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I'm a little confused on how I am supposed to know how many lone pairs and atom will have when drawing expanded octet Lewis Structures. For example on question 3.63, Xe has 2 bound pairs and 3 lone pairs, but on 3.65, Xe can have 4 bound pairs and 2 lone pairs, I am just kind of confused on how to tell how many lone pairs an atom should have in the expanded octet rule.
I believe the number of lone pairs an atom has is dependent on what the atom is bonding with. For instance, in 3.63, Xe has 2 bound pairs and 3 lone pairs because there are 22 valence electrons total. 8 electrons each are given to F, the remaining electrons just become lone pairs (2 pairs of electrons are bonding the Xe and F btw). This ends up totaling 22 valence electrons. Then for 3.65, Xe has 4 bound pairs and 2 lone pairs because there are 36 valence electrons total. 8 electrons each are given to the four Fs, the remaining electrons become lone pairs. This totals 36 valence electrons. So according to what an atom is trying to bond with will determine how many lone pairs it has.
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