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In the case of Boron, the abbreviated electron configuration is [He] 2s 2 2p 1, so the number of valence electrons is 1 (2s 2 2p 1) In the case of Iron, the abbreviated electron configuration is [Ar] 4s 2 3d 6 , so the number of valence electrons is 8 (4s 2 3d 6 )
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To write the electron configuraiton for a cation, start by writing the electron configuration for the neutral atom. Then determine the number of electrons that were added to form the anion. (The addition of a single electron results in a -1 charge, the addition of two electrons results in a -2 charge, and so on.)
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