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

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detailed solution

Correct option is C

The maximum number of $e^{-}$ in any orbital, according to the Pauli exclusion principle, is equal to 2.
Therefore, $1 s^{7}$ is not conceivable.
It's best to choose C.

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If the nitrogen atom had electronic configuration 1s7, it would have energy lower that of normal ground state configuration 1s22s22p3 , because the electrons would be closer to the nucleus. Yet 1s7 is not observed because it violates: