0
$\begingroup$

The energy of an orbital is proportional to its mean radial distance, and since the 3d orbital is much larger it is much higher in energy than the 3s and 3p orbitals

All references from JD Lee Concise inorganic chemistry

However, d orbitals are in general too large and too high in energy to mix completely with s and p orbitals.

In this caption, JD Lee says that the d orbitals find it hard to undergo hybridization because of its too large size and higher energy, but to decrease its energy and size , it employs various solutions, such as:

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 A., but when two 3d orbitals are occupied the distance drops to 1.60 A.

A further small contraction of d orbitals may arise by coupling of the spins of electrons occupying different orbitals. It seems probable that d orbitals do participate in bonding in cases where d orbital contraction occurs.

One of them was that when electrons start filling in d orbitals, contraction of orbitals occurs. When pairing of electrons commences, d orbitals contract further.

But I find this line confusing, because the pairing/filling of electrons in the d orbitals must increase the energy, and thus the size, of d orbitals. But JD LEE states that the opposite happens. Could someone please explain why?

$\endgroup$
  • $\begingroup$ If electrons are excited from 3p to 3d, then that would mean less shielding from the inner orbitals which would serve to contract outer orbitals, I guess? $\endgroup$ – Jan Oct 23 at 9:54
  • $\begingroup$ There is no size of d orbitals, they extend to infinity. Therefore d orbitals cannot get larger and smaller, and they most certainly do not 'employ solutions'. So wherever these numbers come from, one would need more information to judge the statement by JD Lee. I have a very bad feeling, that the argument leads towards hybridisation involving d orbitals; this would be wrong and you can safely assume that the rest is also phishy. $\endgroup$ – Martin - マーチン Oct 29 at 13:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.