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Compare the energies of following sets of quantum numbers for multielectron system.
(A) \(\mathrm{n}=4,1=1\)
(B) \(\mathrm{n}=4, \mathrm{l}=2\)
(C) \(\mathrm{n}=3,1=1\)
(D) \(\mathrm{n}=3,1=2\)
(E) \(\mathrm{n}=4,1=0\)
Choose the correct answer from the options given below :
ChemistryStructure of AtomJEE MainJEE Main 2024 (09 Apr Shift 1)
Options:
  • A \(\text { (B) } > \text { (A) } > \text { (C) } > \text { (E) } > \text { (D) }\)
  • B \(\text { (E) } < (\text { C }) < \text { (D) } < (\text { A }) < (\text { B })\)
  • C \(\text { (E) } > (\text { C }) > (\text { A }) > (\text { D }) > (B)\)
  • D \(\text { (C) } < \text { (E) } < \text { (D) } < \text { (A) } < \text { (B) }\)
Solution:
2407 Upvotes Verified Answer
The correct answer is: \(\text { (C) } < \text { (E) } < \text { (D) } < \text { (A) } < \text { (B) }\)
Energy level can be determined by comparing ( $\mathrm{n}+\ell$ ) values
(A) $\mathrm{n}=4, \ell=1 \Rightarrow(\mathrm{n}+\ell)=5$
(B) $\mathrm{n}=4, \ell=2 \Rightarrow(\mathrm{n}+\ell)=6$
(C) $\mathrm{n}=3, \ell=1 \Rightarrow(\mathrm{n}+\ell)=4$
(D) $\mathrm{n}=3, \ell=2 \Rightarrow(\mathrm{n}+\ell)=5$
(E) $\mathrm{n}=4, \ell=0 \Rightarrow(\mathrm{n}+\ell)=4$
For same value of $(n+\ell)$, orbital having higher value of $n$, will have more energy.
(B) $>(\mathrm{A})>(\mathrm{D})>(\mathrm{E})>(\mathrm{C})$

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