Molecular Orbital Theory

Comparison of Bonding Theories

The table given below explains the major differences between the valence bond theory and molecular orbital theory:

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Molecular orbital theory describes the distribution of electrons in molecules in much the same way that the distribution of electrons in atoms is described using atomic orbitals. Using quantum mechanics, the behavior of an electron in a molecule is still described by a wave function, Ψ, analogous to the behavior in an atom. Just like electrons around isolated atoms, electrons around atoms in molecules are limited to discrete (quantized) energies. The region of space in which a valence electron in a molecule is likely to be found is called a molecular orbital (Ψ²). Like an atomic orbital, a molecular orbital is full when it contains two electrons with opposite spin.

We will consider the molecular orbitals in molecules composed of two identical atoms (H₂ or Cl₂, for example). Such molecules are called homonuclear diatomic molecules. In these diatomic molecules, several types of molecular orbitals occur.

The mathematical process of combining atomic orbitals to generate molecular orbitals is called the linear combination of atomic orbitals (LCAO). The wave function describes the wavelike properties of an electron. Molecular orbitals are combinations of atomic orbital wave functions. Combining waves can lead to constructive interference, in which peaks line up with peaks, or destructive interference, in which peaks line up with troughs as shown in the figure below. In orbitals, the waves are three-dimensional, and they combine with in-phase waves producing regions with a higher probability of electron density and out-of-phase waves producing nodes, or regions of no electron density.

(a) When in-phase waves combine, constructive interference produces a wave with greater amplitude.

(b) When out-of-phase waves combine, destructive interference produces a wave with less (or no) amplitude.

Let us apply this theory to homonuclear diatomic molecules such as hydrogen molecule. Consider the two atoms of hydrogen in the molecule as A and B. Each hydrogen atom has one electron in 1s orbital in ground state. These atomic orbitals may be represented by the wave functions $
\psi_A
$ and $
\psi_B
$ respectively. Then according to LCAO method, the molecular orbitals in the H₂ molecule are given by linear combination (addition or subtraction) of wave functions of the individual atoms, i.e., of $
\psi_A
$ and $
\psi_B
$ as shown below.

$
\begin{aligned}
& \psi_{M O}=\psi_A \pm \psi_B \\
& \psi_b=\psi_A+\psi_B \\
& \psi_a=\psi_A-\psi_B
\end{aligned}
$

The molecular orbital $
\psi_B
$ formed by the addition overlap of atomic orbitals is called bonding molecular orbital (BMO) and the molecular orbital $
\psi_A
$ formed by the subtraction overlap of atomic orbitals is called anti-bonding molecular orbital (ABMO).

The combination of 1s orbitals of hydrogen atoms to form molecular orbitals has been shown in figure below.

The probability density of bonding molecular orbital (BMO) is given by $
\begin{aligned}
& \psi_b^2 \operatorname{or}\left(\psi_A+\psi_B\right)^2 \\
& \text { or }\left(\psi_A^2+\psi_B^2+2 \psi_A \psi_B\right)
\end{aligned}
$, which means that shared electron density $
\left(\psi_A+\psi_B\right)^2
$ is higher than the mere sum of the electron densities of two seperate orbitals $
\left(\psi_A^2+\psi_B^2\right)
$. Thus,addition combination leads to increase in electron density between two nuclei A and B. In other words electron present in BMO experience greater force of attraction thereby lowering the energy of BMO as compared to individual atomic orbitals.

Similarly the probability density of antibonding molecular orbital (ABMO) is given by $
\begin{gathered}
\psi_a^2 \operatorname{or}\left(\psi_A-\psi_B\right)^2 \operatorname{or} \\
\left(\psi_A^2+\psi_B^2-2 \psi_A \psi_B\right)
\end{gathered}
$ which means that the shared electron density $
\left(\psi_A-\psi_B\right)^2
$ is lower than the sum of electron densities of seperate orbitals $
\left(\psi_A^2+\psi_B^2\right)
$. Thus, subtraction combination leads to lowering of electron density in between the nuclei. The electron density, in fact, is concentrated away from the nuclei creating a node in between the nuclei. The electrons present in the ABMO experience repulsive interactions thereby raising the energy of ABMO as compared to atomic orbitals.

Relative energies of atomic orbitals and molecular orbitals in hydrogen molecule as shown in figure below:

It may be noted that energy of the antibonding orbital is raised above the energy of the atomic orbitals that have combined by an amount more than that by which the energy of the bonding orbitals has been lowered. In other words, the destabilisation effect of ABMO is more pronounced than the stabilisation effect of BMO.

