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

The correct option is sp² and 6.

Let us understand the step-by-step reasoning behind the hybridization and the electron count around the central boron atom in boron trifluoride (BF₃):

1. Electronic Configuration of Boron:
Boron (B) is the central atom with an atomic number of 5.
Its ground state electronic configuration is:

$1s^{2}2s^{2}2p^1$

This means boron has 3 valence electrons in its outermost shell (two in the 2s orbital and one in the 2p orbital).

2. Hybridization:
In order to form three covalent bonds with three fluorine (F) atoms, one electron from the 2s orbital is excited and promoted to an empty 2p orbital. This excited state configuration is:

$1s^{2}2s^{1}2p_x^{1}2p_y^1$

These three orbitals (one 2s and two 2p orbitals) mix together to form three equivalent hybrid orbitals. Therefore, the hybridization of the central boron atom is:

${\mathrm{sp}}^2$

The three hybrid orbitals point towards the corners of an equilateral triangle, giving BF₃ a trigonal planar geometry with bond angles of 120°.

3. Number of Electrons around the Central Atom:
Boron shares its 3 valence electrons with 3 fluorine atoms (where each fluorine atom contributes 1 electron for sharing to form a single B-F covalent bond).
Since there are three B-F single bonds, the total number of shared bonding electrons around the central boron atom is:

$3\times 2=6$

Because the central boron atom has only 6 electrons in its valence shell, it has an incomplete octet, making BF₃ a stable but electron-deficient compound.