Problem 1 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The species that has a spin-only magnetic moment of \(5.9 \mathrm{BM}\), is : \(\left(\mathrm{T}_{\mathrm{d}}\right.\) = tetrahedral) (a) \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) (square planar) (b) \(\left[\mathrm{NiCl}_{4}\right]^{2-}\left(\mathrm{T}_{d}\right)\) (c) \(\mathrm{Ni}(\mathrm{CO})_{4}\left(\mathrm{~T}_{\mathrm{d}}\right)\) (d) \(\left[\mathrm{MnBr}_{4}\right]^{2-}\left(\mathrm{T}_{\mathrm{d}}\right)\)

Step-by-Step Solution

Answer

The species is \([\text{MnBr}_4]^{2-}\) (Option d).

Step 1: Understand Spin-Only Magnetic Moment Formula

The spin-only magnetic moment \( \mu \) is given by the formula \( \mu = \sqrt{n(n+2)} \), where \( n \) is the number of unpaired electrons.

Step 2: Calculate Number of Unpaired Electrons for Given Moment

We are given a magnetic moment of 5.9 BM. Using the formula \( \mu = \sqrt{n(n+2)} \), for \( \mu = 5.9 \), solve \( 5.9 = \sqrt{n(n+2)} \) by squaring both sides, giving \( 34.81 = n(n+2) \). Solving for \( n \), we find that \( n \approx 5 \).

Step 3: Analyze Each Complex for Unpaired Electrons

Determine the electron configuration of the central metal ion in each complex: - (a) \([\text{Ni}( ext{CN})_4]^{2-}\) is square planar (low spin); Nickel has 3d\(^8\) configuration, giving 0 unpaired electrons.- (b) \([\text{NiCl}_4]^{2-}\) is tetrahedral; Nickel has 3d\(^8\) configuration, with 2 unpaired electrons.- (c) \(\text{Ni(CO)}_4\) is tetrahedral; Nickel forms a low spin state with 0 unpaired electrons.- (d) \([\text{MnBr}_4]^{2-}\) is tetrahedral; Manganese in this complex gives a 3d\(^5\) configuration, with 5 unpaired electrons.

Step 4: Identify the Correct Option

Among the complexes, only (d) \([\text{MnBr}_4]^{2-}\) has 5 unpaired electrons, matching the calculated \( n \) that corresponds to a spin-only magnetic moment of approximately 5.9 BM as required by the problem.

Key Concepts

Unpaired Electrons

Unpaired electrons play a key role in determining the magnetic properties of a compound. Essentially, these are electrons that do not have another electron paired with them in their atomic or molecular orbitals.
Metals with unpaired electrons exhibit paramagnetism, which means they are attracted to magnetic fields.
The presence of these electrons can be calculated using the formula \( \mu = \sqrt{n(n+2)} \), where \(\mu\) is the spin-only magnetic moment and \(n\) is the number of unpaired electrons.
For example, if a compound has a magnetic moment of 5.9 BM, using the formula we solve \(5.9 = \sqrt{n(n+2)}\).
This calculation reveals \(n\) is approximately 5, meaning the species has 5 unpaired electrons.

• Zero unpaired electrons lead to diamagnetic behavior (no magnetic attraction).
• More unpaired electrons generally increase the magnetic moment, enhancing paramagnetic properties.

Tetrahedral Complexes

Tetrahedral complexes often consist of a central metal ion surrounded by four ligands positioned at the corners of a tetrahedron.
These shapes alter the distribution of electron orbitals due to the arrangement of ligands, which can affect the magnetic characteristics of the complex.
In a tetrahedral field, the splitting of \(d\)-orbitals is different from octahedral or square planar fields, usually having three lower energy orbitals and two higher energy ones.
This smaller energy gap allows for more unpaired electrons, contributing to stronger paramagnetic properties.

• Tetrahedral complexes are high spin, generally leading to more unpaired electrons.
• For example, \([\text{MnBr}_4]^{2-}\) is tetrahedral and has 5 unpaired electrons, matching a high magnetic moment.

Square Planar Complexes

Square planar complexes consist of a metal ion coordinated to four ligands in a plane, forming a square shape.
This configuration tends to result in large differences in energy between \(d\)-orbitals, often causing these complexes to be low spin and have fewer unpaired electrons.
A typical example is the \([\text{Ni(CN)}_4]^{2-}\) complex. Nickel in a square planar setting often adopts a 3d\(^8\) electron configuration without any unpaired electrons, contributing to diamagnetic properties.

• These complexes often have no unpaired electrons, hence a lower or zero magnetic moment.
• Consequently, square planar complexes are typically less magnetic than their tetrahedral counterparts.

This distinct difference in electronic structure explains why square planar complexes might be less magnetic.