# What is a degenerate orbital

I don't really get what degenerate orbitals are, I mean I know that orbitals are degenerate when they have the same energy level, and I know that the energy levels depend on n, but does that mean that 1s is never degenerate and 2s and 2p are always degenerate or am I misunderstanding something, because 2s and 2p have the same principal quantum number n=2, thus same energy levels? Is this correct or am I missing something

• It seems that you are mixing two ideas. If you solve the Schrödinger equation for your system $\hat{H}\Phi_i =E\Phi_i$ you would obtain two kind of solutions. The scalar (eigenvalues) $E$, the energy, and the wavefunction $\Phi$ (eigenfunctions). In the case of the hydrogen Hamiltonian solutions, you will find that there are wavefunctions that have the same energy (they are degenerate), but when other observables are considered (e.g. angular momentum), they render different values. Check your Physical Chemistry textbook.
– PAEP
Jan 19 at 17:00
• Jan 19 at 19:39

The orbits $$2$$s and $$2$$p are degenerate in $$\ce{H}$$ atom. In heavier atoms, these orbitals have different energies. They are not degenerate any more.

• And that's because with heavier atoms the energy levels also depend on $l$ and then f.e. 2p is degenerate but not 2s? Jan 19 at 17:01
• yes, the three 2p orbitals are degenerate Jan 19 at 17:50
• @Maurice I think we're glossing over an important concept here which is that in an atom with a single electron, the degeneracy is across the level. Adding additional electrons breaks the degeneracy because nuclear penetration is different.
– Zhe
Jan 19 at 19:27
• Note that ns and np orbitals are not exactly degenerate in H atom either, but the involved effects of later developed quantum electrodynamics are not addressed by Schroedinger equation. Feb 10 at 15:58

There is multiplicity and degeneracy. In the H atom the n=2 levels the multiplicity is four one 2s, and three 2p, and they are also degenerate in the simplest models. This means that these levels are equal in energy. However, even in the H atom experiments show that this is not strictly true because in the 2p orbital the electron has some orbital angular momentum and as the electron also has angular momentum (due to its spin) these interact by a tiny but measurable amount and the multiplicity remains but the degeneracy is broken. (The interaction is called spin-orbit coupling). The three p orbitals scan also be split in energy, thus removing their degeneracy, by an electric field (Stark effect) or by a magnetic field (Zeeman effect).

Concerning your doubts, I suggest you to approach the problem in a more mathematical fashion. Orbitals for hydrogenoid atoms (or even for molecules) are stationary wavefunctions $$\phi_i$$ that satisfy the time-independent Schroedinger's equation: $$\hat{H}\phi_i = E_i\phi_i$$ Now, if the set of orbitals $${\phi_i}$$ is such that it is possible to have k different $$\phi_ij$$ such that $$E_i$$ is always the same for different j. $$\hat{H}\phi_{ij} = E_i\phi_{ij}\ \ \forall j={1,...,k}$$ We call those orbitals "k-fold degenerate orbitals", i.e. k orbitals with the same energy value (due to the particular simmetry of the molecule/atom under study). Now: for the hydrogen atom there is no difference in the energy of any orbital with the same quantum number $$n$$.

The solution of Schroedinger's equation for the hydrogenoid atom, however, gives us different energies for the orbitals $$\phi_{n,l,m,s}$$ associated with the usual quantum numbers n, l, m, s. For the same n, there exist n-1 orbitals with the secondary quantum number l ranging from 1 to n-1, each set of orbitals with a different l has a different shape and a different energy (in general this holds for approximate orbitals of non-hydrogenoid atoms). For the same n there exist a range of quantum numbers m={-l,..., 0,..., +l} of orbitals with the same shape but different orientation in space, these have the same energy and are therefore $$degenerate$$. The spin number s is not of our concern here, it can assume only half-integer values for any atomic orbital.

In this context:

In the energy levels with n=1 there is only one spin-orbital with $$l=0$$, $$m=0$$ and $$s=\pm\frac{1}{2}$$, meaning that there is one orbital that we commonly call 1s. For levels with n=2 there are two set of possible orbitals: $$n=2,\ l=0,\ m=0,\ s=\pm\frac{1}{2} \implies \ 2s$$ $$n=2,\ l=1,\ m=\{-1,0,+1\},\ s=\pm\frac{1}{2}\ \implies 2p_x,2p_y, 2p_z$$ So 1s and 2s orbitals are nondegenerate while 2p orbitals are 3-fold degenerate. In the same fashion 3p orbitals are 3-fold degenerate and 3d orbitals are 5-fold degenerate ($$l=\{-2,\ -1,\ \ 0,\ +1,\ +2\}$$).

The energy levels for orbitals in mono-electronic species (i.e. atoms or ions with one electron only; for example, hydrogen) are dependent on just the principal quantum number $$n$$. But, for multielectronic species (i.e. atoms or ions with more than one electron), the energy level depends on both the principal quantum number $$n$$ and the azimuthal quantum number $$l$$.

That is,

• For monoelectronic-species:
$$1s<2s=2p<3s=3p=3d<4s=4p=4d=4f\ (\text{Energy comparison})$$

• For multielectronic-species:
$$1s<2s<2p<3s<3p<4s<3d<4p<4d<4f\ (\text{Energy comparison})$$

This energy order for multielectronic species, by the way, is exactly what you derive from the Aufbau principle.

Notice that in the orbital energy level diagram for monoelectronic species (each box corresponds to one orbital), all the orbitals in the $$2s$$ and $$2p$$ subshells, for example, have the same energy, and are thus degenerate. In the orbital energy level diagram for multi-electronic species, however, $$2s$$ and $$2p$$ subshells have different energies and thus all their orbitals "together" are not degenerate.
But, (irrespective of whether the species in question is mono-electronic or multi-electronic) the three orbitals in the $$2p$$ subshell, namely $$2p_x$$, $$2p_y$$ and $$2p_z$$, do have the same energy, and thus we can say that they ARE degenerate, as Gaurav also pointed out. Hopefully, that alleviates your confusion.