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 the $(n+l)$ values are the same,the orbital

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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 the $(n+l)$ values are the same,the orbital with the lower value of $n$ has lower energy.Calculating $(n+l)$ values for each:$(i)$ $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 $(n+l)=4$: $(iv)$ has $n=3$ and $(ii)$ has $n=4$. Thus,$iv For $(n+l)=5$: $(iii)$ has $n=3$ and $(i)$ has $n=4$. Thus,$iii 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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