Describe the orbitals used in bonding and the bond angles in the following compounds:
e. BF₃

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Identify the central atom in BF3. Boron (B) is the central atom, as it is less electronegative than fluorine (F) and can form three bonds.

Determine the hybridization of the central atom. Boron in BF3 forms three sigma bonds with fluorine atoms. To achieve this, boron undergoes sp² hybridization, which involves mixing one s orbital and two p orbitals to form three equivalent sp² hybrid orbitals.

Describe the geometry of the molecule. The sp² hybridization results in a trigonal planar geometry, where the three sp² hybrid orbitals are arranged 120° apart in a plane to minimize electron pair repulsion.

Explain the bond angles. In a trigonal planar geometry, the bond angles between the fluorine atoms are approximately 120°.

Discuss the orbitals involved in bonding. Each B-F bond is formed by the overlap of a boron sp² hybrid orbital with a fluorine p orbital, resulting in a sigma bond.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hybridization

Hybridization is the process by which atomic orbitals mix to form new hybrid orbitals that are suitable for the pairing of electrons to form chemical bonds. In BF3, the boron atom undergoes sp2 hybridization, resulting in three equivalent sp2 hybrid orbitals that are oriented in a planar arrangement.

Bond Angles

Bond angles are the angles formed between adjacent bonds in a molecule, which are influenced by the arrangement of electron pairs around the central atom. In BF3, the bond angles are approximately 120 degrees due to the trigonal planar geometry resulting from the sp2 hybridization of boron.

Molecular Geometry

Molecular geometry refers to the three-dimensional arrangement of atoms in a molecule. BF3 has a trigonal planar geometry, which arises from the arrangement of the three fluorine atoms around the central boron atom, minimizing electron pair repulsion according to VSEPR theory.

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