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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.

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  • $\begingroup$ 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. $\endgroup$ – M.A.R. Aug 25 at 9:58
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There is not much to explain because the concept is wrong. There is no sulphur compounds in which the 3d orbitals (remember: they have an energy level similar to 4s) are involved in bonding in any noteworthy way, shape or form.

From what I believe I may remember about the outdated concept, ‘one d orbital’ refers to cases like $\ce{SO3^2-}$ with sulphur in the $\mathrm{+IV}$ oxidation state while ‘two d orbitals’ would refer to e.g.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.

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  • $\begingroup$ Thank you for your answer. In the first paragraph, you have said the concept of d orbital participation in sulphur compounds is incorrect. I understood the reason you have mention in the same paragraph. In the last paragraph, you have said (or what I understood based on your answer and what was given in my book as the first reason for the participation of d orbitals in bonding) increase in electronegative substituents causes d orbitals to contract in size thereby causing a fall in energy similar to that of the orbitals participating in bonding. $\endgroup$ – Intellex Aug 25 at 3:18
  • $\begingroup$ If the concept is incorrect why must it have an explanation? Sorry to irritate you by this large comment. Kindly reply when you find time. $\endgroup$ – Intellex Aug 25 at 3:19
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    $\begingroup$ @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. $\endgroup$ – Jan Aug 26 at 2:48

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