Beryllium hydride does not obey the octet rule. Isn't this impossible, unless there are more elements comprising the molecule?

  • 4
    $\begingroup$ It's really more like a guideline. ;) $\endgroup$
    – jzx
    Aug 5, 2015 at 22:35

1 Answer 1


Technically, it does not.

$\ce{BeH2}$ forms a network of $\ce{BeH4}$ tetrahedrons sharing corners. Since each hydrogen belongs to two such tetrahedron, the net formula is $\ce{BeH2}$. In this structure, each electron pair belongs to a triple of atoms $\ce{Be-H-Be}$. Such bonding is known as three-centre two-electron bonding, and is fairly common in electron-deficient compounds that would otherwise violate the octet rule. For example, this kind of bonding is essential for boron-hydrogen chemistry.

With that said, unambiguous violations of the octet rule are not unknown. The most obvious and reasonable stable example would be nitric oxide, $\ce{NO}$. However, this example again features strong 'extra' electron delocalization with 'extra' electron shared between the oxygen and nitrogen atoms. The molecule still shows tendency to dimerize at low temperatures. This kind of molecule is more of an exception and not a rule. The exact way such particles are stabilised is highly dependent on the particle.

Of course, if we are talking about transition elements or lanthanides, the situation is more complex, because of the involvement of fairly strongly bonded electrons in d- or f-orbitals. Still, the extra stability of complexes following the equivalent 18-electron rule is clearly observed.


Not the answer you're looking for? Browse other questions tagged or ask your own question.