Identify the orbital having the highest energ | vedclass

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(D) The energy of an orbital is determined by the $(n+l)$ rule. Higher the sum of $(n+l)$ value of an orbital,the higher is its energy.If two orbitals

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$4p$

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(D) The energy of an orbital is determined by the $(n+l)$ rule. Higher the sum of $(n+l)$ value of an orbital,the higher is its energy.If two orbitals have the same value of $(n+l)$,then the orbital with the higher value of $n$ will have more energy.Let us calculate $(n+l)$ for each:$(A)$ $2p$: $n=2, l=1 \implies n+l = 3$$(B)$ $3s$: $n=3, l=0 \implies n+l = 3$$(C)$ $3d$: $n=3, l=2 \implies n+l = 5$$(D)$ $4p$: $n=4, l=1 \implies n+l = 5$Comparing $3d$ and $4p$,both have $(n+l) = 5$. Since $4p$ has a higher principal quantum number $(n=4)$ compared to $3d$ $(n=3)$,$4p$ has the highest energy.

Identify the orbital having the highest energy from the following:

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