Final answer:
The BrF5 molecule has a total of 42 valence electrons, with the bromine atom as the center of the structure having five bonds to fluorine atoms and one lone pair. Its Lewis structure shows four fluorine atoms in equatorial positions and one in an axial position with slightly less than 90° angles.
Step-by-step explanation:
To determine the number of valence electrons in BrF5, we must look at the individual atoms. A bromine atom has seven valence electrons, and each fluorine atom also has seven valence electrons. Therefore, for BrF5, there are a total of 7 (from Br) + 5 × 7 (from each F) = 42 valence electrons.
Now, drawing the corresponding Lewis structure, we start with bromine as the central atom. Since there are five fluorine atoms, BrF5 falls under the AX5E category in VSEPR theory which implies that there are five bonding pairs and one lone pair on the central atom. This gives us a structure with four fluorine atoms in a plane around the bromine in equatorial positions, one fluorine atom and the lone pair of electrons in axial positions. The lone pair causes the Faxial-Br-Fequatorial angles to be slightly less than 90° due to repulsion.
The Lewis structure for BrF5 can be depicted as a central Br atom with single bonds to five F atoms, and one lone pair of electrons indicated typically by a pair of dots.