In an atom,the order of increasing energy of | vedclass

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(D) The order of increasing energy is determined using the $(n+l)$ rule. If two orbitals have the same $(n+l)$ value,the orbital with the lower value

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$iv < ii < iii < i$

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(D) The order of increasing energy is determined using the $(n+l)$ rule. If two orbitals have the same $(n+l)$ value,the orbital with the lower value of $n$ has lower energy.$(i)$ For $n=4, l=1$,$(n+l) = 4+1 = 5$.$(ii)$ For $n=4, l=0$,$(n+l) = 4+0 = 4$.$(iii)$ For $n=3, l=2$,$(n+l) = 3+2 = 5$.$(iv)$ For $n=3, l=1$,$(n+l) = 3+1 = 4$.Comparing the values: $(iv)$ and $(ii)$ have $(n+l) = 4$,where $(iv)$ has a lower $n$ value $(3 $(i)$ and $(iii)$ have $(n+l) = 5$,where $(iii)$ has a lower $n$ value $(3 Combining these,the correct order of increasing energy is $(iv) < (ii) < (iii) < (i)$.

In an atom,the order of increasing energy of electrons with quantum numbers $(i)$ $n=4, l=1$,$(ii)$ $n=4, l=0$,$(iii)$ $n=3, l=2$,and $(iv)$ $n=3, l=1$ is:

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