Give the Lewis Structure of BF4 and Determine the Number of Valence Electrons

To determine the Lewis structure of BF₄, we first need to find the total number of valence electrons available in the molecule.

Boron (B) is in group 13 of the periodic table and has 3 valence electrons, while fluorine (F) is in group 17 and has 7 valence electrons. Since there are four fluorine atoms in BF₄, we can calculate the total number of valence electrons as follows:

• Valence electrons from Boron: 3
• Valence electrons from four Fluorine atoms: 4 × 7 = 28

Adding these together gives us:

3 (from B) + 28 (from 4 F) = 31 valence electrons.

However, when drawing the Lewis structure, we consider the bonding and arrangement of these atoms. In BF₄, the boron atom is the central atom and forms single bonds with each of the four fluorine atoms. Each bond accounts for 2 electrons, so we have:

• 4 single bonds x 2 electrons/bond = 8 electrons used for bonding.

This means we have:

31 total valence electrons – 8 bonding electrons = 23 remaining electrons.

In the Lewis structure, each fluorine atom will complete its octet by having 3 lone pairs of electrons (6 electrons) plus the 1 shared electron from the bond. Therefore, each fluorine will have a complete octet.

After placing the bonds and lone pairs, we find that boron only has 8 electrons around it (4 bonds), which is an exception to the octet rule. The final Lewis structure of BF₄ shows boron at the center with single bonds to four fluorine atoms, which each have 3 lone pairs:

In summary, the Lewis structure of BF₄ is characterized by the central boron atom bonded to four fluorine atoms, and we have determined that there are a total of 31 valence electrons in the molecule.