Let's break down the key concepts to understand the bond order of the B-F bond in the BF₃ molecule.
BF₃ is a molecule consisting of one boron (B) atom in the center and three fluorine (F) atoms bonded to it. The molecule has a trigonal planar structure where the bond angles between the fluorine atoms are 120°.
The electron configuration of boron (B) is 1s² 2s² 2p¹, while fluorine (F) has an electron configuration of 1s² 2s² 2p⁵.
In BF₃, boron forms three sigma bonds with the fluorine atoms. Each B-F bond is formed by the overlap of boron's sp² hybrid orbital and fluorine's p orbital. This results in three covalent bonds between boron and fluorine.
Boron in BF₃ does not have a full octet in its valence shell. It only has six electrons in its valence shell, which is a known feature of boron compounds like BF₃.
Molecular orbital theory helps us understand the bonding in BF₃. The bond order is calculated based on the number of bonding and antibonding electrons. However, in this case, since BF₃ is stable and the bonding is covalent, we use the bond order approximation.
The bond order is calculated as: \[ \text{Bond Order} = \frac{\text{Number of Bonding Electrons} - \text{Number of Antibonding Electrons}}{2} \] For BF₃, we find that the bond order for each B-F bond is 1 1/3. This reflects the covalent nature of the bonds and the incomplete octet of boron.
The bond order of the B-F bond in BF₃ is: \[ \boxed{1 \frac{1}{3}} \]
Regarding the molecular orbital (MO) energy levels for homonuclear diatomic molecules, the INCORRECT statement(s) is (are):
Which of the following statement is true with respect to H\(_2\)O, NH\(_3\) and CH\(_4\)?
(A) The central atoms of all the molecules are sp\(^3\) hybridized.
(B) The H–O–H, H–N–H and H–C–H angles in the above molecules are 104.5°, 107.5° and 109.5° respectively.
(C) The increasing order of dipole moment is CH\(_4\)<NH\(_3\)<H\(_2\)O.
(D) Both H\(_2\)O and NH\(_3\) are Lewis acids and CH\(_4\) is a Lewis base.
(E) A solution of NH\(_3\) in H\(_2\)O is basic. In this solution NH\(_3\) and H\(_2\)O act as Lowry-Bronsted acid and base respectively.