Answer:-

The complex [VCl4] features vanadium (V) in a +4 oxidation state, which means it has lost four electrons from its neutral state. To understand the bonding in this complex, we can use valence bond theory.

Valence Bond Description:

1. Vanadium: In its ground state, vanadium has the electron configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d³. In [VCl4], vanadium has lost its four outermost electrons, resulting in a 3d³ configuration.

2. Chlorine: Each chloride ion (Cl⁻) contributes one electron to the formation of the complex.

Now, let’s describe the bonding in [VCl4] using valence bond theory:

• Vanadium will form four sigma (σ) bonds with the four chloride ions. These sigma bonds will result from the overlap of the half-filled 3d orbitals of vanadium with the 2p orbitals of chlorine atoms.

Box Diagram for the Free Metal Ion: Before forming bonds with chlorine ions, let’s consider the box diagram for the free vanadium ion (V⁴⁺):

• In a box diagram, each box represents an atomic orbital, and the electrons are represented as arrows. In this case, we have three unpaired electrons in the 3d orbitals:

  3d: ↑ ↓ ↑

Hybrid Orbitals: In the valence bond theory, vanadium does not necessarily form hybrid orbitals like carbon or nitrogen. Instead, it utilizes its available atomic orbitals (in this case, the 3d orbitals) to form sigma bonds.

Number of Unpaired Electrons: As mentioned earlier, there are three unpaired electrons in the 3d orbitals of vanadium in [VCl4].

So, in summary, the bonding in [VCl4] involves vanadium forming sigma bonds with four chloride ions using its 3d orbitals. There are three unpaired electrons in the 3d orbitals of vanadium in this complex.