MCQ
The electrons, identified by quantum by numbers $n$ and $l,$ $(i)\, n = 4, l= 1$ $(ii)\, n = 4, l = 0$ $(iii)\, n = 3, l = 2$ $(iv)\,n = 3, l = 1$ can be placed in order of increasing energy, from the lowest to highest, as
  • $(iv) < (ii) < (iii) < (i)$
  • B
    $(ii) < (iv) < (i) < (iii)$
  • C
    $(i) < (iii) < (ii) < (iv)$
  • D
    $(iii) < (i) < (iv) < (ii)$

Answer

Correct option: A.
$(iv) < (ii) < (iii) < (i)$
a
According to afbau rule energy is more for more $(n+l)$ value.

and if $(n+l)$ is Same, then $n$ will less no. has less energy

$(i) \,n+l=0$

$(ii)\,n+l=4$

$( ii) \,n=3, l=2 \quad n+l=5$

$(iv) \,n+l=4$

$(iv) < (ii) < (iii) < (i)$

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