Why isn't Beryllium a noble gas? [closed]








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Somebody answered it before on physics StackExchange, but I couldn't follow the high-level parlance. All I could follow is that it is related to the p subshell. I cannot understand why elements form bonds when they have pairs of spins(If it matters: I may be wrong, but please apprise me about my misconceptions). An example would be Aluminium, though it's not entirely related. It losses all three of its electrons when it can conveniently lose one and have a pair of opposite spin numbers.


Please answer :-)




I interpret this as a more general question of why alkaline earth elements are electropositive metals and not noble gases.


The answer is basically they do not have enough electrons in their valence shells to be strongly stabilizedcas elements. All the electrons below the outer $s^2$ pair are in shells having fewer nodes in their quantum mechanical wavefunctions, and the presence of fewer nodes makes these inner-electron waves so compact they shield almost all of the nuclear charge from the outer pair. So the outer pair of electrons have low effective nuclear charge attracting them, which translates to their own wavefunctions being relatively diffuse and their ionization energies being relatively low.


Helium and, to a lesser extent, beryllium are exceptions because helium has no such inner-shell electrons and beryllium has only one pair, whereas magnesium and heavier $s^2$ elements have multiple pairs and more nodes in their outer-shell wavefunctions. Hence the failure of helium to readily form compounds and the lesser failure of beryllium to form predominantly ionic ones. By the time we get to magnesium we already have the shielded structure described above well-enough developed to enable the ready formation of predominantly ionic compounds, even with the relatively low-electronegativity elements [hydrogen](https://chemistry.stackexchange.com/a/157450/17175) and [boron](https://chemistry.stackexchange.com/a/143849/17175).


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