Why BF3 is a stable compound even if boron is not satisfying the octet rule?
Why BF3 is a stable compound even if boron is not satisfying the octet rule?
here is a phenomenon known as pπ−pπ back bonding taking place in BF3. It may sound a bit complicated, but i can explain in simple terms.
now, when Boron forms bonds with fluorine, boron has 6 electrons and fluorine has 8 electrons. Now fluorine being small in Size and having High Electronegativity (tendency to pull negative charge) has a high amount of charge density (charge per volume) on it, which it prefers to reduce. So, it loses an electron and Boron accepts that electron. Then Fluorine makes a bond with the electron it just lost, forming a double bond. This is known as pπ−pπ back bonding. As a result of this, both boron and fluorine have eight electrons. All three fluorine atoms are capable of this type of bonding, making boron a highly stable molecule.
BF3 is pretty stable. Boron atoms have three valence electrons so it's pretty stable when they form three covalent bonds with fluorine atoms. They are still, however, good Lewis acids and do still readily gain an electron pair in reactions.
Actually BF3 doesn't exist in nature. To decrease the incomplete electron structure, two BF3 molecules attach each other and we have B2F6. With the half bound that is formed this way, the electron structure becomes a semi 8 electron shell, which is more stable.
here is a phenomenon known as pπ−pπ back bonding taking place in BF3. It may sound a bit complicated, but i can explain in simple terms.
now, when Boron forms bonds with fluorine, boron has 6 electrons and fluorine has 8 electrons. Now fluorine being small in Size and having High Electronegativity (tendency to pull negative charge) has a high amount of charge density (charge per volume) on it, which it prefers to reduce. So, it loses an electron and Boron accepts that electron. Then Fluorine makes a bond with the electron it just lost, forming a double bond. This is known as pπ−pπ back bonding. As a result of this, both boron and fluorine have eight electrons. All three fluorine atoms are capable of this type of bonding, making boron a highly stable molecule.
BF3 is pretty stable. Boron atoms have three valence electrons so it's pretty stable when they form three covalent bonds with fluorine atoms. They are still, however, good Lewis acids and do still readily gain an electron pair in reactions.
Actually BF3 doesn't exist in nature. To decrease the incomplete electron structure, two BF3 molecules attach each other and we have B2F6. With the half bound that is formed this way, the electron structure becomes a semi 8 electron shell, which is more stable.
