Chemistry Chemical Bonding MDCAT 2019
PMDC Verified Question 85 of 102
Nitrogen has the atomic number 7. Which of the following electronic configurations is of a Nitrogen atom in ground state?
A
\( 1\text{s}^2, 2\text{s}^2, 2\text{px}^1, 2\text{py}^1, 2\text{pz}^1 \)
B
\( 1\text{s}^2, 2\text{s}^2, 2\text{py}^2, 2\text{pz}^1 \)
C
\( 1\text{s}^2, 2\text{s}^2, 2\text{px}^2, 2\text{py}^1 \)
D
\( 1\text{s}^2, 2\text{s}^2, 2\text{px}^2, 2\text{pz}^1 \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: \( 1\text{s}^2, 2\text{s}^2, 2\text{px}^1, 2\text{py}^1, 2\text{pz}^1 \)
Concept:

According to Hund's Rule, electrons occupy degenerate orbitals singly, with parallel spins, before pairing up.

Formula:

$$ \text{Total Electrons} = 7 $$

Solution:

  • The first two electrons fill the \( 1s \) orbital: \( 1s^2 \).


  • The next two electrons fill the \( 2s \) orbital: \( 2s^2 \).


  • The remaining three electrons must be distributed among the three degenerate \( 2p \) orbitals (\( p_x, p_y, p_z \)).


  • By Hund's Rule, they each take one empty orbital rather than pairing up: \( 2p_x^1, 2p_y^1, 2p_z^1 \).


Why other options are incorrect:

  • Option B, Option C, Option D: These configurations show electrons pairing up in one of the p-orbitals while leaving another empty, which violently violates Hund's Rule of maximum multiplicity.

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