Problem 48 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

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

Step-by-Step Solution

Answer

(d) \([\mathrm{Ni}( ext{CO})_4]\) is diamagnetic, \([\mathrm{NiCl}_4^{2-}]\) and \([\mathrm{Ni}( ext{CN})_4]^{2-}\) are paramagnetic.

Step 1: Determine the Electron Configuration of Ni

Nickel (Ni) has an atomic number of 28. The electron configuration is \[\text{Ni: } [\text{Ar}] \, 3d^8 \, 4s^2.\] When Ni forms a 2+ oxidation state, the electrons are removed from the 4s orbital first, resulting in \[\text{Ni}^{2+}: [\text{Ar}] \, 3d^8\] electron configuration.

Step 2: Analyzing \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\)

In \(\left[\mathrm{Ni}(\ ext{CO})_{4}\right]\), Ni is in the zero oxidation state. CO is a strong field ligand and causes pairing of all 3d electrons. Thus, the electron configuration in presence of CO is \[\text{Ni}( ext{CO})_4: [\text{Ar}] \, 3d^{10} \, 4s^0\] making it diamagnetic with no unpaired electrons.

Step 3: Analyzing \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\)

In \(\left[\mathrm{Ni}( ext{CN})_{4}\right]^{2-}\), Ni is in the +2 oxidation state. CN is also a strong field ligand, which causes pairing of the 3d electrons. The electron configuration becomes \[\text{Ni(CN)}_4^{2-}: [\text{Ar}] \, 3d^{10} \, 4s^0\] and therefore it is diamagnetic.

Step 4: Analyzing \(\left[\mathrm{NiCl}_{4}^{2-}\right]\)

In \(\left[\mathrm{NiCl}_{4}^{2-}\right]\), Ni is in the +2 oxidation state just like in \(\left[\mathrm{Ni}( ext{CN})_{4}\right]^{2-}\). However, Cl is a weak field ligand and does not pair up 3d electrons, leading to \[\text{NiCl}_4^{2-}: [\text{Ar}] \, 3d^8 \, 4s^0\] with two unpaired electrons in the 3d orbitals, making it paramagnetic.

Key Concepts

Crystal Field Theory

Crystal Field Theory is an essential model used in coordination chemistry to understand the interaction between ligands and metal ions within a complex. This theory considers ligands as negative point charges or dipoles, which interact with the d orbitals of the central metal ion.

• In an octahedral complex, the approach of ligands raises the energy of the d orbitals because of the electrostatic repulsion. This interaction splits the degenerate d orbitals into two separate sets: the higher energy \( e_g \) orbitals (\(d_{x^2-y^2}\) and \(d_{z^2}\)) and the lower energy \( t_{2g} \) orbitals (\(d_{xy}\), \(d_{xz}\), and \(d_{yz}\)).
• Ligands are categorized based on their field strength: strong field ligands, like CO and CN⁻, significantly split the d orbitals and thus often lead to the pairing of electrons, forming low-spin complexes. Conversely, weak field ligands like Cl⁻ cause less splitting and usually result in high-spin complexes with unpaired electrons.

Understanding how ligands affect electron pairing is key to explaining the magnetic properties of a complex. Strong field ligands can force electron pairs, reducing magnetic interactions, while weak field ligands allow for more unpaired electrons.

Diamagnetism and Paramagnetism

Magnetism in coordination complexes depends on the electron configuration and the presence of unpaired electrons. This is where the concepts of diamagnetism and paramagnetism come into play.

• Diamagnetic materials have all paired electrons. They create an internal magnetic field that slightly repels an external magnetic field. Thus, when a complex is diamagnetic, it exhibits no net magnetic moment and isn't drawn to a magnetic field. For example, in \([\mathrm{Ni}(\mathrm{CO})_4]\) and \([\mathrm{Ni}(\mathrm{CN})_4]^{2-}\), strong field ligands result in fully paired d electrons, making these complexes diamagnetic.
• Paramagnetic substances have one or more unpaired electrons. These unpaired electrons create tiny magnetic moments that align with an external magnetic field, making paramagnetic complexes attractive to such fields. An example from the original exercise is \([\mathrm{NiCl}_4^{2-}]\), where weak field ligands leave 3d electrons unpaired, resulting in a paramagnetic material.

By assessing whether electrons are paired or unpaired, we can predict the magnetic properties of transition metal complexes, which is crucial in applications ranging from material science to biochemistry.

Electronic Configuration of Transition Metals

Transition metals are characterized by their ability to form various oxidation states due to their d orbital electrons. Understanding the electronic configuration of these metals is crucial in predicting their chemical behavior and properties.

• For the element Nickel (Ni), its neutral state has the configuration \([\text{Ar}] \, 3d^8 \, 4s^2\). When forming compounds, electrons are typically removed starting from the 4s orbital before the 3d orbitals. Thus, in the \( \mathrm{Ni}^{2+} \) ion, the configuration becomes \([\text{Ar}] \, 3d^8\).
• The ligands bound to a transition metal can greatly influence its electronic configuration by causing electron pairings, which subsequently affect the magnetic properties of the complex. Strong ligands like CN⁻ and CO can induce the electron pairing process, converting a high-energy configuration into a lower-energy one, which often results in diamagnetism.
• Alternatively, weak field ligands like Cl⁻ might not significantly alter the electron configuration from what is expected in the high-spin state, often resulting in unpaired electrons and paramagnetism.

A deep understanding of electronic configurations helps in predicting the geometry, reactivity, and physical properties, such as magnetism, of transition metal complexes. This is central to the application and synthesis of coordination compounds in chemistry.