The attachment of a fluoride ion to the boron in BF_3, through a coordinate covalent bond, creates the BF_4- ion. What is the geometric shape of this ion?

${\text{BF}}_{3}$ starts out as a trigonal planar structure, since boron only requires $6$ valence electrons to be stable, thus giving it $3$ electron groups.
The $2 {p}_{z}$ orbital on boron is currently empty. Thus, ${\text{BF}}_{3}$ can accept electrons into that orbital and is a lewis acid.
When a ligand (or atom) donates electrons into that $2 {p}_{z}$ atomic orbital, it distorts the geometry from trigonal planar (${D}_{3 h}$ symmetry) into a tetrahedral structure.
(SIDENOTE: If the atom is also an $\text{F}$, then it is a ${T}_{d}$ symmetry, like methane, but if $\text{L}$ $\ne$ $\text{F}$, then it is a ${C}_{3 v}$ symmetry, like ammonia.)