Question

${{O}^{2-}}$ and $S{{i}^{4+}}$ are isoelectronic ions. If the ionic radius of ${{O}^{2-}}$ is $1\overset{\text{o}}{{\text{A}}}\,$ the ionic radius of $S{{i}^{4+}}$ will be :
(A) $1.4\overset{\text{o}}{{\text{A}}}\,$
(B) $0.41\overset{\text{o}}{{\text{A}}}\,$
(C) $2.8\overset{\text{o}}{{\text{A}}}\,$
(D) $1.5\overset{\text{o}}{{\text{A}}}\,$

Answer 1

Hint: Atomic or ionic radius is inversely proportional to the effective nuclear charge for an isoelectronic species. The effective nuclear charge is directly proportional to the positive charge. Here $S{{i}^{4+}}$ has +4 charge and thus it will have a shorter radius than ${{O}^{2-}}$.

Complete step by step solution:
-Let start with the concept of isoelectronic species. They are the atoms or ions having the same number of electrons or having the same electronic configuration. These species tend to have similar chemical properties. This concept is usually applied to predict rare or unknown compounds based on their resemblance in electronic configuration to known species.
-As we know, the ionic radius is used to describe the size of an ion including cations and anions. A cation (positive ion) is formed by the removal of electrons from the valence shell of an atom. Thus, it will have a lower number of electrons and the same number of protons compared to the parent atom. As a result, the remaining electrons in the valence shell will experience a greater effective nuclear charge and are hence more attracted to the nucleus.
-Similarly, an anion (negative ion) is formed by the addition of electrons to the valence shell. This will result in a greater repulsion between electrons and a decreased effective nuclear charge and this causes the anion to be larger in size than cation.
-We are given two isoelectronic species ${{O}^{2-}}$ and $S{{i}^{4+}}$ and we are asked to find the ionic radius of the ion $S{{i}^{4+}}$. The number of electrons in both species is ten and the number of protons varies. In ${{O}^{2-}}$ there are eight protons and in $S{{i}^{4+}}$ there are fourteen protons.
-As the number of protons in the nucleus increases, the nuclear charge will also increase and thereby the ionic radius will decrease. In the given question $S{{i}^{4+}}$ has a higher number of protons and thus it will have a lower ionic radius than ${{O}^{2-}}$.Its given that the ionic radius of ${{O}^{2-}}$as $1\overset{\text{o}}{{\text{A}}}\,$ and among the options only one option have lower ionic radius than $1\overset{\text{o}}{{\text{A}}}\,$and that is option (B) $0.41\overset{\text{o}}{{\text{A}}}\,$

Therefore, the answer is option (B) $0.41\overset{\text{o}}{{\text{A}}}\,$.

Note: The answer can be found in a simple way also. Keep in mind that the anionic radius of an isoelectronic species will be always greater than the cationic radius. Here ${{O}^{2-}}$ is the anion and $S{{i}^{4+}}$is the cation. Therefore, ${{O}^{2-}}$ will have the higher ionic radius and $S{{i}^{4+}}$will have a lower ionic radius.