Valence Bond Theory (VBT theory): If you believe the Lewis hypothesis explains everything about compounds and molecules, you are wrong. The Valence Bond Theory (VBT Theory) was developed since it failed to explain many concepts. The valence bond theory describes the structure and magnetic properties of several coordination compounds. It explains the structure and magnetic properties of various coordination compounds with the help of postulates, However, there are flaws in this concept as well.

In this article, we have provided detailed information about VBT, Valence Bond Theory postulates, valence bond theory examples, limitations, etc. Continue reading this article to know more about the valence bond theory.

Valence Bond Theory: Overview

The valence bond theory explains the structure and magnetic properties of a wide variety of coordination compounds. The structure of coordination compounds and the bond linkages were explained using the valence bond theory.

According to valence bond theory, a metal atom or ion can use its \(\left({{\text{n}} – 1} \right){\text{d}},{\text{ns}},{\text{np}},{\text{nd}}\) orbitals for hybridisation to yield a set of equivalent orbitals of definite geometry such as octahedral, tetrahedral, square planar, and so on, when it is influenced by ligands. The ligand orbitals that can give electron pairs for bonding can overlap with the hybridised orbitals.

Valence Bond Theory in Coordination Compounds

Pauling mainly developed valence bond treatment of bonding in complexes. It is the simplest of the three theories and explains the structure and magnetic properties of a large number of co-ordinate compounds satisfactorily.

Postulates of Valence Bond Theory

 The important postulates of this theory are as follows:

(i) The central metal ion has several empty orbitals for accommodating electrons donated by the ligands. The number of empty orbitals is equal to the coordination number of the metal ion for the particular complex.
(ii) The metal orbitals and ligand orbitals overlap to form strong bonds. Now we know that the greater the extent of overlapping stronger will be the bond, the more stable the complex will be. To achieve greater stability, the atomic orbitals (\({\text{s}},{\text{p}}\) or \({\text{d}}\)) of the metal ion hybridize to form a new set of equivalent hybridized orbitals with definite directional properties. These hybrid orbitals now overlap with the ligand orbitals to form strong chemical bonds.
(iii) The \({\text{d}}\)-orbitals involved in the hybridisation may be either inner \(\left({{\text{n}} – 1} \right){\text{d}}\)-orbitals or outer \({\text{nd}}\)- orbitals. The complex formed in these two ways is referred to as low spin and high spin complexes, respectively.
(iv) The non-bonding metal electrons occupy the inner \({\text{d}}\)- orbitals which do not participate in hybridisation and thus in bond formation with the ligand.
(v) Each ligand contains a lone pair of electrons.
(vi) A covalent bond is formed by overlapping a vacant hybridized metal orbital and a filled orbital of the ligand. The bond is also sometimes called a co-ordinate bond.
(vii) Magnetic properties of coordination compounds: If after the formation of the complex, no unpaired electron is present, the complex is diamagnetic. If, however some unpaired electrons are present the complex is paramagnetic. Greater the number of unpaired electrons present greater is the paramagnetic character of the complex, that is, greater is the magnetic moment \(\left( {{\rm{\mu = }}\sqrt {{\rm{n}}\left( {{\rm{n + 2}}} \right)} {\rm{BM}}} \right).\) Conversely knowing the magnetic moment of the complex, the number of unpaired electrons present in the complex can be calculated from which the geometry of the complex can be predicted.

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The magnetic moment measurements as illustrated below:

Relation between unpaired electron and magnetic moment

Magnetic moment (Bohr magnetons)	\(0\)	\(1.73\)	\(2.83\)	\(3.87\)	\(4.90\)	\(5.92\)
Number of unpaired electrons	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)

Thus, the knowledge of the magnetic moment can be of great help in ascertaining the type of complexity.

(viii) Under the influence of a strong ligand, the electrons can be forced to pair up against Hund’s rule of maximum multiplicity.

 Let us consider a few examples to illustrate the valency bond theory.

(A) Octahedral Complexes

These complexes result from \({{\text{d}}^2}{\text{s}}{{\text{p}}^3}\) (inner orbital) or \({\text{s}}{{\text{p}}^3}{{\text{d}}^2}\) (outer orbital) hybridisation.

(i) Inner Orbital Complexes

As mentioned above, in these complexes, the d orbitals used are of lower quantum number, i.e., \(\left( {{\text{n}} – 1} \right).\)
Formation of \({\left[ {{\text{Cr}}{{\left({{\text{N}}{{\text{H}}_3}} \right)}_6}} \right]^{3 + }}\) ion: Note that the chromium is in \( + 3\) oxidation state. So first, let us write the electronic configuration of chromium (atomic number \(24\)) and \({\text{C}}{{\text{r}}^{ + 3}}\) ion.
Electronic configuration of \({\text{Cr}} = \left[ {{\text{Ar}}}\right]3{{\text{d}}^5}4{{\text{s}}^1}\)
\(\therefore \) Electronic configuration of \({\text{C}}{{\text{r}}^{ + 3}} = \left[{{\text{Ar}}} \right]3{{\text{d}}^3}\)
Since the \(3{\text{p}}\) orbitals are completely filled, they do not take part in chemical bonding and are called non-bonding orbitals. These are generally omitted in picturising the structure of the complexion where only orbitals involved in bond formation are drawn, thus in the present case, only \(3{\text{d}},4{\text{s}}\) and \(4{\text{p}}\) orbitals are drawn because they take part in chemical bonding. This type of representation will be used throughout the topic.
Thus, the electronic configuration of chromium atom and \({\text{C}}{{\text{r}}^{ + 3}}\) ion may be shortened as below:

