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(B) The energy of an electron in an orbital is determined by the $(n+l)$ rule (also known as the $Aufbau$ principle).$I$. The energy of an orbital inc

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$n = 4, l = 0, m = 0, s = +1/2$

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(B) The energy of an electron in an orbital is determined by the $(n+l)$ rule (also known as the $Aufbau$ principle).$I$. The energy of an orbital increases as the sum of the principal quantum number $(n)$ and the azimuthal quantum number $(l)$ increases.$II$. If two orbitals have the same $(n+l)$ value,the one with the lower $n$ value has lower energy.$III$. Calculating $(n+l)$ for each option:$(A)$ $n=3, l=2 \implies n+l = 3+2 = 5$$(B)$ $n=4, l=0 \implies n+l = 4+0 = 4$$(C)$ $n=4, l=1 \implies n+l = 4+1 = 5$$(D)$ $n=5, l=0 \implies n+l = 5+0 = 5$Since option $(B)$ has the lowest $(n+l)$ value of $4$,it corresponds to the electron with the least energy.

Which electron is associated with the least energy?

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