Draw and explain the Lewis structure of BrF2, state its molecular geometry and its bond angles

To draw the Lewis structure of BrF₂, we start by determining the valence electrons for the central atom, bromine (Br). Bromine has 7 valence electrons. Each fluorine (F) atom has 7 valence electrons as well, and since there are two fluorine atoms, we have a total of 14 valence electrons from fluorine. Therefore, the total number of valence electrons for BrF₂ is:

7 (from Br) + 2 x 7 (from F) = 21 valence electrons.

Next, we place bromine at the center and connect each fluorine with a single bond:

  F
   |
Br--F

Each bond accounts for 2 electrons, so using 2 bonds consumes 4 electrons, leaving us with:

21 – 4 = 17 valence electrons.

Next, we complete the octet of the surrounding fluorine atoms by adding 6 more electrons, 3 pairs, to each fluorine:

    ..   ..
    F:   F:
     |   |
    Br
     

Now, we have used 12 more electrons (6 for each F) for a total of 16. This leaves us with:

17 – 12 = 5 electrons, which we place on bromine in the form of three lone pairs:

   ..   ..
   F:   F:
    |   |
  :Br:
   ..   ..

Ultimately, we have the complete Lewis structure of BrF₂ with two fluorine atoms bonded to a central bromine atom, which also has three lone pairs of electrons.

The molecular geometry of BrF₂ can be described as linear due to the two bonded fluorine atoms and the three lone pairs on the bromine atom. According to the VSEPR theory, lone pairs are positioned in a way that minimizes the repulsion between the surrounding electron pairs. In this case, the three lone pairs occupy equatorial positions in a trigonal bipyramidal arrangement, while the two fluorine atoms are located at opposite ends, resulting in a linear shape.

The bond angles in BrF₂ are approximately 180 degrees because of its linear geometry, where the two F atoms are as far apart as possible from each other due to the repulsive forces of the lone pairs.