Predicting Shape of molecules from TRIGONAL PLANER to OCTAHEDRON ([{VSEPR}]):---# There are three places on the central atom in boron trifluoride (BF 3 ) where valence electrons can be found. Repulsion between these electrons can be minimized by arranging them toward the corners of an equilateral triangle. The VSEPR theory therefore predicts a trigonal planar geometry for the BF 3 molecule, with a F-B-F bond angle of 120 o . BeF2 and BF 3 are both two- dimensional molecules, in which the atoms lie in the same plane. If we place the same restriction on methane (CH 4 ), we would get a square-planar geometry in which the H-C-H bond angle is 90o . If we let this system expand into three dimensions, however, we end up with a tetrahedral molecule in which the H-C-H bond angle is 109 o 28. Repulsion between the five pairs of valence electrons on the phosphorus atom in PF 5 can be minimized by distributing these electrons toward the corners of a trigonal bipyramid . Three of the positions in a trigonal bipyramid are labeled equatorial because they lie along the equator of the molecule. The other two are axial because they lie along an axis perpendicular to the equatorial plane. The angle between the three equatorial positions is 120 o , while the angle between an axial and an equatorial position is 90o . There are six places on the central atom in SF 6 where valence electrons can be found. The repulsion between these electrons can be minimized by distributing them toward the corners of an octahedron . The term octahedron literally means eight sides, but it is the six corners, or vertices, that interest us. To imagine the geometry of an SF 6 molecule, locate fluorine atoms on opposite sides of the sulfur atom along the X , Y , and Z axes of an XYZ coordinate system.
Posted on: Wed, 17 Sep 2014 13:57:25 +0000
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