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Now since two \(3{\text{d}},\) one \(4{\text{s}}\) and three \(4{\text{p}}\) orbitals (all of the nearly equal energy) are to be used for metal-ligand bonding, that is, for accommodating six ligand molecules, they must hybridize to give six \({{\text{d}}^2}{\text{s}}{{\text{p}}^3}\) hybrid orbitals of equal energy so that they give same properties and equal strength to all the metal-ligand bonds.
Thus, resulting six \({{\text{d}}^2}{\text{s}}{{\text{p}}^3}\) hybrid orbitals are occupied by the electron pairs (marked as by \({\text{XX}}\)) donated by six ligand molecules, each donating one lone pair.
Now we know that \({{\text{d}}^2}{\text{s}}{{\text{p}}^3}\) hybridisation leads to octahedral geometry, so the complex \({\left[{{\text{Cr}}{{\left({{\text{N}}{{\text{H}}_3}} \right)}_6}} \right]^{3 + }}\) will be octahedral in shape. Further, since the complexion has three unpaired electrons, it must be paramagnetic, which is found to be so.
Other examples of chromium complexes with a similar structure are \({\left[ {{\text{Cr}}{{\left( {{\text{CN}}} \right)}_6}} \right]^{3 – }}\) and \({\left[{{\text{Cr}}{{\left( {{{\text{H}}_2}{\text{O}}} \right)}_6}} \right]^{3 + }}\)

(ii) Outer Orbital Complexes

In these complexes, \({\text{s,p}}\) and \({\text{d}}\)-orbitals involved in hybridisation belong to the highest quantum number\(\left({\text{n}} \right)\); note the difference from inner complexes where \(\left({{\text{n-1}}} \right){\text{d}},{\text{ns}}\) and \({\text{np}}\) orbitals undergo hybridisation. Complexes formed by the use of inner orbitals are diamagnetic or have reduced Para magnetism. They are called low spin complexes. On the other hand, complexes formed by the use of outer \({\text{nd}}\) orbitals will be paramagnetic. These complexes are called high spin or spin-complexes. The outer orbital complexes have a greater number of unpaired electrons.

Let us take an example to illustrate the formation of outer orbital complexes.

Formation of \({\left[{{\text{Co}}{{\text{F}}_6}} \right]^{3 – }}\):

The electronic configuration of cobalt (atomic number \(27\)) and \({\text{C}}{{\text{o}}^{ + 3}}\) ion present in the given complex are as follows:
Physical experiments show that \({\left[{{\text{Co}}{{\text{F}}_6}} \right]^{3 – }}\) ion has a paramagnetic character corresponding to the four unpaired electrons in \(3{\text{d}}\) orbital. Thus, to get six hybrid orbitals, none of the \(3{\text{d}}\) orbitals is involved in hybridisation. This is also in accordance with the fact that the fluoride ion is a weak ligand and cannot force the pairing up of electrons against Hund’s rule.
To account for the paramagnetic character of the ion, Huggin assumed that \(4{\text{d}}\) (instead of \(3{\text{d}}\)) orbitals are involved in hybridisation. In other words, one \(4{\text{s}},\) there \(4{\text{p}}\) and two \(4{\text{d}}\) orbitals mix to give six equivalent hybrid orbitals which can accept six lone pairs donated by six ligands (i.e., \({{\text{F}}^ – }\) ion) to form \({\left[{{\text{Co}}{{\text{F}}_6}} \right]^{3 – }}\) ion.

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Other examples of outer orbital complexes are \({\left[ {{\rm{Fe}}{{\rm{F}}_6}} \right]^{3 – }},{\left[ {{\rm{Fe}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_6}} \right]^{2 + }},{\left[ {{\rm{Ni}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_6}} \right]^{2 + }},\) etc.

(B) Tetrahedral Complexes

These are formed by \({\text{s}}{{\text{p}}^3}\) hybridisation. Complexes of \({\text{Z}}{{\text{n}}^{2 + }}\) are invariably tetrahedral because they involve \({\text{s}}{{\text{p}}^3}\) hybrid orbitals.
Let us illustrate it by taking \(\left[{{\text{Ni}}{{\left({{\text{CO}}} \right)}_4}} \right].\)

Formation of Nickel tetracarbonyl: Nickel is present in a zero-oxidation state, i.e., in the form of metallic nickel. Its formation can be represented as below:

In the presence of carbon monoxide, ligand rearrangement occurs, and electrons are paired up against Hund’s rule, i.e., the two \(4{\text{s}}\) electrons go to the \(3{\text{d}}\) orbitals to vacate the \(4{\text{s}}\) orbital for electrons donated by the ligand \({\text{CO}}{\text{.}}\) The empty one \(4{\text{s}}\) and three \(4{\text{p}}\) orbitals mix to form four equivalent hybrid orbitals, each of which accepts an electron pair from carbon monoxide molecule forming \({\text{Ni}}{\left({{\text{CO}}}\right)_4}.\)

Thus, nickel tetracarbonyl is tetrahedral and diamagnetic in nature.

