Question

An element has an electron configuration of $1{s^2}2{s^2}2{p^6}3{s^2}$. Upon forming an ion, its new configuration is $1{s^2}2{s^2}2{p^6}$ . Which is the ion formed?

Answer 1

In this question we are clearly given the electronic configuration of the element both in the neutral state and the ionic form. Electronic configuration gives us the no. of electrons present in the atom, from which we can determine the atomic number of the element.

Complete answer:
The configuration given to us is: $1{s^2}2{s^2}2{p^6}3{s^2}$
Total number of electrons present in the atom: $12$
The electronic configuration is the electron distribution in an atom or molecule in atomic or molecular orbitals. It describes each electron to be moving independently in their orbital, in a field created by all other orbitals.
The given element to us has its last electron going into the s orbital, hence we can say that it belongs to the s-block. As there are two valence electrons, it belongs to group 2 or the periodic table. The element given to us is therefore Magnesium.
On ionization of magnesium, it forms Magnesium di-cation. Every atom wants to attain the stable electronic configuration of Noble gas. By losing two electrons, the atom attains stability (Neon Gas configuration) with the electronic configuration of $1{s^2}2{s^2}2{p^6}$. Magnesium loses two electron in its vicinity to the ‘electron acceptors’ and attains the stable electronic configuration of $1{s^2}2{s^2}2{p^6}$ to form $M{g^{ + 2}}$.

Note:
The elements that belong to the group 2, are known as the alkaline earth metals. The elements that belong to this group have an outer electronic configuration of $n{s^2}$. The elements that belong to this group are: Beryllium, Magnesium, Calcium, Strontium and Barium.