Question

Why is $BrF_4^ - $ square planar, whereas $BF_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 $Br$,
$\left[ {Ar} \right]4{s^2}3{d^{10}}4{p^5}$
For $B$,
$\left[ {He} \right]2{s^2}2{p^1}$
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 $BrF_4^ - $ has square planar and $BF_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.