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(B) The energy of an orbital is determined by the $(n+l)$ rule.$A$. $n = 3, l = 0, m = 0 \implies (n+l) = 3 + 0 = 3$ ($3s$ orbital)$B$. $n = 4, l = 0,

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$D > B > C > A$

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(B) The energy of an orbital is determined by the $(n+l)$ rule.$A$. $n = 3, l = 0, m = 0 \implies (n+l) = 3 + 0 = 3$ ($3s$ orbital)$B$. $n = 4, l = 0, m = 0 \implies (n+l) = 4 + 0 = 4$ ($4s$ orbital)$C$. $n = 3, l = 1, m = 0 \implies (n+l) = 3 + 1 = 4$ ($3p$ orbital)$D$. $n = 3, l = 2, m = 1 \implies (n+l) = 3 + 2 = 5$ ($3d$ orbital)Comparing $(n+l)$ values: $D (5) > B (4) = C (4) > A (3)$.For $B$ and $C$,both have $(n+l) = 4$. According to the rule,the orbital with the higher $n$ value has higher energy. Thus,$B (n=4) > C (n=3)$.The decreasing order of energy is $D > B > C > A$.

Arrange the following orbitals in decreasing order of energy:
$A$. $n = 3, l = 0, m = 0$
$B$. $n = 4, l = 0, m = 0$
$C$. $n = 3, l = 1, m = 0$
$D$. $n = 3, l = 2, m = 1$
The correct option for the order is:

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Arrange the following orbitals in decreasing order of energy:$A$. $n = 3, l = 0, m = 0$$B$. $n = 4, l = 0, m = 0$$C$. $n = 3, l = 1, m = 0$$D$. $n = 3, l = 2, m = 1$The correct option for the order is: