Boron trifluoride VSEPR

A boron trifluoride molecule, BF3, has the Lewis structure shown in (5). There are three bonding pairs attached to the central atom and no lone pairs. According to the VSEPR model, as illustrated in Fig. 3.4, the three bonding pairs, and the fluorine atoms they link, lie at the corners of an equilateral triangle. Such a structure is trigonal planar, and all three F—B—F angles are 120° (6). 

Boron trifluoride is a classic Lewis acid. Therefore, an accurate molecular orbital picture of BF3 should show, among other things, an orbital capable of acting as an electron pair acceptor. The VSEPR shape is a planar triangle, consistent with experimental observations. 

Next we will look at the BF3 (boron trifluoride) molecule, known to have planar geometry based on VSEPR. Considering only the valence electrons, the orbital diagram of... 

Boron trifluoride is a Lewis acid, an electron-pair acceptor. Molecular orbital theory of BF3 must provide an orbital capable of acting as such an acceptor to be consistent with this chemical property. The VSEPR-predicted shape is trigonal, consistent with experimental observations. 

Thus, IBr has a nonzero dipole moment, which points from the Br atom to the I atom, (b) Because fluorine is more electronegative than boron, each B—F bond in BF3 (boron trifluoride) is polar, and the three bond moments are equal. According to the VSEPR model, however, BF3 should be trigonal planar and the symmetry of the trigonal planar shape means that the three bond moments exactly cancel one another ... 

The shape of boron trifluoride molecule is planar triangle with bond angles 120°. VSEPR theory predicts sp hybridization of boron atom and a planar triangle shape. 

Boron trifluoride (BF3) has no dipole moment (jm = 0 D). Explain how this observation confirms the geometry of BF3 predicted by VSEPR theory. 

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