If the nitrogen atom had electronic configuration $1s^7$,it would have energy lower than that of the normal ground state configuration $1s^2, 2s^2, 2p^3$ because the electrons would be closer to the nucleus. Yet,$1s^7$ is not observed because it violates:

  • A
    Heisenberg's uncertainty principle
  • B
    Hund's rule
  • C
    Pauli's exclusion principle
  • D
    Bohr postulate of stationary orbit

Explore More

Similar Questions

Identify degenerate orbitals from the following for a hydrogen atom.

There is no difference between a $2p$ and a $3p$ orbital regarding

Identify the elements represented by the following electronic configurations:
$(a)$ $[He] \, 2s^1$
$(b)$ $[Ne] \, 3s^2 \, 3p^3$
$(c)$ $[Ar] \, 4s^2 \, 3d^1$

The quantum number/s needed to describe an electron fully in an atom is/are:

Which of the following species has an equal number of electrons in the outermost shell and the penultimate shell?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo