The relative energies of $ 4s $, $ 4p $ and $ 3d $ orbitals are in the order: · Atomic Structure MDCAT Past Paper MCQ

Chemistry Atomic Structure MDCAT 2012

PMDC Verified Question 93 of 95

The relative energies of $ 4s $, $ 4p $ and $ 3d $ orbitals are in the order:

$ 4p < 4s < 3d $

$ 4p < 3d < 4s $

$ 3d < 4p < 4s $

$ 4s < 3d < 4p $

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Propolis Cognitive Error Autopsy

Official Correct Choice:

Option D: $ 4s < 3d < 4p $

Concept: The energy of an orbital is directly proportional to its $ (n + l) $ value (Aufbau Principle).

Formula: $$ \text{Energy} \propto (n + l) $$

Solution:

For $ 4s $: $ n=4, l=0 \implies n+l = 4 $
For $ 3d $: $ n=3, l=2 \implies n+l = 5 $
For $ 4p $: $ n=4, l=1 \implies n+l = 5 $
Since $ 4s $ has the lowest value, it has the lowest energy.
For $ 3d $ and $ 4p $, both have $ n+l=5 $. When $ (n+l) $ values are identical, the orbital with the lower $ n $ value has lower energy. Thus, $ 3d < 4p $.
Overall order: $ 4s < 3d < 4p $.

Why other options are incorrect: They violate the $ (n+l) $ rule for filling order.

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