Here's why:
* Boron (B) only has 3 valence electrons.
* In BH3, boron forms three single bonds with three hydrogen atoms. This uses up all three of boron's valence electrons.
* Boron only has 6 electrons in its valence shell. This is less than the 8 electrons needed to satisfy the octet rule.
Therefore, boron in BH3 is an example of an atom that does not obey the octet rule. It is actually considered to be electron deficient.
This is a common characteristic of boron compounds, and it contributes to their unique reactivity.
