Name: All About Molecular Orbital Theory
Uploaded: 2019-07-19
Duration: 9 min 36 s
Description: The molecular orbital theory revolutionised the study of chemical bonding by approximating the states of bonded electrons. Let's learn more about this theory.

Question

Show how the

LCAO

approximation gives rise to bonding and anti bonding orbitals. Illustrate your answer with reference to three different diatomic molecules.

Accepted Answer

Electron can be combined as particle or wave. Therefore, an electron in an atom be described as occupying an atomic orbital or by wave function $ψ$ . Electron in a molecule are said to occupy molecular orbitals. Wave function describing a molecular orbital may be obtained by Linear combination of atomic orbitals $(LCAO)$ .

Let us consider two atoms, $A$ and $B$ which have atomic orbitals described by the wave functions $ψ_{(A)}$ and $ψ_{(B)}$ . If the electron cloud of these two atoms overlap then the wave function of the molecule can be obtained by linear combination of atomic orbitals ( $ψ_{(A)}$ and $ψ_{(B)}$ ). Molecular orbital can be represented as $(ψ_{(AB)})$

Electron phase that describe atomic orbitals have positive and negative amplitude. Waves may interact either constructively or destructively. If atomic orbitals overlap in phase, constructive interference occurs in region between the two nuclei which results in bonding orbital.If atomic orbitals overlap out of phase, destructive interference reduce the probability of finding an electron in the region between the nuclei which results in anti bonding orbital.

$H_{2}$ molecule

One electron from each atom, so two electrons in molecule. They occupy the lowest energy molecular orbital which is bonding molecular orbital $σ_{1 s^{2}}$ and anti bonding molecular orbital $σ_{1 s}^{*}$ is empty.

$B_{2}$ molecule

Each boron atom has $5$ electrons .

Configuration: $1 s^{2} 2 s^{2} 2 p^{1}$

So, $B_{2}$ molecule has 10 electrons, which are arranged in molecular orbital as $σ_{1 s^{2}} < σ_{1 s^{2}}^{*} < σ_{2 s^{2}}^{} < σ_{2 s^{2}}^{*} < π_{2 p_{x}^{1}} = π_{2 p_{y}^{1}}$

The inner shell $(1 s)$ does not participate in bonding. The effects of bonding and anti bonding in $σ_{2 s}$ orbitals gets cancel but stabilization occur from the filling of $π_{2 p}$ orbitals and hence a bond is formed and $B_{2}$ exist.

$N_{2}$ molecule

Each nitrogen atom has $7$ electrons.

Configuration: $1 s^{2} 2 s^{2} 2 p^{3}$

So, $N_{2}$ molecule has $14$ electrons. They are arranged in molecular orbitals as $σ_{1 s^{2}} < σ_{1 s^{2}}^{*} < σ_{2 s^{2}} < σ_{2 s^{2}}^{*} < π_{2 p_{x}^{2}} = π_{2 p_{y}^{2}} < σ_{2 p_{z}^{2}}$

Inner shell $(1 s)$ does not participate in bonding. Bonding and anti bonding in $2 s$ level gets cancel. One $σ$ and two $π$ remain. It gives a total of three bonds $(N \equiv N)$ .

J. D. Lee Solutions for Chapter: The Covalent Bond, Exercise 1: Exercise 1

J. D. Lee Chemistry Solutions for Exercise - J. D. Lee Solutions for Chapter: The Covalent Bond, Exercise 1: Exercise 1

Questions from J. D. Lee Solutions for Chapter: The Covalent Bond, Exercise 1: Exercise 1 with Hints & Solutions

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