The electrons,identified by quantum numbers n | vedclass

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(A) According to the $(n+l)$ rule,the higher the value of $(n+l)$,the higher is the energy of the orbital.If the $(n+l)$ values are the same,the orbit

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

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(A) According to the $(n+l)$ rule,the higher the value of $(n+l)$,the higher is the energy of the orbital.If the $(n+l)$ values are the same,the orbital with the lower value of $n$ has lower energy.Calculating $(n+l)$ values:$(i) n=4, l=1 \implies n+l = 5$$(ii) n=4, l=0 \implies n+l = 4$$(iii) n=3, l=2 \implies n+l = 5$$(iv) n=3, l=1 \implies n+l = 4$Comparing the values:For $(iv)$ and $(ii)$,both have $(n+l) = 4$. Since $n=3 For $(i)$ and $(iii)$,both have $(n+l) = 5$. Since $n=3 Combining these,the order of increasing energy is $(iv) < (ii) < (iii) < (i)$.

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

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