- #1
Indranil
- 177
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Homework Statement
As we know in NH3, N full fills its octet with three single bonds with H and a lone pair of electrons and so H. So this molecule is stable. In BF3, F full fills their octets by three single bonds with B but B is unable to fulfill its octet. It got 6 electrons by three single bonds with F. So, B needs 2 more electrons to fulfill its octet. So, BF3 receives the lone pair of electrons from the NH3 and make a coordinate covalent bond between NH3 and BF3, making a new molecule NH3BF3.
So my question is how to calculate the proton numbers in "NH3'' after giving the lone pair of electrons and ''BF3" after receiving the lone pair of electrons?
Homework Equations
As we know in NH3, N full fills its octet with three single bonds with H and a lone pair of electrons and so H. So this molecule is stable. In BF3, F full fills their octets by three single bonds with B but B is unable to fulfill its octet. It got 6 electrons by three single bonds with F. So, B needs 2 more electrons to fulfill its octet. So, BF3 receives the lone pair of electrons from the NH3 and make a coordinate covalent bond between NH3 and BF3, making a new molecule NH3BF3.
So my question is how to calculate the proton numbers in "NH3'' after giving the lone pair of electrons and ''BF3" after receiving the lone pair of electrons?
The Attempt at a Solution
Don't know how to calculate in NH3^+- BF3^-