Question

Explain with suitable examples, Bohr-bury scheme for the distribution of electrons in different orbits in an atom.

Answer 1

The electronic configuration is defined as the distribution of electrons in various energy levels or shells of an atom of the element. distribution of electrons in an atom of an element is given by Bohr-Bury.

The main features of the Bohr-Bury scheme for electronic configuration are as follows:

• Total number of electrons present in a neutral atom is equal to its atomic number(number of protons).
• The maximum number of electrons that can be accommodated in a particular energy level or orbit of an atom is equal to $2{\mathrm{n}}^2$, where $\mathrm{n}$is the number of energy shell.

So, the maximum number of electrons in different energy levels or orbits can be calculated as follows:

(A) For 1st energy level or orbit(K), n =$1$, so, the maximum number of electrons = $2{\mathrm{n}}^2$ = $2*1^2=2$

(B) For the 2nd energy level or orbit(L), n =$2$, so, the maximum number of electrons = $2{\mathrm{n}}^2$= $2*2^2=8$

(C) For the 3rd energy level or orbit(M), n =$3$, so, the maximum number of electrons = $2{\mathrm{n}}^2$= $2*3^2=18$

(D) For the 4th energy level or orbit(N), n =$4$, so, the maximum number of electrons = $2{\mathrm{n}}^2$= $2*4^2=32$

• Electrons will occupy their positions in various energy shells in the increasing order of energy. K shell has the lowest energy, so it will be accommodated first, then L, then M, then N, and so on.
• The outermost shell of an atom cannot accommodate more than 8 electrons, even if it has the capacity to hold more electrons(except the K-shell which can accommodate a maximum of 2 electrons.)

Example

• Chemical symbol of Sodium = $\mathrm{Na}$
• Atomic number = $11$
• Number of protons = Number of electrons = $11$
• Number of electrons present in the K-shell(n=1) of Sodium = $2*1^2=2$
• Number of electrons present in the L-shell(n=2) of Sodium = $2*2^2=8$
• Remaining one electron is present in M-shell.
• Electronic distribution = $$\begin{gathered} \mathrm{K}\mathrm{L}\mathrm{M} \\ 281 \end{gathered}$$