Which of the following sets of quantum number | vedclass

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(D) The energy of an orbital is determined by the $(n + l)$ rule.For option $A$: $n = 4, l = 0$,so $(n + l) = 4 + 0 = 4$ ($4s$ orbital).For option $B$

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$n = 3, l = 2, m = 1, s = +\frac{1}{2}$

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(D) The energy of an orbital is determined by the $(n + l)$ rule.For option $A$: $n = 4, l = 0$,so $(n + l) = 4 + 0 = 4$ ($4s$ orbital).For option $B$: $n = 2, l = 0$,so $(n + l) = 2 + 0 = 2$ ($2s$ orbital).For option $C$: $n = 3, l = 1$,so $(n + l) = 3 + 1 = 4$ ($3p$ orbital).For option $D$: $n = 3, l = 2$,so $(n + l) = 3 + 2 = 5$ ($3d$ orbital).Comparing the $(n + l)$ values: $5 > 4 > 2$. Since the $3d$ orbital has the highest $(n + l)$ value,it possesses the highest energy.

Which of the following sets of quantum numbers belongs to the highest energy orbital?

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