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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 $(n+l)$ values are equal,the orbital with 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 $(n+l)$ values are equal,the orbital with the lower $n$ value has lower energy.Calculating $(n+l)$ 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 $(iv)$ and $(ii)$,$(n+l) = 4$. Since $n(iv) For $(i)$ and $(iii)$,$(n+l) = 5$. Since $n(iii) Combining these,the increasing order of energy is $(iv) < (ii) < (iii) < (i)$.

For the given quantum numbers $n$ and $l$ of electrons:
$(i) n = 4, l = 1$
$(ii) n = 4, l = 0$
$(iii) n = 3, l = 2$
$(iv) n = 3, l = 1$
Arrange them in the increasing order of their energy.

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For the given quantum numbers $n$ and $l$ of electrons:$(i) n = 4, l = 1$$(ii) n = 4, l = 0$$(iii) n = 3, l = 2$$(iv) n = 3, l = 1$Arrange them in the increasing order of their energy.