Verified Solution • 2025–26 Session moderate

QJEEM23Ch385a: JEE Main 2023 Chemistry - Class 11 JEEM23Ch385a Solution: Arrange the following orbitals in decreasing order of energy.…

Arrange the following orbitals in decreasing order of energy.

A. n=3,l=0, m=0\mathrm{n}=3, \mathrm{l}=0, \mathrm{~m}=0

B. n=4,l=0, m=0\mathrm{n}=4, \mathrm{l}=0, \mathrm{~m}=0

C. n=3,l=1, m=0\mathrm{n}=3, \mathrm{l}=1, \mathrm{~m}=0

D. n=3,l=2, m=1\mathrm{n}=3, \mathrm{l}=2, \mathrm{~m}=1

The correct option for the order is :

Multiple Choice Options

  • A B>D>C>A\mathrm{B}>\mathrm{D}>\mathrm{C}>\mathrm{A}
  • B A>C>B>D\mathrm{A}>\mathrm{C}>\mathrm{B}>\mathrm{D}
  • C D>B>A>C\mathrm{D}>\mathrm{B}>\mathrm{A}>\mathrm{C}
  • D D>B>C>A\mathrm{D}>\mathrm{B}>\mathrm{C}>\mathrm{A}

Step-by-Step Verified Solution

(A) n = 3; l = 0; m = 0 ; 3s orbital

(B) n = 4; l = 0; m = 0 ; 4s orbital

(C) n = 3; l = 1; m = 0 ; 3p orbital

(D) n = 3; l = 2; m = 0 ; 3d orbital

As per Hund’s ruleo of energy is given by (n + l) value. If value of (n + l) remains same then energy is given by n only.

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