JEE Main 2023ChemistryStructure of AtomEasyMCQ

JEE Main 2023 — Structure of Atom Question with Solution

JEE Main 2023 (31 Jan Shift 2)

Question

Arrange the following orbitals in decreasing order of energy.

A. n=3, l=0, m=0
B. n=4, l=0, m=0
C. n=3, l=1, m=0
D. n=3, l=2, m=1

The correct option for the order is:

Choose an option

Show full solutionCorrect option: A
Correct answer
AD > B > C > A

Step-by-step explanation

In multi-electronic species, energy is decided on the basis of (n+l) rule.

3s = 3 + 0 + 0= 3

4s = 4 + 0 +0 = 4

3p = 3 + 1 +0 = 4

3d = 3 + 2 + 1 = 6

For 4s and 3p, an orbital with a higher value of n has higher energy.

So increasing the order of energy is 3d>4s>3p>3s.

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About this question

This is a previous-year question from JEE Main 2023, covering the Structure of Atom chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.