Problem 9 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Consider that a \(d^{6}\) metal ion \(\left(\mathrm{M}^{2+}\right)\) forms a complex with aqua ligands, and the spin only magnetic moment of the complex is \(4.90 \mathrm{BM}\). The geometry and the crystal field stabilization energy of the complex is : (a) octahedral and \(-2.4 \Delta_{0}+2 \mathrm{P}\) (b) tetrahedral and \(-0.6 \Delta_{\text {t }}\) (c) octahedral and \(-1.6 \Delta_{0}\) (d) tetrahedral and \(-1.6 \Delta_{\mathrm{t}}+1 \mathrm{P}\)

Step-by-Step Solution

Answer

The geometry is octahedral with CFSE of -1.6 \(\Delta_0\), so the correct answer is (c).

Step 1: Determine the Magnetic Moment Formula

The spin-only magnetic moment (\( \mu \)) can be calculated using the formula: \( \mu = \sqrt{n(n+2)} \), where \( n \) is the number of unpaired electrons. Given that \( \mu = 4.90 \text{ BM} \), we will solve for \( n \).

Step 2: Solve for Unpaired Electrons

Set up the equation from the formula: \( \mu = \sqrt{n(n+2)} = 4.90 \). Solving gives \( n(n+2) = 24 \). Therefore, \( n^2 + 2n - 24 = 0 \). This factors to \( (n-4)(n+6) = 0 \). Thus, \( n = 4 \) as \( n = -6 \) is not physically possible.

Step 3: Analyze the Electronic Configuration

A \( d^6 \) metal ion with 4 unpaired electrons suggests that it fills its orbitals in a high-spin configuration. For a high-spin \( d^6 \) configuration with 4 unpaired electrons, the electron configuration in an octahedral field would be \( t_{2g}^4 e_g^2 \). Thus, the geometry is likely octahedral.

Step 4: Determine Crystal Field Stabilization Energy (CFSE)

For an octahedral high-spin \( d^6 \) configuration (\( t_{2g}^4 e_g^2 \)), the CFSE can be calculated as follows: \( \text{CFSE} = (0.4 \Delta_0 \times 4) - (0.6 \Delta_0 \times 2) = 1.6 \Delta_0 \). This is the stabilization energy relative to a sphere of uniform charge distribution.

Step 5: Match the Given Options with Calculated Values

From the calculated CFSE and geometry, the option that matches an octahedral configuration and a CFSE of \(-1.6 \Delta_0\) is option (c).

Key Concepts

Magnetic Moment

The magnetic moment is a crucial concept in understanding the behavior of complex ions. It refers to the magnitude of a magnetic property or field of a complex due to its electrons' motion and spin. For transition metal complexes, the spin-only magnetic moment can provide insights into the number of unpaired electrons in the d-orbitals.

To calculate the magnetic moment (\( \mu \)) of a complex using the spin-only formula, the equation \( \mu = \sqrt{n(n+2)} \) is applied, where \( n \) is the number of unpaired electrons. This parameter helps in predicting the electronic structure of metal complexes. For example, in our provided scenario, a calculated magnetic moment of \( 4.90 \text{ BM} \) indicates the presence of four unpaired electrons. Such a configuration is typically characteristic of high-spin complexes.

• Magnetic moment is measured in Bohr Magneton (BM).
• The number of unpaired electrons can be deduced from the value of the magnetic moment.
• Magnetic moments are key to determining whether a complex is high-spin or low-spin, hinting towards its electron distribution.

High-Spin Complex

A high-spin complex is a type where the maximum number of unpaired electrons is maintained. In transition metals, this occurs due to the crystal field splitting where the energy gap between the d-orbitals is small compared to the pairing energy. Thus, electrons tend to occupy higher energy orbitals rather than pairing in the lower ones.

Considering the \(d^{6}\) configuration in a high-spin complex, despite the crystal field, the energy difference doesn't encourage pairing of electrons. Therefore, you end up with more unpaired electrons, which subsequently leads to a higher magnetic moment. In an octahedral arrangement for a high-spin \(d^{6}\) complex, the electron configuration could be \(t_{2g}^{4} e_{g}^{2}\), illustrating electrons filling the lower set of orbitals first, before the higher set, to maintain the high-spin situation.

• High-spin complexes have high magnetic moments due to more unpaired electrons.
• The energy considerations favor electron placement in higher energy orbitals rather than electron pairing.
• They commonly occur in metal ions with weak field ligands.

Octahedral Geometry

Octahedral geometry is a common shape for coordination complexes where a central metal atom is surrounded by six ligands positioned symmetrically like the vertices of an octahedron. This geometry influences the splitting pattern of the d-orbitals, giving rise to what is known as crystal field splitting.

In the octahedral field, the d-orbitals split into two sets: \(t_{2g}\) and \(e_{g}\). The \(t_{2g}\) orbitals are of lower energy than the \(e_{g}\) orbitals. The extent of this splitting determines whether an ion forms a high-spin or low-spin complex. For instance, with weak field ligands, the splitting is not significant enough to encourage electron pairing, resulting in high-spin configurations as in the given exercise.

• Symmetrical coordination around the metal leads to the unique properties of the octahedral field.
• The stabilization energy due to the distribution of electrons affects the complex's properties.
• Understanding the octahedral splitting is fundamental for predicting complex geometry and electronic configuration.