Explain Bohr-Bury scheme.
Answer
597k+ views
Hint:The Bohr-Bury scheme describes the arrangement of electrons and the number of electrons that can accommodate in a shell or orbital. One can elaborate electrons that can accommodate in each shell.
Complete step by step answer:1) First of all we will learn the concept of the Bohr-Bury scheme where it states as the maximum number of electrons that could be present in a shell is given by a formula $2{n^2}$.
Where ${\text{n}}$ is the orbit or shell number of that atom.
2) Now if we apply this formula for shells we will get the maximum occupancy as below,
For first orbit i.e. K shell $ = 2{n^2} = 2 \times {1^2} = 2$ electrons.
For second orbit i.e. L shell $ = 2{n^2} = 2 \times {2^2} = 8$ electrons.
For the third orbit i.e. M shell $ = 2{n^2} = 2 \times {3^2} = 18$ electrons.
For fourth orbit i.e. N shell $ = 2{n^2} = 2 \times {4^2} = 32$ electrons.
3) The Bohr-Bury scheme further elaborates as the electrons in an orbit must be filled in a stepwise manner, that is one can not fill the electrons in a higher shell if the lower shell is not fully filled.
4) Electrons revolve around the nucleus in a specific fixed path and that path is called an orbit. There is a certain limit of each orbit that can have a certain number of electrons.
5) The Bohr-bury scheme also elaborates its theory as an electron when revolving in an orbit does not lose energy which means it sustains its energy.
Note:In an atom, there are orbits or shells as K, L, M, N from lower to higher respectively. Each shell or orbit has a subshell named as s, p, d, f which has ${\text{2, 6, 10, 14}}$ electrons occupancy respectively.
Complete step by step answer:1) First of all we will learn the concept of the Bohr-Bury scheme where it states as the maximum number of electrons that could be present in a shell is given by a formula $2{n^2}$.
Where ${\text{n}}$ is the orbit or shell number of that atom.
2) Now if we apply this formula for shells we will get the maximum occupancy as below,
For first orbit i.e. K shell $ = 2{n^2} = 2 \times {1^2} = 2$ electrons.
For second orbit i.e. L shell $ = 2{n^2} = 2 \times {2^2} = 8$ electrons.
For the third orbit i.e. M shell $ = 2{n^2} = 2 \times {3^2} = 18$ electrons.
For fourth orbit i.e. N shell $ = 2{n^2} = 2 \times {4^2} = 32$ electrons.
3) The Bohr-Bury scheme further elaborates as the electrons in an orbit must be filled in a stepwise manner, that is one can not fill the electrons in a higher shell if the lower shell is not fully filled.
4) Electrons revolve around the nucleus in a specific fixed path and that path is called an orbit. There is a certain limit of each orbit that can have a certain number of electrons.
5) The Bohr-bury scheme also elaborates its theory as an electron when revolving in an orbit does not lose energy which means it sustains its energy.
Note:In an atom, there are orbits or shells as K, L, M, N from lower to higher respectively. Each shell or orbit has a subshell named as s, p, d, f which has ${\text{2, 6, 10, 14}}$ electrons occupancy respectively.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Trending doubts
Which is more stable and why class 12 chemistry CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

What are the major means of transport Explain each class 12 social science CBSE

Sulphuric acid is known as the king of acids State class 12 chemistry CBSE

