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

We need to understand the rules for writing the electronic configuration of any element of the periodic table. Electronic configuration is defined as the distribution of electrons into orbitals. The electronic configuration follows certain rules which help us categorize into the respective blocks. If two electrons are filled in the ‘s’ subshell of the first shell, the electronic configuration is noted as $1{s^2}$. The rules including the Aufbau principle, Pauli’s exclusion principle, Hund’s Rule of maximum multiplicity are the three main rules followed in electronic configuration.

Complete step by step answer:
The following must be kept in mind before writing any electronic configuration.
Principle quantum number is defined as the maximum number of electrons that can be accommodated in a shell is calculated by the formula $2{n^2}$ where is the shell number or the energy level characterized by a set of four quantum numbers. When n=1, the maximum number of electrons that can be accommodated are $2{\left( 1 \right)^2} = 2.$
Azimuthal quantum number: The subshells into which electrons are distributed are based on the azimuthal quantum number, denoted by ‘l’. When n=4 the subshells correspond to $l = 0,{\text{ }}l = 1,{\text{ }}l = 2,{\text{ }}and{\text{ }}l = 3$ and are named the s, p, d, and f subshells respectively. The maximum number of electrons that can be accommodated by a subshell is given by the formula $2\left( {2l{\text{ }} + {\text{ }}1} \right).$ So s, p, d and f subshells can accommodate $2,6,10$ and $14$ electrons respectively.
The Aufbau principle: This principle along with the electronic configuration of atoms provides a theoretical foundation for the periodic classification into s-block, p-block, d-block, and f-block elements. It states that in the ground state of an atom or ion, electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels.
Pauli’s exclusion principle: This principle states “no two electrons in the same atom have the same values for all four quantum numbers”.
Hund’s Rule of maximum multiplicity: It states that for a given electron configuration, the lowest energy term is the one with the greatest value of spin multiplicity. This implies that if two or more orbitals of equal energy are available, electrons will occupy them singly before filling them in pairs.
The given element given in the question is $Cu$ whose atomic number is 19. Therefore, the electronic configuration of $Cu$ according to the rules is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^9}$.
But there are exceptions such as in case $Cu,Cr$ etc. This is because Fully-filled orbitals and half-filled orbitals have extra stability. Hence, one of the $4{s^2}$ electrons jumps to the $3{d^9}$ to give it extra stability. Hence, the electronic configuration becomes $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^1}3{d^{10}}$.
However, the electronic configuration of an element can also be written in a short form using the nearest noble gas configuration. Thus, the electronic configuration of $Cu$ is $\left[ {Ar} \right]3{d^{10}}4{s^1}$.
We are to find the electronic configuration of $C{u^{2 + }}$ ion. In this cation, the charge on the atom says that two electrons have been removed from the $Cu$ atom. Hence the electronic configuration becomes $\left[ {Ar} \right]3{d^9}4{s^0}$.

Therefore, option B is correct.

Note: It must be noted that this arrangement of atoms in a diamond unit cell is not the same for all the cubic unit cells. The corner and face-centred atoms may be the same but the contribution of voids may vary depending on the close packing. Note that the arrangement of atoms in diamond makes it one of the hardest substances on earth.