Question

The element Cu with atomic number 29 undergoes chemical reaction to formation with oxidation number +2. Write down the subshell electronic configuration of this ion.

Answer 1

Hint: Aufbau principle is used for the filling of atomic orbitals. This principle says that electrons are filled first at the orbital that possesses a lower energy level. When the lower energy levels are filled, then electrons enter the orbitals that possess higher energy.

Complete Step by Step Solution:
Here, the element given is Copper. Its atomic number is 29. Now, we have to write its electronic configuration. The count of electrons to be filled is 29. So, Copper has the following electronic configuration.
\[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^1}3{d^{10}}\]

Now, we have to write the electronic configuration of the ion of copper which possesses the oxidation number of +2. So, now the electrons to be filled are 27. So, the ion of copper (\[{\rm{C}}{{\rm{u}}^{2 + }}\] ) has the following configuration.
\[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^9}\]

As in copper, the d orbital is fully filled. It attains stability. So, the removal of the two electrons from the d orbital is difficult. Therefore, one electron is removed from the 4s orbital and the other electron is to be removed from the 3d orbital. Therefore, the two ions lost from the copper ion are to be removed from the 4s and 3d orbital.

Hence, \[{\rm{C}}{{\rm{u}}^{2 + }}\] possesses the configuration of \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^9}\].

Note: Students might get confused that electronic configuration of copper is \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^9}\], but it is not correct. The correct electronic configuration is \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^1}3{d^{10}}\]. It is an exception to the electronic configuration of other elements. Ro attain stability, the 3d orbital is filled first and then the 4s orbital is filled.