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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 the electronic configuration is 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.

Its 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 '19 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 '19 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 '19 at 19:58
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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. For example, 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 $\pu{9.3 eV}$, the second requires $\pu{18.2 eV}$, the third requires $\pu{153.9 eV}$, and the fourth requires $\pu{217.7 eV}$. The energy required to remove the first two electrons is in the range of other metals that lose two electrons: For comparison, the first and second ionization energies of $\ce{Mg}$ are $7.6$ and $\pu{15.0 eV}$ while those of $\ce{Ca}$ are $6.1$ and $\pu{11.9 eV}$, respectively. So, $\ce{Be}$ 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 hypothetically happens in physics and mathematics!).

The empty orbitals on $\ce{Be}$ are quite high in energy, i.e., if you stuck an electron into a $\mathrm{p}$ orbital on $\ce{Be}$, 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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  • $\begingroup$ I have added few words to make it better. If you don't like what I have done, please feel free to roll back. $\endgroup$ Aug 22 at 11:32
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    $\begingroup$ I don't remember what it was before, but it sounds good now, so, Thanks! $\endgroup$ Aug 22 at 12:46
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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 '19 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 '19 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. $\endgroup$ Feb 23 '19 at 13:41
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After the electron configuration, the last shell of the beryllium atom has two electrons. In this case, both the valence and valence electrons of beryllium are 2. We know the details about this. The elements that have 1, 2, or 3 electrons in the last shell donate the electrons in the last shell during bond formation.

The elements that form bonds by donating electrons are called cations and beryllium is a cation element. Beryllium donates the electrons of the last shell to form bonds and turns into beryllium ions:

$$\ce{Be^0 – 2e– -> Be^2+} \ \text{or} \ \ce{Be -> Be^2+ + 2e–}$$

The electron configuration of beryllium ion $(\ce{Be^2+})$ is $\mathrm{1s^2}$, which shows that beryllium ion, $\ce{Be^2+}$ has only one shell and that shell has total of two electrons. The electron configuration of $\ce{Be^2+}$ also shows that the beryllium atom has acquired the electron configuration of helium. That is, in this case, the valence of the beryllium-ion is $+2$. Since the last shell of a beryllium-ion has two electrons, the valence electrons of a beryllium-ion are two. Source- https://valenceelectrons.com/category/valence-electrons-of-all-elements/

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