Shapes of Cation and Anion Formed by Autoionization of Bromine Trifluoride

To determine the shapes of the cation ($\text{BrF}_2^+$) and the anion ($\text{BrF}_4^-$) formed by the autoionization of liquid bromine trifluoride ($\text{BrF}_3$), we can apply Valence Shell Electron Pair Repulsion (VSEPR) theory:

1. Shape of the Cation ($\text{BrF}_2^+$)

• Central Atom: Bromine ($\text{Br}$), which belongs to Group 17 and has 7 valence electrons.
• Valence electrons on $\text{Br}$ in $\text{BrF}_2^+$:
  $\text{Valence electrons} = 7 - 1 = 6$
• Bond Pairs: 2 single covalent bonds formed with two fluorine atoms.
• Lone Pairs:
  $\text{Number of lone pairs} = \frac{6 - 2}{2} = 2$
• Steric Number (SN):
  $\text{SN} = \text{Bond Pairs} + \text{Lone Pairs} = 2 + 2 = 4$
• Hybridization: $sp^3$ (Tetrahedral electron-pair geometry)
• Molecular Shape: With 2 bond pairs and 2 lone pairs, the species adopts a bent (or V-shaped) geometry.

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2. Shape of the Anion ($\text{BrF}_4^-$)

• Central Atom: Bromine ($\text{Br}$).
• Valence electrons on $\text{Br}$ in $\text{BrF}_4^-$:
  $\text{Valence electrons} = 7 + 1 = 8$
• Bond Pairs: 4 single covalent bonds formed with four fluorine atoms.
• Lone Pairs:
  $\text{Number of lone pairs} = \frac{8 - 4}{2} = 2$
• Steric Number (SN):
  $\text{SN} = \text{Bond Pairs} + \text{Lone Pairs} = 4 + 2 = 6$
• Hybridization: $sp^3d^2$ (Octahedral electron-pair geometry)
• Molecular Shape: The two lone pairs position themselves at trans (axial) positions to minimize electron-pair repulsions, resulting in a square planar geometry.

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Conclusion

The shapes of the cation and anion are bent and square planar, respectively.

Correct Option: A