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

Short answer

(b) sp³, +1, zero

Step-by-step solution

Determine the Oxidation State of Copper

First, we need to calculate the oxidation state of copper in \(\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{3-}\). Cyanide \(\mathrm{CN}^{-}\) has a charge of \(-1\). There are four \(\mathrm{CN}^{-}\) ligands, so the total negative charge from the ligands is \(-4\). The complex overall has a charge of \(-3\). Let the oxidation state of copper be \(x\). Therefore,\[x + 4(-1) = -3\]Solving this gives:\[x - 4 = -3\]\[x = +1\] Thus, the oxidation state of copper is \(+1\).

Identify Hybridization

Given that the oxidation state of copper is \(+1\) and we are working with \(\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{3-}\), the electron configuration of \(\mathrm{Cu}^{+}\) is \([\mathrm{Ar}] 3d^{10}\). The CN\(^{-}\) ligands are strong field ligands that cause pairing of electrons. This full pairing supports a \(\mathrm{sp}^{3}\) hybridization configuration. Therefore, \([\mathrm{Cu}(\mathrm{CN})_{4}]^{3-}\) shows \(\mathrm{sp}^{3}\) hybridization.

Determine the Number of Unpaired Electrons

After determining the oxidation state, we know \(\mathrm{Cu}^{+}\) has the electron configuration \([\mathrm{Ar}] 3d^{10}\). As cyanide is a strong field ligand and causes pairing, the \(d\)-orbitals will be completely filled, resulting in no unpaired electrons.

Key concepts

Oxidation State Determination

Determining the oxidation state is crucial for understanding a metal's role in a coordination complex. In the complex \([\mathrm{Cu}(\mathrm{CN})_{4}]^{3-}\), we find the oxidation state of copper by considering the charges. Each cyanide ligand (\(\mathrm{CN}^{-}\)) has a charge of \(-1\). Because there are four cyanide ligands, they collectively contribute \(-4\) to the total charge.
The entire complex is charged \(-3\). Since the total charge is the sum of the copper's oxidation state and the ligands' charges, we set it up as an equation:
\[x + 4(-1) = -3\]
Simplifying gives \(x - 4 = -3\) and solving this, we find \(x = +1\). This means copper is in the +1 oxidation state in this complex.
The ability to deduce an element's oxidation state helps predict its chemical behavior, such as its potential to undergo redox reactions and bonding characteristics.

Hybridization in Complexes

Understanding hybridization is key to explaining the geometry of coordination complexes. Hybridization involves the mixing of atomic orbitals to form new hybrid orbitals suitable for bonding. In the case of the complex \([\mathrm{Cu}(\mathrm{CN})_{4}]^{3-}\), copper with a +1 oxidation state has the electron configuration \([\mathrm{Ar}] 3d^{10}\).
Cyanide ligands are known as *strong field ligands*, which means they have a significant ability to induce electron pairing when attached to metals. This leads the copper's 3d orbitals to completely fill with paired electrons, leaving no room for unpaired electrons.
As a result, copper uses its s and p orbitals to form hybrid \(sp^3\) orbitals, which accommodate the bonding with cyanide ligands. This hybridization results in a tetrahedral geometry around the metal ion.

Unpaired Electrons in Transition Metals

Transition metals are often characterized by their d electron configurations which contribute to their chemical behavior. Unpaired electrons influence the magnetic properties and chemical reactivity of these metals. In \([\mathrm{Cu}(\mathrm{CN})_{4}]^{3-}\), we deal with \(\mathrm{Cu}^{+}\) having \([\mathrm{Ar}] 3d^{10}\).
The strong field cyanide ligands trigger the pairing of all available d electrons. Hence, in this specific complex, there are no unpaired electrons. This leads to diamagnetic behavior since all electrons are paired.
Knowing whether electrons are unpaired is important for predicting magnetic properties and potential reactions that the metal may undergo. In practical terms, any experiment regarding the magnetic susceptibility of \(\mathrm{Cu}^{+}\) in this complex would show no response to an external magnetic field.