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(B) The energy of an electronic configuration is determined by the sum of the energies of all the electrons present in the orbitals.For a given princi

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(B) The energy of an electronic configuration is determined by the sum of the energies of all the electrons present in the orbitals.For a given principal quantum number $n=3$,the energy of the orbitals follows the order $3s To find the configuration with the maximum energy,we compare the total number of electrons in the higher energy orbitals ($3p$ and $3d$).Option $A$: $2$ electrons in $3s$,$2$ in $3p$,$0$ in $3d$. Total electrons in $3p+3d = 2$.Option $B$: $2$ electrons in $3s$,$3$ in $3p$,$2$ in $3d$. Total electrons in $3p+3d = 5$.Option $C$: $2$ electrons in $3s$,$3$ in $3p$,$1$ in $3d$. Total electrons in $3p+3d = 4$.Option $D$: $2$ electrons in $3s$,$3$ in $3p$,$1$ in $3d$. Total electrons in $3p+3d = 4$.Comparing the total number of electrons in the higher energy orbitals,Option $B$ has the highest number of electrons in the $3p$ and $3d$ orbitals,thus representing the state with the maximum energy.

Which of the following electronic configurations has the maximum energy?

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