# What is meant by “no d-orbital”?

Sulfur and oxygen belong to the same group. Sulfur has a vacant d-orbital while oxygen has no vacant d-orbital.

What does having "no d-orbital" mean? Orbitals are just the spaces around the atom. How can you say that there is no space around the atom which makes the d-orbital?

There is a bit of a philosophical debate as to whether orbitals exist only when they're populated, or if they're always there. In both oxygen and sulfur, there is no occupied $d$ orbital in the ground state (so both $3d$ orbitals are vacant), but in sulfur the promotion energy of an electron from a $3s$ or $3p$ orbital to a $3d$ orbital is much less than the promotion energy of an electron in the $2s$ or $2p$ orbitals in oxygen to a $3d$ orbital ($2d$ orbitals don't exist, of course). Arguably, this means that sulfur can access its $3d$ orbitals under the right conditions since the promotion energy required is relatively low and could be supplied in chemically relevant situations (so the sulfur $3d$ orbital is accessible).

It is not expected that an oxygen atom could ever populate its $3d$ orbital in a stable substance. It is possible to occupy an oxygen $3d$ orbital for a short while, however, by exciting the atom with a photon of the proper frequency.

• So, we say that sulphur has a vacant d-orbital because the electrons can occupy that vacancy, right? – Rafique Apr 21 '13 at 9:21
• Strictly speaking, vacancy merely implies that the orbital is empty. The 3d orbitals are vacant in both oxygen and sulphur, but they're only accessible in the latter. – Nicolau Saker Neto Apr 21 '13 at 14:06
• Actually 3d orbital could be used, if you write down a Full-CI expansion for the ground state wavefunction, isn't it? – user26143 Dec 1 '13 at 22:16
• @user26143 I don't have a large understanding of computational chemistry, but I have read that calculations can improve if you also include orbitals that aren't formally populated. There are some subtleties which escape me, however. I know it is a point of confusion with calculations performed for "hypervalent" molecules; even though they are calculated to have negligible population of upper d orbitals, including them in the calculation provides lower energies. Hybridization fans will say it's proof of hypervalency, while computational chemists tend to say it's something else entirely. – Nicolau Saker Neto Dec 1 '13 at 22:35
• Sorry, I mean in isolated atom.... – user26143 Dec 1 '13 at 22:38

As Nicolau hints in his answer, orbitals are not regions of space around an atomic nucleus. They are mathematical constructs formulated as wavefunctions describing the properties of electrons, including their energy, angular momentum, and probabilistic distribution in space. The pictures of "orbitals" that you are used to seeing are the probability density functions (the absolute value of the wavefunction squared).

Ultimately, the only way we can know what a probability density function looks like or what the energy of orbital would be is to pretend there is an electron in the orbital and then do the math. So, orbitals without electrons in them might as well not exist. As Nicolau and ManishEarth point out for oxygen, the orbitals in the 3d subshell are basically inaccessible to the electrons in oxygen. Thus those orbitals do not exist.

Oxygen has a d-orbital as well. Just that there are $3s$ and $3p$ orbitals in between, so extra electrons will get filled there first. The $3d$ orbital has too high energy for it to be of any use.

Also, hybridization almost always (the exception is with coordination compounds) happens for orbitals with the same principal quantum number (shell number). So the $3s$ and $3p$ orbitals are not of any use for accomodating extra electrons/bonds.