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(D) The energy of atomic orbitals in multi-electron atoms is determined by the $(n+l)$ rule.$1$. According to the $(n+l)$ rule,the orbital with the lo

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$E < B < A < D < C$

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(D) The energy of atomic orbitals in multi-electron atoms is determined by the $(n+l)$ rule.$1$. According to the $(n+l)$ rule,the orbital with the lower $(n+l)$ value has lower energy.$2$. If the $(n+l)$ values are equal,the orbital with the lower value of $n$ has lower energy.Calculating $(n+l)$ for each orbital:$A: n=3, l=2 \implies n+l = 3+2 = 5$$B: n=4, l=0 \implies n+l = 4+0 = 4$$C: n=6, l=1 \implies n+l = 6+1 = 7$$D: n=5, l=1 \implies n+l = 5+1 = 6$$E: n=2, l=1 \implies n+l = 2+1 = 3$Comparing the $(n+l)$ values: $3 (E) Therefore,the order of increasing energy is $E < B < A < D < C$.

Arrange the following atomic orbitals of multi-electron atoms in order of increasing energy.
$A$. $n=3, l=2, m=+1$
$B$. $n=4, l=0, m=0$
$C$. $n=6, l=1, m=0$
$D$. $n=5, l=1, m=+1$
$E$. $n=2, l=1, m=+1$
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Arrange the following atomic orbitals of multi-electron atoms in order of increasing energy.$A$. $n=3, l=2, m=+1$$B$. $n=4, l=0, m=0$$C$. $n=6, l=1, m=0$$D$. $n=5, l=1, m=+1$$E$. $n=2, l=1, m=+1$Choose the correct answer from the options given below: