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

Vedclass · Accepted Answer

$n = 4, l = 0, m = 0, s = +\frac{1}{2}$

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(B) The energy of an electron in a multi-electron atom is determined by the $(n + l)$ rule. According to this rule,the orbital with the lower $(n + l)$ value has lower energy. If the $(n + l)$ values are equal,the orbital with the lower $n$ value has lower energy. Calculating $(n + l)$ for each option: $A: n = 3, l = 2 \implies n + l = 3 + 2 = 5$ $B: n = 4, l = 0 \implies n + l = 4 + 0 = 4$ $C: n = 4, l = 1 \implies n + l = 4 + 1 = 5$ $D: n = 5, l = 0 \implies n + l = 5 + 0 = 5$ Comparing the values,option $B$ has the lowest $(n + l)$ value of $4$. Therefore,the electron in option $B$ is associated with the least energy.

Which electron is associated with the least energy?

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