The geometry and type of hybridization of Central atom BF3 is
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Answer:
BF3 has a boron atom with three outer-shell electrons in its ground state and three fluorine atoms containing seven outer electrons. Further, if we observe closely, one boron electron is unpaired in the ground state. During the formation of this compound, the 2s orbital and two 2p orbitals hybridize. Only one of the empty p-orbital is left behind as the lone pair. In short, Boron needs 3 hybridised orbitals to make bonds with 3 atoms of F where the 2pz orbitals get overlapped with these hybridised sp2 orbitals and bonds are formed.
Important Points to RememberThe three hybridised sp2 orbitals are usually arranged in a triangular shape.BF3 molecule is formed by bonding between three sp2 orbitals of B and p of 3 F atoms.All the bonds in BF3 are sigma bonds.BF3 Molecular Geometry and Bond Angles
Normally, boron forms monomeric covalent halides which have a planar triangular geometry. This shape is mainly formed by the overlap of the orbitals between the two compounds. To be more precise, the BF3molecular geometry is trigonal planar. It further has symmetric charge distribution on the central atom and is nonpolar.
The bond angle is 120o where all the atoms are in one plane. Each of them also makes an equilateral triangle.
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