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

(d) 3,1, and 0

Step 1: Determine the Oxidation State and Electron Configuration of Cr in [Cr(CN)6]3-

In the complex \([\mathrm{Cr}(\mathrm{CN})_{6}]^{3-}\), the CN is a neutral ligand, so the oxidation state of Cr is +3. Chromium's electron configuration in the elemental state is \([\mathrm{Ar}] 3d^5 4s^1\). In the +3 oxidation state, it becomes \([\mathrm{Ar}] 3d^3\), meaning there are 3 unpaired electrons.

Step 2: Analyze the Electron Configuration of Cu in [CuCl4]2-

For \([\mathrm{CuCl}_4]^{2-}\), Cu is in the +2 oxidation state. The electron configuration of Cu^2+ is \([\mathrm{Ar}] 3d^9\). Typically, in this configuration, there is 1 unpaired electron.

Step 3: Analyze [Cu(NH3)2]+

In \([\mathrm{Cu}({\mathrm{NH}}_{3})_{2}]^{+}\), Cu has an oxidation state of +1. Its electron configuration is \([\mathrm{Ar}] 3d^{10}\), which means all electrons are paired, and thus it has 0 unpaired electrons.

Key Concepts

Electron Configuration

Electron configuration is crucial in understanding the properties and behavior of coordination complexes. It refers to the distribution of electrons in atomic or molecular orbitals, which dictates how these atoms or ions interact and bond with others.

For example, in the complex \([\mathrm{Cr}(\mathrm{CN})_{6}]^{3-}\), we determine that chromium has an electron configuration of \([\mathrm{Ar}] 3d^3\) in its +3 oxidation state. This means that three electrons occupy the d orbital.

Similarly, for \([\mathrm{CuCl}_4]^{2-}\), copper in the +2 state has an electron configuration of \([\mathrm{Ar}] 3d^9\). These configurations not only help predict reactivity and magnetic properties but also guide chemists on how these complexes might be synthesized or manipulated.

• In the elemental state, Cr is \([\mathrm{Ar}] 3d^5 4s^1\).
• For Cu in various states, its configurations adjust to \([\mathrm{Ar}] 3d^{10}\) or \([\mathrm{Ar}] 3d^9\).

Oxidation States

Oxidation state is a theoretical charge often used to describe the number of electrons lost or gained by an atom in a compound. This concept is essential because it influences the electron configuration and the reactivity of a complex.

In coordination complexes like \([\mathrm{Cr}(\mathrm{CN})_{6}]^{3-}\), chromium exhibits an oxidation state of +3. This means chromium has lost three electrons compared to its elemental state.

In \(\left[\mathrm{CuCl}_4\right]^{2-}\), copper is in the +2 oxidation state, indicating it has lost two electrons. Meanwhile, in \([\mathrm{Cu}({\mathrm{NH}}_{3})_{2}]^{+}\), copper is in the +1 state, implying the loss of one electron. Recognizing the oxidation state helps predict the electron distribution and the possible coordination number.

Unpaired Electrons

Unpaired electrons play a pivotal role in determining the magnetic properties of a compound. Compounds with unpaired electrons are typically paramagnetic, meaning they are attracted to magnetic fields.

In \([\mathrm{Cr}(\mathrm{CN})_{6}]^{3-}\), chromium's configuration results in three unpaired electrons, making this complex paramagnetic.

However, in \(\left[\mathrm{CuCl}_4\right]^{2-}\), copper has one unpaired electron because one of the nine electrons in the d orbital is not paired, making it less strongly paramagnetic than the chromium complex.

Lastly, \([\mathrm{Cu}({\mathrm{NH}}_{3})_{2}]^{+}\) features copper in a completely filled \([\mathrm{Ar}] 3d^{10}\) configuration, with no unpaired electrons, making it diamagnetic and not attracted to magnetic fields. Understanding whether a complex has unpaired electrons helps chemists determine its potential applications and behaviors.