Why does beryllium chloride not follow octet rule?

Answer 1

Beryllium is in the 2d period of the modern Periodic Table. Its electron config. is [He] 2s2 and forms the Be2+ ion by losing two elctrons and in this instance can be said to disobey the octet. In some compounds such as the hydride it also disobeys the octet rule. But in many others it does obey the octet rule! For example the difluoride, BeF2, is essentially covalent with four coordinate Be centers linked by flourine bridges.

Answer 2

Be has 2 valence elelectrons. BeCl2 in the solid state does follow the octet rule, it has a chain structure with 4 coordinate tetrahedral beryllium atoms. There are "dative" bonds from Cl bringing the number of electrons around the Be to 8. The hybridusation is not sp3 as the bond angles are not tetrahedral.

In the vapour phase it forms a dimer, approxmate sp2 hybridisation wherether are 6 electrons around the Be.

At high temperature the dimer dissociates to monomeric BeCl2 which is linear. This has only 4 electrons around the Be.

Answer 3

Beryllium chloride does not follow the octet rule because beryllium typically forms compounds where it has only four electrons in its valence shell, not a full octet of eight electrons. This is due to its small atomic size and high charge density, making it prefer to bond using fewer electrons.

Answer 4

It needs an ionic bond, but I need to know why it needs and ioic bond??

Answer 5

They don't have enough electrons