Question

The electronic configuration of $R{b^ + }$ in absence of Aufbau principle, is:
$(A)\,1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^{10}},4{s^2}4{p^6}$
$(B)\,1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^{10}},4{s^2}4{p^4}$
$(C)\,1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^{10}},4{s^2}4{p^3}$
$(D)\,None\,of\,these$

Answer 1

Rubidium is the ${37^{th}}$ element of the modern periodic table with chemical symbol Rb. It is an alkali metal and belongs to the first group. Therefore, like all other alkali metals it readily loses an electron from its outermost orbital. The arrangement of electrons in its atomic orbitals can be determined by Aufbau principle and various other rules.

Complete answer:
The number of electrons in the ground state of an atom will be equal to its atomic number. Rubidium (Rb) with atomic number $37$ will therefore have $37$ electrons in its atomic orbital. Now let us discuss the Aufbau principle. Aufbau principle states that the arrangement of electrons in the ground state occurs in such a way that the atomic orbitals with lower energy will be filled first. The energy of atomic orbitals is determined by the $(n + l)$ rule, where ‘n’ and ‘l’ stand for principal quantum number and azimuthal quantum number respectively. Atomic orbital with lower $(n + l)$ will have lower energy value and thus will be filled first. Let us try to understand this with an example. When we take the case of $3d$ and $4s$ we might think $3d$ will be having lower energy. As we calculate the $(n + l)$ values of each, we get the values as:
$3d = 3 + 2 = 5 \\\ 4s = 4 + 0 = 4$
Thus, we can conclude that $3d$ has higher energy than $4s$ .Now, the order of filling of atomic orbitals in Rb is : $1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s$ and its electronic configuration is:
$Rb = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}4{p^6}5{s^1}$
Rb on losing an electron from its $5s$ becomes $R{b^ + }$ .
$R{b^ + } = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}4{p^6}$
In the absence of Aufbau principle, the electronic configuration of $R{b^ + }$ will be:
$R{b^ + } = \,1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^{10}}4{s^2}4{p^6}$
The electrons are arranged in the increasing order of principal quantum numbers.
Therefore, the right option is $(A)\,1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^{10}},4{s^2}4{p^6}$

Note:
The distribution or arrangement of electrons in its atomic orbital is called electronic configuration. It is governed by the following rules:
Aufbau principle: Electrons fill lower energy atomic orbitals before filling the higher ones.
Hund’s rule: Electrons must singly occupy the degenerate orbitals and must have the same spin before pairing up.
Pauli’s exclusion principle: No two electrons can have the same set of quantum numbers.