# Why does the d orbital size decreases on addition of electrons?

In Concise Inorganic Chemistry by JD Lee (4th edition; adapted by Sudarshan Guha), on page 80 under section 3.7 "The Extent of d-orbital Participation in Molecular Bonding" it is given:

A second factor affecting the size of d orbitals is the number of d orbitals occupied by electrons. If only one 3d orbital is occupied on an S atom, the average radial distance is $$2.46$$Å, but when two 3d orbitals are occupied the distance drops to $$1.60$$Å.

I wish to know the reason why the size of the d orbital drops when additional electrons are added. I think this is counter-intuitive since on addition of an extra electron the interelectronic repulsions increases and confining them to a small volume increases overall energy of the system thereby making it unstable. So, I think the orbital size must increase, but it is happening the opposite way.

Kindly explain the above-mentioned concept.

• Jan's answer below is spot on but I think the author borrowed the same logic from why atomic radii decreases as you move across a period. – M.A.R. Aug 25 '19 at 9:58

From what I believe I may remember about the outdated concept, ‘one d orbital’ refers to cases like $$\ce{SO3^2-}$$ with sulphur in the $$\mathrm{+IV}$$ oxidation state while ‘two d orbitals’ would refer to e.g. $$\ce{SO4^2-}$$ or an oxidation state of $$\mathrm{+VI}$$. In the Lewis formalism, the answer is easily deduced if you draw octet-abiding structures: $$\ce{SO3^2-}$$ will include a formal single positive charge on sulphur while in $$\ce{SO4^2-}$$ sulphur has a formal double positive charge. The higher an oxidation state, the more contacted an atom becomes.
A more correct answer would elaborate on ($$\mathrm{sp}^n$$-type, formed only by sulphur’s s and p atomic orbitals) molecular orbitals and the increasing number of electronegative substituents that need to bond to the central sulphur, reducing its charge density and leading to the same effect.
• @Intellex You misunderstood me. The calculated molecular orbitals for these types of compounds are made up of over 99 % s and p contributions. With these $\mathrm{sp}^n$-type orbitals, you can do the electronegativity discussion and reduce their energies/contact them. While the outer d orbitals as well as any other highly virtual orbitals will also contact and be lowered in energy, this has no effect on the bonding situation as the difference between the energies remains almost the same. The charge density mentioned is sulphur’s overall charge density across all (including core) orbitals. – Jan Aug 26 '19 at 2:48