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I have a rudimentary understanding of orbitals, as in what they are, the shapes (l, I think) and of the principles.

So I was doing the electronic configuration for beryllium, it has 4 electrons. So 1s2, 2s2. Why does berylium lose two electrons to complete its “inner duplet”? I thought that atoms want to gain, lose or share electrons so their orbitals are complete and thus don’t react.

It’s orbitals are complete, is it something to do with the promotion of an electron to a higher energy level, but won’t that be temporary change? Or perhaps there is some fundamental misunderstanding on my part.

(I’m in my IGCSE, this isn’t in our syllabus, just trying to understand, so I hope my ignorance is forgiven)

Edit: I am not familiar with hybridisation but I don’t think it applies to this, unless you are giving a bonding example)

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  • $\begingroup$ 1) Complete shell not "orbitals" 2) Actually Be is tetracoordinate and it's bonds quite covalent. $\endgroup$ – Mithoron Feb 22 at 19:09
  • $\begingroup$ @Mithoron Oh, I see it forms covalent bonds due to its strong pull on its electrons. One example being BeCl2 with chlorine’s dative bond. But aren’t orbitals involved? And your comment answers my question to some extent that it isn’t ionic (thanks for that), but it’s s orbital is complete what reason does it have to form covalent compounds? $\endgroup$ – Serapion Feb 23 at 9:09
  • $\begingroup$ It's p orbitals are accessible so it can just try to get octet (even if sometimes incomplete). It's bonding may be more ionic or more covalent depending on compound, but it's very much not binary thing, but continuum. $\endgroup$ – Mithoron Feb 23 at 19:58
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All main group elements follow the octet rule, which is to bond in such a way as to achieve 8 electrons in their valence shells.

I've never heard of the term "inner duplet", but perhaps this is referring to its desire to complete its valence shell with 8 electrons. Valence electrons are the electrons in the outermost shell of an atom.

For example, Beryllium can bond with oxygen and donates those 2 electrons to oxygen to combine with oxygen, which has 6 valence electrons. After this, Be will have a charge of 2+ and oxygen will have a charge of 2-. These ions will be attracted to each other and bond. This results in them both having completed valence shells with 8 electrons each.

Here is a brief explanation of how the octet rule affects metals: https://socratic.org/questions/how-does-the-octet-rule-affect-metals

Explanation of valence electrons: https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table

I hope this answers your question and I didn't just re-explain something that you already knew :)

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    $\begingroup$ I already know that, beryllium loses two electrons so it has completed the duplet of the first shell and then subsequent attraction to the anion. But when you draw out the orbitals, it doesn’t need to lose 2 electrons since the 2s orbital is complete. I always thought that atoms complete their orbitals to achieve stability which is inertness. $\endgroup$ – Serapion Feb 23 at 8:37
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    $\begingroup$ You seem confused by the usage of inner duplet, beryllium has 4 electrons. Effectively if it were to lose two electrons it’s first shells duplet would be complete, I used air quotes since the 2s orbital is already complete, so it doesn’t need to lose electrons (which I learnt now it forms covalent bonds). But even so it’s s orbitals are complete, I simple can’t understand why it would react? My knowledge on orbitals isn’t very extensive, half knowledge is dangerous they say $\endgroup$ – Serapion Feb 23 at 9:14
  • $\begingroup$ Right, I understand what you are saying about both s orbitals being full. I suppose the only reason it would lose electrons would be to achieve the the state of a noble gas by following the octet rule. $\endgroup$ – Kinesthetic76 Feb 23 at 13:41
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Electrons exist in the outer spaces of elements, away from the nucleus. Orbitals are mathematical descriptions of places in these outer spaces where electrons seem to be when they are there; they can be mathematically manipulated and reorganized but are a way to help you imagine what is really going on.

Manipulating the electrons is done experimentally, not mathematically. E.g., an element can be ionized and the energy required can be measured. Beryllium can suffer the loss of up to four electrons: the first requires 9.3 eV, the second requires 18.2 eV, the third requires 153.9 eV, and the fourth requires 217.7 eV. The energy required to remove the first two electrons is in the range of other metals that lose two electrons (Mg 7.6, 15.0; Ca 6.1, 11.9 eV). So Be can lose 2 electrons with a relatively small amount of energy (which will be supplied by something like empty orbitals and ionic or covalent bonding energy to other atoms). Losing those last two is a big jump and essentially doesn't happen in chemistry (it happens in physics and mathematics!).

The empty orbitals on Be are quite high in energy, i.e., if you stuck an electron into a p orbital on Be, it would be floating so far out from the nucleus that it would be looking for a better place. The idea of orbitals gives you a mental picture of what might happen - whether it really happens or not depends on whether that mathematical picture is true and is the most stable (lowest energy) condition available.

Completing shells or orbitals is a qualitative way of looking at the issue, but numerically, ionization potentials gives you the data you need to develop the picture, and the orbitals (as calculated for the hydrogen atom) are a reasonable way to begin.

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