Question

The electronic configuration of $C{u^{2 + }}$ ion is,
A.$\left[ {Ar} \right]3{d^8}4{s^1}$
B.$\left[ {Ar} \right]3{d^9}4{s^0}$
C.$\left[ {Ar} \right]3{d^7}4{s^2}$
D.$\left[ {Ar} \right]3{d^8}4{s^0}$

Answer 1

Hint: In an atom electrons are filled in orbitals according to the Aufbau principle. The electrons are filled in the increasing order of energy of orbitals. The ultimate aim is to get the most stable electronic configuration.

Complete step by step answer:The atomic number of copper is $29$. It is a d-block element. Hence the last electron fills in the d-orbital. In order to find the electronic configuration of $C{u^{2 + }}$ ion, we first need to find the electronic configuration of Cu atoms. We can get an electronic configuration of $C{u^{2 + }}$ ion, by taking out two electrons from the outermost shell of Cu atom.
Following Aufbau principle the electronic configuration of Cu atom is,
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^1}3{d^{10}}$
Electrons are filled in the increasing order of orbital energy level. It is also the order of increasing (n+l) value of the orbital. $4s$ lies outer to \[3d\]. But $4s$ fills first to \[3d\]. This is because the (n+l) value of $4s$ is less than that of \[3d\].
(n+l) for $4s$$ = 4 + 0 = 4$
(n+l) for \[3d\] $ = 3 + 2 = 5$
But $4s$ contains only one electron. This is an exceptional case. If we fill two electrons in $4s$, then $3d$ will be containing only nine electrons. This is not a stable configuration. A completely full or half full configuration is more stable than partially filled d- configuration.
$C{u^{2 + }}$ contains two electrons less than that of Cu. Hence we need to eliminate two electrons from Cu to get $C{u^{2 + }}$. The electron will be lost from the outermost orbitals. The outer orbital of Cu is $4s$. It contains only one electron. So one electron from $3d$ will also be eliminated to give the electronic configuration of $C{u^{2 + }}$ as,
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^0}3{d^9}$
The noble gas before Cu is Argon. Hence the electronic configuration can also be written as,
$\left[ {Ar} \right]3{d^9}4{s^0}$
Therefore, the correct option is B.

Note:
Before filling the electrons, we should be aware of the extra stability of completely filled or half-filled electronic configuration of d-orbitals. This is applicable to both copper and chromium in the third period.