Question

Which of the following groups has isoelectronic species?
A) ${O^ - },{F^ - },Na,M{g^ + }$
B) ${O^{2 - }},{F^ - },Na,M{g^{2 + }}$
C) ${O^ - },{F^ - },N{a^ + },M{g^{2 + }}$
D) ${O^{2 - }},{F^ - },N{a^ + },M{g^{2 + }}$

Answer 1

Hint: Species that have the same number of electrons are called isoelectronic. Hence, calculate the number of electrons present in each atom or ion and the group that has the same number of electrons will be isoelectronic in nature.

Complete answer:Isoelectronic species are those that have the same number of electrons.
In order to find the isoelectronic group, we need to calculate the number of electrons present in each atom and ion.
We know that Oxygen has eight electrons, ${O^ - }$ will have nine electrons and ${O^{2 - }}$ will have ten electrons.
Similarly, we can calculate for Fluorine. Fluorine has nine electrons and ${F^ - }$ has ten electrons.
Similarly, we know that Sodium has eleven electrons and hence $N{a^ + }$ will have ten electrons.
Finally, we know that Magnesium has twelve electrons so $M{g^ + }$ will have eleven electrons and $M{g^{2 + }}$ will have ten electrons.
Therefore the group having same number of electrons is ${O^{2 - }},{F^ - },N{a^ + },M{g^{2 + }}$ and hence they are said to be isoelectronic.

Additional information: In isoelectronic species, the number of electrons is the same so the size of the atom or ion is determined by the number of protons. Isoelectronic species are expected to show useful predictability and consistency in their properties.

Hence, the correct answer is option (D).

Note: Ions or atoms of different elements that have the same number of electrons but differ from one another in magnitude of nuclear charge are known to be isoelectronic in nature. ${O^{2 - }},{F^ - },N{a^ + },M{g^{2 + }}$ have ten electrons in them and hence are isoelectronic species.