MCQ
The electrons identified by quantum numbers $n$ and $l$ :

$(A)\, n=4,l=1$                        $(B)\, n=4,l=0$

$(C) \,n=3,l=2$                        $(D)\, n=3,l=1$

can be placed in order of increasing energy as :

  • A
    $(C) < (D) < (B) < (A)$
  • $(D) < (B) < (C) < (A)$
  • C
    $(B) < (D) < (A) < (C)$
  • D
    $(A) < (C) < (B) < (D)$

Answer

Correct option: B.
$(D) < (B) < (C) < (A)$
b
$(a) n=4, l=1(p-\text { subshell }),$ so $4 p$

$(b) =4, l=0$ ( $s-$ subshell), so $4 s$

$(c) n=3, l=2(d \text { -subshell), so } 3 d$

$(d) n=3, l=1$ ( $p-$ subshell), so $3 p$

Accroding to the Bohr $(n+l)$ rule, Enery order of the subshell : $3 p<4 s<3 d<4 p$

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