The values of n and l for orbital A are 3 and | vedclass

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Question

(A) For orbital $A$: $n = 3, l = 2$. The orbital is $3d$. The energy is determined by $(n + l)$ rule. $(n + l) = 3 + 2 = 5$. For orbital $B$: $n = 5,

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Energy of $B$ is greater than $A$

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(A) For orbital $A$: $n = 3, l = 2$. The orbital is $3d$. The energy is determined by $(n + l)$ rule. $(n + l) = 3 + 2 = 5$. For orbital $B$: $n = 5, l = 0$. The orbital is $5s$. The energy is determined by $(n + l)$ rule. $(n + l) = 5 + 0 = 5$. Since both orbitals have the same $(n + l)$ value of $5$,the orbital with the lower $n$ value has lower energy. Therefore,$3d$ $(A)$ has lower energy than $5s$ $(B)$,or energy of $B$ is greater than $A$.

The values of $n$ and $l$ for orbital $A$ are $3$ and $2$ respectively. The values of $n$ and $l$ for orbital $B$ are $5$ and $0$ respectively. Compare their energies.

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