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Boron forms a compound with hydrogen having formula BH(₃). Does the compound follow the octet rule?

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Final answer:

No, the compound BH3 does not follow the octet rule because boron only has 3 valence electrons. Instead, boron forms 3 covalent bonds with hydrogen, resulting in a molecule with a trigonal planar shape.

Step-by-step explanation:

The compound BH3 does not follow the octet rule. According to the octet rule, atoms tend to gain, lose, or share electrons in order to achieve a stable configuration with a full outer shell of 8 electrons. However, boron only has 3 valence electrons, so it cannot satisfy the octet rule by sharing or losing electrons because it would result in an incomplete octet.



Instead, in BH3, boron forms 3 covalent bonds with hydrogen, resulting in a molecule with a trigonal planar shape. The boron atom is sp2 hybridized, meaning it forms three hybrid orbitals by mixing one s orbital and two p orbitals. Each hybrid orbital then overlaps with a hydrogen 1s orbital to form a sigma bond.



Since BH3 does not have a complete octet for boron, it is considered an exception to the octet rule. Other elements, like boron and aluminum, also tend to form covalent compounds with incomplete octets.

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