How many valence electrons can an atom have?
Answer 1:
the first subshell has 2, and all the other subshells have 8. Helium is stable.
Explanation:
I’m not exactly sure how in depth you need it to be, but in case you do, each orbital is arranged according to s sub-orbitals, p sub-orbitals, d sub-orbitals, and f sub-orbitals. Each sub-orbital has a fixed amount of electrons that it can hold.
The first orbital has an s orbital only which can hold a maximum of 2 electrons. This is labeled as \(1 s^{a}\), where a is the number of electrons it could potentially have. The number in the label represents which shell it is, the closest to the nucleus is labeled 1, and these increase as the number of shells increase.
Helium: \(1 s^{2}\) maximum in this shell, therefore it is stable.
Hydrogen \(1 s^{1}\), half filled shell, unstable.
When you have 2 shells, what happens is that there are 2 subshells. 1 s subshell and 1 p subshell, however, because you are 2 shells away from the nucleus, then it changes to \(2 s^{a}\). There is also the introduction of the p sub orbital, which can hold a total of 6 electrons. It is labeled as \(2 p^{b}\). Note that the number of electrons (the exponent) for 2s and 2p can be different depending on the element chosen.
carbon: \(1 s^{2} 2 s^{2} 2 p^{2}\)
Oxygen: \(1 s^{2} 2 s^{2} 2 p^{4}\)
I’ll go into more detail if you ask me to but I think this covers it all.
Answer 2:
In atoms before the third period, the number of electrons outside of the noble gas core add up to no more than 8, and that tends to be the valence electrons… But that means nothing regarding the number of valence electrons for atoms past the third period.
The valence electrons are simply those that are most important in chemical bonding, irrespective of whether they are in the highest n or not. Atoms have no regard for any of the rules we make up. They don’t care. The number of valence electrons most used in practice is shown in the oxidation state.
Here are two exceptions to the notion that all the electrons outside the noble gas core are valence…
\(\mathrm{Ni}:[A r] 3 d^{8} 4 s^{2}, 10\) electrons past the noble gas core;
max valence electrons used so far: \(4\) e.g. in \(\mathrm{BaNiO}_{3}\).
\(\mathrm{Rh}:[K r] 4 d^{7} 5 s^{2}, 9\) electrons past the noble gas core;
max valence electrons used so far: \(6\) e.g. in \(\mathrm{RhF}_{6}\).
But of course, not all of these outer-core electrons are used. In \(\mathrm{Ni}\), only \(2\) are usually used, and in \(\mathrm{Rh},\) only \(3\) are usually used. The above are the maximum that have been used.
In principle, more than 8 electrons can be in the highest-energy orbitals for bonding, but actually, at most, 9 valence electrons have so far been used, namely in iridium, while osmium, ruthenium, xenon, and hassium have used at most 8.
Iridium currently holds the record with \(9\) valence electrons used, since \(\mathrm{Ir}^{+9}\) exists in the tetrahedral \(\mathrm{IrO}_{4}^{+}\) polyatomic ion.
Its electron configuration demonstrates that they are from the 6s and 5d orbitals:
\([X e] 4 f^{14} 5 d^{7} 6 s^{2}\)