The atomic numbers of vanadium (V), chromium (Cr), manganese (Mn) and iron (Fe) are, respectively, 23, 24, 25 and 26. Which one of these may be expected to have the highest second ionization enthalpy?
Answer & explanation
Correct answer: option 1
The second ionization enthalpy is the energy required to remove the second electron from an atom. The second electron is removed from the valence shell of the atom, which is already positively charged after the removal of the first electron. This makes it more difficult to remove the second electron, and the second ionization enthalpy is therefore higher than the first ionization enthalpy.
In the case of V, Cr, Mn, and Fe, the second electron is removed from the 3d orbital. The 3d orbital is a relatively stable orbital, and the second ionization enthalpy is therefore high for all of these elements. However, the second ionization enthalpy is highest for Cr because the 3d orbital in Cr is half-filled. This makes the 3d orbital more stable, and it is therefore more difficult to remove the second electron.
Therefore, the answer is (1) Cr has the highest second ionization enthalpy.