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) \(\mathrm{sp}^{3},+1\), zero (c) \(\mathrm{sp}^{3},+2,1\) (d) \(\mathrm{dsp}^{2},+2\), zero

Step-by-Step Solution

Answer

The correct option is (b) \(\mathrm{sp}^{3},+1\), zero.

Step 1: Determine the oxidation state of Copper

In the complex \([\mathrm{Cu}(\mathrm{CN})_{4}]^{3-}\), the cyanide ion \((\mathrm{CN}^{-})\) is a monodentate ligand with a \(-1\) charge. The charge on the complex is \(-3\). Let the oxidation state of Copper (Cu) be \(x\). Thus, the equation for the oxidation state is: \[ x + 4(-1) = -3 \] Solving for \(x\), we find: \[ x = -3 + 4 = +1 \] Therefore, the oxidation state of Copper is \(+1\).

Step 2: Determine the hybridization of Copper

The copper atom in this complex is in the \(+1\) oxidation state, which corresponds to the electronic configuration of \([\mathrm{Ar}] 3d^{10}\). With \(4\) cyanide ligands, there are \(4\) coordination sites to be filled. The cyanide ion being a strong field ligand will pair up the electrons:- Utilizing steric number = 4 suggests a possibility of \(\mathrm{sp}^{3}\) hybridized orbitals, describing a coordination number of 4 with tetrahedral geometry.Thus, the hybridization of copper in this complex is \(\mathrm{sp}^{3}\).

Step 3: Determine the number of unpaired electrons

Copper in \(+1\) oxidation state means it has a \(3d^{10}\) configuration which is completely filled. As a result, all the electrons are paired. Therefore, there are zero unpaired electrons in the complex.

Key Concepts

Oxidation State

The oxidation state of an atom in a coordination complex tells us how many electrons have been lost or gained compared to the neutral atom form. This concept is crucial for understanding the behavior and properties of the complex. In the complex \([\text{Cu}(\text{CN})_{4}]^{3-}\), the cyanide ions, which are monodentate ligands, each have a charge of \-1\. We determine the oxidation state of copper by setting up an equation where \([x + 4(-1) = -3]\). Solving for \([x]\), we find that the oxidation state of copper in this complex is \(+1\). This means copper has lost one electron when forming this complex.

Hybridization

Hybridization describes the mixing of atomic orbitals to form new hybrid orbitals that can form covalent bonds in coordination complexes. For copper in the complex \([\text{Cu}(\text{CN})_{4}]^{3-}\), we need to understand its electronic configuration in the \(+1\) oxidation state. Copper normally holds an electronic configuration of \[\text{[Ar]} \, 3d^{10}\, 4s^0\, 4p^0\]\ once it loses an \([\text{s}]\) electron. With four CN\(^-\) ligands, copper uses sp\(^3\) hybridization to form four new bonds. This hybridization suggests a tetrahedral geometry due to its steric number of four, letting the strong field ligands like CN\(^-\) pair up with d-orbital electrons.

Complex Ions

Complex ions are formed when a central metal atom bonds with one or more ligands—these can be ions or molecules. The entire system usually carries an electric charge. In \([\text{Cu}(\text{CN})_{4}]^{3-}\), the central copper atom is surrounded by four cyanide ions, creating a complex ion with an overall negative three charge. Complex ion geometry and hybridization affect properties such as color, magnetism, and reactivity. These properties arise because the central metal's hybrid orbitals overlap with the ligand orbitals, creating distinct energy states in the d orbitals of the metal.

Unpaired Electrons

The number of unpaired electrons in a complex helps to determine its magnetic properties. Unpaired electrons create a magnetic field, causing the complex to be paramagnetic. However, if all electrons are paired, the complex is diamagnetic. For copper in its \(+1\) oxidation state and a \[\text{3d}^{10}\]\ electron configuration, the electrons are completely filled across its orbitals. As a result, there are no unpaired electrons left in the copper atoms of the complex \([\text{Cu}(\text{CN})_{4}]^{3-}\). This full pairing leads to the complex being diamagnetic, with no magnetic field produced.