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JEE Main 20249 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Structure of Atom

JEE Main 2024 · 9 April · Morning Shift · Chemistry · Structure of Atom

Compare the energies of following sets of quantum numbers for...

Compare the energies of following sets of quantum numbers for multielectron system. (A) (n=\mathrm{n}=4,1=1) (B) (n=\mathrm{n}=4, l=\mathrm{l}=2) (C) (n=\mathrm{n}=3,1=1) (D) (n=\mathrm{n}=3,1=2) (E) (n=\mathrm{n}=4,1=0) Choose the correct answer from the options given below :

Options

  1. A(\text{ (B) } > \text{ (A) } > \text{ (C) } > \text{ (E) } > \text{ (D) })
  2. B(\text{ (E) } < (\text{ C }) < \text{ (D) } < (\text{ A }) < (\text{ B }))
  3. C(\text{ (E) } > (\text{ C }) > (\text{ A }) > (\text{ D }) > (B))
  4. D(\text{ (C) } < \text{ (E) } < \text{ (D) } < \text{ (A) } < \text{ (B) })

Correct answer

D. (\text{ (C) } < \text{ (E) } < \text{ (D) } < \text{ (A) } < \text{ (B) })

Step-by-step solution

Energy level can be determined by comparing ( n+ℓ\mathrm{n}+\ell ) values (A) n=4,ℓ=1⇒(n+ℓ)=5\mathrm{n}=4, \ell=1 \Rightarrow(\mathrm{n}+\ell)=5 (B) n=4,ℓ=2⇒(n+ℓ)=6\mathrm{n}=4, \ell=2 \Rightarrow(\mathrm{n}+\ell)=6 (C) n=3,ℓ=1⇒(n+ℓ)=4\mathrm{n}=3, \ell=1 \Rightarrow(\mathrm{n}+\ell)=4 (D) n=3,ℓ=2⇒(n+ℓ)=5\mathrm{n}=3, \ell=2 \Rightarrow(\mathrm{n}+\ell)=5 (E) n=4,ℓ=0⇒(n+ℓ)=4\mathrm{n}=4, \ell=0 \Rightarrow(\mathrm{n}+\ell)=4
For same value of (n+ℓ)(n+\ell), orbital having higher value of nn, will have more energy. (B) >(A)>(D)>(E)>(C)>(\mathrm{A})>(\mathrm{D})>(\mathrm{E})>(\mathrm{C})
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