Q5 / 90 · JEE Main 2024 (09 Apr Shift 1) · Chemistry

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 :

  • A
    \(\text { (B) } > \text { (A) } > \text { (C) } > \text { (E) } > \text { (D) }\)
  • B
    \(\text { (E) } < (\text { C }) < \text { (D) } < (\text { A }) < (\text { B })\)
  • C
    \(\text { (E) } > (\text { C }) > (\text { A }) > (\text { D }) > (B)\)
  • Correct answer
    D
    \(\text { (C) } < \text { (E) } < \text { (D) } < \text { (A) } < \text { (B) }\)

Solution

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) $>(\mathrm{A})>(\mathrm{D})>(\mathrm{E})>(\mathrm{C})$

Hence, the correct answer is option: 'D'.

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