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 $\ce{Cr}$.
https://www.chemguide.co.uk/atoms/properties/3d4sproblem.html
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)