How to Draw the Lewis Structure for BrF5

To draw the Lewis structure for BrF5 (Bromine Pentafluoride), follow these steps:

1. Count the total number of valence electrons:
   • Bromine (Br) has 7 valence electrons.
   • Each Fluorine (F) atom has 7 valence electrons.
   • Total valence electrons = 7 (Br) + 5 × 7 (F) = 7 + 35 = 42 electrons.
2. Determine the central atom:
   • Bromine (Br) is the central atom because it is less electronegative than Fluorine (F).
3. Draw the skeletal structure:
   • Place the Bromine (Br) atom in the center and surround it with five Fluorine (F) atoms.
4. Distribute the remaining electrons:
   • After forming single bonds between Br and each F, you have used 10 electrons (5 bonds × 2 electrons).
   • Remaining electrons = 42 – 10 = 32 electrons.
   • Distribute these electrons as lone pairs around the Fluorine atoms, giving each F atom 6 additional electrons (3 lone pairs).
5. Check the octet rule:
   • Each Fluorine atom now has 8 electrons (2 in the bond and 6 as lone pairs), satisfying the octet rule.
   • Bromine has 10 electrons (5 bonds × 2 electrons), which is acceptable as it can have an expanded octet.

The final Lewis structure for BrF5 shows Bromine (Br) at the center with five single bonds to Fluorine (F) atoms, and each Fluorine atom has three lone pairs of electrons.