(C) Square Planar Complexes

These are formed by \({\text{ds}}{{\text{p}}^2}\) hybridisation. These complexes tend to be formed when the central ion has only one d orbital available in the inner shell.

Formation of nickelocyanide ion, \({\left[{{\text{Ni}}{{\left({{\text{CN}}} \right)}_4}} \right]^{2 – }}\)

The electronic configuration of the nickel atom and \({\text{N}}{{\text{i}}^{2 + }}\) ion is depicted below.
Since the coordination number of \({\text{Ni}}\) in this complex is four, the configuration of \({\text{N}}{{\text{i}}^{2 + }},\) at first sight, shows that the complex is formed by \({\text{s}}{{\text{p}}^3}\) hybridisation, and it is paramagnetic as it has two unpaired electrons. However, experiments show that the complex is diamagnetic, i.e., it does not have any unpaired electrons. This is possible when the \(3{\text{d}}\) electrons rearrange against Hund’s rule, as shown below. This is also following the fact that the ligand involved here is a strong, i.e., \({\text{C}}{{\text{N}}^ – }\) ion.
Hence, now \({\text{ds}}{{\text{p}}^2}\) hybridisation, involving one \(3{\text{d}},\) one \(4{\text{s}}\) and two \(4{\text{p}}\) orbitals, takes place, leading to four \({\text{ds}}{{\text{p}}^2}\) hybrid orbitals, each of which accepts an electron pair from \({\text{C}}{{\text{N}}^ – }\) ion forming \({\left[{{\text{Ni}}{{\left( {{\text{CN}}} \right)}_4}} \right]^{2 – }}\) ion.

The resulting complex is square planar and is diamagnetic as it has no unpaired electron.

Limitations of Valence Bond Theory

1. Although it provides a satisfactory pictorial representation of the complex qualitatively, it does not provide a quantitative interpretation of the stability of complexes.
2. It does not explain the spectra(colour) of the complexes.
3. It does not predict any distortion in symmetrical complexes, whereas all the copper \(\left({{\text{II}}} \right)\) and titanium \(\left({{\text{III}}} \right)\) complexes are distorted.
4. It does not give any detailed information about the magnetic properties of the complexes. In particular, it cannot explain the temperature-dependent Paramagnetism of the complexes.
5. It does not explain why at one time, the electrons must be arranged against Hund’s rule while at other times, the electronic configuration is not disturbed.
6. It does not provide any satisfactory explanation for the existence of inner orbital and outer orbital complexes.
7. Sometimes the theory requires the transfer of electrons from a lower energy level (Example. \(3{\text{d}}\)) to the higher energy level \(\left({4{\text{p}}} \right),\) which is very much unrealistic in the absence of an energy supplier.
8. Electron spin resonance shows that in \({\text{Cu}}\left({{\text{II}}} \right)\) complexes, the electron is not in the 4p level, and the complex is planar.
9. It cannot explain why certain complexes are more labile than others. Labile complexes are those in which one ligand can be easily displaced by another ligand. On the other hand, inert complexes are those in which displacement of ligands is slow.

Valence Bond Theory: Summary

To explain chemical bonding, the Valence Bond Theory (VBT) looks at the interaction between atoms. It is one of two prominent theories that help to explain how atoms join together. The valence bond theory explains the formation of covalent bonds. It also helps in finding the electronic structure of molecules. Furthermore, by using VBT and hybridisation one can explain the geometry of an atom in a molecule. However, VBT fails to explain the existence of inner orbital and outer orbital complexes.

FAQs on Valence Bond Theory

Q.1: What is the valence bond theory?
Ans: The valence bond theory describes the formation of covalent bonds and the electronic structure of molecules. The hypothesis assumes that electrons occupy individual atoms atomic orbitals inside a molecule and that electrons from one atom are attracted to the nucleus of another atom.

Q.2: State any two limitations of Valence Bond Theory.
Ans: The limitations of Valence bond theory are as follows:
1. It does not explain the spectra(colour) of the complexes.
2. It does not provide any satisfactory explanation for the existence of inner orbital and outer orbital complexes.

Q.3: What are valence bond theory and the concept of hybridisation?
Ans: The geometry of an atom in a molecule can be explained and predicted using valence bond theory and hybridisation. The concept of hybridisation is particularly important for understanding the geometry of organic molecules.

Q.4: What is the purpose of VBT?
Ans: The purpose of VBT is to explain the structure and magnetic properties of a large number of coordination compounds.

Q.5: Who introduced the valence bond theory?
Ans: The valence bond (VB) theory was proposed by two American chemists namely, Linus Pauling and John C. Slater.

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