Conditions for the Combination of Atomic Orbitals

For the atomic orbitals to combine resulting in the formation of molecular orbitals the main conditions are:

• The combining atomic orbitals should have almost same energies.
• The extent of overlap between the atomic orbitals of the two atoms should be large.
• The combining atomic orbitals should have the same symmetry about the molecular axis.

It must be noted that Z-axis is taken as the internuclear axis according to modern conventions.

Related Topics:

• Molecular Geometry
• Energy Level Diagram
• Dipole Moment

Some Solved Examples

Example 1: During the formation of a molecular orbital from atomic orbital, the electron density is :

1) Minimum in the nodal place

2) Maximum in the nodal place

3) Zero in the nodal place

4) Zero on the surface of the lobe

Solution:

Nodal planes are regions around atomic nuclei where the probability of finding an electron is zero.

For example, see in the p-orbitals

Hence, option (3) is correct.

Example 2: The stability of molecular orbital is:

1) Less than atomic orbitals

2) More than atomic orbitals

3) Can’t be predicted

4) None of these.

Solution:

The number of molecular orbitals formed is equal to the number of combining atomic orbitals. When two atomic orbitals combine, two molecular orbitals are formed. One is known as a bonding molecular orbital while the other is called an antibonding molecular orbital.
Hence, the answer is the option (1).

Example 3: The number of molecular orbitals is:

1) Equal to the number of combining atomic orbitals.

2) Not equal to the number of combining atomic orbitals.

3) Equal to twice the number of combining atomic orbitals.

4) None of these.

Solution:

The number of molecular orbitals formed is equal to the number of combining atomic orbitals. When two atomic orbitals combine, two molecular orbitals are formed. One is known as a bonding molecular orbital while the other is called an antibonding molecular orbital.
Hence, the answer is the option (1).

Example 4: Of the species $\mathrm{NO},{\mathrm{NO}}^{+},{\mathrm{NO}}^{2+}$ and ${\mathrm{NO}}^{-}$ the one with the minimum bond strength is

1)NO⁺
2) NO
3)NO²⁺
4)NO^-

Solution:

Bond order of NO²⁺ = 2.5
Bond order of NO⁺ = 3
Bond order of NO = 2.5
Bond order of NO^- = 2

Bond order $\propto$ bond strength

Thus, NO^- has the minimum bond strength.

Hence, the answer is the option (4).

Example 5:

Total number of molecules/species from following which will be paramagnetic is-------------

$\qquad$ $\mathrm{O}_2, \mathrm{O}_2^{+}, \mathrm{O}_2^{-}, \mathrm{NO}, \mathrm{NO}_2, \mathrm{CO}, \mathrm{K}_2\left[\mathrm{NiCl}_4\right]$, $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3, \mathrm{~K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]$ [JEE Main 2025]

Solution:

To determine which molecules or species are paramagnetic, we need to identify those with unpaired electrons.

$\mathrm{O}_2 \rightarrow 2$ unpaired electrons according to Molecular orbital theory.
$\mathrm{O}_2^{+} \rightarrow 1$ unpaired electrons according to Molecular orbital theory.
$\mathrm{O}_2^{-} \rightarrow 1$ unpaired electrons according to Molecular orbital theory.

NO: As a species with an odd number of electrons, NO has unpaired electrons, making it paramagnetic.
$\mathrm{NO}_2$ : This molecule is also an odd-electron species, which means it has unpaired electrons and is paramagnetic.

$\mathrm{K}_2\left[\mathrm{NiCl}_4\right]$ : In this compound, $\mathrm{Ni}^{2+}$ (nickel with a +2 oxidation state) has an electron configuration of $3 d^8$. Given the presence of weak field ligands and a coordination number of 4 , this forms a tetrahedral complex. As a result, it is paramagnetic with 2 unpaired electrons.

Thus, the paramagnetic species in the provided list are $\mathrm{O}_2, \mathrm{O}_2^{+}, \mathrm{O}_2^{-}, \mathrm{NO}, \mathrm{NO}_2$, and $\mathrm{K}_2\left[\mathrm{NiCl}_4\right]$.

Hence, the answer is 6.

Practice More Questions From the Link Given Below:

Summary

The molecular geometry is a three-dimensional arrangement of atoms in a molecule. It is decided by the spatial distribution of electron pairs around the central atom. The VSEPR theory predicts this geometry by minimizing electron pair repulsion. The key factors are bonding pairs, lone pairs, and bond types. The common shapes are linear, bent, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral. For example, CO₂ is linear, while NH₃ has a trigonal pyramidal geometry due to a lone pair at nitrogen.

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