The electrons identified by quantum numbers n | vedclass

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Question

(A) The energy of an orbital is determined by the $(n + l)$ rule. $(i)$ $n = 4, l = 1 \implies n + l = 5$ ($4p$ orbital). $(ii)$ $n = 4, l = 0 \implie

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$iv < ii < iii < i$

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(A) The energy of an orbital is determined by the $(n + l)$ rule. $(i)$ $n = 4, l = 1 \implies n + l = 5$ ($4p$ orbital). $(ii)$ $n = 4, l = 0 \implies n + l = 4$ ($4s$ orbital). $(iii)$ $n = 3, l = 2 \implies n + l = 5$ ($3d$ orbital). $(iv)$ $n = 3, l = 1 \implies n + l = 4$ ($3p$ orbital). For orbitals with the same $(n + l)$ value,the one with the lower $n$ value has lower energy. Comparing $(ii)$ $(n=4)$ and $(iv)$ $(n=3)$,$(iv)$ has lower energy. Comparing $(i)$ $(n=4)$ and $(iii)$ $(n=3)$,$(iii)$ has lower energy. Thus,the order of increasing energy is $(iv) < (ii) < (iii) < (i)$.

The electrons identified by quantum 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

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