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

Question

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

Aicharya · Accepted Answer

Not more than two electrons can be present in same atomic orbital. This is Pauli's exclusion principle. ,

If the nitrogen atom had electronic configuration $1 \mathrm{~s}^{7}$ it would have energy lower than that…

Solution