Question

The isoelectronic pair is:
(A) \[C{{l}_{2}}O,IC{{l}_{2}}^{-}\]
(B) \[C{{l}_{2}}^{-},Cl{{O}_{2}}\]
(C) \[I{{F}_{2}}^{+},{{I}_{3}}^{-}\]
(D) \[Cl{{O}_{2}}^{-},Cl{{F}_{2}}^{+}\]

Answer 1

Hint: Isoelectronic species have the same electron configuration, to find it, the positive charge is subtracted from the total number of electrons and negative charge is added to the total number of electrons.

Complete step by step solution:

Let’s see each of the options given here in the question:

* In \[C{{l}_{2}}O\], we know the atomic number of Chlorine is 17 and oxygen is 8, So, \[C{{l}_{2}}O\] will have a total of 17 × 2 + 8 = 42 electrons. Similarly, in \[IC{{l}_{2}}^{-}\], iodine has 53 electrons and Chlorine has 17, plus there is a negative charge on this compound, thus, the total number of electrons in \[IC{{l}_{2}}^{-}\] are 53 + 17 × 2 + 1 = 88.
As we can see, the number of electrons in both the compounds are not the same, therefore, they are not isoelectronic.
* In \[C{{l}_{2}}^{-}\], we know the atomic number of Chlorine is 17, plus there is a negative charge on the ion, So, \[C{{l}_{2}}^{-}\] will have a total of 17 × 2 + 1 = 35 electrons. Similarly, in \[Cl{{O}_{2}}\], Chlorine has 17 electrons and Oxygen has 8,Thus, the total number of electrons in \[Cl{{O}_{2}}\] are 17 + 8 × 2 = 33.
Here also the number of electrons in both the compounds are not the same, therefore they are not isoelectronic.
* In \[I{{F}_{2}}^{+}\], we know the atomic number of Iodine is 53 and fluorine is 9 plus there is a positive charge on the compound , So, \[I{{F}_{2}}^{+}\] will have a total of 53 + 9 × 2 - 1 = 70 electrons. Similarly, in \[{{I}_{3}}^{-}\], Iodine has 53 electrons, plus there is a negative charge on this ion, thus, the total number of electrons in \[{{I}_{3}}^{-}\] are 53 × 3 + 1 = 160.
The number of electrons in both the compounds are not the same, therefore, they are not isoelectronic.
* In \[Cl{{O}_{2}}^{-}\], we know the atomic number of Chlorine is 17 and oxygen is 8 plus there is a negative charge, So, \[Cl{{O}_{2}}^{-}\] will have a total of 17 + 2 × 8 + 1 = 34 electrons. Similarly, in \[Cl{{F}_{2}}^{+}\], Chlorine has 17 and fluorine has 9, plus there is a positive charge on this compound, thus, the total number of electrons in \[Cl{{F}_{2}}^{+}\] are 17 + 9 × 2 - 1 = 34.
As we can see the number of electrons in both the compounds are the same, therefore they are isoelectronic.

Therefore, from the above statements we can conclude that the correct option is (d).

Note: The importance of the concept lies in identifying related species, as pairs or series. Isoelectronic species can be expected to show useful consistency and predictability in their properties.