This is a quote from my textbook:

The irregular trend in the first ionisation enthalpy of 3d lmetals,can be accounted for by considering that the removal of one electron alters the relative energies of 4s and 3d orbitals. So the unipositive ions have dn configuration with no 4s electrons.There is thus a reorganization energy accompanying ionisation with some gains in exchange energy as number of electrons increases and from transference of s electrons into d orbitals.However the value of Cr is lower because of the absence of any change in d configuration.

What I basically understand from this is that during first ionisation, electron is lost from $\mathrm{4s}$ orbital because it have higher energy and forms uni-positive ion and the remaining one electron is transferred back into $\mathrm{3d}$ lower energy orbital as energy gap between $\mathrm{3d}$ and $\mathrm{4s}$ lowers. This transition of electrons into lower orbital gives a net decrease in ionisation enthalpy (where energy is released as reorganization energy). But I don't understand what they mentioned about $\ce{Cr}$.


In this site when they explain about Vanadium electronic configuration it has been mentioned that Vanadium in its unipositive ion state have 4 electrons in $\mathrm{3d}$ rather than 3 electrons in $\mathrm{3d}$ and 1 electron in $\mathrm{4s}$ which supports the idea that electrons is transferred from $\mathrm{4s}$ to $\mathrm{3d}$.

Is it true that during ionisation there is a transference of electron from $\mathrm{4s}$ to $\mathrm{3d}$ orbital?Why the text mention reorganization and exchange enthalpy when talking about ionisation?

Text source:NCERT textbook ($\mathrm{d}$- and $\mathrm{f}$-blocks, page 215)

  • $\begingroup$ This doesn't answer your whole question, but I think the point with Cr is that it's configuration is already 3d5 4s1 (for reasons explained in the Chemguide article you linked), so when it's ionised, it has no electrons in its 4s orbital and there is no need to reorganise. Compare this with, say, vanadium, which is 3d3 4s2 when neutral. If an electron is removed from the 4s orbital, there's still one electron left in the 4s which is then transferred to the 3d, and there is reorganisation. $\endgroup$ – atbm Feb 14 at 15:15
  • $\begingroup$ Is there some energy release during reorganization $\endgroup$ – Grace Feb 14 at 16:21

As your textbook notes, the relative energy of atomic orbitals is not fixed as you start ionizing your atom. As you ionize, all orbitals drop in energy. However, higher angular momentum orbitals drop more than lower ones, and specifically in your case, the $\mathrm{3d}$ orbitals drop below the $\mathrm{4s}$ orbitals. So consider vanadium, the atom has a $\mathrm{[Ar]4s^23d^3}$ configuration, but after you ionize it, the $\mathrm{3d}$ orbitals drop below the $\mathrm{4s}$ ones and you get $\mathrm{[Ar]3d^4}$ as the electron configuration of the $\ce{V+}$ ion. So your measured ionization energy includes both the energy it takes to remove an electron from $\ce{V}$ and also some reorganization energy. You know there is some reorganization energy since two electrons (both $\mathrm{4s}$ ones) have "moved" but you only removed one.

Now, what about $\ce{Cr}$? Recall that $\ce{Cr}$ is a bit special and has a $\mathrm{[Ar]3d^54s^1}$ configuration when neutral. After we remove an electron (in this case, a $\mathrm{4s}$ one), we have an $\mathrm{[Ar]3d^5}$ configuration for $\ce{Cr+}$. Note that no electrons had to move around besides the one we removed. So there is no reorganization energy in the case of $\ce{Cr}$, as your textbook notes.

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  • $\begingroup$ I read from above mentioned site that 3d orbital have less energy than 4s orbital before ionisation,thus ionisation occur from 4s rather than 3d.And since reorganization provides some stability(due to electron moving from higher to lower orbital)...Cr doesn't get to have this stability so shouldn't that mean a higher ionisation enthalpy for Cr. $\endgroup$ – Grace Feb 15 at 1:37

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