Chemistry Atomic Structure SZABMU 2024
PMDC Verified Question 7 of 95
(Deleted) Electronic configuration of \( _{11}\text{Na}^{23} \) is
A
\( \text{[Ne]} 3s^1 \)
B
\( \text{[Ne]} 3s^2 \)
C
\( \text{[Ne]} 3s^0 \)
D
\( \text{[Ne]} 3s^2 3p \dots \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: \( \text{[Ne]} 3s^1 \)
Concept: Noble gas notation simplifies electronic configurations by replacing inner, fully filled shells with the bracketed symbol of the preceding noble gas.

Formula: None required.

Solution:
  • Sodium (Na) has an atomic number \( Z = 11 \).
  • The nearest preceding noble gas is Neon (Ne), which has \( Z = 10 \).
  • Neon's configuration perfectly covers the core electrons: \( 1s^2, 2s^2, 2p^6 \).
  • Sodium has one remaining valence electron (\( 11 - 10 = 1 \)).
  • This final electron enters the next available energy level, which is the \( 3s \) orbital.
  • The condensed configuration is therefore \( \text{[Ne]} 3s^1 \).


Why other options are incorrect: Option B represents Magnesium (\( Z=12 \)). Option C represents a Sodium ion (\( \text{Na}^+ \)). Option D implies too many electrons.

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