Degenerate orbital

See also: Degenerate energy level

Degenerate orbitals for electrons in an atomic subshell are orbitals at identical energy levels. These are important in physical chemistry because they affect the ways electrons fill atoms (see Hund's rule). For example, all the 3p orbitals have same energy level, and so do all the 5d orbitals. Each orbital is defined as if it lies along a set of x, y, z axes. The 2p orbitals are thus 2px, 2py and 2pz. The 3p orbitals would be the same shape but larger and at higher energy.

Degenerate orbitals play an important role in Molecular Orbital theory. For example, consider the three 2p orbitals. When forming a molecular orbital, the 2px orbital combines with another 2px orbital, the 2py with a 2py and a 2pz with a 2pz. The 2px orbitals combine "head on" forming sigma bonding and antibonding orbitals. The 2py orbitals and the 2pz orbitals combine separately forming 2 pairs of pi bonding and antibonding orbitals. These pi orbitals are degenerate since they have the same energy (different from the sigma orbitals). The sequence for the orbitals is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2.

It is important to note that the symmetry of degenerate orbitals is lost as they come closer to any other molecule because of asymmetric interactions of these orbitals with the electric and magnetic fields of the other molecule. As a result there is a change in their energy levels too and the degeneracy is lost.

Here is a simple quantum mechanical example of degeneracy. Let's consider a particle in a 2-D square box of side a. Inside the box the potential is 0 everywhere while outside it's infinity. This ensures that the particle is confined inside the box. Let's solve the time independent schrodinger equation to get the possible states of the system

    HΨ = EΨ
   - h2 / (8mπ2) [ ∂2 Ψ(x,y) /dx2 +  ∂2 Ψ(x,y) /dy2] + V Ψ(x,y) = E Ψ(x,y)
   - h2 / (8mπ2) [∂2 Ψ(x,y) /dx2 +  ∂2 Ψ(x,y) /dy2 ] = E Ψ(x,y)

(V = 0 inside & Ψ = 0 outside)

This differential euation can be solved using the separation of variables method i.e by substituting:

 Ψ ( x , y) = Ψ (x) Ψ (y)

The following solution is obtained:

Ψ (nx , ny) ( x , y)=(2/a) sin ( nx πx /a ) sin ( ny πy / a)

E = ( h2/ (8 m π2)) (nx 2 + ny 2)

It is obvious from the expression for energy that E (m,n) = E (n,m), so these two states are degenerate states. But had it been a non-square rectangular box E (m,n) ≠ E (n,m). The argument can be extended for a 3-D cubic box. So it can be concluded that degeneracy is just manifestation of symmetry. ...

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