The correct order of increasing energy of ato | vedclass

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(B) According to the $(n+l)$ rule,the energy of an orbital increases with the value of $(n+l)$.For $5p$: $n=5, l=1$,so $(n+l) = 6$.For $6s$: $n=6, l=0

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$5p < 6s < 4f < 5d$

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(B) According to the $(n+l)$ rule,the energy of an orbital increases with the value of $(n+l)$.For $5p$: $n=5, l=1$,so $(n+l) = 6$.For $6s$: $n=6, l=0$,so $(n+l) = 6$.For $4f$: $n=4, l=3$,so $(n+l) = 7$.For $5d$: $n=5, l=2$,so $(n+l) = 7$.If $(n+l)$ values are equal,the orbital with the lower $n$ value has lower energy.Comparing $5p$ and $6s$: both have $(n+l) = 6$,but $5p$ has a lower $n$ value,so $5p Comparing $4f$ and $5d$: both have $(n+l) = 7$,but $4f$ has a lower $n$ value,so $4f Thus,the correct order is $5p < 6s < 4f < 5d$.

The correct order of increasing energy of atomic orbitals is

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