Frequently Asked Question

The smaller ionic radius of Ni2+ compared to Zr4+ can be attributed to several factors:

1. Effective Nuclear Charge (Zeff): Ni2+ has a higher Zeff than Zr4+ due to the loss of two electrons. This increased Zeff results in a stronger electrostatic attraction between the nucleus and the remaining electrons, causing the electrons to be held more tightly and reducing the ionic radius.

2. Lanthanide Contraction: Zr4+ belongs to the 4d transition series, while Ni2+ belongs to the 3d transition series. As you move from left to right across a transition series, the atomic number increases, leading to an increase in the number of electrons in the d orbitals. However, the shielding effect of the d electrons is not sufficient to completely offset the increasing nuclear charge, resulting in a gradual decrease in the ionic radii known as the lanthanide contraction. This effect is more pronounced for the 4d transition series, which Zr4+ belongs to, compared to the 3d transition series, where Ni2+ is located.

3. Crystal Field Stabilization Energy (CFSE): In octahedral coordination complexes, Ni2+ exhibits a high CFSE due to the stabilization of its d electrons in the t2g orbitals. This stabilization contributes to the stability of the Ni2+ ion and influences its ionic radius. Zr4+, on the other hand, has a lower CFSE, which makes its ionic radius less affected by this factor.

Therefore, the combination of a higher Zeff, lanthanide contraction, and CFSE effects contributes to the smaller ionic radius of Ni2+ compared to Zr4+.