The atomic numbers of vanadium chromium, manganese and iron are respectively 23, 24, 25 and 26. Which one of these may be expected to show highest second ionization enthalpy 

Atomic number of chromium is 24 and its electronic configuration is $\left[\mathrm{Ar}\right]3{\mathrm{d}}^{5}4{\mathrm{s}}^1.$ After loosing one electron, the configuration becomes $\left[\mathrm{Ar}\right]3{\mathrm{d}}^5.$ After losing one electron, Cr achieves a stable half-filled configuration, therefore its second ionization energy is very high.