained by an atom. This value is helpful in understanding electron transfer processes in chemical reactions.
For instance, in the compound \([ ext{Cr}( ext{H}_2 ext{O})_6]^{2+}\), chromium holds a +2 oxidation state. This means that it has lost two electrons from its neutral atom configuration. Similarly, cobalt in \([ ext{CoCl}_4]^{2-}\), iron in \([ ext{Fe}( ext{H}_2 ext{O})_6]^{2+}\), and manganese in \([ ext{Mn}( ext{H}_2 ext{O})_6]^{2+}\) also have a +2 oxidation state.

• Oxidation states determine electron configuration.
• Critical in identifying electron gain/loss in reactions.

Understanding the oxidation states helps in predicting the magnetic properties of these complexes.

Electron Configuration

Electron configuration refers to the arrangement of electrons in an atom's electron shells. It plays an essential role in determining an atom's chemical behavior and properties.
When an atom loses electrons and forms an ion, like in transition metals, its electron configuration typically changes. For example, we've seen that \([ ext{Cr}( ext{H}_2 ext{O})_6]^{2+}\) has a chromium ion with a configuration that resembles 3d\(^4\), having lost 2 electrons from its previous state.

• Cr: Loses electrons to become 3d\(^4\)
• Co: Becomes 3d\(^7\)
• Fe: Assumes configuration 3d\(^6\)
• Mn: Changes to 3d\(^5\)

These altered configurations directly influence the number of unpaired electrons within these ions, which is critical for further analyzing their magnetic behavior.

Unpaired Electrons

Unpaired electrons are electrons that are alone in an orbital and play a pivotal role in the magnetic properties of a substance. The presence of unpaired electrons is responsible for a material's paramagnetic property, meaning that it gets attracted to a magnetic field.
Examining the complexes provided, we determine the unpaired electrons as follows:

• \([ ext{Cr}( ext{H}_2 ext{O})_6]^{2+}\) has 4 unpaired electrons.
• \([ ext{CoCl}_4]^{2-}\) has 3 unpaired electrons.
• \([ ext{Fe}( ext{H}_2 ext{O})_6]^{2+}\) also contains 4 unpaired electrons.
• \([ ext{Mn}( ext{H}_2 ext{O})_6]^{2+}\) holds 5 unpaired electrons.

These unpaired electrons are essential to calculate the magnetic moment of the complexes using a specific formula. A larger number of unpaired electrons usually indicates a stronger magnetic response.

Magnetic Moment Formula

The magnetic moment of a metal complex provides a quantitative measure of its magnetic properties. A common formula used in such calculations is \( \mu = \sqrt{n(n+2)} \), where \(n\) represents the number of unpaired electrons.
This formula helps predict how these complexes behave in a magnetic field. For example, both \([ ext{Cr}( ext{H}_2 ext{O})_6]^{2+}\) and \([ ext{Fe}( ext{H}_2 ext{O})_6]^{2+}\) have 4 unpaired electrons; hence, their calculated magnetic moment is: \[\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9 \mu_B\] Utilizing this formula, we see that they share approximately the same magnetic moment, indicating similar magnetic properties.

• The greater the number of unpaired electrons, the higher the magnetic moment.
• Magnetic moments provide insights into spin and electron arrangement.

This calculation highlights why understanding unpaired electrons and oxidation states in complexes is so crucial to predicting magnetic behaviors.