Courses for Kids
Free study material
Offline Centres
Store Icon

Write the electronic configuration of the following elements.
(A) Sulphur (Z = 16)
(B) Krypton (Z = 36)

Last updated date: 14th Apr 2024
Total views: 33.6k
Views today: 0.33k
MVSAT 2024
33.6k+ views
Hint: While we write electron configurations, a standardized notation procedure is followed in which the energy level and the type of orbital are written first, then the number of electrons present in the orbital is written in superscript. Follow Aufbau’s and Hund’s principle for proper arrangement of electrons in orbitals.

Complete step by step answer:
Let us first establish what the electronic configuration of an atom really is before moving onto answering the given question.
The electron configuration is a representation of the arrangement of electrons distributed among the shells and subshells in the orbitals of an atom; and is mostly used for describing the electronic arrangement in the orbitals of an atom in its ground state.

Let us now come to answering the given questions.
Given that Atomic Number of Sulphur = 16
No. of electrons in 1st orbit = 2 (Maximum Capacity is 2)
No. of electrons in 2nd orbit = 8 (Maximum Capacity is 8)
No. of electrons in 3rd orbit = 16-(2+8) = 6 (Maximum Capacity is 8)

1st orbit has only 1s orbital. So, we can write it as $1{{s}^{2}}$ .
The 2nd orbit has 2s and 2p orbitals. As we know s orbital can hold maximum 2 electrons and p orbital can hold maximum 6 electrons, so we can write it as $2{{s}^{2}}2{{p}^{6}}$ .
The 3rd orbit has 3s, 3p and 3d orbitals. Now we know that the order of energy in the 3rd shell is as follows:
Energy of 3s < Energy of 3p < Energy of 3d

Therefore, 3s is filled at first followed by 3p and 3d as per Aufbau's principle. So, we can write it as $3{{s}^{2}}3{{p}^{4}}$ . Now since all the electrons are accommodated within 3p , 3d will remain empty. Moreover, in the 3p orbital all the triply degenerate orbitals get 1 electron to begin with, following which pairing occurs according to Hunds' rule.
So, final electronic configuration of Sulphur is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{4}}$ .

Similarly, in the electronic configuration of Krypton, the total number of electrons is 36, so a fourth orbital will also be used. Its electronic configuration is as follows: $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{10}}4{{s}^{2}}4{{p}^{6}}$ .

Note: Remember that Hund’s maximum spin multiplicity implies only for the orbitals that have the same amount of energy. That means you need not to put electrons in the 4p orbital of Krypton until the 4s orbital is full.