The ground state electron configuration of the nitrogen atom is A. 1s22s22px1 B. 1s22s22px1py1 C. 1s22s22px1py1pz1 D. 1s22s 2

Question

BlackTom AI · Accepted Answer

Nitrogen has atomic number 7, so it contains 7 electrons to place into orbitals according to the Aufbau principle, Pauli exclusion principle, and Hund’s rule.

1) Relevant concepts
- Aufbau principle: fill lowest-energy orbitals first (1s → 2s → 2p).
- Pauli exclusion principle: each orbital can hold at most two electrons with opposite spins.
- Hund’s rule: for degenerate orbitals (the three 2p orbitals), place one electron in each orbital with parallel spins before pairing.

2) Step-by-step electron placement
- Electrons to place: 7.
- Fill 1s: 1s can hold 2 → place 2 electrons. Remaining electrons = 7 − 2 = 5.
- Fill 2s: 2s can hold 2 → place 2 electrons. Remaining electrons = 5 − 2 = 3.
- Fill 2p: three degenerate 2p orbitals (2p_x, 2p_y, 2p_z) share the remaining 3 electrons.

3) 2p occupancy using Hund’s rule
- With 3 electrons and three degenerate p orbitals, Hund’s rule directs one electron in each orbital with parallel spin: 2p_x^1, 2p_y^1, 2p_z^1.

4) Why option C is correct
- Option C explicitly lists 1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1, which matches the filling sequence and Hund’s rule for nitrogen’s three 2p electrons. This yields the correct total of 7 electrons and the lowest-energy (ground) arrangement.

5) Why the other options are incorrect
- A (1s^2 2s^2 2p_x^1): shows only one 2p electron (total 5 electrons) — incomplete.
- B (1s^2 2s^2 2p_x^1 2p_y^1): shows two 2p electrons (total 6 electrons) — still incomplete.
- D (1s^2 2s^2): lists only the filled 1s and 2s shells (total 4 electrons) — far fewer than nitrogen’s 7.

Conclusion: Using Aufbau, Pauli, and Hund’s rules produces 1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1, so the correct answer is C (matches the provided answer).

The ground state electron configuration of the nitrogen atom is A. 1s22s22px1 B. 1s22s22px1py1 C. 1s22s22px1py1pz1 D. 1s22s 2Single choice

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