The correct decreasing order of energy for th | vedclass

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(A) According to the $(n+\ell)$ rule,the energy of an orbital is determined by the sum of its principal quantum number $(n)$ and azimuthal quantum num

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$D > B > C > A$

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(A) According to the $(n+\ell)$ rule,the energy of an orbital is determined by the sum of its principal quantum number $(n)$ and azimuthal quantum number $(\ell)$.For $A$: $n+\ell = 3+0 = 3$ ($3s$ orbital)For $B$: $n+\ell = 4+0 = 4$ ($4s$ orbital)For $C$: $n+\ell = 3+1 = 4$ ($3p$ orbital)For $D$: $n+\ell = 3+2 = 5$ ($3d$ orbital)Rules for energy:$1$. Higher $(n+\ell)$ value means higher energy.$2$. If $(n+\ell)$ values are equal,the orbital with the higher $n$ value has higher energy.Comparing the values:$D$ $(n+\ell=5)$ has the highest energy.Between $B$ and $C$,both have $(n+\ell)=4$. Since $B$ has $n=4$ and $C$ has $n=3$,$B$ has higher energy than $C$.$A$ $(n+\ell=3)$ has the lowest energy.Therefore,the decreasing order of energy is $D > B > C > A$.

The correct decreasing order of energy for the orbitals having the following set of quantum numbers:
$A$. $n=3, \ell=0, m=0$
$B$. $n=4, \ell=0, m=0$
$C$. $n=3, \ell=1, m=0$
$D$. $n=3, \ell=2, m=1$

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The correct decreasing order of energy for the orbitals having the following set of quantum numbers:$A$. $n=3, \ell=0, m=0$$B$. $n=4, \ell=0, m=0$$C$. $n=3, \ell=1, m=0$$D$. $n=3, \ell=2, m=1$