Question

How would you write the electronic configuration for the following ions?
${{\text{K}}^{ + 1}}$
${\text{B}}{{\text{r}}^{ - 1}}$
${\text{A}}{{\text{l}}^{ + 3}}$
${\text{A}}{{\text{s}}^{ - 3}}$?

Answer 1

The arrangement of electrons in the energy levels around the nucleus of an atom is known as the electronic configuration. The electrons fill up in the energy levels according to the Aufbau’s principle. To solve this we must know the atomic numbers of given elements.

Complete solution:
We know that the Aufbau’s principle states that in the ground state of the atoms, the orbitals are filled with electrons in order of the increasing energies. The order of energy of different orbitals in an atom is as follows:
$1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d$ and so on.
The maximum capacity of s-orbital is 2 electrons, the maximum capacity of p-orbital is 6 electrons, the maximum capacity of d-orbital is 10 electrons and the maximum capacity of f-orbital is 14 electrons.
For ${{\text{K}}^{ + 1}}$:
The atomic number of ${\text{K}}$ i.e. potassium is 19. Thus, the electronic configuration of ${\text{K}}$ is as follows:
$1{s^2}\,2{s^2}\,2{p^6}\,3{s^2}\,3{p^6}\,4{s^1}$
We know that the potassium atom by losing 1 electron forms ${{\text{K}}^{ + 1}}$ ions. The one electron is lost from the valence orbitals of the potassium atom. One electron is lost from the 4s orbital. Thus, the electronic configuration of ${{\text{K}}^{ + 1}}$ ion is as follows:
$1{s^2}\,2{s^2}\,2{p^6}\,3{s^2}\,3{p^6}\,4{s^0}$
For ${\text{B}}{{\text{r}}^{ - 1}}$:
The atomic number of ${\text{Br}}$ i.e. bromine is 35. Thus, the electronic configuration of ${\text{Br}}$ is as follows:
$1{s^2}\,2{s^2}\,2{p^6}\,3{s^2}\,3{p^6}\,4{s^2}\,3{d^{10}}\,4{p^5}$
We know that the bromine atom by gaining 1 electron forms ${\text{B}}{{\text{r}}^{ - 1}}$ ions. The one electron is gained in the valence orbitals of the bromine atom. One electron is gained in the 4p orbital. Thus, the electronic configuration of ${\text{B}}{{\text{r}}^{ - 1}}$ ion is as follows:
$1{s^2}\,2{s^2}\,2{p^6}\,3{s^2}\,3{p^6}\,4{s^2}\,3{d^{10}}\,4{p^6}$
For ${\text{A}}{{\text{l}}^{ + 3}}$:
The atomic number of ${\text{Al}}$ i.e. aluminium is 13. Thus, the electronic configuration of ${\text{Al}}$ is as follows:
$1{s^2}\,2{s^2}\,2{p^6}\,3{s^2}\,3{p^1}$
We know that the potassium atom by losing 3 electrons forms ${\text{A}}{{\text{l}}^{ + 3}}$ ions. The three electrons are lost from the valence orbitals of the aluminium atom. One electron is lost from the 3p orbital and two electrons are lost from the 3s orbital. Thus, the electronic configuration of ${\text{A}}{{\text{l}}^{ + 3}}$ ion is as follows:
$1{s^2}\,2{s^2}\,2{p^6}\,3{s^0}\,3{p^0}$
For ${\text{A}}{{\text{s}}^{ - 3}}$:
The atomic number of ${\text{As}}$ i.e. arsenic is 33. Thus, the electronic configuration of ${\text{As}}$ is as follows:
$1{s^2}\,2{s^2}\,2{p^6}\,3{s^2}\,3{p^6}\,4{s^2}\,3{d^{10}}\,4{p^3}$
We know that the arsenic atom by gaining 3 electron forms ${\text{A}}{{\text{s}}^{ - 3}}$ ions. The three electrons are gained in the valence orbitals of the arsenic atom. Three electrons are gained in the 4p orbital. Thus, the electronic configuration of ${\text{A}}{{\text{s}}^{ - 3}}$ ion is as follows:
$1{s^2}\,2{s^2}\,2{p^6}\,3{s^2}\,3{p^6}\,4{s^2}\,3{d^{10}}\,4{p^6}$

Note: The electrons fill the orbitals according to the Aufbau’s principle. The Aufbau’s principle states that in the ground state of the atoms, the orbitals are filled with electrons in order of the increasing energies. The order of energy of different orbitals in an atom is as follows:
$1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d$ and so on.
The maximum number of electrons that can be accommodated is s-orbital are 2, p-orbital are 6, d-orbital are 10 and f-orbital is 14.