Chemistry Transition Elements MDCAT 2017
PMDC Verified Question 54 of 68
Scandium has atomic number 21; which one will be its electronic configuration?
A
\( 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3s^{6}, 3d^{3} \)
B
\( 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6}, 4s^{2}, 3d^{1} \)
C
\( 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6}, 4s^{2}, 4p^{1} \)
D
\( 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6}, 4s^{1}, 4p^{2} \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: \( 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6}, 4s^{2}, 3d^{1} \)
Concept:

Scandium (\(Z=21\)) is the first element of the 3d transition series. Its electron configuration follows the standard Aufbau sequence up to 21 electrons.

Solution:

  • The first 18 electrons fill standard shells: \(1s^2 \; 2s^2 \; 2p^6 \; 3s^2 \; 3p^6\) (Argon core).


  • The remaining 3 electrons enter the next available energy levels. According to the \(n+l\) rule, the \(4s\) orbital fills before the \(3d\) orbital.


  • Thus, 2 electrons go into \(4s\) (\(4s^2\)), and the last electron enters \(3d\) (\(3d^1\)).


  • The full correct configuration is: \(1s^{2} \; 2s^{2} \; 2p^{6} \; 3s^{2} \; 3p^{6} \; 4s^{2} \; 3d^{1}\).


Why other options are incorrect:

  • Option A contains a typo (\(3s^6\)) and incorrect total electron placement.


  • Option C places the final electron in \(4p\) instead of \(3d\).


  • Option D skips standard \(n+l\) filling orders by putting electrons in \(4p\) prematurely.

Quality & Fidelity Assurance: Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.

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