MCQ
Given below are the quantum numbers for $4$ electrons.

$A. \;n =3, l=2, m _{1}=1, m _{ s }=+1 / 2$

$B.\; n =4, l=1, m _{1}=0, m _{ s }=+1 / 2$

$C. \;n =4, l=2, m _{1}=-2, m _{ s }=-1 / 2$

$D. \;n =3, l=1, m _{1}=-1, m _{ s }=+1 / 2$

The correct order of increasing energy is

  • A
    $D < B < A < C$
  • $D < A < B < C$
  • C
    $B < D < A < C$
  • D
    $B < D < C < A$

Answer

Correct option: B.
$D < A < B < C$
b
Energy order of subshell decided by $( n +\lambda)$ rule.

$A \Rightarrow 3 d \Rightarrow n +1=5$

$B \Rightarrow 4 p \Rightarrow n +\lambda=5$

$C \Rightarrow 4 d \Rightarrow n +\ell \Rightarrow 6$

$D \Rightarrow 3 s \Rightarrow( n +\ell)=4$

$D < A < B < C$

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