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

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(D) The order of increasing energy is determined by the $(n+l)$ rule.If two orbitals have the same value of $(n+l)$, 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 by the $(n+l)$ rule.If two orbitals have the same value of $(n+l)$, 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 $(n+l) = 4$. Since (iv) has lower $n$, $(iv) $(i)$ and (iii) have $(n+l) = 5$. Since (iii) has lower $n$, $(iii) Thus, the increasing order of 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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