Question

If the nitrogen atom has electronic configuration 1s7

,

it would have energy lower than that of the normal

ground state configuration 1s22s22p3

, because the

electrons would be closer to the nucleus.Yet 1s7

is

not observed because it violates

(1) Heisenberg’s uncertainty principle

(2) Hund’s rule

(3) Pauli’s exclusion principle

(4) Bohr’s postulate of stationary orbits


(Note:- Only genuine answer otherwise I will report you)​

Answer 1

HERE IS THE ATTACHMENT SIS ✨⚡❄️

Answer 2

Answer:

(3)Pauli's exclusion principle . The principle states that every orbital must have 2 electrons and that too with the same spin. s orbital means magnetic quantum number is zero, hence it can accommodate only two electrons with+1/2 and -1/2 spins and not 7 electrons.

Formula to find the no of the preferred orientations of electrons in a 3D space ( magnetic quantum number ) is given by the formula 2l+1 where l is the azumthal quantum number.

Here orbital given is s so l value is 0.

Therefore, preferred orientations of electrons can be calculated as 2l+1

=2×0+1

=1

Therfore in 1 orientations the maximum possible no of electrons that can accommodate is 2.

So s can carry 2 electrons and not 7.

Hope this helps. The concept should be crystal clear.

Hope this helps