Question

Among the following, the set of isoelectronic ions is:
(A) $N{{a}^{+}}$,$M{{g}^{+2}}$,${{F}^{-}}$,$C{{l}^{-}}$
(B) $N{{a}^{+}}$, $C{{a}^{+2}}$,${{F}^{-}}$,${{O}^{-}}$
(C) $N{{a}^{+}}$, $M{{g}^{+2}}$,${{F}^{-}}$, ${{O}^{2-}}$
(D) $N{{a}^{+}}$,${{K}^{+}}$,${{S}^{2-}}$, $C{{l}^{-}}$

Answer 1

Hint: Firstly we have to understand the definition of isoelectronic ions. On the basis of the definition we can answer this question which will be the easiest way approaching these kinds of questions.

Complete step by step answer:
Isoelectronic ions are the cations or anions of different atoms, these ions have the same number of electrons and same configuration.
-Option A is $N{{a}^{+}}$, $M{{g}^{+2}}$, ${{F}^{-}}$, $C{{l}^{-}}$. Let’s check out electronic configuration of $N{{a}^{+}}$, $M{{g}^{+2}}$, ${{F}^{-}}$, $C{{l}^{-}}$ in order to check out the set of isoelectronic ions. If the electronic configuration of all the compounds are the same then this is the correct answer.

Isoelectronic ions	Number of electrons	Electronic configuration
$N{{a}^{+}}$	10	$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$
$M{{g}^{+2}}$	10	$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$
${{F}^{-}}$	10	$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$
$C{{l}^{-}}$	10	$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$

-From the above mentioned table we can observe that all ions have the same configuration and same number of electrons. Thus, option A is the correct answer.
-Option B is $N{{a}^{+}}$, $C{{a}^{+2}}$, ${{F}^{-}}$, ${{O}^{-}}$. Let’s check out electronic configuration of $N{{a}^{+}}$, $C{{a}^{+2}}$, ${{F}^{-}}$, ${{O}^{-}}$ in order to check out the set of isoelectronic ions.

Isoelectronic ions	Number of electrons	Electronic configuration
$N{{a}^{+}}$	10	$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$
$C{{a}^{+2}}$	22	$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}$
${{F}^{-}}$	10	$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$
${{O}^{-}}$	9	$1{{s}^{2}}2{{s}^{2}}2{{p}^{5}}$

-Here we can see that the number of electrons in $C{{a}^{+2}}$ and ${{O}^{-}}$is 22 and 9, accordingly the electronic configuration is also different. Hence this is not the set of isoelectronic ions. As two ions have different numbers of electrons as well as the electronic configuration. Therefore, this option is the wrong answer.
-Option C is $N{{a}^{+}}$, $M{{g}^{+2}}$, ${{F}^{-}}$, ${{O}^{2-}}$. Let’s check out electronic configuration of $N{{a}^{+}}$, $M{{g}^{+2}}$, ${{F}^{-}}$, ${{O}^{2-}}$ in order to check out the set of isoelectronic ions.

Isoelectronic ions	Number of electrons	Electronic configuration
$N{{a}^{+}}$	10	$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$
$M{{g}^{+2}}$	10	$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$
${{F}^{-}}$	10	$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$
${{O}^{2-}}$	10	$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$

-If the electronic configuration of all the compounds are the same then this is the correct answer. Yes, we can observe that all the ions in this option have the same number of electrons and the same electronic configuration. Therefore, this is the correct answer.
-Option D is $N{{a}^{+}}$, ${{K}^{+}}$,${{S}^{2-}}$, $C{{l}^{-}}$. Let’s check out electronic configuration of $N{{a}^{+}}$,${{K}^{+}}$,${{S}^{2-}}$, $C{{l}^{-}}$ in order to check out the set of isoelectronic ions.

Isoelectronic ions	Number of electrons	Electronic configuration
$N{{a}^{+}}$	10	$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$
${{K}^{+}}$	18	$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}$
${{S}^{2-}}$	18	$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}$
$C{{l}^{-}}$	10	$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$

We can see that in this set there is lots of variation in electronic configuration. Therefore, this option is the wrong answer.

Thus, option A and C are the correct answer.

Note: The number of electrons can be easily calculated if we know its atomic number. For example: The atomic number of Na is 11. If it has a positive charge on it then, it means it lost the electron. If it has a negative charge on it, then it means it gains electrons. Therefore, $N{{a}^{+}}$ has 10 electrons and $N{{a}^{-}}$has 9 electrons.