Question

Which of the following pairs has the same size?
A. $${\text{Z}}{{\text{n}}^{{\text{2 + }}}}$$, $${\text{H}}{{\text{f}}^{{\text{4 + }}}}$$
B. $${\text{F}}{{\text{e}}^{{\text{2 + }}}}$$, $${\text{N}}{{\text{i}}^{{\text{4 + }}}}$$
C. $${\text{Z}}{{\text{n}}^{{\text{4 + }}}}$$, $${\text{T}}{{\text{i}}^{{\text{4 + }}}}$$
D. $${\text{Z}}{{\text{r}}^{{\text{4 + }}}}$$, $${\text{H}}{{\text{f}}^{{\text{4 + }}}}$$

Answer 1

Hint: The shielding effect is also known as screening effect. This is the phenomenon which occurs when the nucleus tends to reduce the force of attraction on the valence electrons because of the presence of electrons present in the inner shell. The electrons which are present at the outermost shell experience the force of attraction from the nucleus and the effect of the repulsion is experienced by them through inner electrons.

Complete step by step answer:
So in this question the $${\text{H}}{{\text{f}}^{{\text{4 + }}}}$$and $${\text{Z}}{{\text{r}}^{{\text{4 + }}}}$$belongs to the group IVB. These two cations have same size as the cation $${\text{H}}{{\text{f}}^{{\text{4 + }}}}$$has same size as that of $${\text{Z}}{{\text{r}}^{{\text{4 + }}}}$$because of the addition of the 14 lanthanide elements before it in which the electrons are added into the f subshell which are been shielded poorly by the outer electrons and de to which the size get contracts.
So these have the same size because of the poor shielding effect. Whereas the other pairs do not have the same size as one of the ions in the pair, more electrons in comparison to others.

So, the correct answer is Option D.

Note: The lanthanide can be corroded or could be brittle if they get contaminated with other metals and non metals. They are also magnetic in nature. They tend to form trivalent compounds. These elements have different reaction tendencies which depend on their basicity. They are lustrous and have silver appearance.