The electrons,identified by quantum numbers n | vedclass

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(A) According to the $(n+l)$ rule,the energy of an orbital increases as the value of $(n+l)$ increases. If two orbitals have the same $(n+l)$ value,th

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

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(A) According to the $(n+l)$ rule,the energy of an orbital increases as the value of $(n+l)$ increases. If two orbitals have the same $(n+l)$ value,the one with the lower $n$ value has lower energy.Calculating $(n+l)$ for each:$(i) n = 4, l = 1 \implies n+l = 4+1 = 5$$(ii) n = 4, l = 0 \implies n+l = 4+0 = 4$$(iii) n = 3, l = 2 \implies n+l = 3+2 = 5$$(iv) n = 3, l = 1 \implies n+l = 3+1 = 4$Comparing the values:For $(ii)$ and $(iv)$,$(n+l) = 4$. Since $(iv)$ has a lower $n$ value $(3 For $(i)$ and $(iii)$,$(n+l) = 5$. Since $(iii)$ has a lower $n$ value $(3 Combining these,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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