Question

Given,$B{e^{2 + }}$ is isoelectronic with which of the following ions?
(A) ${H^ + }$
(B) $L{i^ + }$
(C) $N{a^ + }$
(D) $M{g^ + }$

Answer 1

The species which have the same number of electrons in their structure are called isoelectronic species. The atomic number of H, Li, Na and Mg are 1, 3, 11 and 12 respectively.

Complete answer:
Isoelectronic species are the species which have the same number of electrons in their structure. So, we will count the number of electrons and compare with that of the given species in order to find the answer.
- Be is an element whose atomic number is 4 and its ground state electronic configuration is $1{s^1}2{s^2}$ . Here, the ion given is $B{e^{2 + }}$. So, its electronic configuration is $1{s^2}$.
- H is an element with atomic number 1. Its ionic form ${H^ + }$ does not have any electrons in its structure and so it is also called proton.
- The atomic number of Li is 3. It has 3 electrons in its ground state and so its electronic configuration is $1{s^2}2{s^1}$. $L{i^ + }$ has one electron less than its ground state. So, its electronic configuration is $1{s^2}$.
- Atomic number of Na is 11. Its ground state electronic configuration is $1{s^2}2{s^2}2{p^6}3{s^1}$. So, the electronic configuration of $N{a^ + }$ is $1{s^1}2{s^2}2{p^6}$.
- Atomic number of Mg is 12. Its ground state electronic configuration is $1{s^2}2{s^2}2{p^6}3{s^2}$. The electronic configuration of $M{g^{2 + }}$ is $1{s^2}2{s^2}2{p^6}$.
- So, we can say that the number of electrons in $B{e^{2 + }}$ and $L{i^ + }$ will be the same.

Therefore, the correct answer to this question is (B).

Note:
Remember that when any atom loses electrons, the electrons are removed from the orbitals which are more far from the nucleus. You can simply subtract the positive charge form the atomic number of the atom in order to find their number of electrons.