Introduction

In this particular article Tetrahedral and square planar complexes,we are going to discuss about tetrahedral complexes with their example. We will also discuss octahedral complexes along with there example in detail.

Tetrahedral complexes

In tetrahedral complexes, the metal cation is either sp³ or sd³ hybridized. Let us consider some example to illustrate the hybridization and geometry.

1. {Ni Cl₄}²⁻ion :

In this complex ion, the oxidation state of Ni is +2 and its valence shell electronic configuration is 3d⁸. Magnetic moment measurements indicate that it is paramagnetic corresponding to two unpaired electrons. Since Cl⁻ is a weak ligand, therefore, no pairing of electrons will occur in 3d orbitals. None of the five 3d- orbitals is vacant. Vacant 4s and 4p orbitals combine to give four sp³ hybrid orbitals because {Ni Cl₄}²⁻ is a tetrahedral complex ion. These four hybrid orbitals form bonds with four ligands by sharing four pairs of electrons, one pair from each of the four ligands.

2. Ni (CO)₄ ion :

In this compound oxidation state of Nickel is zero and its valence shell electronic configuration is 3d⁸ 4s². Magnetic moment measurements indicate that Ni(CO)₄ is diamagnetic i.e., it has no unpaired electrons. CO is a strong ligand and has the tendency to pair up the 3d- electrons. And the two electrons of 4s orbitals shift into one of the five 3d- orbitals. Thus there is no vacant 3d- orbitals. Thus the vacant orbitals available for hybridization are 4s and 4p to give four sp³ hybrid orbitals.

3. MnO⁻₄ ion :

In this complex ion, the oxidation state of Mn is +7. The valence shell electronic configuration of Mn is 3d⁵ 4s². In Mn⁷⁺ ion all the five 3d, 4s orbitals are vacant. The vacant 4s and three 3d- orbitals combine to give four sd³ -hybrid orbitals. Thus Mn⁷⁺ ion is sd³ -hybridized in this complex ion as shown below :

Tetrahedral and square planar complexes

In Mn⁷⁺ ion energy of vacant 4s orbital becomes lower than that of vacant 3d- orbitals

In sd³ – hybridization vacant dxy, dyz, and dzx orbitals are involved.

4. Cr₂O₇²⁻ ion :

In this complex ion, the oxidation state of Cr is + 6. Velence shell electronic configuration of Cr is 3d⁵ 4s¹. Cr⁶⁺ ion has no electron in 3d- orbitals. The 4s – and three 3d – vacant orbitals of each Cr⁶⁺ ion combine to give four sd³ – hybrid orbitals. Thus each Cr⁶⁺ ion in Cr₂O₇²⁻ ion is sd³- hybridized. One of the seven oxide ions shares both the Cr⁶⁺ ions.

Square planar complexes

In square planar complexes, the central metal cation is dsp² – hybridized. The dsp² – hybrid orbitals point towards the four corners of a square.

Let us consider some examples to discuss the hybridization in square planar complexes.

(1) {Ni (CN)₄}²⁻ ion :

In this complex ion, the oxidation state of Ni is +2 and its velence shell electronic configuration is 3d⁸. Magnetic moment measurements indicate that this complex ion is diamagnetic i.e., it has no unpaired electrons. Since CN⁻ is a strong ligand, therefore these ligands cause to pair up the two unpaired electrons in one d-orbitals resulting in a vacant 3d orbital. This vacant 3d – orbitals get hybridized with the vacant 4s and two of the 4p- orbitals to give four dsp² – hybrid orbitals. These hybrid orbitals form bonds to the ligands by accepting four pairs of ligand electrons, one pair from each of the four ligands.

(2) {Cu (NH₃)₄}²⁺ ion :

In this ion, the oxidation state of copper is +2 and its velence shell electronic configuration is 3d⁹. Magnetic moment measurements indicate that this complex ion is paramagnetic corresponding to the presence of one unpaired electron. There is no possibility of the pairing of electrons by the ligands even the ligand is strong. Because there is only one unpaired electron in 3d – orbitals. Since coordination number of Cu² ion in {Cu (NH₃)₄}²⁺ is 4, therefore according to VBT, Cu²⁺ ion should be sp³ – hybridized and the structure is tetrahedral.

 

But according to esr and X- rays structure determination, the structure of {Cu(NH₃)₄}²⁺ is found to be square planar. Thus, to make the Cu²⁺ ion dsp² hybridized. And it is considered that the unpaired electron in the 3dx²-y² orbital is to be promoted to 4pz orbital as shown.

In the above electronic configuration, the unpaired electrons are present in the higher energy 4pz orbitals and are expected to be lost easily.  i.e., {Cu (NH₃)₄}²⁺ ion may be easily oxidized to {Cu(NH₃)₄}³⁺.

But, experiments have shown that {Cu (NH₃)₄}³⁺ ion does not exist i.e., oxidation of {Cu (NH₃)₄}²⁺ to {Cu (NH₃)₄}³⁺ is not possible.

Finally, Huggin suggested that in square planar {Cu (NH₃)₄}³⁺ ion, Cu²⁺ ion is sp²d hybridized as shown below.

In this configuration one of the three 4p orbitals, 4pz orbital remains unhybridized i.e., does not participate in hybridization because pz orbitals lie above and below the plane of the ion. {Cu(py)₂}²⁺,{Cu(en)₂}²⁺,{Cu(CN)₄}²⁻  complex ions are square planar and Cu²⁺ ion is sp²d hybridized.

The conclusion of Tetrahedral and square planar complexes

In order to discuss various information about the octahedral and tetrahedral complexes, we are now able to understand all the basic information of Tetrahedral and square planar complexes.

 

 

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