Bromine trifluoride autoionizes to form BrF_2^ and BrF_4^ . The...

The autoionization of bromine trifluoride is given by the reaction:

$2\text{BrF}_3 \rightleftharpoons \text{BrF}_2^{\oplus} + \text{BrF}_4^{\ominus}$

For the cation $\text{BrF}_2^{\oplus}$:
The central atom Br has $7$ valence electrons. Due to the positive charge, it has $7 - 1 = 6$ electrons available.
It forms $2$ single bonds with F atoms, leaving $4$ electrons, which constitute $2$ lone pairs.
Steric number = $2$ (bond pairs) + $2$ (lone pairs) = $4$.
The hybridization is $sp^3$. With $2$ bond pairs and $2$ lone pairs, the shape is bent (or V-shaped).

For the anion $\text{BrF}_4^{\ominus}$:
The central atom Br has $7$ valence electrons. Due to the negative charge, it has $7 + 1 = 8$ electrons available.
It forms $4$ single bonds with F atoms, leaving $4$ electrons, which constitute $2$ lone pairs.
Steric number = $4$ (bond pairs) + $2$ (lone pairs) = $6$.
The hybridization is $sp^3d^2$. To minimize repulsion, the $2$ lone pairs occupy opposite (trans) positions in the octahedral geometry, resulting in a square planar shape.

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

Answer: bent, square planar