Question

Electronic configuration of sodium is:
A. $1{s^2}2{s^2}2{p^5}3{s^2}$
B. $1{s^2}2{s^2}3{p^5}4{s^6}$
C. $1{s^2}2{s^2}2{p^7}3{s^1}$
D. $1{s^2}2{s^2}2{p^6}3{s^1}$

Answer 1

The number of electrons to be distributed is equal to the atomic number and should be distributed from lower energy to higher energy orbitals using aufbau principle.

Complete step by step answer:
The atomic number of sodium is 11 i.e. 11 electrons need to be distributed in its orbitals starting from the lowest and then going to the higher energy shells. Each orbital contains 2 electrons and the number of orbitals of each type - l to + l is decided by the angular quantum number, 1. which has values from 0 to $(n - 1)$ where n is the principal quantum number. The number of values of $'l'$ give us the number of types of orbitals that a principal shell can have.
Distributing electrons in the first shell, i.e. n = 1
For n = 1, l = 0 and the number of orbitals is $- l$ to $+ l$ i.e. 0. So, the first shell has only one orbital which is called 1s and it will have 2 electrons.
Therefore, $1{s^2}.....$is the distribution of the first two electrons of sodium. Remaining 9 electrons need to be distributed to higher shells.
Distributing electrons in the second shell, i.e. n = 2
For n = 2,l = 0,1
The number of orbitals for l = 0 is 1 and in this case will be called 2s and in case of l = 1, the number of orbitals will be 3 and will be referred to as $2{p_x},{\text{ }}2{p_y},{\text{ }}and{\text{ }}2{p_z}$ separately and collectively as 2p. These correspond to the values from $- l{\text{ }}to{\text{ }}+ l{\text{ }}as{\text{ }} - 1,{\text{ }}0,{\text{ }} + 1$ . These three 2p orbitals will contain 2 electrons each and 6 electrons in total.
Therefore, the distribution of electrons in the second shell for sodium is: $2{s^2}2{p^6}$
The remaining 1 electron needs to go to third shell
Distributing electrons in the second shell, i.e., n = 3
For $n = {\text{ }}3,{\text{ }}l{\text{ }} = {\text{ }}0,{\text{ }}1,{\text{ }}2$
For $n{\text{ }} = {\text{ }}3{\text{ }}and{\text{ }}l{\text{ }} = {\text{ }}0$, only one orbital called 3s is possible and it can accommodate the remaining 1 electron. The other orbitals (corresponding to $l{\text{ }} = {\text{ }}1{\text{ }}and{\text{ }}l{\text{ }} = 2$) in the third shell remain empty and need not be mentioned.
Therefore, the distribution of electrons in the third shell for sodium is: $3{s^1}$ .
Hence, the electronic configuration or distribution of electrons of Sodium, obtained by combining the distributions obtained in the above three steps, is: $1{s^2}2{s^2}2{p^6}3{s^1}$ .

Therefore, the correct answer is D.

Note: Electrons should always be distributed from lower energy shells to higher energy shells. One should always keep track of the remaining number of electrons to be distributed and once the required number of electrons are distributed, the remaining higher energy shells and their orbitals remain empty and need not be mentioned.