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Question: \[{O^{2 - }}\] and \[S{i^{4 + }}\] are isoelectronic ions. If the ionic radius of \[{O^{2 - }}\] is ...

O2{O^{2 - }} and Si4+S{i^{4 + }} are isoelectronic ions. If the ionic radius of O2{O^{2 - }} is 1Ao1{A^o}, the ionic radius of Si4+S{i^{4 + }} will be:
A.1.4 Ao1.4{\text{ }}{A^o}
B.0.41 Ao0.41{\text{ }}{A^o}
C.2.8 Ao2.8{\text{ }}{A^o}
D.1.5 Ao1.5{\text{ }}{A^o}

Explanation

Solution

Isoelectronic ions are the ones which have a similar electronic configuration. To find the radius of isoelectronic species we first need to calculate the ratio of atomic number to number of electrons. The larger the ratio smaller is the size of the ion.

Complete step by step answer:
Let’s start with discussing the isoelectronic ions for better understanding of the question. Isoelectronic ions are the ones which have a similar electronic configuration.
In the question we are given 22 species which are O2{O^{2 - }} and Si4+S{i^{4 + }} both are having the similar electronic configuration which is 1s22s22p61{s^2}2{s^2}2{p^6}. We are given with the radius of O2{O^{2 - }} ion which is 1Ao1{A^o}, we need to find the radius of Si4+S{i^{4 + }}. To find the radius of isoelectronic species we first need to calculate the ratio of atomic number to number of electrons. The larger the ratio smaller is the size of the ion. So the ratio for O2{O^{2 - }} is given below,
O2ze=810=0.8{O^{2 - }} \Rightarrow \dfrac{z}{e} = \dfrac{8}{{10}} = 0.8, the atomic number zz of OO is 88 and the number of electron is 8+2=108 + 2 = 10.
Similarly for Si4+S{i^{4 + }}, the ratio will be
Si4+ze=1410=1.4S{i^{4 + }} \Rightarrow \dfrac{z}{e} = \dfrac{{14}}{{10}} = 1.4, the atomic number zz of SiSi is 1414 and the number of electron in Si4+S{i^{4 + }} is144=1014-4 = 10.
Since the ratio of Si4+S{i^{4 + }} is greater than O2{O^{2 - }}, its size will be smaller than O2{O^{2 - }}, hence the size of Si4+S{i^{4 + }} will be smaller than O2{O^{2 - }}. So, the size of Si4+S{i^{4 + }} will be less than 1Ao1{A^o}.
\therefore The answer to this question is option B. 0.41Ao0.41{A^o}.

Note:
The reason behind the reducing size as the ratio between atomic number to number of electron increases is that, the atomic number of any species remain the same as it’s the number of proton present in it but the number of electron can be changed, so if the atom accepts electron due to electron-electron repulsion in outermost orbit the size of electron increases whereas if the atoms are donated then the inward attractive force of the atom increases as there are more proton than electron results in smaller size.