Question

The first ionization enthalpies of 5d elements is higher than that of the corresponding 3d and 4d elements. This is best explained by:

• A. Greater extent of metallic bonding in 5d elements
• B. Greater effective nuclear charge in the case of 5d elements due to the poor shielding of valence electrons by the 4f electrons
• C. Neither a nor b
• D. Both a and b

Answer 1

The correct option is B Greater effective nuclear charge in the case of 5d elements due to the poor shielding of valence electrons by the 4f electrons

In general, among the s, p d and f orbitals, what is the decreasing order of magnitude of shielding effect on the valence electrons?
For the same principal quantum number, s orbitals electrons do a far better job at shielding the valence electrons from the nucleus than the p orbital electrons. Similarly, the shielding effect of p orbitals screen the valence electrons better than the d orbital electrons and so on.
s > p > d > f (magnitude of shielding effect on valence electrons for the same n)
5d elements have much more nuclear charge and the weakest shielding of valence electrons. Hence their elements experience the maximum electrostatic attraction from the nucleus and thus also have the highest first ionization energies among 3d 4d and 5d elements. Do note that ionization enthalpy is the amount of energy required to remove the most loosely bound valence electron of an isolated gaseous atom to form a cation. So metallic bonding does not come into considerations here!