Problem 105 - A Complete Resource Book in Chemistry for JEE Main

Question

In the complex \(\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{3-}\) the hybridization state, oxidation state and number of unpaired electrons are (a) \(\mathrm{dsp}^{2},+1,1\) (b) sp \(^{3},+1\), zero (c) \(\mathrm{sp}^{3},+2,1\) (d) \(\mathrm{dsp}^{2},+2\), zero

Step-by-Step Solution

Answer

The hybridization state, oxidation state, and number of unpaired electrons are sp\(^{3}, +1\), and zero.

Step 1: Determine the oxidation state of Cu

In the complex \([\mathrm{Cu}(\mathrm{CN})_{4}]^{3-}\), each \(\mathrm{CN}^-\) ligand carries a charge of \(-1\). Therefore, their total charge contribution is \(-4\). Let the oxidation state of copper be \(x\). Equating the total charge to the given charge of the complex, we get:\[x - 4 = -3\]Solving for \(x\), we find:\[x = +1\]Thus, the oxidation state of Cu is \(+1\).

Step 2: Find the hybridization state of Cu

For copper, with an oxidation state of \(+1\) in this complex, the electronic configuration before hybridization is:\([ ext{Ar}] \, 3d^{10} \, 4s^0\)Since CN is a strong field ligand, it causes pairing of electrons, and no d-electrons are unpaired. The configuration remains \(3d^{10}\), suggesting the available orbitals are \(s\) and \(p\) for hybridization with no participation of \(3d\). The complex geometry and ligand number (4) suggest \(sp^3\) hybridization, as it forms a tetrahedral complex.

Step 3: Determine the number of unpaired electrons

Given copper is in the \(+1\) oxidation state, its electronic configuration is \(3d^{10}\). With \(3d^{10}\), all d-electrons are paired, meaning there are zero unpaired electrons in the complex.

Key Concepts

Hybridization

In coordination chemistry, understanding the hybridization state is crucial because it dictates the shape and bonding characteristics of a metal complex. For the complex \([\text{Cu} (\text{CN})_4]^{3-}\), let's examine the hybridization of the copper ion. Copper, in its +1 oxidation state, has the electronic configuration of:\[[\text{Ar}] \, 3d^{10} \, 4s^0\]Since cyanide (CN\(^-\)) is a strong field ligand, it causes the d electrons to pair up, resulting in no available \(d\) orbitals to participate in hybridization.This leaves the available \(s\) and \(p\) orbitals to form bonds with the four CN\(^-\) ligands. Therefore, the complex is configured with \(sp^3\) hybridization. This hybridization type corresponds to a tetrahedral geometry, which is typical when four ligands are involved. It is fascinating how hybridization explains the spatial arrangement of the atoms in coordination complexes.

Oxidation State

Calculating the oxidation state of a metal within a complex is a key step in understanding its chemical behavior and bonding nature. For the complex \([\text{Cu} (\text{CN})_4]^{3-}\), copper's oxidation state needs to be determined.Here's how it's calculated:

• Each CN\(^-\) ligand carries a charge of \(-1\).
• With four CN\(^-\) ligands, their collective charge is \(-4\).
• The overall charge of the complex is \(-3\).
• Let the oxidation state of copper be \(x\).

The equation to determine the oxidation state is:\[x - 4 = -3\]Solving for \(x\) gives:\[x = +1\]Therefore, the copper ion in this complex holds an oxidation state of \(+1\). Understanding the oxidation state can also give insights into the electron distribution and the magnetic properties of the complex.

Unpaired Electrons

The number of unpaired electrons in a metal complex influences its magnetic properties and reactivity. In the case of \([\text{Cu} (\text{CN})_4]^{3-}\), determining the number of unpaired electrons involves looking at the electronic configuration of copper in its \(+1\) oxidation state.Copper in \(+1\) state has the configuration:\[[\text{Ar}] \, 3d^{10} \, 4s^0\]Here, all d electrons are completely paired up due to the full 3d subshell. CN\(^-\), being a strong field ligand, ensures there is no unpairing of electrons.Consequently, there are zero unpaired electrons in the complex, making it diamagnetic. This lack of unpaired electrons affects how the complex interacts with magnetic fields and specific reactions involving electron transfer.