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(A) According to the $(n+l)$ rule, the energy of an orbital increases as the value of $(n+l)$ increases. If two orbitals have the same $(n+l)$ value,

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$ns < (n-2)f < (n-1)d < np$

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(A) According to the $(n+l)$ rule, the energy of an orbital increases as the value of $(n+l)$ increases. If two orbitals have the same $(n+l)$ value, the one with the lower $n$ value has lower energy.For a given principal quantum number $n$, the energy order is determined by the $(n+l)$ values:$1$. For $ns$: $n+0 = n$$2$. For $(n-2)f$: $(n-2)+3 = n+1$$3$. For $(n-1)d$: $(n-1)+2 = n+1$$4$. For $np$: $n+1$Comparing $(n-2)f$, $(n-1)d$, and $np$, all have $(n+l) = n+1$. Among these, the one with the lowest $n$ value has the lowest energy. Thus, $(n-2)f Combining these, the correct order is $ns < (n-2)f < (n-1)d < np$.

Which of the following represents the correct general order of increasing energy for orbitals in a multi-electron atom?

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