The order of increasing energies of the orbit | vedclass

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Question

(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.For

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$3s < 3p < 4s < 3d < 4p$

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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.For $3s$: $n=3, l=0$,so $(n+l) = 3+0 = 3$.For $3p$: $n=3, l=1$,so $(n+l) = 3+1 = 4$.For $4s$: $n=4, l=0$,so $(n+l) = 4+0 = 4$.For $3d$: $n=3, l=2$,so $(n+l) = 3+2 = 5$.For $4p$: $n=4, l=1$,so $(n+l) = 4+1 = 5$.If $(n+l)$ values are equal,the orbital with the lower $n$ value has lower energy.Comparing $3p$ $(n=3)$ and $4s$ $(n=4)$,$3p Comparing $3d$ $(n=3)$ and $4p$ $(n=4)$,$3d Thus,the increasing order of energy is $3s < 3p < 4s < 3d < 4p$.

The order of increasing energies of the orbitals follows:

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