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 Electronic Configurations

Identify the electronic configurations of the central metal ions in each complex. For each metal ion in its oxidation state: - \(\text{Cr}^{2+}\) in \(\left[\text{Cr}(\text{H}_2\text{O})_6\right]^{2+}\) has the electronic configuration of \([\text{Ar}] 3d^4\).- \(\text{Co}^{2+}\) in \(\left[\text{CoCl}_4\right]^{2-}\) has the electronic configuration of \([\text{Ar}] 3d^7\).- \(\text{Fe}^{2+}\) in \(\left[\text{Fe}(\text{H}_2\text{O})_6\right]^{2+}\) has the electronic configuration of \([\text{Ar}] 3d^6\).- \(\text{Mn}^{2+}\) in \(\left[\text{Mn}(\text{H}_2\text{O})_6\right]^{2+}\) has the electronic configuration of \([\text{Ar}] 3d^5\).

Step 2: Calculate the Unpaired Electrons

For each electronic configuration identified:- \(\text{Cr}^{2+}\): \(3d^4\) configuration has 4 unpaired electrons.- \(\text{Co}^{2+}\): \(3d^7\) configuration has 3 unpaired electrons.- \(\text{Fe}^{2+}\): \(3d^6\) configuration has 4 unpaired electrons.- \(\text{Mn}^{2+}\): \(3d^5\) configuration has 5 unpaired electrons.

Step 3: Calculate the Spin-Only Magnetic Moment

Calculate the magnetic moment \((\mu)\) using the formula:\[ \mu = \sqrt{n(n+2)} \] where \(n\) is the number of unpaired electrons:- For \(\text{Cr}^{2+}\) and \(\text{Fe}^{2+}\) with 4 unpaired electrons: \[ \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9 \mu_B \]

Step 4: Match the Complexes With the Same Magnetic Moment

Find which complexes have the same magnetic moment based on their unpaired electrons:- \(\text{Cr}^{2+}\) and \(\text{Fe}^{2+}\) both have 4 unpaired electrons, thus they have the same spin-only magnetic moment.

Key Concepts

Electronic Configurations

Understanding electronic configurations is key to grasping how the magnetic properties of coordination complexes emerge. The electronic configuration of a metal ion is influenced by its oxidation state and its position in the periodic table. For example,

• The complex \(\left[\text{Cr}(\text{H}_2\text{O})_6\right]^{2+}\) has a central \(\text{Cr}^{2+}\) ion. In its +2 oxidation state, this chromium ion has an electronic configuration of \([\text{Ar}] 3d^4\).
• For \(\left[\text{Fe}(\text{H}_2\text{O})_6\right]^{2+}\), the metal \(\text{Fe}^{2+}\) ion adopts the configuration \([\text{Ar}] 3d^6\).
• The \(\left[\text{Mn}(\text{H}_2\text{O})_6\right]^{2+}\) has \(\text{Mn}^{2+}\) with the configuration \([\text{Ar}] 3d^5\).
• Lastly, the complex \(\left[\text{CoCl}_4\right]^{2-}\) features a \(\text{Co}^{2+}\) ion configured as \([\text{Ar}] 3d^7\).

Using these configurations, we can predict the number of unpaired electrons, which directly affects the magnetic properties of these complexes.

Unpaired Electrons

Unpaired electrons play a pivotal role in determining the magnetic moment of coordination complexes. The more unpaired electrons present, the higher the magnetic moment. In each 3d transition metal ion, electrons fill according to Hund's Rule, which states that electrons prefer to occupy separate orbitals of the same energy before pairing up.

Let's look at the examples from the previous electronic configurations:

• \(\text{Cr}^{2+}\) with \(3d^4\) has four unpaired electrons as each electron occupies a separate orbital.
• \(\text{Fe}^{2+}\) in \(3d^6\) also has four unpaired electrons.
• \(\text{Mn}^{2+}\), configured \(3d^5\), reaches its half-filled stability with five unpaired electrons.
• \(\text{Co}^{2+}\) in the \(3d^7\) state has three unpaired electrons, as electrons start pairing up after the half-filled stage is surpassed.

These unpaired electrons are the main drivers behind the magnetic characteristics of the complexes and help in calculating their magnetic moments.

Spin-Only Magnetic Moment Calculation

The concept of spin-only magnetic moment is crucial for understanding the magnetic behavior of metal complexes, particularly those adhered to by weak field ligands which do not significantly affect the electron configurations of the metal ions. The spin-only magnetic moment can be calculated using the formula:\[\mu = \sqrt{n(n+2)}\]where \(n\) denotes the number of unpaired electrons. This formula emerges from the notion that each unpaired electron contributes a unit magnetic moment due to its intrinsic spin. Let's dive into a calculation:

For complexes with four unpaired electrons, such as those containing \(\text{Cr}^{2+}\) or \(\text{Fe}^{2+}\), the spin-only magnetic moment is calculated as follows:\[\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9 \mu_B\]where \(\mu_B\) is the Bohr magneton, the unit for magnetic moments. This measurement highlights that both these complexes share the same magnetic moment, based only on their electron spins. Thus, accurately determining the number of unpaired electrons allows us to predict and compare the magnetic moments of various coordination complexes efficiently.