4d, 5p, 5f and 6p orbitals are arranged in th | vedclass

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(A) To determine the energy order,we use the $(n+l)$ rule:For $4d$: $n=4, l=2$,so $(n+l) = 4+2 = 6$.For $5p$: $n=5, l=1$,so $(n+l) = 5+1 = 6$.For $5f$

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$5f > 6p > 5p > 4d$

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(A) To determine the energy order,we use the $(n+l)$ rule:For $4d$: $n=4, l=2$,so $(n+l) = 4+2 = 6$.For $5p$: $n=5, l=1$,so $(n+l) = 5+1 = 6$.For $5f$: $n=5, l=3$,so $(n+l) = 5+3 = 8$.For $6p$: $n=6, l=1$,so $(n+l) = 6+1 = 7$.Comparing the $(n+l)$ values: $8 (5f) > 7 (6p) > 6 (4d, 5p)$.For orbitals with the same $(n+l)$ value,the one with the higher $n$ value has higher energy. Thus,$5p (n=5) > 4d (n=4)$.The final decreasing order of energy is $5f > 6p > 5p > 4d$.

$4d, 5p, 5f$ and $6p$ orbitals are arranged in the order of decreasing energy. The correct option is

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