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(D) The energy of an orbital in a multi-electron atom is determined by the $(n+l)$ rule.According to this rule,the orbital with the higher $(n+l)$ val

Vedclass · Accepted Answer

$n=5, l=1, m=0, s=+\frac{1}{2}$

Vedclass · Answer

(D) The energy of an orbital in a multi-electron atom is determined by the $(n+l)$ rule.According to this rule,the orbital with the higher $(n+l)$ value has higher energy.If two orbitals have the same $(n+l)$ value,the one with the higher principal quantum number $(n)$ has higher energy.Calculating $(n+l)$ for each option:$A$: $n=5, l=0 \implies n+l = 5+0 = 5$$B$: $n=4, l=2 \implies n+l = 4+2 = 6$$C$: $n=4, l=1 \implies n+l = 4+1 = 5$$D$: $n=5, l=1 \implies n+l = 5+1 = 6$Comparing options $B$ and $D$,both have $(n+l) = 6$. Since option $D$ has a higher $n$ value $(n=5)$ compared to option $B$ $(n=4)$,option $D$ has the highest energy.

For a multi-electron atom,the highest energy level among the following is

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