# Why aren't elements with two valence shell electrons stable?

We all know that hydrogen becomes stable by completing duplet and helium is stable since its valence shell has completed the duplet

Still, other elements don't behave like that. I want to know that why aren't elements like magnesium, beryllium etc, with two valence electrons be unreactive like helium atom, or why can't other elements become stable by completing duplet in the valence shell ?

• Titles are not supposed to be clear enough, like book titles are not supposed to tell the book story. They should just give a good hint what it is about. OTOH, the question body is supposed to provide good question elaboration, not just rephrasing the title. That includes results of textbook and online research. Otherwise such question can become a closure candidate. Oct 4, 2020 at 11:23
• @Poutnik okay I will edit it. Thanks for response. 🙂 Oct 4, 2020 at 11:26
• Stable is a very poor word here. The alkaline earths are all perfectly stable elements. Do you mean "unreactive"? Oct 4, 2020 at 11:32
• @Ian Bush yes , by stable I meant unreactive. Oct 4, 2020 at 11:33
• @Ankit do you know about orbitals? Oct 4, 2020 at 12:06

$$\mathrm{s}$$ electrons of alkali metals and alkali earth metals perceive relatively low central electrostatic charge, because of shielding the kernel by inner electrons. They are also farther from the kernel, compared to 1s electrons of hydrogen and helium, what overall leads to low ionization energy of such electrons and high reactivity of such atoms.

Helium and hydrogen have no inner electrons to shield the kernel and their 1s orbitals are smaller, compared to respective higher s orbitals So their ionization energy, especially for helium, is high, and reactivity reciprocally low.

For these reasons and the ones more described below, there is significant decreasing trend in ionization energy of ns electrons of group 1 and2 elements ( values approximated):

$$\ce{H}$$ $$\pu{13.6 eV}$$
$$\ce{He}$$ $$\pu{24.7 eV}$$
$$\ce{Li}$$ $$\pu{5.3 eV}$$
$$\ce{Be}$$ $$\pu{9.3 eV}$$
$$\ce{Na}$$ $$\pu{5.1 eV}$$
$$\ce{Mg}$$ $$\pu{7.8 eV}$$
$$\ce{K}$$ $$\pu{4.3 eV}$$
$$\ce{Ca}$$ $$\pu{6.1 eV}$$

$$\mathrm{p}$$ electrons of chalkogens and halogens perceive relatively high central electrostatic force, leading to the high ionization energy of such electrons and relatively high electron affinity and because that also high reactivity of such atoms.

Classical case of electrostatics

There is the well known rule from electrostatics, that if there is a system of charges with perfect spherical symmetry, it acts toward its neighbourhood like if it were a point-like charge with the effective charge equal to the sum of charges.

If we have such a system with point-like central big positive charge and negatively charged spherical layer with spherically symmetric charge distribution, the central positive charge is shielded/screened by the outer negative charge of the layer, so the system neighborhood perceives much lower central charge than it really is, i.e. the difference of both charges.

Multielectron atomic quantum models

Statistical distribution of electron density around atomic kernels follows wave functions called orbitals(*). The electron density and related charge distribution is not therefore spherically symmetric. Orbitals also spatially overlap and go sideways, what is further worsening the screening of the kernel charge.

The overall decreasing of screening efficiency goes in order s, p, d, f. Also, electrons from inner "shells"/"subshells" perform better screening then electrons with the same shell or subshell.

It has the consequence the overall kernel screening and electron energy depends on particular atomic electron configuration. For more you can read about Slater's rules.

Atoms with a new principal quantum level just started being filled, like alkali and alkali earth metals, have the screening by inner electrons particularly good. Therefore these $$\mathrm{ns}$$ electrons perceive particularly low effective kernel charge.

This has few consequences:

• The last occupied s orbital is particularly big - see atomic radii.

• The electrons in the last occupied s orbital have particularly high energy ( compared to other atoms ).

• High electron energies lead to favourable chemical bonding, releasing energy, for covalent bond cases like $$\ce{R-Mg-X}$$ or $$\ce{R-Li}$$

• High electron energies lead to easy ionization due low ionization energy , which can be more than compensated by lattice ionic energy of solid salts, or by hydration energy in water solutions.

Similarly, elements with just few missing p electrons like halogens or chalkogens have for these electrons poor kernel shielding. This leads to a high effective kernel charge and significant amount of energy released by capturing extra electron to form anion or polar covalent bond. The related term is electron affinity.

(*) Note that the term orbital has 3 major, related but distinguished meanings:

• orbital(1) - a wave function as the particular solution of the Schroedinger's quantum wave equation

• orbital(2) as a quantum state of an electron, belonging to orbital(1)

• orbital(3) as a 3D region(or 3D surface/shape), describing the region of significant probability of electron occurance (or its border threshold value), belonging to orbital(1,2)

• I think you got this upside-down: "For such system is said the central inner charges are shielded/screened by the outer charges." Oct 5, 2020 at 14:20
• @BuckThorn Perhaps I have misformulated it. I meant the kernel is for outer electrons partially screened (off?)/shielded by the inner electrons. I will check it. Oct 5, 2020 at 14:23
• @BuckThorn I think it is right. Note that that paragraph as about a classical and symmetric scenario. If you have e.g. a central point charge +2e and spherically symmetric charged layer -1e, it would act for the neighborhood as if it was a point charge +e, as +2e is partially screened off by the negative layer. Oct 5, 2020 at 14:28
• I'm a tad confused. Isn't your point later in the answer that inner electrons shield outer (valence) s electrons rendering the nuclear charge effectively less, increasing reactivity and lower ionization energy? Oct 5, 2020 at 14:30
• @Buckthorn the first paragraph is about classical spherically symmetric scenario: Central pointlike positive, spherical negative layer. For neighborhood it would be like central-layer summary effective charge. the2nd paragrapha is about atoms, with lack of sph. symmetry and orbital overlapping. There is partial screening of the kernel by inner electrons for the outer electrons. this screening is good for s electrons of K or Ca, but bad for p electrons of O or F. Oct 5, 2020 at 14:34

Woah, that answer above me is very complicated. Im sure that it is also right, but let's explore a simpler way to look at this. Instead of thinking of elements becoming unreactive when their octet/duplet is filled, lets think about unreactivity in terms of noble gases. If we look at the noble gases on the periodic table, we can see that they all have 8 valence electrons (except for He). You can verify this by writing the electron configuration or looking at the group number. That is why we say that atoms want to fill their octet to become unreactive, because that would generally give them noble gas configuration. Now, this is not always true. Instead, we should really say that atoms want to become isoelectronic with a noble gas, which means that they have the same electron configuration as the noble gas. Now back to the question, why does only Hydrogen need 2 valence e- to become unreactive? Well, its because it only needs to have 2 valence electrons to become isoelectronic with the nearest noble gas, helium. This wouldn't work with other elements, which need to have 8 valence electrons to become unreactive, since all other noble gases besides He have 8 valence electrons. I am just an AP chemistry student, so I don't exactly have any credentials, so feel free to correct me, but this made sense to me. Have a nice day.