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The electronic configuration of nitrogen ($Z = 7$) in the ground state is:
A1s$^2$ 2s$^2$ 2p$^3$, with three unpaired p electrons here
B1s$^2$ 2s$^2$ 2p$^4$, with one paired and two unpaired ones
C1s$^2$ 2s$^2$ 2p$^2$, missing one electron from nitrogen
D1s$^2$ 2p$^5$, skipping the 2s subshell entirely here
Answer & Solution
Correct answer: A. 1s$^2$ 2s$^2$ 2p$^3$, with three unpaired p electrons here
N has 7 electrons: 1s$^2$ 2s$^2$ 2p$^3$, with three unpaired p electrons (Hund).
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