Problem 107 - A Complete Resource Book in Chemistry for JEE Main

Question

Among \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right],\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) and \(\left[\mathrm{NiCl}_{4}\right]^{2}\) (a) \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) and \(\left[\mathrm{NiCl}_{4}\right]^{2-}\) are diamagnetic and \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) is paramagnetic (b) \(\left[\mathrm{NiCl}_{4}\right]^{2}\) and \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) are diamagnetic and \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) is paramagnetic (c) \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) and \(\left[\mathrm{NiCN}_{4}\right]^{2}\) are diamagnetic and \([\mathrm{Ni}(\mathrm{Cl})]^{2-}\) is paramagnetic (d) \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) is diamagnetic and \(\left[\mathrm{NiCl}_{4}\right]^{2}\) and \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) are paramagnetic

Step-by-Step Solution

Answer

(c) [Ni(CO)_4] and [NiCN_4]^(2-) are diamagnetic; [NiCl_4]^(2-) is paramagnetic.

Step 1: Understanding the Coordination Complexes

We are given three coordination complexes: Ni(CO)_4, Ni(CN)_4}^(2-), and NiCl_4}^(2-). We need to classify each as either diamagnetic or paramagnetic based on their electron configuration.

Step 2: Determine Electron Configuration of Nickel

Nickel has an atomic number of 28, so its electronic configuration is [Ar] 3d^8 4s^2. In complexes, nickel will typically use its 3d, 4s, and 4p orbitals depending on the ligands.

Step 3: Analyze Ni(CO)_4 Complex

CO is a strong field ligand, leading to a low-spin complex. Nickel in Ni(CO)_4 is in the 0 oxidation state. The configuration for nickel in this complex becomes 3d^{10}, with paired electrons making it diamagnetic.

Step 4: Analyze [Ni(CN)_4]^(2-) Complex

CN^- is also a strong field ligand, causing pairing of electrons as a low-spin complex. Nickel's oxidation state is +2, yielding [Ar] 3d^8. The two paired electrons in each orbital mean it remains diamagnetic.

Step 5: Analyze [NiCl_4]^(2-) Complex

Cl^- is a weak field ligand resulting in a high-spin complex. Nickel's oxidation state is +2, with the configuration [Ar] 3d^8. The remaining two unpaired electrons make this complex paramagnetic.

Step 6: Conclude about Magnetic Properties

From the above analysis, Ni(CO)_4 and [Ni(CN)_4]^(2-) are diamagnetic due to their paired electrons, while [NiCl_4]^(2-) is paramagnetic due to unpaired electrons.

Key Concepts

Magnetic Properties of Complexes

Coordination complexes exhibit varied magnetic properties based on their electron arrangements. These properties can categorize complexes as either paramagnetic or diamagnetic. A paramagnetic substance has one or more unpaired electrons, which align with magnetic fields, causing attraction. In contrast, diamagnetic substances have all their electrons paired, resulting in a slight repulsion from magnetic fields.

For example, \([\text{Ni}(\text{CO})_4]\) and \([\text{Ni} (\text{CN})_4]^{2-}\) demonstrate diamagnetism since all their electrons are paired within their orbitals. Conversely, \([\text{NiCl}_4]^{2-}\) exhibits paramagnetism because it has unpaired electrons due to its electron configuration.

Understanding the spin state of the complex, whether high-spin or low-spin, is crucial for predicting its magnetic behavior. High-spin complexes have more unpaired electrons, while low-spin complexes often have them paired.

Electron Configuration

The electron configuration of a metal in a coordination complex is fundamental in determining its chemistry and magnetic properties. Nickel, with an atomic number of 28, typically has the electron configuration \[ [\text{Ar}] 3d^8 4s^2 \]. However, when participating in coordination complexes, the electron configuration of nickel adjusts based on the nature of the ligand.

When \([\text{Ni}(\text{CO})_4]\) is formed, nickel is in the 0 oxidation state, leading to the electron configuration \[3d^{10}\]. In this state, all electrons are paired, explaining its diamagnetic nature. On the other hand, for \([\text{Ni}(\text{CN})_4]^{2-}\), nickel is in a +2 oxidation state, maintaining the configuration \[ [\text{Ar}] 3d^8 \]. Despite having more unpaired electrons in its neutral state, the presence of strong-field ligands forces electron pairing, making the complex diamagnetic.

For \[\text{NiCl}_4]^{2-}\], the complex retains unpaired electrons due to the weak-field nature of the chloride ligands, leading to a paramagnetic behavior with the configuration \[ [\text{Ar}] 3d^8 \]. These nuances in electron configuration underscore the importance of ligand field strength in influencing a complex's properties.

Strong and Weak Field Ligands

In coordination chemistry, ligands are classified into strong and weak field based on their ability to split the d-orbitals of the central metal ion. This splitting affects the electron arrangement and therefore the spin state and magnetic properties of the complex.

Strong field ligands, such as CO and CN⁻, cause significant splitting of the d-orbitals. This effect often leads to a low-spin configuration, where electrons are forced to pair up, despite any higher energy required to do so. In the complexes \([\text{Ni}(\text{CO})_4]\) and \([\text{Ni}(\text{CN})_4]^{2-}\), the strong field nature of CO and CN⁻ causes electron pairing, resulting in diamagnetism.

Weak field ligands like Cl⁻ lead to minimal splitting of the d-orbitals, resulting in a situation where electrons remain unpaired, occupying higher-energy orbitals. Hence, in \([\text{NiCl}_4]^{2-}\), the presence of unpaired electrons characterizes it as paramagnetic.

Knowledge of whether a ligand is strong or weak field helps predict the electronic structure, and thus the physical properties, including color and magnetism of the resulting complexes.