MCQ
Compare the energies of following sets of quantum numbers for multielectron system.

$(A)$ $n=4,1=1$   $(B)$ $\mathrm{n}=4,1=2$    $(C)$ $\mathrm{n}=3,1=1$     $(D)$ $\mathrm{n}=3,1=2$     $(E)$ $n=4,1=0$

Choose the correct answer from the options given below:

  • A
    $(\mathrm{B})>(\mathrm{A})>(\mathrm{C})>\text { (E) }>\text { (D) }$
  • B
    $(\mathrm{E})>(\mathrm{C})<\text { (D) }<\text { (A) }<\text { (B) }$
  • C
    ${ (E) }>\text { (C) }>\text { (A) }>\text { (D) }>\text { (B) }$
  • $(\mathrm{C})<\text { (E) }<\text { (D) }<\text { (A) }<\text { (B) }$

Answer

Correct option: D.
$(\mathrm{C})<\text { (E) }<\text { (D) }<\text { (A) }<\text { (B) }$
d
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) $>$ (A) $>$ (D) $>$ (E) $>$ (C)$

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