Question

Why is $Br{F}_4^{-}$ square planar, whereas $B{F}_4^{-}$ is tetrahedral?

Answer 1

Hint: The geometry of square planar molecules describes the stereochemistry adopted by chemical compounds. The d-orbital splitting diagram for square planar $\left(D_{4h}\right)$transition metal complexes is derived from the general octahedral $\left(O_h\right)$splitting diagram.

Complete answer:
In this question, bromine has $7$ valence electrons in its ground state electronic configuration whereas boron has only $3$ valence electrons in its ground state electronic configuration as given below:
For $Br$,
$\left[Ar\right]4s^{2}3d^{10}4p^5$
For $B$,
$\left[He\right]2s^{2}2p^1$
In these electronic configurations, we can see that the valence shell of bromine contains $7$ electrons, $2$ from the $s$ subshell and $5$ in the $p$ subshell.
In the first case when $Br$ is the central atom bonded with four $F$ atoms, the four of the seven electrons of bromine form a bond with $F$ atom. Here this shows that three electrons do not take part in bond formation. As there is a negative charge in the compound this indicates the additional electrons which will be paired with one of the three lone electrons.

Here you can see that four $$F$$ will align in a square plane and the lone pairs are on either side of the plane.
In the case of $B{F_4}^{-}$, boron only has three valence electrons, so three $F$ will bond with boron and the fourth $F$will occupy the open $p$ orbital of $B$. Here the extra electron shown in red is not present in the ground state configurations that provide the negative charge.

Thus, by these diagrams it is clear that $Br{F}_4^{-}$ has square planar and $B{F}_4^{-}$ has a tetrahedral shape.

Note:
In square planar, the constituent atoms surround the central atom which form the corners of a square on the same plane. Whereas in the tetrahedral, on the centre of the four substituents, there the central atom forming the corners of the tetrahedron.