I'm bit confused on how to find the number of valence electrons. For example, Boron ($\ce{B) has the electronic configuration as: $\mathrm{1s^2 2s^2 2p^1}$. So what would be its number of valence electrons : 1 or 3. According to me the answer should the 1 as the last orbital ($\mathrm{p}$ orbital) has 1 electron. Am I wrong? Please guide me if I am wrong.
1 Answer
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Any electrons present in the electron configuration of the element in question, but not present in the nearest noble gas with an atomic number below that of the element, is a valence electron.
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1$\begingroup$ Not quite correct. For example manganese (Mn) has configuration $\ce{[Ar]\ 4s^2\ 3d^5}$ and hence 7 valence electrons. $\endgroup$– MaxWMay 9, 2020 at 17:27
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$\begingroup$ Fair point @MaxW, I will edit. $\endgroup$– user93414May 9, 2020 at 17:30
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$\begingroup$ I'd use the KISS principle here. // The valence electrons are the electrons that can participate in bonding. For n=2 there can be $\ce{sp^3}$ hybridization so all the 2s and 2p electrons are valence electrons. // If you want to elaborate more certainly some paragraphs would help. $\endgroup$– MaxWMay 9, 2020 at 17:46