My Math Forum What would be the number of electrons in a spectroscopic notation?

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 October 25th, 2019, 10:00 PM #1 Senior Member     Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus What would be the number of electrons in a spectroscopic notation? I found this a bit "trivial" question regarding quantum numbers but I'm still confused over the given alternatives in the answer. Could it be that whoever posed the question made a mistake or made the question in a silly way? The problem is as follows: How many electrons are in the following spectroscopic notation $4d^{9}$?$\begin{array}{ll} 1.&4e^{-}\\ 2.&e^{-}\\ 3.&13e^{-}\\ 4.&5e^{-}\\ 5.&8e^{-}\\ \end{array}$ If I am not mistaken I believe that the answer would be just $9$ as it is strictly stated over the notation for the d-subshell. But this doesn't appear within the alternatives. Upon inspecting the "answer" it still doesn't make much sense as well since $4d^{9}$ is not stable but instead $4d^{8}$ is but for that to happen this atom from which no further information is given has not lost any electrons, needless to say that I cannot assume anything more from the given information, my only guess it is that it could be referring to silver, but again $4d^{9}$ is not the correct configuration for that atom. What would be the answer or is it just that the question was incorrectly stated or the alternatives aren't right?

 Tags electrons, notation, number, spectroscopic

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