If the given four electronic configurations:( | vedclass

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(A) According to the $(n+l)$ rule:$1$. The orbital with a lower $(n+l)$ value has lower energy.$2$. If the $(n+l)$ values are the same, the orbital wi

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

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

If the given four electronic configurations:
$(i)$ $n=4, l=1$
$(ii)$ $n=4, l=0$
$(iii)$ $n=3, l=2$
$(iv)$ $n=3, l=1$
are arranged in order of increasing energy, then the order will be:

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The following sets of quantum numbers represent four electrons in an atom:
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The sequence representing the increasing order of energy is:

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If the given four electronic configurations:$(i)$ $n=4, l=1$$(ii)$ $n=4, l=0$$(iii)$ $n=3, l=2$$(iv)$ $n=3, l=1$are arranged in order of increasing energy, then the order will be: