Number of 90 Degree Bond Angles in Bromine Pentafluoride

To determine the number of $90^\circ$ bond angles ($\text{F}-\text{Br}-\text{F}$) in bromine pentafluoride ($\text{BrF}_5$), we use Valence Shell Electron Pair Repulsion (VSEPR) theory:

1. Determination of Steric Number and Hybridization

• Central Atom: Bromine ($\text{Br}$) has $7$ valence electrons.
• Monovalent Atoms: $5$ Fluorine ($\text{F}$) atoms.
• Steric Number ($\text{SN}$):
  $\text{SN} = \frac{1}{2} \left( \text{Valence electrons of Br} + \text{Monovalent atoms} \right) = \frac{1}{2} (7 + 5) = 6$

A steric number of $6$ corresponds to $sp^3d^2$ hybridization and an octahedral electron-pair geometry.

2. Geometry and Lone Pairs

• Bond Pairs ($\text{bp}$): $5$
• Lone Pairs ($\text{lp}$): $6 - 5 = 1$

With 5 bond pairs and 1 lone pair, the molecular geometry of $\text{BrF}_5$ is square pyramidal.

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3. Analysis of Bond Angles

Case I: Ideal Octahedral Geometry (Ignoring Distortion)

In an un-distorted square pyramidal structure derived from an octahedron:

1. Equatorial-Equatorial angles: The 4 equatorial Fluorine atoms form a square plane around the central $\text{Br}$ atom. The angle between adjacent equatorial fluorine atoms ($\text{F}_{\text{eq}}-\text{Br}-\text{F}_{\text{eq}}$) is:
   $4 \times 90^\circ = 4 \text{ angles}$
2. Axial-Equatorial angles: The 1 axial Fluorine atom is perpendicular to the equatorial plane, forming an angle of $90^\circ$ with each of the 4 equatorial fluorine atoms ($\text{F}_{\text{ax}}-\text{Br}-\text{F}_{\text{eq}}$):
   $4 \times 90^\circ = 4 \text{ angles}$

$\text{Total ideal } 90^\circ \text{ angles} = 4 + 4 = 8$

Case II: Actual VSEPR Geometry (Considering Repulsions)

According to VSEPR theory:
$\text{Lone Pair - Bond Pair (lp-bp) repulsion} > \text{Bond Pair - Bond Pair (bp-bp) repulsion}$

• The lone pair on the $\text{Br}$ atom exerts a strong repulsive force, pushing the four equatorial $\text{F}$ atoms away from itself (towards the axial $\text{F}$ atom).
• This causes the axial-equatorial bond angles ($\text{F}_{\text{ax}}-\text{Br}-\text{F}_{\text{eq}}$) to decrease from $90^\circ$ to approximately $84.8^\circ$.
• Due to this distortion, the four equatorial atoms are no longer in the same plane as the bromine atom, making all equatorial-equatorial angles slightly different from $90^\circ$ as well.

Thus, due to real VSEPR distortion, 0 bond angles are strictly equal to $90^\circ$.

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Conclusion

Depending on whether ideal octahedral geometry or distorted VSEPR geometry is considered:

• Number of $90^\circ$ angles in ideal geometry = 8
• Number of $90^\circ$ angles in distorted real geometry = 0

Hence, the number of $90^\circ$ $\text{F}-\text{Br}-\text{F}$ angles is 0 or 8.