The energy is lowest for the orbital :

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Question

(A) According to the $(n+l)$ rule,the energy of an orbital is determined by the sum of its principal quantum number $(n)$ and azimuthal quantum number

Vedclass · Accepted Answer

$4s$

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(A) According to the $(n+l)$ rule,the energy of an orbital is determined by the sum of its principal quantum number $(n)$ and azimuthal quantum number $(l)$.For $4s$: $n=4, l=0$,so $(n+l) = 4+0 = 4$.For $4p$: $n=4, l=1$,so $(n+l) = 4+1 = 5$.For $3d$: $n=3, l=2$,so $(n+l) = 3+2 = 5$.For $4f$: $n=4, l=3$,so $(n+l) = 4+3 = 7$.Comparing the $(n+l)$ values,$4s$ has the lowest value of $4$. If $(n+l)$ values are equal,the orbital with the lower $n$ value has lower energy. Since $4s$ has the lowest $(n+l)$ value,it has the lowest energy.

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