Ionization enthalpy is the energy required to remove the electron from the outer most orbit.
Electronic configuration :
V(23)=[Ar]3d34S2 Cr(24)=[Ar]‌‌3d5‌‌4S1 Mn(25)=[Ar]3d54S2 Fe(26)=[Ar]‌‌3d6‌‌4S2 After first ionization enthalpy the electronic configuration will be,
V(23)=[Ar]3d34S1Cr(24)=[]3d5Mn(25)=[.Fe(26)=[Ar]3d64S1 Now in 2 nd ionization enthalpy one more electron will be removed. And after second ionization enthalpy electronic configuration will be,
V(23)=[.Cr(24)=[Ar]3d4Mn(25)=[]3d5Fe(26)=[Ar]3d6 For
V(23),Mn(25) and
Fe(26) electron is removed from
4S orbital but for
Cr(24) electron is removed from
3d orbital.
We know distance of
4S>3d from the nucleus. So, the attraction on the electrons of 45 shell is less compared to
3d shell by the nucleus. So, the removed of electron will be easier for the electrons of
4S shell than
3d shell.
So, for
Cr(24) second ionization is more than other as electron is removed from
3d shell.
And for
Cr(24) outer most shell "
3d′′ was half filled after 1 st ionization enthalpy, so
3d shell was stable. So removal of electron will be difficult from stable shell. That is why also
2nd ionization enthalpy of
Cr(24) is higher compare to other given atoms.