BF₃ is planar and electron deficient compound. Hybridization and number of electrons around the central atom, respectively are :

To determine the hybridization of the central boron atom in BF₃, we first count the regions of electron density around boron.

Boron forms three sigma bonds with three fluorine atoms. There are no lone‑pair regions on boron, because boron contributes only three valence electrons and each B–F bond uses one electron from boron and one from fluorine. Hence, the total number of electron‑density regions (bonding pairs) around boron is:

$3\; \backslash text\{\; bonding\; pairs\}$

Three regions correspond to an sp² hybridization (one s orbital + two p orbitals → three sp² hybrids). The remaining unhybridized p orbital stays empty, which explains the electron‑deficient nature of BF₃.

Next, we count the total electrons that are present around the central atom (the bonded electrons). Each B–F sigma bond contains two electrons, so the total number of electrons surrounding boron is:

$3\; \backslash text\{\; bonds\}\; \backslash times\; 2\; \backslash text\{\; electrons\; per\; bond\}\; =\; 6\; \backslash text\{\; electrons\}$

Therefore, the hybridization of boron in BF₃ is sp² and the number of electrons around the central atom is 6.