# What is the electron configuration for the oxide (${O^{ - 2}}$) ion? A $1{s^2}2{s^2}2{p^6}$ B $1{s^2}2{p^6}3{s^2}$ C $1{s^2}2{s^2}2{p^2}$ D $1{s^2}2{s^2}2{p^4}$

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Hint: Electronic distribution of an atom/ion is an arrangement in a space of the electrons around the center of the mass(nucleus) of the atom or ion among the orbits or shells. This electronic distribution defines the chemical properties of the atoms.

Complete step by step solution:
There are some rules or principles by which an electronic distribution can be written. Those rules are
A. Aufbau principle: This principle states that electrons occupy the lower energy level before occupying the higher energy level. For example: 2s shell is filled before 2p shell. According to the Aufbau principle the energy of the orbitals depends upon the ${\text{(L + S)}}$ value. Higher the value of ${\text{(L + S)}}$ higher will be the energy of the orbital and vice-versa.
B. Pauli's exclusion principle: Any two electrons from a particular electronic distribution cannot have the same values of four quantum numbers. At least one of them should be different.
C. Hund's rule of maximum multiplicity: Electrons will occupy singly the orbitals with the same energy before filling them in pairs.
Therefore, according to these rules the electronic configuration of ${{\text{O}}^{{\text{ - 2}}}}$ can be written.
${O^{-2}}$is formed from oxygen by gaining two electrons to its last shell. Now the electronic configuration of oxygen with 8 electrons is, $1{s^2}2{s^2}2{p^4}$.
After gaining two electrons the total electron would be 10. So, the electronic configuration of ${O^{- 2}}$is $1{s^2}2{s^2}2{p^6}$.

The correct option is A.

Note: This electronic distribution defines the chemical properties of the atoms. Higher the value of ${\text{(L + S)}}$ higher will be the energy of the orbital and vice-versa. Remember these three rules of electronic distribution of an atom/ion.