Question

The radius of an isoelectronic series:
A. Decreases with decreasing nuclear charge
B. Decrease with increase in effective nuclear charge
C. Same for all
D. First increases then decreases

Answer 1

In the isoelectric species, the size of the ions depends on the relative number of protons and electrons present. The more the number of protons, the more is the effective nuclear charge and thus, the lesser is the diameter of the ion or atom.

Complete step by step answer:
Here we are asked to find a pattern in the radius of species in an isoelectronic series. First of all, we need to understand what the isoelectronic series means. Isoelectronic series is a series or a group of such atoms or ions which have the same number of electrons in their shells.
For example: ${O^{2 - }}$, ${F^ - }$, $N{a^ + }$, $M{g^{2 + }}$ have the same number of electrons and thus, they form an isoelectronic series.

Species	Radius	Number of Protons	Number of Electrons	Electronic Configuration
${O^{2 - }}$	126	8	10	$1{s^2}2{s^2}2{p^6}$
${F^ - }$	119	9	10	$1{s^2}2{s^2}2{p^6}$
$N{a^ + }$	116	11	10	$1{s^2}2{s^2}2{p^6}$
$M{g^{2 + }}$	86	12	10	$1{s^2}2{s^2}2{p^6}$

It is clear from the above table that ${O^{2 - }}$, ${F^ - }$, $N{a^ + }$, $M{g^{2 + }}$ form an isoelectronic series as the number of electrons is same in all these species i.e. 10. But, as in go down the series the number of protons is continuously increasing with the decrease in the radius of the species. The reason is the number of electrons is the same, thus the electrons provide a similar amount of shielding effect in all the species, but the number of protons is increasing as the atomic number increases from ${O^{2 - }}$ to $M{g^{2 + }}$. Thus, the effective nuclear charge is more on the atoms or ions having a greater number of protons, and 12 protons in $M{g^{2 + }}$ will bind the outermost electrons more tightly than the 8 protons of ${O^{2 - }}$. Therefore, species with higher atomic numbers will show higher effective nuclear charge and their radius will be lesser.

So, the correct answer is Option B.

Note: The effective nuclear charge is different from just nuclear charge i.e. the number of protons. The term "effective" is used because of the repelling effect of inner-layer electrons, there is a considerable shielding effect by the electron to prevent the electron in the outermost shell to experience the nuclear pull.