To draw the Lewis dot structure for BF3, we start by identifying the valence electrons of the atoms involved. Boron (B) has 3 valence electrons, and each fluorine (F) has 7 valence electrons. Since there are three fluorine atoms, we have a total of:
3 (from B) + 3 × 7 (from F) = 3 + 21 = 24 valence electrons.
In the Lewis structure, boron will be the central atom surrounded by three fluorine atoms. Each fluorine atom will form a single bond with boron, using up 6 of the 24 valence electrons (2 electrons per bond). The remaining 18 valence electrons will be allocated to the fluorine atoms, giving each of them three lone pairs of electrons:
Now, let’s summarize the requested information:
- a) Number of bond pairs: There are 3 bond pairs in BF3, corresponding to the three B-F bonds.
- b) Number of lone pairs: There are no lone pairs on the central boron atom. Each fluorine has 3 lone pairs, but these are not on the boron.
- c) Molecular geometry: The molecular geometry of BF3 is trigonal planar. This geometry arises because the three B-F bonds are arranged around the boron atom at 120-degree angles to minimize electron repulsion.
- d) Hybridization of the central atom: The hybridization of the central boron atom in BF3 is sp2. This is due to the involvement of one s orbital and two p orbitals in the bonding configuration.
In summary, the Lewis dot structure of BF3 shows a central boron atom bonded to three fluorine atoms, with specific characteristics regarding bonds, lone pairs, molecular geometry, and hybridization.