Answer:
The central atom boron in BCl₄⁻ ion is sp³ hybridized and has 4 sp³ hybrid orbitals.
Explanation:
Boron is an element which belongs to group 13 of periodic table.
According to the Valence Bond Theory, in BCl₄⁻ ion, one 2s and three 2p orbitals of boron hybridize to form four sp³ hybrid orbitals, which overlap with four 2p orbitals of chlorine to form BCl₄⁻ anion.
Therefore, the central atom boron in BCl₄⁻ ion is sp³ hybridized and has 4 sp³ hybrid orbitals.
