Step-by-step derivation:

1. Determine the central atom and valence electrons. In the molecule $BrF_5$, Bromine (Br) is the central atom as it is less electronegative than Fluorine (F). Bromine is in Group 17 of the periodic table, so it has 7 valence electrons.

2. Calculate the number of bonding pairs and lone pairs.

   • The central Br atom forms single covalent bonds with five F atoms. Therefore, there are 5 bonding pairs (bp).
   • The number of electrons used in bonding is 5.
   • The number of non-bonding electrons on Br is (Total valence electrons) - (Electrons used in bonding) = 7 - 5 = 2.
   • These 2 non-bonding electrons form one lone pair (lp).
   • So, the central Br atom has 5 bonding pairs and 1 lone pair.

3. Determine the VSEPR formula and electron geometry.

   • The total number of electron domains (steric number) around the central atom is the sum of bonding pairs and lone pairs: 5 bp + 1 lp = 6.
   • The VSEPR formula for $BrF_5$ is $AX_5E_1$, where A is the central atom, X are the surrounding atoms, and E is the lone pair.
   • For a steric number of 6, the electron pairs arrange themselves in an octahedral geometry to minimize repulsion.

4. Determine the molecular geometry.

   • In the octahedral arrangement, there are 6 positions for the electron pairs. One position is occupied by the lone pair, and the other five are occupied by the bonding pairs with Fluorine atoms.
   • The resulting molecular shape is square pyramidal. The four F atoms form the base of the pyramid in the equatorial plane, and one F atom is at the apex (axial position). The lone pair occupies the position opposite to the axial F atom.

5. Analyze the bond angles based on VSEPR theory.

   • VSEPR theory states that the order of repulsion between electron pairs is: lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair.
   • In an ideal octahedral geometry (like in $SF_6$), all bond angles are exactly 90° or 180°.
   • However, in $BrF_5$, the presence of the lone pair distorts this ideal geometry.
   • The lone pair exerts a stronger repulsion on the adjacent bonding pairs than another bonding pair would. The lone pair repels the four equatorial Br-F bonds more strongly than the axial Br-F bond repels them.
   • This stronger repulsion pushes the four equatorial F atoms slightly upwards, away from the lone pair and towards the axial F atom.
   • As a result, the angles between the axial F and the equatorial F atoms (F$_{axial}$-Br-F$_{equatorial}$) are compressed and become less than 90°. (Experimental value is ~84.8°).
   • Similarly, the distortion affects the angles between adjacent equatorial F atoms (F$_{equatorial}$-Br-F$_{equatorial}$). Because the Br atom is pushed slightly out of the plane of the four equatorial F atoms, these angles are also no longer exactly 90°.
   • Therefore, due to the repulsion from the lone pair, all the F-Br-F angles that would be 90° in a perfect octahedron are distorted.

6. Conclusion. Based on the VSEPR theory, there are no F-Br-F angles that are exactly 90° in the $BrF_5$ molecule.

The number of 90 degree F-Br-F angles is 0.