Question

describe, in term of shielding and effective nuclear charge, the following phenomena:

1. why, for main group elements, do atomic radii decrease going from left to right across the periodic table?

2. why, for transition metals, do atomic radii generally stay constant going from left to right across the periodic table?

3. why, in an isolectronic series, do the ions with the more positive chrage have the smallest radius?

Answer 1

First remember that shielding effect and effective nuclear charge work against each other. Increase in nuclear charge tend to decrease atomic radius and increase in shielding tend to increase atomic radius.

1. As we move from left to right across the periodic table , both nuclear charge and shielding increases. But increase in nuclear charge dominate over increase in shielding. As a result atomic radius of main group elements decreases as we move from left to right across periodic table.

2. We know that transition element have d-orbitals. Shielding effect of d-orbitals are quite poor. So, nucleus attract the electron of d-orbital with same effective nuclear charge. Hence, all the valence electrons are almost same distance away from the nucleus. As we move from left to right across transition metals , atomic radius remain almost same due to poor shielding effect of d-orbitals.

3. More positive ion simply means more number of protons than electron . So, effective nuclear charge of more positive ion is greater. Due to this greater effective nuclear charge , isoelectronic species with more positive ion have lower atomic radius.