Identify the orbital having the lowest energy | vedclass

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(A) The energy of an orbital is determined by the $(n+l)$ rule,where $n$ is the principal quantum number and $l$ is the azimuthal quantum number. The

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$2p$

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(A) The energy of an orbital is determined by the $(n+l)$ rule,where $n$ is the principal quantum number and $l$ is the azimuthal quantum number. The orbital with the lowest $(n+l)$ value has the lowest energy. If $(n+l)$ values are equal,the orbital with the lower $n$ value has lower energy.Calculating $(n+l)$ for each:$2p: n=2, l=1 \implies n+l = 3$$3s: n=3, l=0 \implies n+l = 3$$3d: n=3, l=2 \implies n+l = 5$$4p: n=4, l=1 \implies n+l = 5$Comparing $2p$ and $3s$,both have $(n+l) = 3$. Since $2p$ has a lower $n$ value $(n=2)$ compared to $3s$ $(n=3)$,$2p$ has the lowest energy.

Identify the orbital having the lowest energy from the following.

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