# Why can't beryllium react with hydrogen gas directly to give hydride?

I read in a textbook that all alkali metal and alkaline earth metal can react with hydrogen directly to give hydride except for beryllium. I want to know the reason why because I know that $\ce{BeH2}$ exists in a polymeric state.

• See this paper: pubs.acs.org/doi/pdf/10.1021/ja00078a057 Apr 9, 2021 at 7:51
• @NilayGhosh this q never got an answer that could stand up to criticism. Maybe summarize key findings from the reference in an answer that will work? Sep 6 at 19:54
• @OscarLanzi Please go ahead and make into an answer. I can see you had deleted your answer 4 years ago due to lack of reference(?) Sep 10 at 7:27
• That paper shows evidence for BeH, BeH2, and several gaseous oligomers under heroic conditions. BeH is endothermic deltaG = 67 kcal. that combined with the strong bond energies of H2 and crystal lattice of Ge direct reaction would be kinetically and thermally difficult. The answer seems to be they do react if all the activation energies are removed at a low temperature to prevent decomposition. Sep 21 at 3:48

As described in Ref. 1, the proposed reaction is endothermic and thus unlikely to be favored. In contrast the other alkaline earth metals give exothermic reactions that ought to proceed at least at low enough temperature. So then the question becomes why does beryllium give an endothermic reaction?

One clue lies in the unusually high melting and boiling points of this metal versus other alkaline-earth metals, which do react directly with hydrogen. From the respective Wikipedia articles we render the following melting and boiling points for the first three alkaline earth metals:

Beryllium: 1287°C, 2469°C

Magnesium: 650°C, 1091°C

Calcium: 842°C, 1484°C

This implies that metallic Beryllium has much stronger bonds than other metallic alkaline earth elements, and a reaction of beryllium metal with hydrogen is thereby hindered by those stronger elemental bonding.

In this respect beryllium resembles not the other alkaline earth metals or even post-transition metals in Group 12, but higher-melting transition metals such as titanium or iron. The lack of a reaction between beryllium and hydrogen thereby fits a more general pattern: high-melting metals, which require breaking strong bonds to form salt-like compounds, do not do so with hydrogen. They either fail to react directly with hydrogen at all (unless they are somehow atomized first, as in Ref. 1), like beryllium or iron; or else they instead form interstitial hydrides that allow most of the metal-metal bonding to be preserved like titanium.

Beryllium is also not alone in being atypically hard to melt or boil. The same thing happens with boron versus the rest of Group 13 and carbon versus the rest of Group 14. In all these cases the small size of the electronic core, which is only a $$1s$$ subshell, allows strong and deep overlap in the delocalized bonds between the atoms, whereas the larger cores of the later-period congeners limits overlap and makes the heavier elements potentially more reactive.

References

1. Thomas J. Tague Jr. and Lester Andrews (1993). "Reactions of beryllium atoms with hydrogen: Matrix infrared spectra of novel product molecules". J. Am. Chem. Soc. 115, 25, 12111–12116. https://doi.org/10.1021/ja00078a057
• It really is fairly simple the product is not particularly stable and reacting from the elements has a large activation energy to supply Ge and H atoms. Ge and H atoms react nicely at temperatures near 0 K, the activation energy is supplied elsewhere. Sep 23 at 1:56
• More is involved than activation energy. Breaking the strong bonds implied by the high melting point is also a thermodynamic barrier. Sep 23 at 10:53
• That is what the activation energy is composed of. Add in a strong H-H bond and finally an endothermic compound why does anyone suspect this reaction to happen at ambient temperatures Sep 24 at 3:36

The explanation lies in the structure of the BeH 2 molecule. Since the Be atom is small, it is always in the center of the BeH 2 molecule. This Be atom is surrounded by a lot of H atoms, and it is hard for the H atoms to be close to each other, just like to have water molecules come close to each other. This explains why BeH 2 is a polymeric state; it is very difficult for H atoms to have H-H contacts.

This is not the case for sodium and potassium. When we study the sodium atom, we see that the sodium atom is very small. It is small enough to have the possibility to have neighboring sodium atoms. In fact, one sodium atom can have 3 neighboring sodium atoms. In other words, one sodium atom has a coordination number of 3. This makes Na+ a cation and Na a neutral atom. The same situation happens for K+. The coordination number of K+ is 4. As a result, the K atoms are bigger than the Be atoms, and are not as enclosed as Be atoms in BeH 2. This is one of the reasons why BeH 2 is an insulating solid. The other reason is that it is polymeric, just like water.

Compare to hydrogen, the H 2 molecule is diatomic. This means that each hydrogen atom is shared by two H atoms. In other words, the H atoms in H 2 have a coordination number of 2. This means that when we have H 2 molecules in a molecule, the H atoms will have less difficulty in approaching each other.

• I don't fully agree though, maybe you can provide evidence from papers to prove your point! Thanks! :3 (I think it has to do with Be's oxidation potential) Sep 22 at 5:30