Problem 115 - A Complete Resource Book in Chemistry for JEE Main

Question

In the complexes \(\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{3},\left[\mathrm{CuCl}_{4}\right]^{2-},\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+} .\) The number of unpaired electrons are respectively (a) 1,3 and 0 (b) 3,2 and 1 (c) 3,2 and 0 (d) 3,1 and 0

Step-by-Step Solution

Answer

The answer is (d) 3,1 and 0.

Step 1: Determine the Electronic Configuration of Chromium in \([\text{Cr(CN)}_6]^3\)

The charge on the complex is 3+. The CN ligand is a strong field ligand, causing \([\text{Cr}^3+]\) to adopt a low-spin configuration. The electronic configuration of \([\text{Cr}^3+]\) is \([\text{Ar}] 3d^3\). In the low-spin state, all three electrons are paired because the \(3d\) orbitals are split and filled in the lower energy orbitals.

Step 2: Evaluate the Configuration for Copper in \([\text{CuCl}_4]^{2-}\)

The copper ion that forms \([\text{CuCl}_4]^{2-}\) is \([\text{Cu}^{2+}]\). Chloride is a weak field ligand, so it cannot pair electrons. The electronic configuration of \([\text{Cu}^{2+}]\) is \([\text{Ar}] 3d^9\). There is one unpaired electron in the \(3d\) orbitals.

Step 3: Identify the Unpaired Electrons in \([\text{Cu(NH}_3)_2]^+\)

In \([\text{Cu(NH}_3)_2]^+\), Cu is present as \([\text{Cu}^+]\). Ammonia, NH_3, is considered a strong field ligand. The electronic configuration of \([\text{Cu}^+]\) is \([\text{Ar}] 3d^{10}\). Since all electrons in \(3d^{10}\) are paired, there are no unpaired electrons in \([\text{Cu(NH}_3)_2]^+\).

Key Concepts

Electronic Configuration

The electronic configuration of transition metal ions like those found in coordination complexes is essential in understanding their chemical behavior. To determine the electronic configuration of a metal ion, we must first consider its oxidation state within the complex. For example, in the complex \([ ext{Cr(CN)}_6]^{3-}\), chromium has a 3+ oxidation state, meaning it loses three electrons. The neutral chromium atom has an electronic configuration of \([ ext{Ar}] 3d^5 4s^1\). By removing three electrons, as needed for \([ ext{Cr}^{3+}]\), we get an electronic configuration of \([ ext{Ar}] 3d^3\). For copper, in \([ ext{CuCl}_4]^{2-}\), the oxidation state of Cu is 2+; thus, the electronic configuration becomes \([ ext{Ar}] 3d^9\). In the complex \([ ext{Cu(NH}_3)_2]^+\), Cu is in the +1 oxidation state, leading to an electronic configuration of \([ ext{Ar}] 3d^{10}\). Understanding these configurations is crucial as they dictate the magnetic properties and reactivity of the complexes.

Ligand Field Theory

Ligand Field Theory (LFT) helps explain the behavior of transition metal complexes in terms of electron distribution in orbitals. LFT considers how ligands surrounding a metal ion affect the energy of its d orbitals. Example: \([ ext{Cr(CN)}_6]^{3-}\) - Cyanide is a strong field ligand, which means it creates a large energy gap between the high and low energy sets of d orbitals, often resulting in a low-spin configuration. - This influences \([ ext{Cr}^{3+}]\) to have its electrons fully paired in the lower energy d orbitals, leading to no unpaired electrons. Weak vs. Strong Field Ligands- Strong field ligands, like CN⁻ and NH₃, cause a large splitting in the d orbital energies and often lead to low-spin states. - Weak field ligands, like Cl⁻, cause smaller splitting and typically result in high-spin states. LFT provides a deeper understanding of the color, magnetic properties, and stability of coordination compounds.

Unpaired Electrons

The presence of unpaired electrons in a metal complex is vital for determining the magnetic properties of the compound. When d orbitals have unpaired electrons, the complex is paramagnetic and will interact with magnetic fields. In the complex \([ ext{Cr(CN)}_6]^{3-}\), chromium is in a low-spin state with a configuration of \([ ext{Ar}] 3d^3\), resulting in paired electrons, therefore exhibiting no magnetism. For \([ ext{CuCl}_4]^{2-}\), Cu has one unpaired electron in its \([ ext{Ar}] 3d^9\) configuration, causing the complex to be weakly paramagnetic. In \([ ext{Cu(NH}_3)_2]^+\), Cu is \([ ext{Ar}] 3d^{10}\) entirely paired and exhibits diamagnetism due to the lack of unpaired electrons. Knowing the number and arrangement of unpaired electrons allows prediction of the compound's response to magnetic fields.

Transition Metal Complexes

Transition metal complexes are formed when transition metals bind to surrounding molecules known as ligands. These complexes can display diverse chemical and physical properties. Core Characteristics of Transition Metal Complexes:

• Variable Oxidation States: Transition metals often exhibit multiple oxidation states, allowing them to form various complexes with different properties.
• Color: The different splitting of d orbitals by various ligands often imparts vivid colors to these complexes.
• Catalytic Properties: Many transition metal complexes implement roles in catalysis due to their ability to lend and accept electrons easily.
• Coordination Number and Geometry: The shape and number of ligands surrounding the metal influence the properties and reactivity of the complex.

In complexes like \([ ext{Cr(CN)}_6]^{3-}\) and \([ ext{CuCl}_4]^{2-}\), the coordination environment and type of ligands explain their distinct characteristics. By using concepts like oxidation states and ligand field theory, the versatile nature of these complexes can be appreciated and utilized in various fields, from materials science to bioinorganic chemistry.