The correct order of energy for orbitals in a | vedclass

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(D) According to the $(n+l)$ rule,the energy of an orbital increases as the value of $(n+l)$ increases. For a given value of $(n+l)$,the orbital with

Vedclass · Accepted Answer

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

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(D) According to the $(n+l)$ rule,the energy of an orbital increases as the value of $(n+l)$ increases. For a given value of $(n+l)$,the orbital with a lower value of $n$ has lower energy. Comparing the orbitals: $ns$: $(n+0) = n$ $np$: $(n+1)$ $(n-1)d$: $(n-1+2) = n+1$ $(n-2)f$: $(n-2+3) = n+1$ Among orbitals with the same $(n+l)$ value of $(n+1)$,the order of increasing energy is $np Thus,the overall energy order is $ns < np < (n-1)d < (n-2)f$.

The correct order of energy for orbitals in a multi-electron atom is:

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