In a multi-electron atom, which of the following orbitals described by the three quantum numbers will have the same energy in the absence of magnetic acid and electric fields? (a) $$n = 1, l = 0, m = 0$$ (b) $$n = 2, l = 0, m = 0$$ (c) $$n = 2, l = 1, m = 1$$ (d) $$n = 3, l = 2, m = 1$$ (e) $$n = 3, l = 2, m = 0$$

For a hydrogen atom the energy of an orbital depends only on the principal quantum number $$n$$, so all the orbitals having the same $$n$$ are degenerate. In a multi-electron atom the electron-electron repulsions break this degeneracy: the energy of an orbital is now a function of both $$n$$ and the azimuthal quantum number $$l$$ (as summarised by the $$n+l$$ rule).

However, in the absence of any external magnetic or electric field the energy still does not depend on the magnetic quantum number $$m$$, because all the different orientations of a given subshell experience the same electrostatic environment inside the atom. Therefore:

• Orbitals with the same pair $$(n,l)$$ but different $$m$$ values remain degenerate.
• Orbitals that differ in either $$n$$ or $$l$$ will, in general, have different energies in a multi-electron atom.

Let us examine the list:

(a) $$n=1,\; l=0,\; m=0$$ — a 1s orbital
(b) $$n=2,\; l=0,\; m=0$$ — a 2s orbital
(c) $$n=2,\; l=1,\; m=1$$ — a 2p orbital
(d) $$n=3,\; l=2,\; m=1$$ — a 3d orbital (one particular orientation)
(e) $$n=3,\; l=2,\; m=0$$ — the same 3d subshell but a different orientation

Only (d) and (e) share the identical quantum numbers $$n=3$$ and $$l=2$$; they differ only in $$m$$. Hence they possess the same energy in a multi-electron atom when no external fields are applied. All the other pairs differ either in $$n$$ or in $$l$$, so their energies are different.

Therefore the orbitals that have the same energy are (d) and (e).

Option D which is: (d) and (e)