Question

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?

Answer 1

`"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 `2p_z` orbital on boron is currently empty. Thus, `"BF"_3` can accept electrons into that orbital and is a lewis acid.

When a ligand (or atom) donates electrons into that `2p_z` atomic orbital, it distorts the geometry from trigonal planar (`D_(3h)` symmetry) into a tetrahedral structure.

(SIDENOTE: If the atom is also an `"F"`, then it is a `T_d` symmetry, like methane, but if `"L"` `ne` `"F"`, then it is a `C_(3v)` symmetry, like ammonia.)

You can tell that it's tetrahedral because there are no lone-pair electron groups and there are exactly four electron groups around boron.